{"msg":"应用数学","code":200,"data":{"currentIndex":null,"examId":null,"examTime":null,"questionList":[{"id":"794933553929408513","title":"<p>某电子商务公司要从A地向B地的用户发送一批价值90000元的货物。从A地到B地有水、陆两条路线。走陆路时比较安全，其运输成本为10000元；走水路时一般情况下的运输成本只要7000元，不过一旦遇到暴风雨天气，则会造成相当于这批货物总价值的10%的损失。根据历年情况，这期间出现暴风雨天气的概率为1/4，那么该电子商务公司 （） 。</p>","analyze":"<p>这是一个不确定性决策问题，其决策树如下图所示：</p><p><img style=\"max-width:100%;height:auto\" title=\"1b79ed441136b1c53133892f36f631cc.jpg\" alt=\"1.jpg\" src=\"https://image.chaiding.com/ruankao/1b79ed441136b1c53133892f36f631cc.jpg?x-oss-process=style/ruankaodaren\"/></p><p>由于该问题本身带有不确定因素，因此实际的运输成本不能预先确定。不过，对掌握一定概率分布的不确定性问题，该电子商务公司可以通过计算数学期望值进行比较决策，而不是盲目碰运气或一味害怕、躲避风险。</p><p>根据上述决策树，走水路时，成本为7000元的概率为75%，成本为16 000元的概率为25%，因此走水路的期望成本为（7000×75%+16000×25%=9250。走陆路时，其成本确定为10 000元。因此，走水路的期望成本小于走陆路的成本，所以应该选择走水路。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933555808456705"],"itemList":[{"id":"794933555808456705","questionId":"794933553929408513","content":"应选择走水路","answer":1,"chooseValue":"A"},{"id":"794933555829428225","questionId":"794933553929408513","content":"应选择走陆路","answer":0,"chooseValue":"B"},{"id":"794933555846205441","questionId":"794933553929408513","content":"难以选择路线","answer":0,"chooseValue":"C"},{"id":"794933555867176961","questionId":"794933553929408513","content":"可以随机选择路线","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933605754228737","title":"<p>某厂生产某种电视机，销售价为每台2500元，去年的总销售量为25000台，固定成本总额为250万元，可变成本总额为4000万元，税率为16%，则该产品年销售量的盈亏平衡点为（）台。(只有在年销售量超过它时才能有盈利)。</p>","analyze":"<p>设盈亏平衡点上的销量是x台，销售总额为25000×2500=6250万，其中可变成本占比4000/6250=64%，由此得出下面等式：</p><p>2500x=2500000+2500x×64%+2500x×16%，解得x=5000。即，盈亏平衡点上的销量是5000台。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933607612305409"],"itemList":[{"id":"794933607612305409","questionId":"794933605754228737","content":"5000","answer":1,"chooseValue":"A"},{"id":"794933607633276929","questionId":"794933605754228737","content":"10000","answer":0,"chooseValue":"B"},{"id":"794933607654248449","questionId":"794933605754228737","content":"15000","answer":0,"chooseValue":"C"},{"id":"794933607675219969","questionId":"794933605754228737","content":"20000","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933494252851201","title":"<p><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">某博览会每天8:00开始让观众通过各入口处检票进场，8:00前已经有很多观众在排队等候。假设8:00后还有不少观众均匀地陆续到达，而每个入口处对每个人的检票速度都相同。根据以往经验，若开设8个入口，则需要60分钟才能让排队观众全部入场；若开设10个入口，则需要40分钟才能消除排队现象。为以尽量少的入口数确保20分钟后消除排队现象，博览会应在8:00和8:20开设的入口数分别为（ ）</span><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p>","analyze":"<p>设早上8点时已有S人在排队等候，以后每分钟新来m人，每个人口处每分钟进场n人，则</p><p>JS+60m=8*60n （1），</p><p>S+40m=10*40n （2），</p><p>（1）（2）两式相减得m=4n，而S=240n。</p><p>若要在20分钟由K个入口消除排队，则S+20m=20Kn，则K=16。</p><p>即8:00时，若开设16个入口，就可以在20分钟消除排队现象。</p><p>由于8:20后，每分钟新来m=4n人，所以应设4个入口，参观者就可以随来随进。</p><p>参考答案：C</p><p><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span><br/></p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933496526163969"],"itemList":[{"id":"794933496484220929","questionId":"794933494252851201","content":"12,2","answer":0,"chooseValue":"A"},{"id":"794933496505192449","questionId":"794933494252851201","content":"14,4","answer":0,"chooseValue":"B"},{"id":"794933496526163969","questionId":"794933494252851201","content":"16,4","answer":1,"chooseValue":"C"},{"id":"794933496551329793","questionId":"794933494252851201","content":"18,6","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933477542744065","title":"<p><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"> </span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">线性规划问题不可能</span></span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>52）</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">。</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">试题解析：</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">线性规划问题的可行解区是一个凸集。如果线性规划问题存在两个最优解，则连接这两个解点的线段上所有的点都必然是可行解。</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">参考答案：（</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">52）C</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933479769919489"],"itemList":[{"id":"794933479740559361","questionId":"794933477542744065","content":"没有最优解","answer":0,"chooseValue":"A"},{"id":"794933479757336577","questionId":"794933477542744065","content":"只有一个最优解","answer":0,"chooseValue":"B"},{"id":"794933479769919489","questionId":"794933477542744065","content":"只有2个最优解","answer":1,"chooseValue":"C"},{"id":"794933479786696705","questionId":"794933477542744065","content":"有无穷多个最优解","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933615115915265","title":"<p>某地天然气输送管线网络图如下，每段管线旁的数字表示输气能力(单位：万立方米/小时）根据该图，从源S到目的地T的最大输气能力为 （66） 万立方米/小时。<br/></p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/fae9dd71acb076f79b04b853fa19554c.jpg?x-oss-process=style/ruankaodaren\" title=\"fae9dd71acb076f79b04b853fa19554c.jpg\" alt=\"111.jpg\"/></p><br/>","analyze":"<p>如下图所示：</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/bd267d9eba1f7f52602bb1b8919425d3.jpg?x-oss-process=style/ruankaodaren\" title=\"bd267d9eba1f7f52602bb1b8919425d3.jpg\" alt=\"111.jpg\"/></p><p>①u3000从S节点出发，经过A节点的流量，可以分为AT、ABT两条线路，流量均为2。AT、AB、BT的流量减去2后均为0，因此把AT、AB、BT均标记为断开。由于A节点已经没有了到达其他节点的线路，可以认为SA线路所有流量最大为4。</p><p>②u3000从S节点出发，经过B节点的流量，经BCT线路即可用完，流量为3。把SB、BC、CT的流量减去3后，SB、CT均为0，标记为断开，BC剩余1。SB线路最大流量3已用完。</p><p>③u3000同理，SCDT线路的流量为4，SC最大流量4已用完，SC标记为断开，CD剩余1，DT剩余2。</p><p>④u3000SDT线路的流量为1，把SD的流量用完。</p><p>综上可得，所有线路的最大输气能力可达4+3+4+1=12。</p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933617062072321"],"itemList":[{"id":"794933617015934977","questionId":"794933615115915265","content":"6","answer":0,"chooseValue":"A"},{"id":"794933617032712193","questionId":"794933615115915265","content":"8","answer":0,"chooseValue":"B"},{"id":"794933617049489409","questionId":"794933615115915265","content":"10","answer":0,"chooseValue":"C"},{"id":"794933617062072321","questionId":"794933615115915265","content":"12","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933516516216833","title":"<p><br/></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">某批发站准备向甲、乙、丙、丁四家小商店供应</span>5箱商品。批发站能取得的利润（单位：百元）与分配的箱数有关（见下表）。</span></p><p class=\"MsoNormal\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/8dd31e307245c2ba8ee2149c7e9f5b41.png?x-oss-process=style/ruankaodaren\"/></p><p class=\"MsoNormal\"><span style=\"font-family:宋体\">批发站为取得最大总利润，应分配</span><span style=\"font-family: 宋体; font-size: 10.5pt; text-indent: 0pt;\">__(57)__。</span><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 21pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">该批发站如将</span>5箱都分配给1家，则最大总利润为9百元（给乙5箱)；</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">如分配给</span>2家（1-4箱或2-3箱)，则最大总利润分别为12或13百元；</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">如分配给</span>3家（1-1-3箱)，则最大总利润为15百元；</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">如分配给</span>3家（1-2-2箱），则最大总利润为16百元（给甲、丙各2箱，给丁1箱); </span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">如分配给</span>4家（1-1-1-2箱)，则最大总利润为16百元（给甲、乙、丁各1箱，给丙2箱)。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">因此，该批发站有两种最优分配方案能取得最大利润</span>16百元。这两种方案中，都需要给丙分配2箱。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"font-family: 宋体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:宋体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">57）C </span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933518445596673"],"itemList":[{"id":"794933518399459329","questionId":"794933516516216833","content":"给甲、丙各1箱","answer":0,"chooseValue":"A"},{"id":"794933518424625153","questionId":"794933516516216833","content":"给乙2箱","answer":0,"chooseValue":"B"},{"id":"794933518445596673","questionId":"794933516516216833","content":"给丙2箱","answer":1,"chooseValue":"C"},{"id":"794933518470762497","questionId":"794933516516216833","content":"给丁2箱","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933568156487681","title":"<p>S公司开发一套信息管理软件，其中一个核心模块的性能对整个系统的市场销售前景影响极大，该模块可以采用S公司自己研发、采购代销和有条件购买三种方式实现。S公司的可能利润（单位万元）收入如下表。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/e33c6a9baf0be3a40eb30ffd86f844e4.jpg?x-oss-process=style/ruankaodaren\" title=\"e33c6a9baf0be3a40eb30ffd86f844e4.jpg\" alt=\"1.jpg\"/></p><p>按经验，此类管理软件销售50万套，20万套，5万套和销售不出的概率分别为15%，25%，40%和20%，则S公司应选择 （ ） 方案。</p>","analyze":"<p>自己研发的可能利润值为：</p><p>45000015%+200000250%-5000040%-15000020%=67500 </p><p>采购代销的可能利润值为：</p><p>6500015%+6500025%+6500040%+6500020%=65000 </p><p>有条件购买的可能利润值为：25000015%+10000025%=62500 </p><p>因此，S公司应选择A方案以获得最高可能利润。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933570073284609"],"itemList":[{"id":"794933570073284609","questionId":"794933568156487681","content":"自己研发","answer":1,"chooseValue":"A"},{"id":"794933570090061825","questionId":"794933568156487681","content":"采购代销","answer":0,"chooseValue":"B"},{"id":"794933570106839041","questionId":"794933568156487681","content":"有条件购买","answer":0,"chooseValue":"C"},{"id":"794933570123616257","questionId":"794933568156487681","content":"条件不足无法选择","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933594278612993","title":"<p>某网络公司为了扩大华北市场，希望在北京举行一个展销会，会址打算选择在北京市内的A、B、C3个会场之一。获利情况与天气有关。通过天气预报了解到展销会当日天气为晴、多云、雨的概率，收益和会场租赁费用如下表所示，那么该公司所做的决策是（65）。</p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/ef29411ac507658e412b09011f88677e.jpg?x-oss-process=style/ruankaodaren\" title=\"ef29411ac507658e412b09011f88677e.jpg\" alt=\"1.jpg\"/>","analyze":"<p>期望货币价值(Expected Monetary Value，EMV)，又称为预期货币值、风险暴露值、风险期望值；是为了确定一项投机的期望货币价值，计算每一种可能出现的结果的货币收益（或损失）与其出现的概率相乘以后的和；</p><p>比如：一件商品有70%的概率可以获利20元，有30%的概率会导致损失30元，此时便可算出EMV=0.7*20+(-30)*0.3=5；</p><p>期望货币价值是定量风险分析的一种技术，常和决策树一起使用，它是将特定情况下可能的风险造成的货币后果和发生概率相乘，包含了现金和风险的考虑；</p><p>机会的EMV通常表示为正值，而威胁的EMV则表示为负值。</p><p>本题中：</p><p>选择A会场的期望货币值：4×0.25+6×0.5+1×0.25-3.6=0.65(万元)。</p><p>选择B会场的期望货币值：5×0.25+4×0.5+1.6×0.25-3.2=0.45(万元)。</p><p>选择C会场的期望货币值：6×0.25+2×0.5+1.2×0.25-3=-0.20(万元)。</p><p>由于0.65＞0.45＞-0.20，因此从货币期望值最大决策考虑，建议把会场设在A处。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933596161855489"],"itemList":[{"id":"794933596161855489","questionId":"794933594278612993","content":"选择A会场","answer":1,"chooseValue":"A"},{"id":"794933596191215617","questionId":"794933594278612993","content":"选择B会场","answer":0,"chooseValue":"B"},{"id":"794933596224770049","questionId":"794933594278612993","content":"选择C会场","answer":0,"chooseValue":"C"},{"id":"794933596254130177","questionId":"794933594278612993","content":"难以选择会场","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933620870500353","title":"<p>甲、乙、丙、丁四个任务分配在A．B．C．D四台机器上执行，每台机器执行一个任务，所需的成本（单位:百元）如表1所示。最低的总成本是 （68） 。<br/></p><p>表1 习题用表</p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/27726a81e1000f9638ab7f08a56f9e92.png?x-oss-process=style/ruankaodaren\" title=\"27726a81e1000f9638ab7f08a56f9e92.png\" alt=\"image.png\" width=\"690\" height=\"131\"/>","analyze":"<p>1.使各行各列中都出现0元素，每行元素都减去该行的最小元素；每列元素中减去该列的最小元素。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/56718269084405eaa846933373375229.jpg?x-oss-process=style/ruankaodaren\" title=\"56718269084405eaa846933373375229.jpg\" alt=\"111.jpg\"/>2.试指派</p><p>1）从含0元素最少的行开始，给该行中的0元素加圈。然后划去◎所在列的其它0元素，记作?，依次进行到最后一行。</p><p>2）从含0元素最少的列开始（画?的不计在内），给该列中的0元素加圈；然后划去◎所在行的0元素，记作?，依次进行到最后一列。</p><p>3）反复进行，直到所有0元素都已圈出和划掉为止。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/dbe92fa7fde0b548b7013e8cec11fcdb.jpg?x-oss-process=style/ruankaodaren\" title=\"dbe92fa7fde0b548b7013e8cec11fcdb.jpg\" alt=\"111.jpg\"/>如果得到n个◎（本题为4），则得到最优解，算法结束。</p><p>本题中，这里得不到4个0，但观察元素（3，2）值为1，元素（1，1），（2，3），（3，2），（4，4）的总和值最小，因此这是最优解之一。</p><p>回到原题，元素（1，1），（2，3），（3，2），（4，4）的总和值为22。</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933622816657409"],"itemList":[{"id":"794933622778908673","questionId":"794933620870500353","content":"20","answer":0,"chooseValue":"A"},{"id":"794933622795685889","questionId":"794933620870500353","content":"21","answer":0,"chooseValue":"B"},{"id":"794933622816657409","questionId":"794933620870500353","content":"22","answer":1,"chooseValue":"C"},{"id":"794933622837628929","questionId":"794933620870500353","content":"23","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933513630535681","title":"<p><br/></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">下面的网络图表示从城市</span>A到城市B运煤的各种路线。各线段上的数字表示该线段运煤所需的费用（百元/车）。城市A有三个装货点，城市B有三个卸货点，各点旁标注的数字表示装/卸煤所需的费用（百元/车）。根据该图，从城市A的一个装卸点经过一条路线到城市B的一个卸货点所需的装、运、卸总费用至少为</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>56）</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">（百元</span>/车）。</span></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/84ecb9ee8560918864e98182758ff0ef.png?x-oss-process=style/ruankaodaren\"/></span></p>","analyze":"<p><span style=\"font-family:宋体\">从</span><span style=\"font-family: 宋体; font-size: 10.5pt; text-indent: 21pt;\">A线出发经过中间5点可以到达B线。首先，很容易计算并标注各条路线从第5点到达B线并卸货的最少费用，可将其标注在相应的点旁。据此就容易计算并标注从第4点到达B线并卸货的最少费用，并将其标注在相应的点旁，依次类推。</span><br/></p><p><span style=\"font-family: 宋体; font-size: 10.5pt; text-indent: 21pt;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/e4eb89cdb2d1fbe5ce436eea4fca92ce.png?x-oss-process=style/ruankaodaren\"/></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">从</span>A的下端出发，向上、上、下、上、上、下到达B的中间点，总费用=3+（2+2+3+2+2+3) + 2=19 (百元/车）最少。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"font-family: 宋体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:宋体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">56）A</span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p><p><span style=\"font-family: 宋体; font-size: 10.5pt; text-indent: 21pt;\"><br/></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 21pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933515555721217"],"itemList":[{"id":"794933515555721217","questionId":"794933513630535681","content":"19","answer":1,"chooseValue":"A"},{"id":"794933515572498433","questionId":"794933513630535681","content":"20","answer":0,"chooseValue":"B"},{"id":"794933515589275649","questionId":"794933513630535681","content":"21","answer":0,"chooseValue":"C"},{"id":"794933515606052865","questionId":"794933513630535681","content":"22","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933556781535233","title":"<p>某工厂生产两种产品S和K，受到原材料供应和设备加工工时的限制。单件产品的利润、原材料消耗及加工工时如下表。为获得最大利润，S应生产 （ ） 件。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/1971a99714f22bd40b4efdccbc22c548.jpg?x-oss-process=style/ruankaodaren\" title=\"1971a99714f22bd40b4efdccbc22c548.jpg\" alt=\"1.jpg\"/></p><br/>","analyze":"<p>设利润为Z，为了获得最大利润，S应生产X1件，K应生产X2件。对该问题求解最优方案可以由下列数学模型描述： </p><p>Max Z=12X1+16X2</p><p>10X1+20X2≤120</p><p>8X1+8X2≤80 </p><p>X1≥0，X2≥0</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933558710915073"],"itemList":[{"id":"794933558681554945","questionId":"794933556781535233","content":"7","answer":0,"chooseValue":"A"},{"id":"794933558710915073","questionId":"794933556781535233","content":"8","answer":1,"chooseValue":"B"},{"id":"794933558731886593","questionId":"794933556781535233","content":"9","answer":0,"chooseValue":"C"},{"id":"794933558757052417","questionId":"794933556781535233","content":"10","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933591460040705","title":"<p>在军事演习中，张司令希望将部队尽快从A地通过公路网(如图所示)运送到F地：</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/65d4159a6f87436ca2948f44d363469f.jpg?x-oss-process=style/ruankaodaren\" title=\"65d4159a6f87436ca2948f44d363469f.jpg\" alt=\"1.jpg\"/></p><p>图中标出了各路段上的最大运量(单位：千人/小时)。根据该图可以算出，从A地到F地的最大运量是（64）千人/小时。</p>","analyze":"<p>该题解题关系是需要将图中节点的输入/输出流量调整平衡，因为只有输入/输出流量平衡才能表现出真实的运量。</p><p>如图所示，对于节点E，他的输出运力为15，而所有输入运力之和为14，则E的最大真实运力，只能达到14，所以将E的输出运力修改为14。对于D节点，其输出运力和为7，而输入运力为8，则需要平衡为7。节点B也需要调，但情况比较复杂，我们需要综合分析B的输出运力与C的输出运力，分析可知，当B到C的运力调整为1时，既能达到节点运力的平衡，又能使运力最大，所以应调整为1。当完成这些调整之后，可轻易得出结论，最大运力为22。</p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/452f40b508be8053a926355f773f0a80.jpg?x-oss-process=style/ruankaodaren\" title=\"452f40b508be8053a926355f773f0a80.jpg\" alt=\"1.jpg\"/>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933593351671809"],"itemList":[{"id":"794933593322311681","questionId":"794933591460040705","content":"20","answer":0,"chooseValue":"A"},{"id":"794933593339088897","questionId":"794933591460040705","content":"21","answer":0,"chooseValue":"B"},{"id":"794933593351671809","questionId":"794933591460040705","content":"22","answer":1,"chooseValue":"C"},{"id":"794933593368449025","questionId":"794933591460040705","content":"23","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933531884146689","title":"<p><span style=\"text-indent: 20.55pt; font-family: New; font-size: 10.5pt;\"><span style=\"font-family:宋体\">袋子里有</span></span><span style=\"text-indent: 20.55pt; font-family: 宋体; font-size: 10.5pt;\">5</span><span style=\"text-indent: 20.55pt; font-family: New; font-size: 10.5pt;\">0<span style=\"font-family:宋体\">个乒乓球，其中</span><span style=\"font-family:New\">20</span><span style=\"font-family:宋体\">个黄球，</span><span style=\"font-family:New\">30</span><span style=\"font-family:宋体\">个白球。现在两个人依次不放回地从袋子中取出一个球，第二个人取出黄球的概率是</span></span><span style=\"text-decoration:underline;\"><span style=\"font-family: New; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（ </span></span></span><span style=\"text-decoration:underline;\"><span style=\"font-family: New; font-size: 10.5pt;\"><span style=\"font-family:宋体\">）</span></span></span><span style=\"text-indent: 20.55pt; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">。</span></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"text-indent: 20.55pt;\"><span style=\"font-family: New; font-size: 10.5pt;\"><span style=\"font-family:宋体\">第二个人取出黄球的概率是</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">2/5*19/49+3/5*20/49=2/5<span style=\"font-family:宋体\">。</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 20.7pt; text-indent: 0pt;\"><span style=\"font-family: Calibri; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:黑体\">参考答案</span></span><span style=\"font-family: Calibri; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"> C </span></p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933533792555009"],"itemList":[{"id":"794933533750611969","questionId":"794933531884146689","content":"1/5","answer":0,"chooseValue":"A"},{"id":"794933533767389185","questionId":"794933531884146689","content":"3/5","answer":0,"chooseValue":"B"},{"id":"794933533792555009","questionId":"794933531884146689","content":"2/5","answer":1,"chooseValue":"C"},{"id":"794933533813526529","questionId":"794933531884146689","content":"4/5","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933507443937281","title":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"> </span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">设甲乙丙三人独立解决某个问题的概率分别为</span>0.45、0.55、0.6，则三人一起解决该问题的概率约为</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>53）</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">。</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 21pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">根据题意，三人一起无法解决该问题的概率为（</span>1-0.45) x (1-0.55) x (1-0.6)=0.099。所以，三人一起能解决该问题的概率为1-0.099=0.901。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">另一种解题思路是：甲解决了该问题的</span>0.45部分，余下0.55部分没有解决。此时，乙能解决其中的0.55部分，即乙能解决总体的0.55x0.55=0.3025部分。甲乙共解决了45+0.3025=0.7525部分，余下0.2475部分没有解决。丙在其中解决了0.6,即丙解决了总体的0.2475x0.6=0.1485部分。甲乙丙三人共解决了问题0.7525+0.1485=0.901部分。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"font-family: 宋体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:宋体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">53）D </span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933509780164609"],"itemList":[{"id":"794933509717250049","questionId":"794933507443937281","content":"0.53","answer":0,"chooseValue":"A"},{"id":"794933509738221569","questionId":"794933507443937281","content":"0.7","answer":0,"chooseValue":"B"},{"id":"794933509759193089","questionId":"794933507443937281","content":"0.8","answer":0,"chooseValue":"C"},{"id":"794933509780164609","questionId":"794933507443937281","content":"0.9","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933632434196481","title":"<p>X、Y、Z是某企业的三个分厂，每个分厂每天需要同一种原料20吨，下图给出了邻近供应厂A、B、C的供应运输路线图，每一段路线上标明了每天最多能运输这种原料的吨数。根据该图可以算出，从A、B、C三厂每天最多能给该企业运来这种原料共 （65） 吨。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/95dc9b70674b019ec1c9e722dd70af36.jpg?x-oss-process=style/ruankaodaren\" title=\"95dc9b70674b019ec1c9e722dd70af36.jpg\" alt=\"1111.jpg\"/></p>","analyze":"<p>①u3000AMX线路，最大流量为10，随后AM=15-10=5，MX=10-10=0，将MX标记为断开。</p><p>②u3000AMY线路，最大流量为5，随后AM=5-5=0，MY=10-5=5，将AM标记为断开。</p><p>③u3000BMY线路，最大流量为5，随后BM=10-5=5，MY=5-5=0，将MY标记为断开。</p><p>④u3000BNY线路，最大流量为10，随后BN=10-10=0，NY=30-10=20，将BN标记为断开。</p><p>⑤u3000CNY线路，最大流量为20，随后CN=20-20=0，NY=20-20=0，将CN、NY标记为断开。</p><p>综上可得，所有线路的最大输气能力可达10+5+5+10+20=50。</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933634313244673"],"itemList":[{"id":"794933634292273153","questionId":"794933632434196481","content":"45","answer":0,"chooseValue":"A"},{"id":"794933634313244673","questionId":"794933632434196481","content":"50","answer":1,"chooseValue":"B"},{"id":"794933634330021889","questionId":"794933632434196481","content":"55","answer":0,"chooseValue":"C"},{"id":"794933634346799105","questionId":"794933632434196481","content":"60","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933559671410689","title":"<p>S公司开发一套信息管理软件，其中一个核心模块的性能对整个系统的市场销售前景影响极大，该模块可以采用S公司自己研发、采购代销和有条件购买三种方式实现。S公司的可能利润（单位万元）收入如下表。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/7761901c50cfff59adf03976df291687.jpg?x-oss-process=style/ruankaodaren\" title=\"7761901c50cfff59adf03976df291687.jpg\" alt=\"1.jpg\"/></p><p>按经验，此类管理软件销售50万套，20万套，5万套和销售不出的概率分别为15%，25%，40%和20%，则S公司应选择 （ ） 方案。</p>","analyze":"<p>自己研发的可能利润值为：</p><p>45000015%+200000250%-5000040%-15000020%=67500 </p><p>采购代销的可能利润值为：</p><p>6500015%+6500025%+6500040%+6500020%=65000 </p><p>有条件购买的可能利润值为：25000015%+10000025%=62500 </p><p>因此，S公司应选择A方案以获得最高可能利润。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933561550458881"],"itemList":[{"id":"794933561550458881","questionId":"794933559671410689","content":"自己研发","answer":1,"chooseValue":"A"},{"id":"794933561567236097","questionId":"794933559671410689","content":"采购代销","answer":0,"chooseValue":"B"},{"id":"794933561579819009","questionId":"794933559671410689","content":"有条件购买","answer":0,"chooseValue":"C"},{"id":"794933561596596225","questionId":"794933559671410689","content":"条件不足无法选择","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933551098253313","title":"<p>在如下线性约束条件下：2x+3y<=30；x+2y>=10；x>=y；x>=5；y>=0，目标函数2x+3y的极小值为（ ）。</p>","analyze":"<p>根据题意，画出可行区域，如图虚线阴影部分。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/4b48cf41cddd71bd6669e6282988b3b4.jpg?x-oss-process=style/ruankaodaren\" title=\"4b48cf41cddd71bd6669e6282988b3b4.jpg\" alt=\"1.jpg\"/></p><p>显然，x=5与x+2y=10相交处时有最小值，联立得x=5，y=2.5，因此2x+3y最小值为10+7.5=17.5。</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933552977301505"],"itemList":[{"id":"794933552956329985","questionId":"794933551098253313","content":"16.5","answer":0,"chooseValue":"A"},{"id":"794933552977301505","questionId":"794933551098253313","content":"17.5","answer":1,"chooseValue":"B"},{"id":"794933552998273025","questionId":"794933551098253313","content":"20","answer":0,"chooseValue":"C"},{"id":"794933553019244545","questionId":"794933551098253313","content":"25","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933542592204801","title":"<p><br/></p><p><br/></p><p><br/></p><p>设三个煤场 A1、A2、A3 分别能供应煤 7、12、11 万吨，三个工厂 B1、 B2、B3 分别需要煤 10、10、10 万吨，从各煤场到各工厂运煤的单价（百元／吨）见下表方框内的数字。只要选择最优的运输方案，总的运输成本就能降到（ ）百万元。</p><br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/992fbdd613cf4939986151ce9dfc7245.jpg?x-oss-process=style/ruankaodaren\" title=\"992fbdd613cf4939986151ce9dfc7245.jpg\" alt=\"1.jpg\"/>","analyze":"<p>优先考虑单位运价最小的供应业务，最大限度满足其供应量。可供应的煤量用完的煤场或者需求已经全部满足的工厂，将不再考虑。然后继续按照上述方法安排运输方案，直到得到一个运费最小的方案。结果为：2*10+10+7+3=40（百万元）</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933544488030209"],"itemList":[{"id":"794933544467058689","questionId":"794933542592204801","content":"30","answer":0,"chooseValue":"A"},{"id":"794933544488030209","questionId":"794933542592204801","content":"40","answer":1,"chooseValue":"B"},{"id":"794933544504807425","questionId":"794933542592204801","content":"50","answer":0,"chooseValue":"C"},{"id":"794933544521584641","questionId":"794933542592204801","content":"61","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933504138825729","title":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">某部门邀请</span>3位专家对12个项目进行评选，每个专家选了5个项目。评选的结果中，有a个项目被3人都选中，有b个项目被2个选中，有c个项目被1人选中，有2个项目无人选中。据此，可以推断</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>52）</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">。</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt;\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">试题分析</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"text-indent: 20.7pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">根据题意：</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"text-indent: 20.7pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">a+b+c=12-2=10①</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"text-indent: 20.7pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">3a+2b+c=3X5=15②</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"text-indent: 20.7pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">由</span>2①- ②可得c-a=5。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">a=0时，c=5，b=5，c=a+b；</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">a=1时，c=6，b=3，c＞a+b；</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">a=2时，c=7，b=1，c＞a+b；</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">a＞2时，c＞7，a+c至少为11，与a+b+c=10矛盾。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">根据上述情况，可以推断供选答案</span>D是正确的。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"font-family: 宋体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:宋体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">52）D </span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933506407944193"],"itemList":[{"id":"794933506357612545","questionId":"794933504138825729","content":"a>2","answer":0,"chooseValue":"A"},{"id":"794933506374389761","questionId":"794933504138825729","content":"b>5","answer":0,"chooseValue":"B"},{"id":"794933506391166977","questionId":"794933504138825729","content":"b为偶数","answer":0,"chooseValue":"C"},{"id":"794933506407944193","questionId":"794933504138825729","content":"c≥a+b","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933487462273025","title":"<p><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">某公司拟将</span>5百万元资金投放下属A、B、C三个子公司（以百万元的倍数分配投资），各子公司获得部分投资后的收益如下表所示（以百万元为单位）。该公司投资的总收益至多为</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>56）</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">百万元。</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/149cb38b3fe49266dbedd9bd3ca04c3b.png?x-oss-process=style/ruankaodaren\"/></span></span></p>","analyze":"<p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/1d67b5d0b12084c9535c2354112ddc6b.png?x-oss-process=style/ruankaodaren\"/></p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933489886580737"],"itemList":[{"id":"794933489836249089","questionId":"794933487462273025","content":"4.8","answer":0,"chooseValue":"A"},{"id":"794933489853026305","questionId":"794933487462273025","content":"5","answer":0,"chooseValue":"B"},{"id":"794933489869803521","questionId":"794933487462273025","content":"5.2","answer":0,"chooseValue":"C"},{"id":"794933489886580737","questionId":"794933487462273025","content":"5.5","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933500850491393","title":"<p><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">某风险投资公司拥有的总资金数为</span>25，分期为项目P1、P2、P3、P4投资，各项目投资情况如下表所示。</span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">若</span>P1和P3分别申请资金数1和2，则公司资金管理处</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>51）</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">。</span></span><br/></p><p><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/9d459ffac1cd305fa6b0214dc8cc677d.png?x-oss-process=style/ruankaodaren\"/></span></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 20.7pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">因为在图</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">中</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">的情况下，公司资金管理处为</span>P3分配资金2个单位后，能保证项目P3得到所需的最大资金完成项目，归还资金，使得公司的可用资金为8, 而项目P1、P2、P4的尚需资金分别为4、7、6,均小于可用资金，故为项目P3进行投资，投资后公司资金周转状态是安全的。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"font-family: 宋体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:宋体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"> （51）B</span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933503077666817"],"itemList":[{"id":"794933503060889601","questionId":"794933500850491393","content":"只能先为项目P1进行投资，因为投资后公司资金周转状态是安全的","answer":0,"chooseValue":"A"},{"id":"794933503077666817","questionId":"794933500850491393","content":"只能先为项目P3进行投资，因为投资后公司资金周转状态是安全的","answer":1,"chooseValue":"B"},{"id":"794933503094444033","questionId":"794933500850491393","content":"可以同时为项目P1、P3进行投资，因为投资后公司资金周转状态是安全的","answer":0,"chooseValue":"C"},{"id":"794933503111221249","questionId":"794933500850491393","content":"不能先为项目P3进行投资，因为投资后公司资金周转状态是不安全的","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933565325332481","title":"<p>某IT企业计划对一批新招聘的技术人员进行岗前脱产培训，培训内容包括编程和测试两个专业，每个专业要求在基础知识、应用技术和实际训练三个方面都得到提高。根据培训大纲，每周的编程培训可同时获得基础知识3学分、应用技术7学分以及实际训练10学分；每周的测试培训可同时获得基础知识5学分、应用技术2学分以及实际训练7学分。企业要求这次岗前培训至少能完成基础知识70学分，应用技术86学分，实际训练185学分。以上说明如下表所示。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/0e819c1cb1798195e4c9c8544e039bd4.jpg?x-oss-process=style/ruankaodaren\" title=\"0e819c1cb1798195e4c9c8544e039bd4.jpg\" alt=\"1.jpg\"/>那么这样的岗前培训至少需要 （ ） 周时间才能满足企业的要求。</p>","analyze":"<p>设编程需要x周，测试需要y周，则下列方程式成立。</p><p>基础知识：3x+ 5y= 70</p><p>应用技术：7x+ 2y=86</p><p>实际训练：10x+ 7y=185</p><p>解上述方程式，得出的x=15、y=5之和20就是需要的值。</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933567229546497"],"itemList":[{"id":"794933567191797761","questionId":"794933565325332481","content":"15","answer":0,"chooseValue":"A"},{"id":"794933567212769281","questionId":"794933565325332481","content":"18","answer":0,"chooseValue":"B"},{"id":"794933567229546497","questionId":"794933565325332481","content":"20","answer":1,"chooseValue":"C"},{"id":"794933567246323713","questionId":"794933565325332481","content":"23","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933497562157057","title":"<p><br/></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">某风险投资公司拥有的总资金数为</span>25，分期为项目P1、P2、P3、P4投资，各项目投资情况如下表所示。公司的可用资金数为</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>50）</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">。</span></span></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/0d02b7d1685c470f854d807a6d125225.png?x-oss-process=style/ruankaodaren\"/></span></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 21pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">企业的总资金数是</span>25,企业资金管理处为项目P1、P2、P3、P4已投资的资金总数=5+5+6+7=23,故可用资金数为</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">25-23=</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">2。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 21pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\"><span style=\"font-family: 宋体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:宋体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">50）C</span></p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933499806109697"],"itemList":[{"id":"794933499768360961","questionId":"794933497562157057","content":"0","answer":0,"chooseValue":"A"},{"id":"794933499785138177","questionId":"794933497562157057","content":"1","answer":0,"chooseValue":"B"},{"id":"794933499806109697","questionId":"794933497562157057","content":"2","answer":1,"chooseValue":"C"},{"id":"794933499822886913","questionId":"794933497562157057","content":"3","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933618005790721","title":"<p>各种线性规划模型都可以将其标准化。线性规划模型标准形式的特点不包括 （67） 。</p>","analyze":"<p>线性规划模型标准形式有如下四个特点：</p><p>1）目标函数是求最大值或最小值</p><p>2）约束条件都是线性等式</p><p>3）决策变量均为非负</p><p>4）约束条件等式的右端常数是非负数</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933619939364865"],"itemList":[{"id":"794933619910004737","questionId":"794933618005790721","content":"目标函数达到最大化（或最小化）","answer":0,"chooseValue":"A"},{"id":"794933619926781953","questionId":"794933618005790721","content":"约束条件都是线性等式","answer":0,"chooseValue":"B"},{"id":"794933619939364865","questionId":"794933618005790721","content":"约束条件等式的右端常数是0","answer":1,"chooseValue":"C"},{"id":"794933619956142081","questionId":"794933618005790721","content":"所有的决策变量均为非负","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933480822689793","title":"<p><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">某石油管理公司拥有下图所示的输油管道网。其中有</span>6个站点，标记为①～⑥。站点①是唯一的供油站。各站点之间的箭线表示输油管道和流向。箭线边上标注的数字表示该管道的最大流量（单位：百吨／小时）。据此可算出，从站点①到达站点⑥的最大流量为</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>54）</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">百吨</span>/小时</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/774507db52a9da9aa0daef3ce7d164e5.png?x-oss-process=style/ruankaodaren\"/></span></p>","analyze":"<p><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">试题解析：</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">本题考查应用数学基础知识。</span><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"text-indent: 21pt; line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">从站点</span>①到⑥有多条线路。显然，每条线路上的最大流量等于该线路上各段管道最大流量的最小值。站点①到⑥的最大总流量等于所有线路最大流量之和。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">我们可以先从流量较大的线路开始计算。例如，线路</span>①②④⑥的最大流量为min(10，5，11)=5。线路①③⑤⑥的最大流量为min(6，8，7)=6。除去这两条线路的流量后，剩余流量的图示如下：</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/5dfd054bcbeed37583520d95a0115d1d.png?x-oss-process=style/ruankaodaren\"/></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">根据此图，线路</span>①②③④⑥的最大流量为min(5，4，5，6)=4。除去该线路上的流量后得:</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/9fad8b896f1b363cad3a8c8ce0de4852.png?x-oss-process=style/ruankaodaren\"/></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">根据此图，线路</span>①②⑤⑥的最大流量为min(1，3，1)=1。除去该线路上的流量后，从①到⑥已不连通，也就不再有剩余流量。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">汇总后，最大总流量可以达到</span>5+6+4+1=16(百吨/小时</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">)</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">。上述实现最大流量的方法是：</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/55468330799ada72517508dff05b9bfb.png?x-oss-process=style/ruankaodaren\"/><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><br/></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">该图中，各管道的实际流量都不超过其最大流量。除起点和终点外，所有站点的进油量等于其出油量。</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">虽然解答此题可以有多种选择线路的方案，但计算得到的最大总流量值都是</span>-致的。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">由于上述解题过程中，管道</span>⑤一④尚未用到，因此，该管道的关闭并不会影响最大总流量值。其他路段管道的关闭是否会影响总流量值呢？</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">为了保持总流量值为</span>16，从①出发的两段管道必须满负苟运输。管道①一②的流量10被分散到②一③、②一④、②一⑤三条管道，关闭其中任何一条管道都将达不到流量为10。同时，管道②一③的流量至少为2。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">同样，为保持最大总流量，管道</span>①一③的流量为6，管道③一⑤显然不能关闭。假设管道③一④关闭，则管道④一⑥的流量至多为8，到达站点⑥的流量至多为15。所以为保持最大总流量，管道③一④不能关闭。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">为保持到达站点</span>⑥的总流量为16，显然管道④一⑥和⑤一⑥任何一个都不能关闭。从而，只有管道⑤一④的关闭对最大总流量没有影响。</span></p><p class=\"MsoNormal\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">参考答案：（</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">54）</span><span style=\"font-family: 宋体; font-size: 10.5pt;\">C</span></p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933483096002561"],"itemList":[{"id":"794933483062448129","questionId":"794933480822689793","content":"14","answer":0,"chooseValue":"A"},{"id":"794933483079225345","questionId":"794933480822689793","content":"15","answer":0,"chooseValue":"B"},{"id":"794933483096002561","questionId":"794933480822689793","content":"16","answer":1,"chooseValue":"C"},{"id":"794933483116974081","questionId":"794933480822689793","content":"18","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933588582748161","title":"张先生向商店订购某一商品，每件定价100元，共订购60件，张先生对商店经理说：“如果你肯减价，每减价1元，我就多订购3件”，商店经理算了一下，如果减价4%，由于张先生多订购，仍可获得原来一样多的总利润。请问这件商品的成本是 （63） 元。","analyze":"<p>设商品的成本为x，张先生总共订购60件，则不降价前的总利润是60(100-x)，降价4%，也就是降到96元后的总利润是72(96-x)，两种情况下的总利润相等，则可以得到以下算式：</p><p>60(100-x)=72(96-x)，x=76</p><p>故：商品的成本价为76元。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933590457602049"],"itemList":[{"id":"794933590457602049","questionId":"794933588582748161","content":"76","answer":1,"chooseValue":"A"},{"id":"794933590491156481","questionId":"794933588582748161","content":"80","answer":0,"chooseValue":"B"},{"id":"794933590520516609","questionId":"794933588582748161","content":"75","answer":0,"chooseValue":"C"},{"id":"794933590549876737","questionId":"794933588582748161","content":"85","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933597172682753","title":"某家具厂有方木材90 m3，木工板600 m3，生产书桌和书柜所用材料数量及利润如下表所示。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/d54ea4d2262679f881db4d311d224e3b.jpg?x-oss-process=style/ruankaodaren\" title=\"d54ea4d2262679f881db4d311d224e3b.jpg\" alt=\"1.jpg\"/>在生产计划最优化的情况下，最大利润为 （64） 元。","analyze":"<p>设书桌x张，书柜y张，则</p><p>0.1x+0.2y90，即x+2y900</p><p>2x+y600</p><p>求MAX（80x+120y）</p><p>显然，x=100，y=400时可以求得最大值。</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933599106256897"],"itemList":[{"id":"794933599039148033","questionId":"794933597172682753","content":"54000","answer":0,"chooseValue":"A"},{"id":"794933599072702465","questionId":"794933597172682753","content":"55000","answer":0,"chooseValue":"B"},{"id":"794933599106256897","questionId":"794933597172682753","content":"56000","answer":1,"chooseValue":"C"},{"id":"794933599135617025","questionId":"794933597172682753","content":"58000","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933580714233857","title":"张先生向商店订购某一商品，每件定价100元，共订购60件，张先生对商店经理说：“如果你肯减价，每减价1元，我就多订购3件”，商店经理算了一下，如果减价4%，由于张先生多订购，仍可获得原来一样多的总利润。请问这件商品的成本是 （） 元。","analyze":"<p>设商品的成本为x，张先生总共订购60件，则不降价前的总利润是60(100-x)，降价4%，也就是降到96元后的总利润是72(96-x)，两种情况下的总利润相等，则可以得到以下算式：</p><p>60(100-x)=72(96-x)</p><p>x=76</p><p>故：商品的成本价为76元。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933582593282049"],"itemList":[{"id":"794933582593282049","questionId":"794933580714233857","content":"76","answer":1,"chooseValue":"A"},{"id":"794933582631030785","questionId":"794933580714233857","content":"80","answer":0,"chooseValue":"B"},{"id":"794933582660390913","questionId":"794933580714233857","content":"75","answer":0,"chooseValue":"C"},{"id":"794933582685556737","questionId":"794933580714233857","content":"85","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933490922573825","title":"<p><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">已知</span>17个自然数（可有重复）的最小值是30，平均值是34，中位数是35，所有各数到38的距离之和比到35的距离之和多5，由此可以推断，这17个数中只有1个</span><span style=\"text-decoration:underline;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>57）</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">。</span></span><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p>","analyze":"<p><span style=\"font-family:宋体\">由于这</span><span style=\"font-family: 宋体; font-size: 10.5pt;\">17个数的中位数是35，所以肯定其中有1个数就是35，左边8个数小于或等于35，右边8个数大于或等于35。以所有各数到35的距离之和为基础，考察各数到38的距离之和的变化。左边和中间共9个数，每个数到38的距离都比到35的距离增加3，共增加27。因此，右边8个数，从离35转到离38的距离之和，应减少27-5=22。</span><br/></p><p class=\"MsoNormal\" style=\"text-indent: 21pt; line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">设右边</span>8个数中，有x个35，y个36，z个37，w个38或38以上。而35、36、37、38以上，对35和38的距离变化分别是+3、+1、-1、-3。所以应该有：</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">3x+y-z-3w=-22，x+y+z+w=8，x、y、z、w都是0〜8之间的整数。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">两式相加得</span>2w-x+z=15，再减前式得w-2x-y=7。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">w只能为7(若w=8，则x=y=zK)，上式不成立).，从而x=y=0，z=l。即17个数中，只有1个37(至此已完成本题解答)，没有36，中位数35的右边没有重复的35。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"text-indent: 21pt; line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">中位数</span>35以及右边的8个数(1个37，7个至少38)到34的距离之和至少为32。由于这17个数的平均值为34，因此，小于34的各数与34的距离之和也应该不少于32(如果左边8数中含有35，则该和数还应该更多)。由于17个数的最小值为30，它与34的距离为4，因此中位数左边8个数必须都是30。也就是说，17个数中，35也只有1个，并没有34，而30则有8个。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">由于中位数左边</span>8个数30与34的距离之和恰好等于32，因此35以及右边8个数与34的距离之和也必须正好等于32。因此35右边除了1个37外，其他只能是7个38。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"text-indent: 21pt; line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">这样就推断出，这</span>17个数只能是：8个30，1个35，1个37，7个38。</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">参考答案：（</span></span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">57）</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\">D</span><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933493212663809"],"itemList":[{"id":"794933493166526465","questionId":"794933490922573825","content":"30","answer":0,"chooseValue":"A"},{"id":"794933493183303681","questionId":"794933490922573825","content":"34","answer":0,"chooseValue":"B"},{"id":"794933493195886593","questionId":"794933490922573825","content":"36","answer":0,"chooseValue":"C"},{"id":"794933493212663809","questionId":"794933490922573825","content":"37","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933526171504641","title":"<p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/fea1e41b9adc05d961e577b4b26e9d30.png?x-oss-process=style/ruankaodaren\"/></p>","analyze":"<p><span style=\"font-family: 宋体; background-color: #FFFFFF; font-size: 10.5pt; text-indent: 20.7pt;\">穷举法</span><br/></p><p class=\"MsoNormal\" style=\"text-indent:20.7000pt;\"><span style=\"font-family: Calibri; font-size: 10.5pt; background: #FFFFFF;\"><span style=\"font-family:宋体\">从</span>A<span style=\"font-family:宋体\">到</span><span style=\"font-family:Calibri\">B </span><span style=\"font-family:宋体\">挨个数，找最小。</span></span><span style=\"font-family: Calibri; font-size: 10.5pt; background: #FFFFFF;\"></span></p><p class=\"MsoNormal\" style=\"text-indent:20.7000pt;\"><span style=\"font-family: 黑体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:黑体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span>B</span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933528042164225"],"itemList":[{"id":"794933528021192705","questionId":"794933526171504641","content":"4","answer":0,"chooseValue":"A"},{"id":"794933528042164225","questionId":"794933526171504641","content":"5","answer":1,"chooseValue":"B"},{"id":"794933528063135745","questionId":"794933526171504641","content":"7","answer":0,"chooseValue":"C"},{"id":"794933528079912961","questionId":"794933526171504641","content":"6","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933629569486849","title":"<p>某家具厂有方木材90m³，木工板600m³，生产书桌和书柜所用材料数量及利润如下表所示。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/ce009fbf91a140de5836ab93873d8cd3.png?x-oss-process=style/ruankaodaren\" title=\"ce009fbf91a140de5836ab93873d8cd3.png\" alt=\"image.png\" width=\"474\" height=\"115\"/><br/></p><p></p><p>在生产计划最优化的情况下，最大利润为 （） 元。</p><br/>","analyze":"<p>设书桌x张，书柜y张，则</p><p>0.1X+0.2y≤90，即x+2y≤900</p><p>2x+y≤600</p><p>求MAX（80x+120y）</p><p>显然，x=100，y=400时可以求得最大值。</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933631473700865"],"itemList":[{"id":"794933631435952129","questionId":"794933629569486849","content":"54000","answer":0,"chooseValue":"A"},{"id":"794933631456923649","questionId":"794933629569486849","content":"55000","answer":0,"chooseValue":"B"},{"id":"794933631473700865","questionId":"794933629569486849","content":"56000","answer":1,"chooseValue":"C"},{"id":"794933631490478081","questionId":"794933629569486849","content":"58000","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933484144578561","title":"<p><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">某石油管理公司拥有下图所示的输油管道网。其中有</span>6个站点，标记为①～⑥。站点①是唯一的供油站。各站点之间的箭线表示输油管道和流向。箭线边上标注的数字表示该管道的最大流量（单位：百吨／小时）。据此可算出，</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\">当管道</span><span style=\"text-decoration: underline;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">（</span>55）</span></span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">关闭维修时管道网仍可按该最大流量值向站点</span>⑥供油。<br/><img style=\"max-width:100%;height:auto\" title=\"6cf866b7428982452824621a1863075f.png\" alt=\"3131.png\" src=\"https://image.chaiding.com/ruankao/6cf866b7428982452824621a1863075f.png?x-oss-process=style/ruankaodaren\"/></span><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">试题解析：</span></span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">本题考查应用数学基础知识。</span></span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%; text-indent: 21pt;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">从站点</span>①到⑥有多条线路。显然，每条线路上的最大流量等于该线路上各段管道最大流量的最小值。站点①到⑥的最大总流量等于所有线路最大流量之和。</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">我们可以先从流量较大的线路开始计算。例如，线路</span>①②④⑥的最大流量为min(10，5，11)=5。线路①③⑤⑥的最大流量为min(6，8，7)=6。除去这两条线路的流量后，剩余流量的图示如下：</span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/c9bb1dcc1e8987d959b2b55db7d25db1.png?x-oss-process=style/ruankaodaren\"/></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">根据此图，线路</span>①②③④⑥的最大流量为mm(5，4，5，6)=4。除去该线路上的流量后得:</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"text-align: center; line-height: 150%;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/34a19900e3ddc04011b7e83aacac9523.png?x-oss-process=style/ruankaodaren\"/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">根据此图，线路</span>①②⑤⑥的最大流量为min(1，3，1)=1。除去该线路上的流量后，从①到⑥已不连通，也就不再有剩余流量。</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">汇总后，最大总流量可以达到</span>5+6+4+1=16(百吨/小时</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\">)</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">。上述实现最大流量的方法是：</span></span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"text-align: center; line-height: 150%;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/113d06fa50a73849ec634ac80daf3c5c.png?x-oss-process=style/ruankaodaren\"/></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">该图中，各管道的实际流量都不超过其最大流量。除起点和终点外，所有站点的进油量等于其出油量。</span></span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">虽然解答此题可以有多种选择线路的方案，但计算得到的最大总流量值都是</span>-致的。</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">由于上述解题过程中，管道</span>⑤一④尚未用到，因此，该管道的关闭并不会影响最大总流量值。其他路段管道的关闭是否会影响总流量值呢？</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">为了保持总流量值为</span>16，从①出发的两段管道必须满负苟运输。管道①一②的流量10被分散到②一③、②一④、②一⑤三条管道，关闭其中任何一条管道都将达不到流量为10。同时，管道②一③的流量至少为2。</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">同样，为保持最大总流量，管道</span>①一③的流量为6，管道③一⑤显然不能关闭。假设管道③一④关闭，则管道④一⑥的流量至多为8，到达站点⑥的流量至多为15。所以为保持最大总流量，管道③一④不能关闭。</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">为保持到达站点</span>⑥的总流量为16，显然管道④一⑥和⑤一⑥任何一个都不能关闭。从而，只有管道⑤一④的关闭对最大总流量没有影响。</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\" style=\"line-height: 150%;\"><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family: 宋体;\">参考答案：</span></span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"> <span style=\"font-family: 宋体;\">（</span>55）</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"> D</span><span style=\"line-height: 150%; font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933486489194497"],"itemList":[{"id":"794933486401114113","questionId":"794933484144578561","content":"②→③","answer":0,"chooseValue":"A"},{"id":"794933486430474241","questionId":"794933484144578561","content":"②→⑤","answer":0,"chooseValue":"B"},{"id":"794933486459834369","questionId":"794933484144578561","content":"③→④","answer":0,"chooseValue":"C"},{"id":"794933486489194497","questionId":"794933484144578561","content":"⑤→④","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933583599915009","title":"<p>载重量限24吨的某架货运飞机执行将一批金属原料运往某地的任务。待运输的各箱原料的重量、运输利润如下表所示。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/d8986139e2c48cccb3e1f514c0d8499b.jpg?x-oss-process=style/ruankaodaren\" title=\"d8986139e2c48cccb3e1f514c0d8499b.jpg\" alt=\"1.jpg\"/></p><p>经优化安排，该飞机本次运输可以获得的最大利润为（）千元。</p>","analyze":"在重量有限制的条件下，为取得最大的利润，显然应优先选择装载“利润重量比”大的货物。先列出每箱货物的利润/重量比如下表所示。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/23716742f4c1ea9bea120bbc6a350772.jpg?x-oss-process=style/ruankaodaren\" title=\"23716742f4c1ea9bea120bbc6a350772.jpg\" alt=\"1.jpg\"/><p>根据利润重量比优先原则，应先装第4箱、第6箱货物。重量已达到16吨，离最大载重量还差8吨，只能再装第1箱，或第3箱，或第5箱。为取得最大利润，再装第1箱更好。</p><p>所以最优方案是装运箱号为1、4、6的三箱，总利润为3+4+3=10千元。</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933585470574593"],"itemList":[{"id":"794933585453797377","questionId":"794933583599915009","content":"11","answer":0,"chooseValue":"A"},{"id":"794933585470574593","questionId":"794933583599915009","content":"10","answer":1,"chooseValue":"B"},{"id":"794933585491546113","questionId":"794933583599915009","content":"9","answer":0,"chooseValue":"C"},{"id":"794933585512517633","questionId":"794933583599915009","content":"8","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933562515148801","title":"<p>某厂需要购买生产设备生产某种产品，可以选择购买四种生产能力不同的设备，市场对该产品的需求状况有三种（需求量较大、需求量中等、需求量较小）。厂方估计四种设备在各种需求状况下的收益由下表给出，根据收益期望值最大的原则，应该购买 （ ） 。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/49a11026db2dc1b0f668b35c65efed38.jpg?x-oss-process=style/ruankaodaren\" title=\"49a11026db2dc1b0f668b35c65efed38.jpg\" alt=\"1.jpg\"/></p>","analyze":"<p>设备1收益期望值为：50*0.3+20*0.4+（-20）*0.3=17</p><p>设备2收益期望值为：30*0.3+25*0.4+（-10）*0.3=16</p><p>设备3收益期望值为：25*0.3+30*0.4+（-5）*0.3=18</p><p>设备4收益期望值为：10*0.3+10*0.4+10*0.3=10</p><p>因此，根据收益期望值最大的原则，应该购买设备3。</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933564398391297"],"itemList":[{"id":"794933564369031169","questionId":"794933562515148801","content":"设备1","answer":0,"chooseValue":"A"},{"id":"794933564381614081","questionId":"794933562515148801","content":"设备2","answer":0,"chooseValue":"B"},{"id":"794933564398391297","questionId":"794933562515148801","content":"设备3","answer":1,"chooseValue":"C"},{"id":"794933564415168513","questionId":"794933562515148801","content":"设备4","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933545435942913","title":"<p>用一辆载重量为 10 吨的卡车装运某仓库中的货物（不用考虑装车时货物的大小），这些货物单件的重量和运输利润如下表。适当选择装运一些货物各若干件，就能获得最大总利润（ ）元。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/1c5566ce2a1ad20faa7aea0fe89d7889.jpg?x-oss-process=style/ruankaodaren\" title=\"1c5566ce2a1ad20faa7aea0fe89d7889.jpg\" alt=\"1.jpg\"/></p>","analyze":"<p>如果装的货均是单位利润最高的那些货物，并且装满10吨，则能获得最大总利润，如下表所示。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/ec3dab6ff2be7a997ac6cab42637b788.jpg?x-oss-process=style/ruankaodaren\" title=\"ec3dab6ff2be7a997ac6cab42637b788.jpg\" alt=\"1.jpg\"/></p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933547356934145"],"itemList":[{"id":"794933547302408193","questionId":"794933545435942913","content":"530","answer":0,"chooseValue":"A"},{"id":"794933547319185409","questionId":"794933545435942913","content":"534","answer":0,"chooseValue":"B"},{"id":"794933547335962625","questionId":"794933545435942913","content":"536","answer":0,"chooseValue":"C"},{"id":"794933547356934145","questionId":"794933545435942913","content":"538","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933528994271233","title":"<p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/c8cc5e7b08a085d2caf55ac7582a7a94.png?x-oss-process=style/ruankaodaren\"/></p>","analyze":"<p><span style=\"font-family:宋体\">设甲生产</span><span style=\"font-family: 宋体; font-size: 10.5pt; text-indent: 0pt;\">X</span><span style=\"font-family:宋体\">套，乙生产</span><span style=\"font-family:Calibri\">Y</span><span style=\"font-family:宋体\">套，则有：</span><span style=\"font-family:Calibri\">2X+3Y</span><span style=\"font-family:宋体\">≤</span><span style=\"font-family:Calibri\">14</span><span style=\"font-family:宋体\">；</span><br/></p><p class=\"MsoNormal\" style=\"margin-left:20.7000pt;text-indent:0.0000pt;mso-char-indent-count:0.0000;\"><span style=\"mso-spacerun:'yes';font-family:宋体;mso-ascii-font-family:Calibri;mso-hansi-font-family:Calibri;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;\"> X<span style=\"font-family:宋体\">≤</span><span style=\"font-family:Calibri\">2</span><span style=\"font-family:宋体\">；</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;mso-ascii-font-family:Calibri;mso-hansi-font-family:Calibri;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left:20.7000pt;text-indent:0.0000pt;mso-char-indent-count:0.0000;\"><span style=\"mso-spacerun:'yes';font-family:宋体;mso-ascii-font-family:Calibri;mso-hansi-font-family:Calibri;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;\"> Y<span style=\"font-family:宋体\">≤</span><span style=\"font-family:Calibri\">4</span><span style=\"font-family:宋体\">；</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;mso-ascii-font-family:Calibri;mso-hansi-font-family:Calibri;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left:20.7000pt;text-indent:0.0000pt;mso-char-indent-count:0.0000;\"><span style=\"mso-spacerun:'yes';font-family:宋体;mso-ascii-font-family:Calibri;mso-hansi-font-family:Calibri;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">同时要满足利润最大，只有</span>X<span style=\"font-family:宋体\">取</span><span style=\"font-family:Calibri\">1</span><span style=\"font-family:宋体\">，</span><span style=\"font-family:Calibri\">Y</span><span style=\"font-family:宋体\">取</span><span style=\"font-family:Calibri\">4</span><span style=\"font-family:宋体\">时利润最大是</span><span style=\"font-family:Calibri\">14</span><span style=\"font-family:宋体\">万元。</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;mso-ascii-font-family:Calibri;mso-hansi-font-family:Calibri;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left:20.7000pt;text-indent:0.0000pt;mso-char-indent-count:0.0000;\"><span style=\"mso-spacerun:'yes';font-family:黑体;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;background:#7F7F7F;mso-shading:#7F7F7F;\"><span style=\"font-family:黑体\">参考答案</span></span><span style=\"mso-spacerun:'yes';font-family:Calibri;mso-fareast-font-family:宋体;mso-bidi-font-family:'Times New Roman';font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"> B</span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933530885902337"],"itemList":[{"id":"794933530864930817","questionId":"794933528994271233","content":"产甲2套，乙3套","answer":0,"chooseValue":"A"},{"id":"794933530885902337","questionId":"794933528994271233","content":"生产甲1套，乙4套","answer":1,"chooseValue":"B"},{"id":"794933530902679553","questionId":"794933528994271233","content":"生产甲3套，乙4套","answer":0,"chooseValue":"C"},{"id":"794933530923651073","questionId":"794933528994271233","content":"生产甲4套，乙2套","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933577866301441","title":"某工程埋设线缆，将中央控制室W与A～F共6个控制点相连通，各控制点的位置及距离(单位：千米)如下图所示。可使所埋设线缆总长最短的方案个数和最短距离分别是（）。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/75664cecfc84e65442a657320979c6c7.jpg?x-oss-process=style/ruankaodaren\" title=\"75664cecfc84e65442a657320979c6c7.jpg\" alt=\"1.jpg\"/>","analyze":"<p>要使如图所示埋设的电缆总长最短，可以有以下两个方案。</p><p>方案1：(WA、WB、WC、WED、EF)，本方案所埋设线缆总长度为18+17+16+16+16+18=101千米。</p><p>方案2：(WA、WCB、WED、EF)，本方案所埋设线缆总长度为18+16+17+16+16+18=101千米。</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933579757932545"],"itemList":[{"id":"794933579741155329","questionId":"794933577866301441","content":"1个，102千米","answer":0,"chooseValue":"A"},{"id":"794933579757932545","questionId":"794933577866301441","content":"2个，101千米","answer":1,"chooseValue":"B"},{"id":"794933579774709761","questionId":"794933577866301441","content":"3个，102千米","answer":0,"chooseValue":"C"},{"id":"794933579791486977","questionId":"794933577866301441","content":"3个，100千米","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933612259594241","title":"<p>某乡8个小村（编号为1~8）之间的距离如表1（单位：km）。1号村离水库最近，为 4km，从水库开始铺设水管将各村连接起来，最少需要铺设 （65） 长的水管（为便于管理和维修，水管分叉必须设在各村处）。<br/></p><p>表1 习题用表</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/e535fa70c07b49b080b53058f98b6521.jpg?x-oss-process=style/ruankaodaren\" title=\"e535fa70c07b49b080b53058f98b6521.jpg\" alt=\"111.jpg\"/></p><br/>","analyze":"<p>此题采用最小生成树法。</p><p>把各村之间的水管从最小的距离开始连接。</p><p>①u30000.5km，把7与8连接起来；</p><p>②u30000.8km，把1与4、6与7连接起来；</p><p>③u30001.0km，把2与3、2与5、3与8、4与8连接起来。而6与8之间如果连接，就会形成闭环，所以不连接6与8。</p><p>如下图所示：</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/328dd808ff795bd0c20cb5ad616dd912.jpg?x-oss-process=style/ruankaodaren\" title=\"328dd808ff795bd0c20cb5ad616dd912.jpg\" alt=\"111.jpg\"/></p><p>至此，全部小村均已连接，并且没有形成闭环，连接完成。水管总长度为：</p><p>4+0.8+1.0+0.8+0.5+1.0+1.0+1.0=10.1</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933614184779777"],"itemList":[{"id":"794933614138642433","questionId":"794933612259594241","content":"6.1km","answer":0,"chooseValue":"A"},{"id":"794933614159613953","questionId":"794933612259594241","content":"11.3km","answer":0,"chooseValue":"B"},{"id":"794933614184779777","questionId":"794933612259594241","content":"10.1km","answer":1,"chooseValue":"C"},{"id":"794933614205751297","questionId":"794933612259594241","content":"16.8km","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933548271292417","title":"<p>某企业准备将4个工人甲、乙、丙、丁分配在A、B、C、D共4个岗位。每个工人由于技术水平不同，在不同岗位上每天完成任务所需的工时见下表。适当安排岗位，可使4个工人以最短的总工时（ ）全部完成每天的任务。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/c6ff273aca9889c4f3fd52305f6997b3.jpg?x-oss-process=style/ruankaodaren\" title=\"c6ff273aca9889c4f3fd52305f6997b3.jpg\" alt=\"1.jpg\"/></p><br/>","analyze":"经分析表中处于左下到右上对角线的位置，4值相加最少，即4+4+3+3=14","multi":0,"questionType":1,"answer":"B","chooseItem":["794933550158729217"],"itemList":[{"id":"794933550141952001","questionId":"794933548271292417","content":"13","answer":0,"chooseValue":"A"},{"id":"794933550158729217","questionId":"794933548271292417","content":"14","answer":1,"chooseValue":"B"},{"id":"794933550171312129","questionId":"794933548271292417","content":"15","answer":0,"chooseValue":"C"},{"id":"794933550183895041","questionId":"794933548271292417","content":"16","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933608736378881","title":"<p>某家具厂有方木材90 m3，木工板600 m3，生产书桌和书柜所用材料数量及利润如下表所示。</p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/e6cc8d2dbd59bb6420e6c70ff4b9d7d6.png?x-oss-process=style/ruankaodaren\" title=\"e6cc8d2dbd59bb6420e6c70ff4b9d7d6.png\" alt=\"image.png\"/><br/><p>在生产计划最优化的情况下，最大利润为 （） 元。</p>","analyze":"<p>设书桌x张，书柜y张，则</p><p>0.1x+0.2y90，即x+2y900</p><p>2x+y600</p><p>求MAX（80x+120y）</p><p>显然，x=100，y=400时可以求得最大值。</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933610984525825"],"itemList":[{"id":"794933610942582785","questionId":"794933608736378881","content":"54 000","answer":0,"chooseValue":"A"},{"id":"794933610963554305","questionId":"794933608736378881","content":"55 000","answer":0,"chooseValue":"B"},{"id":"794933610984525825","questionId":"794933608736378881","content":"56 000","answer":1,"chooseValue":"C"},{"id":"794933611005497345","questionId":"794933608736378881","content":"58 000","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933523340349441","title":"<p><span style=\"font-family:宋体\">同时抛掷</span><span style=\"font-family: 宋体; font-size: 10.5pt;\"> 3 </span><span style=\"font-family:宋体\">枚均匀的硬币，恰好有两枚正面向上的概率为（ ）。</span></p><p class=\"MsoNormal\"><br/></p>","analyze":"<p><br/></p><p class=\"MsoNormal\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">每枚硬币都有正反两面，三枚硬币共有</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">8<span style=\"font-family:宋体\">种情况，</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">两枚正面向上</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">的情况有：正正反、正反正、反正正</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">,3<span style=\"font-family:宋体\">种。</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p><p class=\"MsoNormal\"><span style=\"font-family: 宋体; font-size: 10.5pt;\"> </span><span style=\"font-family: 黑体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:黑体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">B </span></p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933525211009025"],"itemList":[{"id":"794933525194231809","questionId":"794933523340349441","content":"1/4","answer":0,"chooseValue":"A"},{"id":"794933525211009025","questionId":"794933523340349441","content":"3/8","answer":1,"chooseValue":"B"},{"id":"794933525227786241","questionId":"794933523340349441","content":"1/2","answer":0,"chooseValue":"C"},{"id":"794933525244563457","questionId":"794933523340349441","content":"1/3","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933575047729153","title":"某系统集成项目为满足某种软件产品的市场需求，拟提出自主开发、部分研发任务外包和外购3个方案。假设各方案中销路好的概率为0.3，销路一般的概率为0.5，销路差的概率为0.2。不同销路的损益值如下表所示。假设该项目经营期为10年，那么该项目所作的决策最可能是（）。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/d557f0ac85069269c5b94352e95602dc.jpg?x-oss-process=style/ruankaodaren\" title=\"d557f0ac85069269c5b94352e95602dc.jpg\" alt=\"1.jpg\"/>","analyze":"<p>自主开发方案的期望值计算方法为：</p><p>(80×0.3+60×0.5+50×0.2)×10-300=340(万元)。</p><p>部分研发任务外包方案的期望值计算方法为：</p><p>(30×0.3+20×0.5+15×0.2)×10-80=140(万元)。</p><p>外购方案的期望值计算方法为(50×0.3+40×0.5+25×0.2)×10-160=240(万元)。</p><p>由于340＞240＞140，因此从货币期望值最大决策角度考虑，建议该项目选择自主开发方案。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933576901611521"],"itemList":[{"id":"794933576901611521","questionId":"794933575047729153","content":"选择自主开发方案","answer":1,"chooseValue":"A"},{"id":"794933576922583041","questionId":"794933575047729153","content":"选择部分研发任务外包方案","answer":0,"chooseValue":"B"},{"id":"794933576939360257","questionId":"794933575047729153","content":"选择外购方案","answer":0,"chooseValue":"C"},{"id":"794933576960331777","questionId":"794933575047729153","content":"条件不足，无法得出结论","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933510690328577","title":"<p><br/></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">某厂准备生产甲、乙、丙三种产品，生产每件产品所需的</span>A、B两种原料数量，能获得的利润，以及工厂拥有的原料数量如下表：</span></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/963f3cde8ddedde0bb2ba0a4fda8d931.png?x-oss-process=style/ruankaodaren\"/></span></p><p class=\"MsoNormal\"><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span><span style=\"text-indent: 0pt; font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">根据该表，只要安排好生产计划，就能获得最大利润</span></span><span style=\"text-indent: 0pt;text-decoration:underline;\"><span style=\"font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span>54）</span></span><span style=\"text-indent: 0pt; font-family: 宋体; line-height: 150%; font-size: 10.5pt;\"><span style=\"font-family:宋体\">万元。</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><br/></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p>","analyze":"<p><br/></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">设该厂计划生产甲</span>x件，乙y件，丙z件，则有线性规划模型：</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">Max S=3x+4y+z</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">6x+5y+3z≤45</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">3x+5y+4z≤30</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">xyz</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;color:#FF0000;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">≥</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\">0</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">线性规划问题的最优解必然在可行解区的顶点处达到。</span> </span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">因此，在</span>x=5, y=3, z=0时能获得最大利润27万元。</span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"><span style=\"font-family:宋体\">可以通过坐标图解答。</span></span><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"></span></p><p class=\"MsoNormal\" style=\"margin-left: 0pt; text-indent: 0pt; line-height: 150%;\"><span style=\"mso-spacerun:'yes';font-family:宋体;line-height:150%;font-size:10.5000pt;mso-font-kerning:1.0000pt;\"> </span><span style=\"font-family: 宋体; font-size: 10.5pt; background: #7F7F7F;\"><span style=\"font-family:宋体\">参考答案</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">：</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\"><span style=\"font-family:宋体\">（</span></span><span style=\"font-family: 宋体; font-size: 10.5pt;\">54）C</span><span style=\"font-family: 宋体; font-size: 10.5pt;\"></span></p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933512703594497"],"itemList":[{"id":"794933512661651457","questionId":"794933510690328577","content":"25","answer":0,"chooseValue":"A"},{"id":"794933512682622977","questionId":"794933510690328577","content":"26","answer":0,"chooseValue":"B"},{"id":"794933512703594497","questionId":"794933510690328577","content":"27","answer":1,"chooseValue":"C"},{"id":"794933512720371713","questionId":"794933510690328577","content":"28","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933623747792897","title":"<p><strong>请作答第<span style=\"color: red\">1</span>空。</strong></p><p>某软件项目的活动图如下图所示，其中顶点表示项目里程碑，链接顶点的边表示包含的活动，边上的数字表示活动的持续时间（天）。完成该项目的最少时间为（）天。由于某种原因，现在需要同一个开发人员完成BC和BD，则完成该项目的最少时间为（）天。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/c0189b44e089e482e577e6d06b0eef3a.jpg?x-oss-process=style/ruankaodaren\" title=\"c0189b44e089e482e577e6d06b0eef3a.jpg\" alt=\"1111.jpg\" width=\"566\" height=\"183\"/></p><br/>","analyze":"<p>1、关键路径为ABCEFJ 和ABDGFJ ，18天。</p><p>2、BC持续时间3天，BD持续时间2天，由一人完成，则可以把先完成BD，再完成BC，则BC持续时间作为5天，则关键路径为ABCEFJ， 20天。</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933625735892993"],"itemList":[{"id":"794933625685561345","questionId":"794933623747792897","content":"11","answer":0,"chooseValue":"A"},{"id":"794933625735892993","questionId":"794933623747792897","content":"18","answer":1,"chooseValue":"B"},{"id":"794933625777836033","questionId":"794933623747792897","content":"20","answer":0,"chooseValue":"C"},{"id":"794933625815584769","questionId":"794933623747792897","content":"21","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933600045780993","title":"<p><strong>请作答第<span style=\"color: red\">1</span>空。</strong></p><p>某企业需要采用甲、乙、丙三种原材料生产Ⅰ、Ⅱ两种产品。生产两种产品所需原材料数量、单位产品可获得利润以及企业现有原材料数如表所示：</p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/a56c1f3791b16a86344169b1097154cb.png?x-oss-process=style/ruankaodaren\" title=\"a56c1f3791b16a86344169b1097154cb.png\" alt=\"image.png\"/><br/><p>则公司可以获得的最大利润是（52）万元。取得最大利润时，原材料（53）尚有剩余。</p>","analyze":"<p>设生产的产品 I为 x 吨，产品II 为 y 吨，则： </p><p>1x+1y≦4 </p><p>4x+3y≦12 </p><p>1x+3y≦6 </p><p>解上述方程可知，x=2,y=4/3.因此，最大利润是：9*2+12*4/3=34 原料“甲”还剩余。</p>","multi":0,"questionType":1,"answer":"B","chooseItem":["794933601937412097"],"itemList":[{"id":"794933601916440577","questionId":"794933600045780993","content":"21","answer":0,"chooseValue":"A"},{"id":"794933601937412097","questionId":"794933600045780993","content":"34","answer":1,"chooseValue":"B"},{"id":"794933601954189313","questionId":"794933600045780993","content":"39","answer":0,"chooseValue":"C"},{"id":"794933601966772225","questionId":"794933600045780993","content":"48","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933635261157377","title":"<p><strong>请作答第<span style=\"color: red\">1</span>空。</strong></p><p>某厂拥有三种资源 A．B．C生产甲、乙两种产品。生产每吨产品需要消耗的资源、可以获得的利润见下表。日前，该厂拥有资源 A．资源 B 和资源 C 分別为 12 吨，7 吨和 12 吨。根据上述说明，适当安排甲、乙两种产品的生产量，就能获得最大总利润 （67）百万元 。如果资源 A 不再受限，则该厂总利润为 （68）百万元 。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/4c9580d67cdeeb26aaf603120f5e25ac.jpg?x-oss-process=style/ruankaodaren\" title=\"4c9580d67cdeeb26aaf603120f5e25ac.jpg\" alt=\"1111.jpg\" width=\"553\" height=\"141\"/></p>","analyze":"<p>本题采用图解法。</p><p>设甲产品生产x吨，乙商品生产y吨，根据题意，得到约束条件如下：</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/c3e274a8b7affbeeda4158a4dd485c14.jpg?x-oss-process=style/ruankaodaren\" title=\"c3e274a8b7affbeeda4158a4dd485c14.jpg\" alt=\"1111.jpg\" width=\"172\" height=\"90\"/></p><p>根据约束条件，得到对应的坐标图具体如图1所示。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/50dbdfdc4356c48bb8a291f7267528d9.jpg?x-oss-process=style/ruankaodaren\" title=\"50dbdfdc4356c48bb8a291f7267528d9.jpg\" alt=\"1111.jpg\" width=\"418\" height=\"471\"/></p><p>符合条件的区域是左下角的深色区，深色区的顶点分别为（0,0）, （0,6）, （4,0）, （2.5,4.5）, （2,5）,将这5组数值代入表达式3x+2y。可得顶点（2.5,4.5）得到最大值N=16.5。</p><p>因此，最大利润为16.5百万元。</p><p>如果生产计划只受资源 B 和 C 的约束，顶点（7,0）才是3x+2y的最大值点，如下图。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/e0bce1c146759c1c250cb09fe3391a3a.jpg?x-oss-process=style/ruankaodaren\" title=\"e0bce1c146759c1c250cb09fe3391a3a.jpg\" alt=\"1111.jpg\"/></p><p>此时，该最大值N为21。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933637127622657"],"itemList":[{"id":"794933637127622657","questionId":"794933635261157377","content":"16.5","answer":1,"chooseValue":"A"},{"id":"794933637144399873","questionId":"794933635261157377","content":"17","answer":0,"chooseValue":"B"},{"id":"794933637165371393","questionId":"794933635261157377","content":"19","answer":0,"chooseValue":"C"},{"id":"794933637186342913","questionId":"794933635261157377","content":"20","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933626729943041","title":"<p><strong>请作答第<span style=\"color: red\">2</span>空。</strong></p><p>某软件项目的活动图如下图所示，其中顶点表示项目里程碑，链接顶点的边表示包含的活动，边上的数字表示活动的持续时间（天）。完成该项目的最少时间为（）天。由于某种原因，现在需要同一个开发人员完成BC和BD，则完成该项目的最少时间为（）天。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/c0189b44e089e482e577e6d06b0eef3a.jpg?x-oss-process=style/ruankaodaren\" title=\"c0189b44e089e482e577e6d06b0eef3a.jpg\" alt=\"1111.jpg\" width=\"566\" height=\"183\"/></p><br/>","analyze":"<p>1、关键路径为ABCEFJ 和ABDGFJ ，18天。</p><p>2、BC持续时间3天，BD持续时间2天，由一人完成，则可以把先完成BD，再完成BC，则BC持续时间作为5天，则关键路径为ABCEFJ， 20天。</p>","multi":0,"questionType":1,"answer":"C","chooseItem":["794933628621574145"],"itemList":[{"id":"794933628592214017","questionId":"794933626729943041","content":"18","answer":0,"chooseValue":"A"},{"id":"794933628608991233","questionId":"794933626729943041","content":"11","answer":0,"chooseValue":"B"},{"id":"794933628621574145","questionId":"794933626729943041","content":"20","answer":1,"chooseValue":"C"},{"id":"794933628638351361","questionId":"794933626729943041","content":"21","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933602910490625","title":"<p><strong>请作答第<span style=\"color: red\">2</span>空。</strong></p><p>某企业需要采用甲、乙、丙三种原材料生产Ⅰ、Ⅱ两种产品。生产两种产品所需原材料数量、单位产品可获得利润以及企业现有原材料数如表所示：</p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/a56c1f3791b16a86344169b1097154cb.png?x-oss-process=style/ruankaodaren\" title=\"a56c1f3791b16a86344169b1097154cb.png\" alt=\"image.png\"/><br/><p>则公司可以获得的最大利润是（52）万元。取得最大利润时，原材料（53）尚有剩余。</p>","analyze":"<p>设生产的产品 I为 x 吨，产品II 为 y 吨，则： </p><p>1x+1y≦4 </p><p>4x+3y≦12 </p><p>1x+3y≦6 </p><p>解上述方程可知，x=2,y=4/3.因此，最大利润是：9*2+12*4/3=34 原料“甲”还剩余。</p>","multi":0,"questionType":1,"answer":"A","chooseItem":["794933604781150209"],"itemList":[{"id":"794933604781150209","questionId":"794933602910490625","content":"甲","answer":1,"chooseValue":"A"},{"id":"794933604802121729","questionId":"794933602910490625","content":"乙","answer":0,"chooseValue":"B"},{"id":"794933604823093249","questionId":"794933602910490625","content":"丙","answer":0,"chooseValue":"C"},{"id":"794933604844064769","questionId":"794933602910490625","content":"乙和丙","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"794933638096506881","title":"<p><strong>请作答第<span style=\"color: red\">2</span>空。</strong></p><p>某厂拥有三种资源 A．B．C生产甲、乙两种产品。生产每吨产品需要消耗的资源、可以获得的利润见下表。日前，该厂拥有资源 A．资源 B 和资源 C 分別为 12 吨，7 吨和 12 吨。根据上述说明，适当安排甲、乙两种产品的生产量，就能获得最大总利润 （67）百万元 。如果资源 A 不再受限，则该厂总利润为 （68）百万元 。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/4c9580d67cdeeb26aaf603120f5e25ac.jpg?x-oss-process=style/ruankaodaren\" title=\"4c9580d67cdeeb26aaf603120f5e25ac.jpg\" alt=\"1111.jpg\" width=\"553\" height=\"141\"/></p>","analyze":"<p>本题采用图解法。</p><p>设甲产品生产x吨，乙商品生产y吨，根据题意，得到约束条件如下：</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/c3e274a8b7affbeeda4158a4dd485c14.jpg?x-oss-process=style/ruankaodaren\" title=\"c3e274a8b7affbeeda4158a4dd485c14.jpg\" alt=\"1111.jpg\" width=\"172\" height=\"90\"/></p><p>根据约束条件，得到对应的坐标图具体如图1所示。<br/><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/50dbdfdc4356c48bb8a291f7267528d9.jpg?x-oss-process=style/ruankaodaren\" title=\"50dbdfdc4356c48bb8a291f7267528d9.jpg\" alt=\"1111.jpg\" width=\"418\" height=\"471\"/></p><p>符合条件的区域是左下角的深色区，深色区的顶点分别为（0,0）, （0,6）, （4,0）, （2.5,4.5）, （2,5）,将这5组数值代入表达式3x+2y。可得顶点（2.5,4.5）得到最大值N=16.5。</p><p>因此，最大利润为16.5百万元。</p><p>如果生产计划只受资源 B 和 C 的约束，顶点（7,0）才是3x+2y的最大值点，如下图。</p><p><img style=\"max-width:100%;height:auto\" src=\"https://image.chaiding.com/ruankao/e0bce1c146759c1c250cb09fe3391a3a.jpg?x-oss-process=style/ruankaodaren\" title=\"e0bce1c146759c1c250cb09fe3391a3a.jpg\" alt=\"1111.jpg\"/></p><p>此时，该最大值N为21。</p>","multi":0,"questionType":1,"answer":"D","chooseItem":["794933640000720897"],"itemList":[{"id":"794933639946194945","questionId":"794933638096506881","content":"18","answer":0,"chooseValue":"A"},{"id":"794933639967166465","questionId":"794933638096506881","content":"19","answer":0,"chooseValue":"B"},{"id":"794933639983943681","questionId":"794933638096506881","content":"20","answer":0,"chooseValue":"C"},{"id":"794933640000720897","questionId":"794933638096506881","content":"21","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239464606224385","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某石油管理公司拥有下图所示的输油管道网。其中有6个站点，标记为①～⑥。站点①是唯一的供油站。各站点之间的箭线表示输油管道和流向。箭线边上标注的数字表示该管道的最大流量（单位：百吨／小时）。据此可算出，从站点①到达站点⑥的最大流量为（54）百吨/小时，而且当管道（55）关闭维修时管道网仍可按该最大流量值向站点⑥供油。<br><img alt=\"\" width=\"460\" height=\"134\" src=\"https://image.chaiding.com/ruankao/a6ccb8cf40230e3c1afdc48dbe164df7.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学基础知识。<br>从站点①到⑥有多条线路。显然，每条线路上的最大流量等于该线路上各段管道最大流量的最小值。站点①到⑥的最大总流量等于所有线路最大流量之和。<br>我们可以先从流量较大的线路开始计算。例如，线路①②④⑥的最大流量为min(10，5，11)=5。线路①③⑤⑥的最大流量为min(6，8，7)=6。除去这两条线路的流量后，剩余流量的图示如下：<br><img alt=\"\" width=\"379\" height=\"119\" src=\"https://image.chaiding.com/ruankao/6eac3780bcca019c4022a9cbc3f62070.jpg?x-oss-process=style/ruankaodaren\"><br>根据此图，线路①②③④⑥的最大流量为mm(5，4，5，6)=4。除去该线路上的流量后得:<br><img alt=\"\" width=\"394\" height=\"126\" src=\"https://image.chaiding.com/ruankao/c7bab18a3ec514e9000ad37e2529e57b.jpg?x-oss-process=style/ruankaodaren\"><br>根据此图，线路①②⑤⑥的最大流量为min(1，3，1)=1。除去该线路上的流量后，从①到⑥已不连通，也就不再有剩余流量。<br>汇总后，最大总流量可以达到5+6+4+1=16(百吨/小时(。上述实现最大流量的方法是：<br><img alt=\"\" width=\"387\" height=\"127\" src=\"https://image.chaiding.com/ruankao/d94b4711ef205f0ec900d508a7609a2c.jpg?x-oss-process=style/ruankaodaren\"><br>该图中，各管道的实际流量都不超过其最大流量。除起点和终点外，所有站点的进油量等于其出油量。<br>虽然解答此题可以有多种选择线路的方案，但计算得到的最大总流量值都是-致的。<br>由于上述解题过程中，管道⑤一④尚未用到，因此，该管道的关闭并不会影响最大总流量值。其他路段管道的关闭是否会影响总流量值呢？<br>为了保持总流量值为16，从①出发的两段管道必须满负苟运输。管道①一②的流量10被分散到②一③、②一④、②一⑤三条管道，关闭其中任何一条管道都将达不到流量为10。同时，管道②一③的流量至少为2。<br>同样，为保持最大总流量，管道①一③的流量为6，管道③一⑤显然不能关闭。假设管道③一④关闭，则管道④一⑥的流量至多为8，到达站点⑥的流量至多为15。所以为保持最大总流量，管道③一④不能关闭。<br>为保持到达站点⑥的总流量为16，显然管道④一⑥和⑤一⑥任何一个都不能关闭。从而，只有管道⑤一④的关闭对最大总流量没有影响。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239465642217473"],"itemList":[{"id":"796239465533165569","questionId":"796239464606224385","content":" ②→③","answer":0,"chooseValue":"A"},{"id":"796239465579302913","questionId":"796239464606224385","content":" ②→⑤","answer":0,"chooseValue":"B"},{"id":"796239465617051649","questionId":"796239464606224385","content":" ③→④","answer":0,"chooseValue":"C"},{"id":"796239465642217473","questionId":"796239464606224385","content":" ⑤→④","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239461603102721","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某石油管理公司拥有下图所示的输油管道网。其中有6个站点，标记为①～⑥。站点①是唯一的供油站。各站点之间的箭线表示输油管道和流向。箭线边上标注的数字表示该管道的最大流量（单位：百吨／小时）。据此可算出，从站点①到达站点⑥的最大流量为（54）百吨/小时，而且当管道（55）关闭维修时管道网仍可按该最大流量值向站点⑥供油。<br><img alt=\"\" width=\"460\" height=\"134\" src=\"https://image.chaiding.com/ruankao/6131f574e943692f03470999364a4c87.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学基础知识。<br>从站点①到⑥有多条线路。显然，每条线路上的最大流量等于该线路上各段管道最大流量的最小值。站点①到⑥的最大总流量等于所有线路最大流量之和。<br>我们可以先从流量较大的线路开始计算。例如，线路①②④⑥的最大流量为min(10，5，11)=5。线路①③⑤⑥的最大流量为min(6，8，7)=6。除去这两条线路的流量后，剩余流量的图示如下：<br><img alt=\"\" width=\"379\" height=\"119\" src=\"https://image.chaiding.com/ruankao/9b1f3e643b019cba9a7b90445529fcf8.jpg?x-oss-process=style/ruankaodaren\"><br>根据此图，线路①②③④⑥的最大流量为mm(5，4，5，6)=4。除去该线路上的流量后得:<br><img alt=\"\" width=\"394\" height=\"126\" src=\"https://image.chaiding.com/ruankao/85e5da65aeb464c463eaa9f0f921e183.jpg?x-oss-process=style/ruankaodaren\"><br>根据此图，线路①②⑤⑥的最大流量为min(1，3，1)=1。除去该线路上的流量后，从①到⑥已不连通，也就不再有剩余流量。<br>汇总后，最大总流量可以达到5+6+4+1=16(百吨/小时(。上述实现最大流量的方法是：<br><img alt=\"\" width=\"387\" height=\"127\" src=\"https://image.chaiding.com/ruankao/5346f8ab39863b1f031a245da250f4a7.jpg?x-oss-process=style/ruankaodaren\"><br>该图中，各管道的实际流量都不超过其最大流量。除起点和终点外，所有站点的进油量等于其出油量。<br>虽然解答此题可以有多种选择线路的方案，但计算得到的最大总流量值都是-致的。<br>由于上述解题过程中，管道⑤一④尚未用到，因此，该管道的关闭并不会影响最大总流量值。其他路段管道的关闭是否会影响总流量值呢？<br>为了保持总流量值为16，从①出发的两段管道必须满负苟运输。管道①一②的流量10被分散到②一③、②一④、②一⑤三条管道，关闭其中任何一条管道都将达不到流量为10。同时，管道②一③的流量至少为2。<br>同样，为保持最大总流量，管道①一③的流量为6，管道③一⑤显然不能关闭。假设管道③一④关闭，则管道④一⑥的流量至多为8，到达站点⑥的流量至多为15。所以为保持最大总流量，管道③一④不能关闭。<br>为保持到达站点⑥的总流量为16，显然管道④一⑥和⑤一⑥任何一个都不能关闭。从而，只有管道⑤一④的关闭对最大总流量没有影响。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239462588764161"],"itemList":[{"id":"796239462525849601","questionId":"796239461603102721","content":" 14","answer":0,"chooseValue":"A"},{"id":"796239462559404033","questionId":"796239461603102721","content":" 15","answer":0,"chooseValue":"B"},{"id":"796239462588764161","questionId":"796239461603102721","content":" 16","answer":1,"chooseValue":"C"},{"id":"796239462618124289","questionId":"796239461603102721","content":" 18","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239453516484609","title":"某博览会每天8:00开始让观众通过各入口处检票进场，8:00前已经有很多观众在排队等候。假设8:00后还有不少观众均匀地陆续到达，而每个入口处对每个人的检票速度都相同。根据以往经验，若开设8个入口，则需要60分钟才能让排队观众全部入场；若开设10个入口，则需要40分钟才能消除排队现象。为以尽量少的入口数确保20分钟后消除排队现象，博览会应在8:00和8:20开设的入口数分别为（）。","analyze":"本题考查应用数学基础知识。<br>设早上8点时已有S人在排队等候，以后每分钟新来m人，每个人口处每分钟进场n人，则JS+60m=8*60n，S+40m=10*40n，两式相减得m=4n，而S=240n。<br>若要在20分钟由K个入口消除排队，则S+20m=20Kn，则K=16。<br>即8:00时，若开设16个入口，就可以在20分钟消除排队现象。<br>由于8:20后，每分钟新来m=4n人，所以应设4个入口，参观者就可以随来随进。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239454456008705"],"itemList":[{"id":"796239454414065665","questionId":"796239453516484609","content":" 12,2","answer":0,"chooseValue":"A"},{"id":"796239454435037185","questionId":"796239453516484609","content":" 14,4","answer":0,"chooseValue":"B"},{"id":"796239454456008705","questionId":"796239453516484609","content":" 16,4","answer":1,"chooseValue":"C"},{"id":"796239454476980225","questionId":"796239453516484609","content":" 18,6","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239456423137281","title":"某团队希望在未来18天内串行选做若干个作业。供选各作业所需的实施时间（天数）、截止时间（最迟必须在指定的数天内完工）以及利润见下表： <br><img alt=\"\" width=\"459\" height=\"95\" src=\"https://image.chaiding.com/ruankao/d703cc60608c416d93716b0bb57c82b6.jpg?x-oss-process=style/ruankaodaren\"><br>该团队只要能适当选择若干个作业依次实施，就能获得最大利润（）万元。","analyze":"本题考查应用数学基础知识。<br>为在规定的时间内获得最大利润，应尽量选做“利润/所需时间”较大的作业。<br><img alt=\"\" width=\"560\" height=\"115\" src=\"https://image.chaiding.com/ruankao/933f0521a1e05bdf3141b329f99fed35.jpg?x-oss-process=style/ruankaodaren\"><br>按“利润/天”从大到小排列得:<br><img alt=\"\" width=\"553\" height=\"104\" src=\"https://image.chaiding.com/ruankao/2d2a61be4edaa3cc417f04d1cdec03a1.jpg?x-oss-process=style/ruankaodaren\"><br>前5个作业T2、T3、T9、T7、T5的实施总时间为18天，但考虑到截止时间，应优先安排截止时间早的作业。依次安排T3(第1~3天)、T5(第4~10天)、T2(第11~13天)、T7(第14~16天)后，不能选T9，改选T4(第17、18天)。所以最大利润为5+8+6+4+2—25万元。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239457387827201"],"itemList":[{"id":"796239457324912641","questionId":"796239456423137281","content":" 23","answer":0,"chooseValue":"A"},{"id":"796239457358467073","questionId":"796239456423137281","content":" 24","answer":0,"chooseValue":"B"},{"id":"796239457387827201","questionId":"796239456423137281","content":" 25","answer":1,"chooseValue":"C"},{"id":"796239457421381633","questionId":"796239456423137281","content":" 26","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239467563208705","title":"某部门聘请了 30位专家评选去年最优秀项目，甲、乙、丙、丁四个项目申报参选。各位专家经过仔细考察后都在心目中确定了各自对这几个项目的排名顺序，如下表：<br><img alt=\"\" width=\"572\" height=\"128\" src=\"https://image.chaiding.com/ruankao/7ffae2a326d9528c150528594ab5377b.jpg?x-oss-process=style/ruankaodaren\"><br>其中，有3人将甲排在第1，将乙排在第4，将丙排在第2，将丁排在第3；依次类推。<br>如果完全按上表投票选择最优秀项目；那么显然，甲项目能得票9张，乙项目能得票8张，丙项目能得票7张，丁项目能得票6张，从而可以选出优秀项目甲。但在投票前，丙项目负责人认为自己的项目评上的希望不大，宣布放弃参选。这样，投票将只对甲、乙、丁三个项目进行，而各位专家仍按自己心目中的排名（只是删除了项目丙）进行投票。投票的结果是评出了优秀项目（58）。","analyze":"“选举”是数学在社会科学\"中的重要应用领域之一。<br>本题是“选举”造论中典型的例子之一。该例子考查在选举过程中，次要项的退出是否会对优势项产生影响。<br>按照题中所列各位专家心目中对各项目的排名，甲是优势项目，乙是次优项目，丙难胜出，丁是最差的。<br>丙退出后，每位专家对各项目的排名顺序没有变化，只需要将排在丙后面的项目丁提前一位，如下表：<br><img alt=\"\" width=\"568\" height=\"101\" src=\"https://image.chaiding.com/ruankao/d9dccd6f629ad00d29abe29e1073e832.jpg?x-oss-process=style/ruankaodaren\"><br>按上表投票，甲项目可得3+6=9票，乙项目可得3+5+2=10票，丁项目可得5+2+4=11票。因此，投票结果选出的优秀项目是项目丁。<br>这个例子说明了，投票制度的混沌性。劣势项目的退出居然对优势项目产生了颠覆性的影响。原来最差的项目居然变成了最优秀的项目。该例子也说明了用简单的数学规则难以很好地描述真实社会。由于社会的复杂性，完全公正的选举规则并不存在。在数学工作者看来，局部社会可能不完美，好像这是粗糙的错误。但正是这种不完美，体现了社会的迷人之处。没有终极真理，需要人们永远探索。这正是社会最伟大的完美！","multi":0,"questionType":1,"answer":"C","chooseItem":["796239468569841665"],"itemList":[{"id":"796239468481761281","questionId":"796239467563208705","content":" 甲","answer":0,"chooseValue":"A"},{"id":"796239468523704321","questionId":"796239467563208705","content":" 乙","answer":0,"chooseValue":"B"},{"id":"796239468569841665","questionId":"796239467563208705","content":" 丁","answer":1,"chooseValue":"C"},{"id":"796239468615979009","questionId":"796239467563208705","content":" 乙和丁","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239450484002817","title":"某公司测试部门共有40名员工，需要测试三类构件，分别是界面构件、算法构件和数据构件。在测试过程中，要求每位测试人员至少测试1类构件，最多测试2类构件。对于任意的测试任务分配方式，至少有一种构件种类完全一致的测试任务，其测试人员不少于（56）名。","analyze":"本题考查应用数学基础知识。<br>设界面构件、算法构件和数据构件分别为A、B、C三类，每个人至少测试一类构件，最多测试两类构件，这意味着每个人的测试必是A、B、C、AB、BC、AC这6种情况之一。因此，如有6个测试人员，则每个人的测试类别可能都不同。如有7个以上测试人员，则必然会出现测试种类相同的情况。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239451394166785"],"itemList":[{"id":"796239451394166785","questionId":"796239450484002817","content":" 7","answer":1,"chooseValue":"A"},{"id":"796239451423526913","questionId":"796239450484002817","content":" 8","answer":0,"chooseValue":"B"},{"id":"796239451452887041","questionId":"796239450484002817","content":" 9","answer":0,"chooseValue":"C"},{"id":"796239451486441473","questionId":"796239450484002817","content":" 10","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239438471516161","title":"在信息系统中，为防止数据偶发性错误，在数字代码上增设校验位是检测错误的常用手段。设计的原则是：查错功能强，增加存储量不多，便于自动计算校验位上的值，便于自动进行校验。<br>例如，第二代身份证号共18位，其中左17位是数字代码，末位是校验位。<br>设i(i=1，…，18)表第二代身份证号从右到左的编号，A<sub><sup>i</sup></sub>(i=2，…，18)表示身份证号第i位上的数字，则A<sub>1</sub>校验位上的数字可以按如下方法计算（注意所有计算均在模11 下进行)：<br><img alt=\"\" width=\"318\" height=\"76\" src=\"https://image.chaiding.com/ruankao/6b352deded1f33c53334770e2b667fe2.jpg?x-oss-process=style/ruankaodaren\"><br>如果A<sub>1</sub>=10，则以“X”表示。<br>从以上算法可知，对18位身份证号A<sub>1</sub>(i=1，…，18)进行校验的方法是验证：<br><img alt=\"\" width=\"124\" height=\"51\" src=\"https://image.chaiding.com/ruankao/7df7661a00a8a114443e51b25ac196b9.jpg?x-oss-process=style/ruankaodaren\"><br>是否等于（）。","analyze":"<img alt=\"\" width=\"499\" height=\"75\" src=\"https://image.chaiding.com/ruankao/4a9de0929dda7c5118f411c7797cd368.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"B","chooseItem":["796239439419428865"],"itemList":[{"id":"796239439390068737","questionId":"796239438471516161","content":" 0","answer":0,"chooseValue":"A"},{"id":"796239439419428865","questionId":"796239438471516161","content":" 1","answer":1,"chooseValue":"B"},{"id":"796239439448788993","questionId":"796239438471516161","content":" 2","answer":0,"chooseValue":"C"},{"id":"796239439478149121","questionId":"796239438471516161","content":" 10","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239447455715329","title":"某公司拟将5百万元资金投放下属A、B、C三个子公司（以百万元的倍数分配投资），各子公司获得部分投资后的收益如下表所示（以百万元为单位）。该公司投资的总收益至多为（）百万元。<br><img alt=\"\" width=\"413\" height=\"130\" src=\"https://image.chaiding.com/ruankao/91d3db7c05f065a400ad5c97d770685b.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考査应用数学基础知识。<br>将5百万资金依次分配给A、B、C子公司。在分配过程中，若以待分配的子公司和剩余的资金数标记结点名，可以绘制如下的网络图：<br><img alt=\"\" width=\"420\" height=\"284\" src=\"https://image.chaiding.com/ruankao/9beb1fe4445126cdaea3a46a9b1aba91.jpg?x-oss-process=style/ruankaodaren\"><br>从每条箭线的两端可以看出对应的投资分配，箭线上还可以标出相应的收益值。从始结点到终结点的多条路径中，收益总和最大的路径就是分配的最优方案。因此，可采用倒推计算方法寻找最长路径：先分别标记C5~C0到终结点0的收益，再分别计算从BC5~BC0到终结点0的最优路径和最大收益，最后再计算从始结点ABC5到终结点0的最优路径和最大收益。<br>如果结点很多，则以图为思考背景，以表格做实际计算，更为方便。<br>第1步，分配给C的各种情况，其路径和收益显然是直接的：<br><img alt=\"\" width=\"556\" height=\"83\" src=\"https://image.chaiding.com/ruankao/38f0d318b626daf145581a29c3e3f407.jpg?x-oss-process=style/ruankaodaren\"><br>第2步，对B的分配，需要计算各路径分段收益求和，并比较取大:<br><img alt=\"\" width=\"544\" height=\"65\" src=\"https://image.chaiding.com/ruankao/598573786d08fe058fc5b97496c54ad1.jpg?x-oss-process=style/ruankaodaren\"><br>第3步，对A的分配，需要计算各路径分段收益求和，并比较取大:<br><img alt=\"\" width=\"547\" height=\"49\" src=\"https://image.chaiding.com/ruankao/2043e8800f5695e9cbcdabaa72d8a1c4.jpg?x-oss-process=style/ruankaodaren\"><br>总之，ABC5—BC4—C3—0属于最优路径，总收益可以达到最大值5.5百万元。<br>也就是说，最优方案中，应分配1百万元给A(收益1.2百万元(，分配1百万元给B(收益0.8百万元)，分配3百万元给C(收益3.5百万元)。最大总收益为5.5百万元。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239448453959681"],"itemList":[{"id":"796239448361684993","questionId":"796239447455715329","content":" 4.8","answer":0,"chooseValue":"A"},{"id":"796239448391045121","questionId":"796239447455715329","content":" 5","answer":0,"chooseValue":"B"},{"id":"796239448424599553","questionId":"796239447455715329","content":" 5.2","answer":0,"chooseValue":"C"},{"id":"796239448453959681","questionId":"796239447455715329","content":" 5.5","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239444402262017","title":"平面坐标系内，有直线L1：y=ax和直线L2：y=bx(a>b>0)，动点（1，0)沿逆时针方向绕原点做如下运动：先沿垂直方向到达直线L1，再沿水平方向到达直线L2，又沿垂直方向到达直线L1，再沿水平方向到达直线L2，…，依次交替沿垂直和水平方向到达直线L1和12。这样的动点将（59）。","analyze":"动点的初始位置是（1，0)，首先会到达直线L1上的点（1，a)，然后到达直线L2上的点（-a/b，a)，再到达直线L1上的点（-a/b，-a<sup>2</sup>/b)，再到达直线L2上的点（a<sup>2</sup>/b<sup>2</sup>，-a<sup>2</sup>/b)，然后到达x轴上的点（a<sup>2</sup>/b<sup>2</sup>,0)。即动点绕一圈后，从x轴上的点1，达到了点a<sup>2</sup>/b<sup>2</sup>。由于a>b>0，因此动点在向外漂移，再绕一圈后将到达点a<sup>4</sup>/b<sup>4</sup>，绕n圈后将到达a<sup>2n</sup>/b<sup>2n</sup>。当n→∞时，动点将发散到无限。<br>显然，当a=b时，动点将沿矩形边界稳定地转圈；当0<a<b时，动点将收敛于原点。<br>这个问题是功能耦合系统动态变化的简例。机器系统、有机体系统、生态系统或社会系统都是复杂的功能耦合系统，有些功能随变量的增长而增长，有些功能则随变量的增长而減少（一般不是线性的）。在持续动态变化中，某些系统则会收敛于某种状态；有些系统则会发散到无穷：有些系统则会持续地稳定波动（周期性娓荡）；有些系统则会呈现非线性波动。通过简例观察动态系统的状态变化，是一种思维方法，也是表述某种哲理的方法。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239445350174721"],"itemList":[{"id":"796239445316620289","questionId":"796239444402262017","content":" 收敛于原点","answer":0,"chooseValue":"A"},{"id":"796239445350174721","questionId":"796239444402262017","content":" 发散到无穷","answer":1,"chooseValue":"B"},{"id":"796239445379534849","questionId":"796239444402262017","content":" 沿矩形边界稳定地转圈","answer":0,"chooseValue":"C"},{"id":"796239445408894977","questionId":"796239444402262017","content":" 随机运动","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239441483026433","title":"已知17个自然数（可有重复）的最小值是30，平均值是34，中位数是35，所有各数到38的距离之和比到35的距离之和多5，由此可以推断，这17个数中只有1个（）。","analyze":"本题考查应用数学基础知识。<br>由于这17个数的中位数是35，所以肯定其中有1个数就是35，左边8个数小于或等于35，右边8个数大于或等于35。<br>以所有各数到35的距离之和为基础，考察各数到38的距离之和的变化。<br>左边和中间共9个数，每个数到38的距离都比到35的距离增加3，共增加27。因此，右边8个数，从离35转到离38的距离之和，应减少27-5=22。<br>设右边8个数中，有x个35，y个36，z个37，w个38或38以上。而35、36、37、38以上，对35和38的距离变化分别是+3、+1、-1、-3。所以应该有：<br>3x+y-z-3w=-22，x+y+z+w=8，x、y、z、w都是0~8之间的整数。<br>两式相加得2w-x+z=15，再减前式得w-2x-y=7。<br>w只能为7(若w=8，则x=y=zK)，上式不成立).，从而x=y=0，z=l。即17个数中，只有1个37(至此已完成本题解答)，没有36，中位数35的右边没有重复的35。<br>中位数35以及右边的8个数(1个37，7个至少38)到34的距离之和至少为32。<br>由于这17个数的平均值为34，因此，小于34的各数与34的距离之和也应该不少于32(如果左边8数中含有35，则该和数还应该更多)。由于17个数的最小值为30，它与34的距离为4，因此中位数左边8个数必须都是30。也就是说，17个数中，35也只有1个，并没有34，而30则有8个。<br>由于中位数左边8个数30与34的距离之和恰好等于32，因此35以及右边8个数与34的距离之和也必须正好等于32。因此35右边除了1个37外，其他只能是7个38。<br>这样就推断出，这17个数只能是：8个30，1个35，1个37，7个38。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239442456104961"],"itemList":[{"id":"796239442384801793","questionId":"796239441483026433","content":" 30","answer":0,"chooseValue":"A"},{"id":"796239442409967617","questionId":"796239441483026433","content":" 34","answer":0,"chooseValue":"B"},{"id":"796239442430939137","questionId":"796239441483026433","content":" 36","answer":0,"chooseValue":"C"},{"id":"796239442456104961","questionId":"796239441483026433","content":" 37","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233974866268161","title":"某地天然气输送管线网络图如下，每段管线旁的数字表示输气能力（单位：万立方米/小时）。根据该图，从源S到目的地T的最大输气能力为（）万立方米/小时。<br> <img src=\"https://image.chaiding.com/ruankao/84f7a7899ed6716fe21e307ebf9a22b8.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"300\" height=\"200\" title=\"\" align=\"\">","analyze":"题考查最大流量问题。<br> 可以按步骤依次抽取最大流量路径的流量，抽取过程如下所示：<br> 【可以有多种不同的抽取方案，都能取得最大流量值】<br> <img src=\"https://image.chaiding.com/ruankao/579abb2a3abaf72d353414494fbb072d.jpg?x-oss-process=style/ruankaodaren\" width=\"300\" height=\"233\" title=\"\" align=\"\" alt=\"\"><br> <img src=\"https://image.chaiding.com/ruankao/febd19e8c3e74392285e1540fcc6605e.jpg?x-oss-process=style/ruankaodaren\" width=\"300\" height=\"233\" title=\"\" align=\"\" alt=\"\"><br> <img src=\"https://image.chaiding.com/ruankao/67014a52bc87d69c0971b19569ce3c78.jpg?x-oss-process=style/ruankaodaren\" width=\"300\" height=\"233\" title=\"\" align=\"\" alt=\"\"><br> <img src=\"https://image.chaiding.com/ruankao/dea814234faadced7feeb39f4e1052ef.jpg?x-oss-process=style/ruankaodaren\" width=\"300\" height=\"233\" title=\"\" align=\"\" alt=\"\"><br> <img src=\"https://image.chaiding.com/ruankao/781836915c9216b59d3017b54acd1322.jpg?x-oss-process=style/ruankaodaren\" width=\"300\" height=\"233\" title=\"\" align=\"\" alt=\"\"><br> 总共抽取流量：3+2+2+1+1=9。 <p> 综上可得，最大输出能力为9万立方米/小时。 </p>","multi":0,"questionType":1,"answer":"C","chooseItem":["796233976007118849"],"itemList":[{"id":"796233975919038465","questionId":"796233974866268161","content":" 4","answer":0,"chooseValue":"A"},{"id":"796233975960981505","questionId":"796233974866268161","content":" 8","answer":0,"chooseValue":"B"},{"id":"796233976007118849","questionId":"796233974866268161","content":" 9","answer":1,"chooseValue":"C"},{"id":"796233976049061889","questionId":"796233974866268161","content":" 10","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239485191868417","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>根据近几个月的数据统计，某车次火车到站晚点时间t (分钟）的概率分布密度函数可用函数<img alt=\"\" width=\"100\" height=\"17\" src=\"https://image.chaiding.com/ruankao/366f32c1661da9b5fd4e9b2c3535260a.jpg?x-oss-process=style/ruankaodaren\">)来描述，因此可以计算出其中的待定系数k= (54), 晚点超过5分钟的概率为（55）。","analyze":"本题考查应用数学（概率统汁）知识。<br>本题中，某次列车的晚点时间t是随机变量<img alt=\"\" width=\"50\" height=\"17\" src=\"https://image.chaiding.com/ruankao/cdbdb229c5e52a1566cc21a90cb788e3.jpg?x-oss-process=style/ruankaodaren\">，其分布密度函数f(t)意味着晚点时间在<img alt=\"\" width=\"45\" height=\"15\" src=\"https://image.chaiding.com/ruankao/79eee8f9237bc89ac86da2a9e519bd1c.jpg?x-oss-process=style/ruankaodaren\">这个时间段内的概率为<img alt=\"\" width=\"35\" height=\"17\" src=\"https://image.chaiding.com/ruankao/9402bc946d649ee5769148e44346eb15.jpg?x-oss-process=style/ruankaodaren\">。由于总概率为1，因此<br><img alt=\"\" width=\"150\" height=\"43\" src=\"https://image.chaiding.com/ruankao/e06f0a4a1ce921f1047613f85b9c44f4.jpg?x-oss-process=style/ruankaodaren\"><br>从而k=0.003。晚点时间超过5分钟的概率为<br><img alt=\"\" width=\"180\" height=\"30\" src=\"https://image.chaiding.com/ruankao/d68bdb87f0ee1a7fad2ce7d3ab3af85b.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"A","chooseItem":["796239486265610241"],"itemList":[{"id":"796239486265610241","questionId":"796239485191868417","content":" 0.003","answer":1,"chooseValue":"A"},{"id":"796239486282387457","questionId":"796239485191868417","content":" 0.03","answer":0,"chooseValue":"B"},{"id":"796239486303358977","questionId":"796239485191868417","content":" 0.3","answer":0,"chooseValue":"C"},{"id":"796239486320136193","questionId":"796239485191868417","content":" 3","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239472374075393","title":"假设一个I/O系统只有一个磁盘，每秒可以接收50个I/O请求，磁盘对每个I/O请求服务的平均时间是10ms,则I/O请求队列的平均长度是（39）个请求。","analyze":"磁盘的I/O请求是一个随机过程，请求事件达到的时间间隔具有泊松分布的概率学特征。根据Little定律，平均队列长度=达到速率×平均等待时间。其中平均等待时间=平均服务时间X服务器利用率/(1-服务器利用率）<br>而服务器利用率=到达速率×平均服务时间，所以平均队列长度=服务器利用率×服务器利用率/(1-服务器利用率）<br>根据本题给出的相关数据，服务器利用率=1/50x0.01=0.5,因此平均队列长度等于0.5。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239473397485569"],"itemList":[{"id":"796239473363931137","questionId":"796239472374075393","content":" 0","answer":0,"chooseValue":"A"},{"id":"796239473397485569","questionId":"796239472374075393","content":" 0.5","answer":1,"chooseValue":"B"},{"id":"796239473426845697","questionId":"796239472374075393","content":" 1","answer":0,"chooseValue":"C"},{"id":"796239473456205825","questionId":"796239472374075393","content":" 2","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239488421482497","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>根据近几个月的数据统计，某车次火车到站晚点时间t (分钟）的概率分布密度函数可用函数<img alt=\"\" width=\"100\" height=\"17\" src=\"https://image.chaiding.com/ruankao/7ecc2ad8d6596efa1e9202bec8d8a4a4.jpg?x-oss-process=style/ruankaodaren\">)来描述，因此可以计算出其中的待定系数k= (54), 晚点超过5分钟的概率为（55）。","analyze":"本题考查应用数学（概率统汁）知识。<br>本题中，某次列车的晚点时间t是随机变量<img alt=\"\" src=\"https://image.chaiding.com/ruankao/68c5e7056f28c1c9042e22d5bdd05930.jpg?x-oss-process=style/ruankaodaren\" height=\"15\" width=\"42\">，其分布密度函数f(t)意味着晚点时间在<img alt=\"\" width=\"40\" height=\"13\" src=\"https://image.chaiding.com/ruankao/47452bd7bb2c2f7b5fbda3064ddec271.jpg?x-oss-process=style/ruankaodaren\">个时间段内的概率为<img alt=\"\" width=\"40\" height=\"20\" src=\"https://image.chaiding.com/ruankao/97e7d59ba051ff854fe82a4bd9505472.jpg?x-oss-process=style/ruankaodaren\">。由于总概率为1，因此<br><img alt=\"\" width=\"150\" height=\"43\" src=\"https://image.chaiding.com/ruankao/63ce8db8f41c01391235c29b6a4c0071.jpg?x-oss-process=style/ruankaodaren\"><br>从而k=0.003。晚点时间超过5分钟的概率为<br><img alt=\"\" width=\"180\" height=\"30\" src=\"https://image.chaiding.com/ruankao/4bc1809516f6b6c0e7c7168838e8618e.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"C","chooseItem":["796239489419726849"],"itemList":[{"id":"796239489344229377","questionId":"796239488421482497","content":" 1/32","answer":0,"chooseValue":"A"},{"id":"796239489381978113","questionId":"796239488421482497","content":" 1/16","answer":0,"chooseValue":"B"},{"id":"796239489419726849","questionId":"796239488421482497","content":" 1/8","answer":1,"chooseValue":"C"},{"id":"796239489453281281","questionId":"796239488421482497","content":" 1/4","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239491395244033","title":"设甲乙丙三人独立解决某个问题的概率分别为0.45、0.55、0.6，则三人一起解决该问题的概率约为（53）。","analyze":"本题考查数学应用（概率）能力。<br>根据题意，三人一起无法解决该问题的概率为（1-0.45) x (1-0.55) x (1-0.6)=0.099。所以，三人一起能解决该问题的概率为1-0.099=0.901。<br>另一种解题思路是：甲解决了该问题的0.45部分，余下0.55部分没有解决。此时，乙能解决其中的0.55部分，即乙能解决总体的0.55x0.55=0.3025部分。甲乙共解决了45+0.3025=0.7525部分，余下0.2475部分没有解决。丙在其中解决了0.6,即丙解决了总体的0.2475x0.6=0.1485部分。甲乙丙三人共解决了问题0.7525+0.1485=0.901部分。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239492364128257"],"itemList":[{"id":"796239492297019393","questionId":"796239491395244033","content":" 0.53","answer":0,"chooseValue":"A"},{"id":"796239492317990913","questionId":"796239491395244033","content":" 0.7","answer":0,"chooseValue":"B"},{"id":"796239492338962433","questionId":"796239491395244033","content":" 0.8","answer":0,"chooseValue":"C"},{"id":"796239492364128257","questionId":"796239491395244033","content":" 0.9","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239475406557185","title":"有一名患者胸部长了一个肿瘤，医院X光检查结果呈阳性。据统计，胸部肿瘤为良性的概率为99%。对良性肿瘤，X光检查的正确率（呈阴性的概率）为90%；对恶性肿瘤，X光检査的正确率（呈阳性的概率）为80%。因此，可推算出该患者患恶性肿瘤的概率是（54）。","analyze":"我们可以将胸部肿瘤的检查情况画出概率树如下：<br><img alt=\"\" width=\"519\" height=\"147\" src=\"https://image.chaiding.com/ruankao/14f57bedc477d6e9a009a3bb3492b3cb.jpg?x-oss-process=style/ruankaodaren\"><br>该树的根为“胸部肿瘤”，其性质99%的概率为良性的，1%的概率为恶性的。对于良性肿瘤，X光检查的结果，90%的概率为阴性，10%的概率为阳性；对于恶性肿瘤，X光检查的结果，80%的概率为阳性，20%的概率为阴性。<br>从“胸部肿瘤”到“X光检查结果呈阳性”的路径有以下两条：<br>胸部肿瘤→良性→X光检査结果呈阳性<br>胸部肿瘤→恶性→X光检查结果呈阳性<br>前一条路径的概率等于其各段概率之积，为99%×10%=0.099。<br>后一条路径的概率等于其各段概率之积，为1%×80%=0.008。<br>从全概率公式可知道，对于胸部肿瘤，X光检查结果呈阳性的总概率的等于所有各条路径的概率之和，所以为0.099+0.008=0.107=10.7%。<br>如果已经知道X光检查结果呈阳性，那么从前一条路径过来（属于良性）的概率为：<br>0.099/(0.099+0.008)≈0.925=92.5%<br>从后一条路径过来（属于恶性）的概率为：<br>0.008/(0.099+0.008)≈0.075=7.5%<br>这个问题的结论常出乎大家的意料，即使医生也非常惊讶。这是著名的“反问题错乱 ”（confhsion of the inverse)现象。<br>对于患某种重病的概率很低的情况，当患者检查结果偏离正常值时，这种结果在医学上称为假阳性，还需要采用其他手段才能确诊。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239476446744577"],"itemList":[{"id":"796239476413190145","questionId":"796239475406557185","content":" 0.8%","answer":0,"chooseValue":"A"},{"id":"796239476446744577","questionId":"796239475406557185","content":" 7.5%","answer":1,"chooseValue":"B"},{"id":"796239476476104705","questionId":"796239475406557185","content":" 80%","answer":0,"chooseValue":"C"},{"id":"796239476505464833","questionId":"796239475406557185","content":" 75%","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235417354522625","title":"某学校希望通过问卷调查了解学生考试作弊的真实情况。若直接在问卷调查中问：“你作弊了吗？”，极少有入真实做答。为此，专家设计的问卷调查表中包括两个问题：①你是男生吗？②你作弊了吗？而每个学生需要利用给自己配发的电子随机选题器选择一题并回答“是”或“否”。学校按照学生实际的男女比例，随机选择了60名男生和40名女生参与匿名答题，而电子随机选题器选择题1和题2的概率相同。学生们认为，此次调查不但匿名，还不透露自己选择了哪题，因此都如实做答。最后，学校回收到35份回答“是”，65份回答“否”，因此计算出考试作弊的比例大致为（59）。","analyze":"本题考查应用数学基础知识（概率统计）。<br>根据题意画出概率图如下（设作弊的比例为X）:<br><img alt=\"\" width=\"338\" height=\"127\" src=\"https://image.chaiding.com/ruankao/a096916fb2fe3f56a99df04ea0521eb5.jpg?x-oss-process=style/ruankaodaren\"><br>则回答“是”的比例等于0.5*0.6+0.5x=0.35，因此x=0.35*2-0.6=0.1。","multi":0,"questionType":1,"answer":"A","chooseItem":["796235418361155585"],"itemList":[{"id":"796235418361155585","questionId":"796235417354522625","content":" 10%","answer":1,"chooseValue":"A"},{"id":"796235418390515713","questionId":"796235417354522625","content":" 15%","answer":0,"chooseValue":"B"},{"id":"796235418419875841","questionId":"796235417354522625","content":" 20%","answer":0,"chooseValue":"C"},{"id":"796235418440847361","questionId":"796235417354522625","content":" 25%","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235395086962689","title":"某市场上某种零件由甲、乙、丙、丁四厂供货，供货数量之比为4:3:2:1。各厂产品的合格率分别为99%、98%、97.5%和95%。某抽检员发现了一件次品，它属于（52）厂的概率最大。","analyze":"本题考查应用数学基础知识（概率）。<br>先根据题意画出概率图如下：<br><img alt=\"\" width=\"399\" height=\"207\" src=\"https://image.chaiding.com/ruankao/22af92c37fb0904a47673db0d30d4c88.jpg?x-oss-process=style/ruankaodaren\"><br>总次品率=0.4*1%+0.3*2%+0.2*2.5%+0.1*5%<br>=0.004+0.006+0.005+0.005=0.02<br>该次品属于甲厂的概率=0.004/0.02=20%<br>该次品属于乙厂的概率=0.006/0.02=30%<br>该次品属于丙厂的概率=0.005/0.02=25%<br>该次品属于丁厂的概率=0.005/0.02=25%","multi":0,"questionType":1,"answer":"B","chooseItem":["796235396034875393"],"itemList":[{"id":"796235395997126657","questionId":"796235395086962689","content":" 甲","answer":0,"chooseValue":"A"},{"id":"796235396034875393","questionId":"796235395086962689","content":" 乙","answer":1,"chooseValue":"B"},{"id":"796235396060041217","questionId":"796235395086962689","content":" 丙","answer":0,"chooseValue":"C"},{"id":"796235396089401345","questionId":"796235395086962689","content":" 丁","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233978213322753","title":"根据历史数据和理论推导可知，某应用中，随机变量s的分布密度函数为f（x）=3x<sup>2</sup>，（0<x<1）。这意味着，当Δx充分小时，随机变量s落在区间（x，x+Δx）内的概率约等于f（x）Δx。为此，开发该应用的仿真系统时，可用（）来模拟该随机变量，其中，r1.r2.r3...为计算机逐个产生的、均匀分布在（0，1）区间内的互相独立的伪随机数。","analyze":"（ 0 ，1 ）区间内的分布密度函数3x<sup>2</sup>。意味着随着x的增大出现的概率也增大。显然，对于min（r1 ,r2 , r3 ） ，出现较小的数值的概率更大些；r1 *r2*r3（两个小于1 的数相乘会变得更小）也会这样。对于随机变量（r1+r2+r3）/2 ， 出现中等大小数值的概率更大一些，出现较大的或较小值的概率会小一些，其分布密度函数会呈凸型。只有max（r1 ,r2 ,r3） ，出现较大数值的概率更大些。","multi":0,"questionType":1,"answer":"A","chooseItem":["796233979287064577"],"itemList":[{"id":"796233979287064577","questionId":"796233978213322753","content":" max（r1，r2，r3）","answer":1,"chooseValue":"A"},{"id":"796233979324813313","questionId":"796233978213322753","content":" min（r1，r2，r3）","answer":0,"chooseValue":"B"},{"id":"796233979358367745","questionId":"796233978213322753","content":" r1*r2*r3","answer":0,"chooseValue":"C"},{"id":"796233979396116481","questionId":"796233978213322753","content":" （r1+r2+r3）/3","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239510743568385","title":"已知某山区六个乡镇C1，C2,…，C6之间的公路距离（公里数）如下表:<br><img alt=\"\" width=\"545\" height=\"160\" src=\"https://image.chaiding.com/ruankao/26b8474fcb25312095072516b1db62b9.jpg?x-oss-process=style/ruankaodaren\"><br>其中符号“∞“表示两个乡镇之间没有直通公路。乡镇C1到C3虽然没有直通公路， 但可以经过其他乡镇达到，根据上表，可以算出C1到C3最短的路程为（57）公里。","analyze":"本题考查图论应用基础知识。<br>根据题中给出的6个乡镇间的公路距离表，可以绘制距离图如下：<br><img alt=\"\" width=\"394\" height=\"152\" src=\"https://image.chaiding.com/ruankao/f4ba0f36e73c8f484ec93a13ac82a8df.jpg?x-oss-process=style/ruankaodaren\"><br>结点①、②、…、⑥分别表示这6个乡镇，结点之间的连线表示有公路直接通达， 连线上的数字表示公里数。从图可以看出，乡镇①到③没有直通公路，但可以通过其他乡镇达到。显然，路径①-⑤-④-③总的里程数45公里是最短的。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239511712452609"],"itemList":[{"id":"796239511649538049","questionId":"796239510743568385","content":" 35","answer":0,"chooseValue":"A"},{"id":"796239511683092481","questionId":"796239510743568385","content":" 40","answer":0,"chooseValue":"B"},{"id":"796239511712452609","questionId":"796239510743568385","content":" 45","answer":1,"chooseValue":"C"},{"id":"796239511746007041","questionId":"796239510743568385","content":" 50","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239502606618625","title":"已知A、B、……、I 九人比赛结果排名（没有并列名次）的部分情况如下图：<br><img alt=\"\" width=\"127\" height=\"105\" src=\"https://image.chaiding.com/ruankao/5809247bda709315ac5ee4b8f7a26c0e.jpg?x-oss-process=style/ruankaodaren\"><br>图中的箭头表示“排名前于”，例如D—>A表示D排名前于A。<br>根据上图中表示的部分排名情况，可以推断，第3名可能是（55）。","analyze":"本题考查应用数学基础知识。<br>根据题中的箭头图画出如下的网络图：<br><img alt=\"\" width=\"281\" height=\"128\" src=\"https://image.chaiding.com/ruankao/7b43ce7e0591d2c80684fa7dddfe6345.jpg?x-oss-process=style/ruankaodaren\"><br>从上图看出，D排名在其他所有人之前，所以D —定是第1名。由于只有ESA仅排在D之后，因此第2名只可能是E或A (G之前有D、E、H)。<br>如果E是第2名，则第3名可能是H、F或A(B之前有DEA, B不可能是第3名）；如果A是第2名，则第3名必是E (B之前有DEA, B不可能是第3名）；<br>因此，第3名只可能是A、E、F或H。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239503533559809"],"itemList":[{"id":"796239503533559809","questionId":"796239502606618625","content":" A、E、F 或 H","answer":1,"chooseValue":"A"},{"id":"796239503583891457","questionId":"796239502606618625","content":" B、F 或 H","answer":0,"chooseValue":"B"},{"id":"796239503613251585","questionId":"796239502606618625","content":" F 或 H","answer":0,"chooseValue":"C"},{"id":"796239503638417409","questionId":"796239502606618625","content":" B、F、H 或 G","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239523162902529","title":"某乡规划了村村通公路网建设方案连接其所属6个村，每两个村之间至多只有一条公路相连，各条公路互不重叠。因此，各村所连接的公路条数形成一个6数序列。以下4个序列中，除(56)外都是不可能的。","analyze":"本题考查应用数学（图论）知识。<br>每条公路在序列中都被计算两次，因此，6数序列的总和应是偶数。5,4,3,3,2,2中各数之和为奇数，所以不可能。5,5,4,3,2,1中的前两数5表示有两个村与其他各村都有公路相连，因此不可能存在只有1条公路的村，也不可能。供5,4,4,3,1,1中最后1 村只有1条公路，而第1村与其他各村都相连，因此这两个村之间有公路连接。不算这两村及其间的公路后，形成5个村和5数序列4,4,4,3,1。该序列中，既然前3村中每村都与其他4村都相连，那么，每个村的公路数至少为3,所以也是不可能的。<br>5,4,4,3,2,2是可能的，如下图，其中各村公路的条数为：A-5,B-2,C-3,D-4,E-2,F-4。<br><img alt=\"\" width=\"300\" height=\"140\" src=\"https://image.chaiding.com/ruankao/aec9837c79e5e8939db21d23b3a8f0a0.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796239524295364609"],"itemList":[{"id":"796239524215672833","questionId":"796239523162902529","content":" 5,4,3,3,2,2","answer":0,"chooseValue":"A"},{"id":"796239524240838657","questionId":"796239523162902529","content":" 5,5,4,3,2,1","answer":0,"chooseValue":"B"},{"id":"796239524270198785","questionId":"796239523162902529","content":" 5,4,4,3,1,1","answer":0,"chooseValue":"C"},{"id":"796239524295364609","questionId":"796239523162902529","content":" 5,4,4,3,2,2","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239507782389761","title":"已知有6个村A~F，相互间的道路距离（单位:里）如下图所示。计划在其中某村建一所学校。据统计，各村希望来上学的学生人数分别为50、40、60、20、70、90。为使全体学生上学所走的总距离最短，学校应建在（58）村。<br><img alt=\"\" width=\"277\" height=\"136\" src=\"https://image.chaiding.com/ruankao/c297b7b401a437251d02b6410704c271.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学（图论）知识。<br>根据题意，各村之间的最短距离（单位：里）可列表如下:<br><img alt=\"\" src=\"https://image.chaiding.com/ruankao/b828a1145b295de899ed856be61027a0.jpg?x-oss-process=style/ruankaodaren\" height=\"120\" width=\"372\"><br>各村到候选校村的学生人里数列表如下:<br><img alt=\"\" width=\"400\" height=\"127\" src=\"https://image.chaiding.com/ruankao/8664d0721ebf844f93922bd502487b9e.jpg?x-oss-process=style/ruankaodaren\"><br>从表中看出，学生人里数总和最少的是各村到A村980人里，因此学校应建在A村。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239508696748033"],"itemList":[{"id":"796239508696748033","questionId":"796239507782389761","content":" A","answer":1,"chooseValue":"A"},{"id":"796239508734496769","questionId":"796239507782389761","content":" B","answer":0,"chooseValue":"B"},{"id":"796239508768051201","questionId":"796239507782389761","content":" E","answer":0,"chooseValue":"C"},{"id":"796239508805799937","questionId":"796239507782389761","content":" F","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239513721524225","title":"开发商需要在某小区9栋楼房之间敷设自来水管道，使各楼都能连通，又能使总成本最低。经勘察，各楼房之间敷设管道的路径和成本（单位：千元）如下图所示。<br><img alt=\"\" width=\"211\" height=\"170\" src=\"https://image.chaiding.com/ruankao/9f78cfa40ee83a4bf0b7016d5a1839f1.jpg?x-oss-process=style/ruankaodaren\"><br>该项目的总成本至少需要（59）千元。","analyze":"本题考查应用数学基础知识。<br>该题可用图论中的最小支撑树算法来求解。<br>最小支撑树算法的核心思想是先确定最小成本的一段（如有多段，则可任选一段)，该段已将两个点连接；在余下未连接的点中，选择1点使其与已连接的点具有最小成本 (如有多点，则可任选一点）；继续这样做，直到所有的点都已经连接。<br>虽然完成连接的总成本最低的方案可有多种，但它们的总成本都一定是相等的。例如，总成本最低的方案之一为：<br><img alt=\"\" width=\"170\" height=\"121\" src=\"https://image.chaiding.com/ruankao/7397a3d763d1ecb2ae7c409a69dc496e.jpg?x-oss-process=style/ruankaodaren\"><br>该项目的总成本需要13千元。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239514631688193"],"itemList":[{"id":"796239514631688193","questionId":"796239513721524225","content":" 13","answer":1,"chooseValue":"A"},{"id":"796239514648465409","questionId":"796239513721524225","content":" 14","answer":0,"chooseValue":"B"},{"id":"796239514669436929","questionId":"796239513721524225","content":" 15","answer":0,"chooseValue":"C"},{"id":"796239514682019841","questionId":"796239513721524225","content":" 16","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239527428509697","title":"某学校运动会准备安排8个项目（命名为A，B，…，H)的决赛，16个团队（编号为1，2，…，16)参加决赛的项目如下表（*表示相应的团队将参加相应的决赛)：<br><img alt=\"\" width=\"595\" height=\"256\" src=\"https://image.chaiding.com/ruankao/b456fb6b9889ca009a516bd4cb9e3113.jpg?x-oss-process=style/ruankaodaren\"><br>运动会组委会希望妥善安排这8个项目决赛顺序的方案，使每个团队不会连续参加两场决赛。针对上表情况，这样的方案（57）。(提示：可在平面上将每个项目用一个点表示，在两个项目之间，只要有同一团队都参加，则在相应点之间用线连接）","analyze":"本题考查考生在数学应用方面的能力。<br>用图的方法解决此类问题比较直观。<br>在平面上将每个项目用一个节点表示。每一团队参加的多个项目，在相应点之间都用线连接（己有连线时不用重复画）。即，每两个项目，如有团队都参加，就在相应两点之间画连线（如图（a）),表示这两个项目不能接续安排。为清晰起见，我们根据图（a）再画一张连线状态相反的图（如图（b）)。同样8个点表示8个项目，但图（a）中凡是两点之间有连线的地方，图（b）中就没有连线；图（a）中凡是两点之间无连线的地方，图（b）中就有连线。因此，图（b）中的每条连线表示相应的两端项目是可以接续安排的。这样，只要在图（b）中找到一条连线通路，正好将这8个点依次不重复地全都连接起来，就形成一种可行的安排方案。<br><img alt=\"\" width=\"535\" height=\"232\" src=\"https://image.chaiding.com/ruankao/4c0799160446a5a48e3ca20d3b27b066.jpg?x-oss-process=style/ruankaodaren\"><br>从图b可以看出，依次连接这8个项目的通路可以有多条，例如：<br>E-D-A-F-B-G-C-H<br>G-B-F-C-E-D-A-H<br>F-C-D-E-A-E-B-G<br>……<br>上述每一条通路表示一种安排方案。按照其中任一方案，各团队都不会连续参加两场决赛。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239529198505985"],"itemList":[{"id":"796239529089454081","questionId":"796239527428509697","content":" 不存在","answer":0,"chooseValue":"A"},{"id":"796239529123008513","questionId":"796239527428509697","content":" 只有1个","answer":0,"chooseValue":"B"},{"id":"796239529160757249","questionId":"796239527428509697","content":" 共有2个","answer":0,"chooseValue":"C"},{"id":"796239529198505985","questionId":"796239527428509697","content":" 多于2个","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235179487154177","title":"X、Y、Z 是某企业的三个分厂，每个分厂每天需要同一种原料20吨，下图给出了邻近供应厂A、B、C的供应运输路线图，每一段路线上标明了每天最多能运输这种原料的吨数。根据该图可以算出，从A、B、C三厂每天最多能给该企业运来这种原料共（ ）吨。<br><img width=\"234\" height=\"178\" alt=\"\" src=\"https://image.chaiding.com/ruankao/d4768c558f2a43422f0459fe49f47f38.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学（运筹学--最大流）基础知识。<br>逐步画出原料供应路线及其运输量如下：<br>A-M-X 10吨<br>C-N-Z 20吨 Z已满足<br>B-N-Z-Y-X 10吨 X已满足<br>此时，各条路径上的剩余流量如下：<br><img width=\"335\" height=\"224\" alt=\"\" src=\"https://image.chaiding.com/ruankao/77e755c40467ddd6928fa099d50710ed.jpg?x-oss-process=style/ruankaodaren\"><br>A-M-Y 5 吨<br>B-M-N-Y 10 吨<br>总之，每天最多供应（运输）55吨。","multi":0,"questionType":1,"answer":"C","chooseItem":["796235180451844097"],"itemList":[{"id":"796235180409901057","questionId":"796235179487154177","content":" 45","answer":0,"chooseValue":"A"},{"id":"796235180430872577","questionId":"796235179487154177","content":" 50","answer":0,"chooseValue":"B"},{"id":"796235180451844097","questionId":"796235179487154177","content":" 55","answer":1,"chooseValue":"C"},{"id":"796235180468621313","questionId":"796235179487154177","content":" 60","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234695049236481","title":"下表记录了六个结点A、B、C、D、E、F之间的路径方向和距离。从A到F的最短距离是（ ）。<br> <img alt=\"\" src=\"https://image.chaiding.com/ruankao/bb38a12a0b9257a2607ec770be8bb50c.jpg?x-oss-process=style/ruankaodaren\" width=\"505\" height=\"155\">","analyze":"本题考查应用数学（运筹学-图论）基础知识。 <br>按照表中的数据，画图如下。<br><img width=\"471\" height=\"158\" alt=\"\" src=\"https://image.chaiding.com/ruankao/06408e0ba069f1b2dd720db19f0e9074.jpg?x-oss-process=style/ruankaodaren\"><br>从E到F的最短距离=15。<br>从D到F的最短距离=min{D-E-F，D-F}=min{14+15，17}=17。<br>从C到F的最短距离=min{C-D-F，C-E-F，C-F}=min{14+17，17+15,22}=22。<br>从B到F的最短距离=min{B-C-F，B-D-F，B-E-F，B-F}=min{13+22，16+17,21+15,29}=29。 <br>从A到F的最短距离=min{A-B-F，A-C-F，A-D-F，A-E-F，A-F}<br>=min{11+29，16+22,24+17，36+15,54}=38。<br>最短路径为A-C-F，最短距离为38。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234695984566273"],"itemList":[{"id":"796234695984566273","questionId":"796234695049236481","content":" 38","answer":1,"chooseValue":"A"},{"id":"796234696005537793","questionId":"796234695049236481","content":" 40","answer":0,"chooseValue":"B"},{"id":"796234696030703617","questionId":"796234695049236481","content":" 44","answer":0,"chooseValue":"C"},{"id":"796234696051675137","questionId":"796234695049236481","content":" 46","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234937278681089","title":"下表记录了六个结点A、B、C、D、E、F之间的路径方向和距离。从A到F的最短距离是（ ）。<br><img alt=\"\" src=\"https://image.chaiding.com/ruankao/dd6a14c28760b9a828933ee793f52b7f.jpg?x-oss-process=style/ruankaodaren\" height=\"173\" width=\"410\">","analyze":"本题考查应用数学（运筹学-图论）基础知识。 <br>按照表中的数据，画图如下。<br><img width=\"471\" height=\"158\" alt=\"\" src=\"https://image.chaiding.com/ruankao/06408e0ba069f1b2dd720db19f0e9074.jpg?x-oss-process=style/ruankaodaren\"><br>从E到F的最短距离=15。<br>从D到F的最短距离=min{D-E-F，D-F}=min{14+15，17}=17。<br>从C到F的最短距离=min{C-D-F，C-E-F，C-F}=min{14+17，17+15,22}=22。<br>从B到F的最短距离=min{B-C-F，B-D-F，B-E-F，B-F}=min{13+22，16+17,21+15,29}=29。 <br>从A到F的最短距离=min{A-B-F，A-C-F，A-D-F，A-E-F，A-F}<br>=min{11+29，16+22,24+17，36+15,54}=38。<br>最短路径为A-C-F，最短距离为38。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234938306285569"],"itemList":[{"id":"796234938306285569","questionId":"796234937278681089","content":" 38","answer":1,"chooseValue":"A"},{"id":"796234938352422913","questionId":"796234937278681089","content":" 40","answer":0,"chooseValue":"B"},{"id":"796234938398560257","questionId":"796234937278681089","content":" 44","answer":0,"chooseValue":"C"},{"id":"796234938423726081","questionId":"796234937278681089","content":" 46","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235176475643905","title":"已知八口海上油井(编号从1#到8#) 相互之间的距离(单位：海里)如下表所示，其中1#油井离海岸最近为5海里。现从海岸开始铺设输油管道，经1#油井将这些油井都连接起来，管道的总长度至少为( ) 海里(为便于计量和维修，管道只能在油井处分叉)。<br><img alt=\"\" width=\"520\" height=\"182\" src=\"https://image.chaiding.com/ruankao/2225bcf1b1818254315091c4bfd0d1df.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学（运筹学-最小支撑树）基础知识。<br>从1#到{2#，3#，4#，5#，6#，7#，8#}的最短距离为（1#，5#) =0.5海里。<br>从{1#，5#}到{2#，3#, 4#, 6#, 7#，8#}的最短距离为（5#，4#) =0.7海里。<br>从{1#，4#，5#}到{2#，3#, 6#, 7#，8#}的最短距离为（5#，8#) =0.8海里。<br>从{1#，4#，5#，8#}到{2#，3#, 6#, 7#}的最短距离为（8#，7#) =0.5海里。<br>从{1#，4#, 5#，8#, 7#}到{2#，3#, 6#}的最短距离为（7#，6#) =0.6海里。<br>从{1#，4#，5#，8#，7#, 6#}到{2#，3#}的最短距离为（8#, 3#)=1.0海里。<br>从{1#，4#，5#，8#，7#, 6#, 3#}到2#的最短距离为(8#, 3#)=0.9海里。<br><img width=\"550\" height=\"129\" alt=\"\" src=\"https://image.chaiding.com/ruankao/1cb3a75383d54c1e0778d235091b7157.jpg?x-oss-process=style/ruankaodaren\"><br>因此，从海岸开始连接8口油井的总距离=5+0.5+0.7+0.8+0.5+0.6+1.0+0.9=10海里。","multi":0,"questionType":1,"answer":"C","chooseItem":["796235177452916737"],"itemList":[{"id":"796235177390002177","questionId":"796235176475643905","content":" 5","answer":0,"chooseValue":"A"},{"id":"796235177419362305","questionId":"796235176475643905","content":" 9","answer":0,"chooseValue":"B"},{"id":"796235177452916737","questionId":"796235176475643905","content":" 10","answer":1,"chooseValue":"C"},{"id":"796235177473888257","questionId":"796235176475643905","content":" 11","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234448579350529","title":"某乡8个小村（编号为1~8)之间的距离如下表（单位：km)。1号村离水库最近，为5km,从水库开始铺设水管将各村连接起来，最少需要铺设（55）长的水管（为便于管理和维修，水管分叉必须设在各村处)。<br> <img src=\"https://image.chaiding.com/ruankao/1c8c8044b07ec9debdf7c94e984d0e4b.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"650\" height=\"224\" title=\"\" align=\"\">","analyze":"本题考查应用数学-运筹学-图论应用的基础知识。<br> 为解决这类问题，可以按最短距离逐村铺设水管进行连接。<br> 从水库到①村先铺设水管，距离为5km。<br> 离①村最近的④村距离为1km，因此铺设水管①-④。<br> 离①、④村最近的为⑧村，④-⑧距离为1km，因此铺设水管④-⑧。<br> 离①、④、⑧村最近的为⑦村，⑦-⑧距离为0.5km，因此铺设水管⑧-⑦。<br> 离①、④、⑦、⑧村最近的为⑥村，⑦-⑥距离为0.8km，因此铺设水管⑦-⑥。<br> ②、③、⑤村中，离①、④、⑥、⑦、⑧村最近的为③村，⑧-③距离为1km，因此铺设水管⑧-③。<br> ②、⑤村中，离①、③、④、⑥、⑦、⑧村最近的为②村，③-②距离为1km，因此铺设水管③-②。<br> ⑤村离①、②、③、④、⑥、⑦、⑧村最近的为②村，②-⑤距离为1km，因此铺设水管②-⑤。至此，所有8村均已与水库连接，如下图：<br> <img width=\"246\" height=\"68\" src=\"https://image.chaiding.com/ruankao/a1cacb89235875d1da9713d3442714fc.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 因此，从水库开始连接各村水管的最小总长度为：5+5X1+0.5+0.8=11.3km。<br> 这种解决方法，虽然连接方式可能不唯一，但最小总长度是确定的。","multi":0,"questionType":1,"answer":"B","chooseItem":["796234449556623361"],"itemList":[{"id":"796234449514680321","questionId":"796234448579350529","content":" 6.3km","answer":0,"chooseValue":"A"},{"id":"796234449556623361","questionId":"796234448579350529","content":" 11.3km","answer":1,"chooseValue":"B"},{"id":"796234449594372097","questionId":"796234448579350529","content":" 11.8km","answer":0,"chooseValue":"C"},{"id":"796234449632120833","questionId":"796234448579350529","content":" 16.8km","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233968251850753","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某乡有7个小山村A~G，村与村之间原有小路可加宽修建公路的线路如下图所示（路边的数字表示路长的公里数）。为实现村村通公路，修建公路总长至少（55）公里。若在（56）村新建一所中学，则可以使人们从离它最远的村到该校所走的优化路程最短。<br> <img src=\"https://image.chaiding.com/ruankao/77a555490416c834476d1dbea0bb7490.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"300\" height=\"248\" title=\"\" align=\"\">","analyze":"本题考查的是最小生成树相关问题。<br> 可以根据普里姆算法，选择某个顶点出发，每次查找离当前集合最近的结点，直到遍历所有结点一次且仅一次，并且无回路，此时找到的就是最小生成树。假设从F顶点出发，此时离F最近的是D结点【找到FD边】，离｛F，D｝最近的是E结点【找到DE边】，离｛F，D，E｝最近的是G结点【找到EG边】，离｛F，D，E，G｝最近的是C结点【找到EC边】，离｛F，D，E，G，C｝最近的是A结点（此时有2条路径AC、AD可选择，都是最小生成树的结果），接下来离｛F，D，E，G，C，A｝结点最近的是B顶点【找到GB边】。最终形成最小生成树可以有两种形态，如下所示： <p> <img src=\"https://image.chaiding.com/ruankao/a97e46138a5c7080e315e06b397e5f9e.jpg?x-oss-process=style/ruankaodaren\" width=\"250\" height=\"208\" title=\"\" align=\"\" alt=\"\"><br> <img src=\"https://image.chaiding.com/ruankao/cd44181cde0401b9707a9ee79a6e08f2.jpg?x-oss-process=style/ruankaodaren\" width=\"250\" height=\"208\" title=\"\" align=\"\" alt=\"\"><br> 综上可知，修建公路总长至少13.8公里。<br> 问题2：此处可以代入选项进行验证。<br> 如果学校建在A结点，此时B结点离学校最远，有9有公里。<br> 如果学校建在C结点，此时B结点离学校最远，有7.3有公里。<br> 如果学校建在D结点，此时B结点离学校最远，有7有公里。<br> 如果学校建在E结点，此时B结点离学校最远，有5.5有公里。此时从离它最远的村走到该小所走的优化路径最短。 </p>","multi":0,"questionType":1,"answer":"A","chooseItem":["796233969266872321"],"itemList":[{"id":"796233969266872321","questionId":"796233968251850753","content":" 13.8","answer":1,"chooseValue":"A"},{"id":"796233969296232449","questionId":"796233968251850753","content":" 14.3","answer":0,"chooseValue":"B"},{"id":"796233969329786881","questionId":"796233968251850753","content":" 14.8","answer":0,"chooseValue":"C"},{"id":"796233969359147009","questionId":"796233968251850753","content":" 15.3","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233971431133185","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某乡有7个小山村A~G，村与村之间原有小路可加宽修建公路的线路如下图所示（路边的数字表示路长的公里数）。为实现村村通公路，修建公路总长至少（55）公里。若在（56）村新建一所中学，则可以使人们从离它最远的村到该校所走的优化路程最短。<br> <img src=\"https://image.chaiding.com/ruankao/fab3ca73cf2c659f7d5f4e94fca6e988.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"300\" height=\"248\" title=\"\" align=\"\">","analyze":"本题考查的是最小生成树相关问题。<br> 可以根据普里姆算法，选择某个顶点出发，每次查找离当前集合最近的结点，直到遍历所有结点一次且仅一次，并且无回路，此时找到的就是最小生成树。假设从F顶点出发，此时离F最近的是D结点【找到FD边】，离｛F，D｝最近的是E结点【找到DE边】，离｛F，D，E｝最近的是G结点【找到EG边】，离｛F，D，E，G｝最近的是C结点【找到EC边】，离｛F，D，E，G，C｝最近的是A结点（此时有2条路径AC、AD可选择，都是最小生成树的结果），接下来离｛F，D，E，G，C，A｝结点最近的是B顶点【找到GB边】。最终形成最小生成树可以有两种形态，如下所示： <p> <img src=\"https://image.chaiding.com/ruankao/db868ac1ebd470a94b11cb07de1490d1.jpg?x-oss-process=style/ruankaodaren\" width=\"250\" height=\"208\" title=\"\" align=\"\" alt=\"\"><br> <img src=\"https://image.chaiding.com/ruankao/c20b7173002554b6996c3919d7a3255d.jpg?x-oss-process=style/ruankaodaren\" width=\"250\" height=\"208\" title=\"\" align=\"\" alt=\"\"><br> 综上可知，修建公路总长至少13.8公里。<br> 问题2：此处可以代入选项进行验证。<br> 如果学校建在A结点，此时B结点离学校最远，有9有公里。<br> 如果学校建在C结点，此时B结点离学校最远，有7.3有公里。<br> 如果学校建在D结点，此时B结点离学校最远，有7有公里。<br> 如果学校建在E结点，此时B结点离学校最远，有5.5有公里。此时从离它最远的村走到该小所走的优化路径最短。 </p>","multi":0,"questionType":1,"answer":"D","chooseItem":["796233972622315521"],"itemList":[{"id":"796233972588761089","questionId":"796233971431133185","content":" A","answer":0,"chooseValue":"A"},{"id":"796233972601344001","questionId":"796233971431133185","content":" C","answer":0,"chooseValue":"B"},{"id":"796233972613926913","questionId":"796233971431133185","content":" D","answer":0,"chooseValue":"C"},{"id":"796233972622315521","questionId":"796233971431133185","content":" E","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234940445380609","title":"某小区有七栋楼房①~⑦（见下图），各楼房之间可修燃气管道路线的长度（单位：百米）已标记在连线旁。为修建连通各个楼房的燃气管道，该小区内部煤气管道的总长度至少为（ ）百米。<br><img alt=\"\" src=\"https://image.chaiding.com/ruankao/24c0a8589517a79d11fde9db2cecd804.jpg?x-oss-process=style/ruankaodaren\" height=\"184\" width=\"344\">","analyze":"本题考查应用数学（运筹学-图论）基础知识。<br>首先选择最短距离的路线③⑥修建管道（长度为2)。<br>其余五个楼房到己通管道的楼房③⑥距离最短的路线为③⑦，确定修建③⑦管道 (长度为3)。<br>尚未接通的四个楼房到已接通楼房③⑥⑦的最短路线为②⑥，确定修建管道②⑥(长度为4)。<br>尚未接通的楼房①④⑤到已接通的楼房②③⑥⑦的最短路线为①②，确定修建管道①②（长度为3)。<br>尚未接通的楼房④⑤到己接通的楼房①②③⑥⑦的最短路线为④⑦，确定修建管道④⑦（长度为5)。<br>尚未接通的楼房⑤到己接通的楼房①②③④⑥⑦的最短路线为②⑤，确定修建管道②⑤（长度为6)。<br>现在，全部楼房已接通（如图），需要修建的管道总长度为2+3+4+3+5+6=23 (百米)。<br>一般来说，修建总长最短管道的方案可能不唯一，但最短总长度是一致的。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234941506539521"],"itemList":[{"id":"796234941506539521","questionId":"796234940445380609","content":" 23","answer":1,"chooseValue":"A"},{"id":"796234941535899649","questionId":"796234940445380609","content":" 25","answer":0,"chooseValue":"B"},{"id":"796234941561065473","questionId":"796234940445380609","content":" 27","answer":0,"chooseValue":"C"},{"id":"796234941590425601","questionId":"796234940445380609","content":" 29","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239536395931649","title":"某部门邀请3位专家对12个项目进行评选，每个专家选了5个项目。评选的结果中，有a个项目被3人都选中，有b个项目被2个选中，有c个项目被1人选中，有2个项目无人选中。据此，可以推断（52）。","analyze":"本题考查数学应用（方程求解分析）能力。<br>根据题意，a，b，c都是非负整数，a+b+c=12-2=10①，3a+2b+c=3X5=15②。由2①- ②可得c-a=5。<br>a=0时，c=5，b=5，c=a+b；<br>a=1时，c=6，b=3，c＞a+b；<br>a=2时，c=7，b=1，c＞a+b；<br>a＞2时，c＞7，a+c至少为11，与a+b+c=10矛盾。<br>根据上述情况，可以推断供选答案D是正确的。<br>(按汉语常规，a、b、c应均是正整数，a=0的情况不存在，此时应有结论c＞a+b)","multi":0,"questionType":1,"answer":"D","chooseItem":["796239537780051969"],"itemList":[{"id":"796239537683582977","questionId":"796239536395931649","content":" a>2","answer":0,"chooseValue":"A"},{"id":"796239537721331713","questionId":"796239536395931649","content":" b>5","answer":0,"chooseValue":"B"},{"id":"796239537750691841","questionId":"796239536395931649","content":" b为偶数","answer":0,"chooseValue":"C"},{"id":"796239537780051969","questionId":"796239536395931649","content":" c≥a+b","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239636337807361","title":"线性规划问题的数学模型通常由(53)组成。","analyze":"本题考查应用数学基础知识。<br>许多实际应用问题常需要求出一组决策变量的值，这些变量应满足一定的约束条件，并使某个函数达到极大（或极小）值。这个函数就称为目标函数。<br>实际问题中的变量一般都是非负的。如果约束条件是一组线性的不等式（或等式)，目标函数也是线性的，那么这种问题就称为线性规划问题。<br>例如，如下的数学模型就是典型的线性规划问题：<br><img alt=\"\" width=\"284\" height=\"73\" src=\"https://image.chaiding.com/ruankao/a5e2f3292bf9789e9ede6c1909d01039.jpg?x-oss-process=style/ruankaodaren\"><br>因此，线性规划问题的数学模型通常由线性目标函数、线性约束条件和变量非负条件组成。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239637285720065"],"itemList":[{"id":"796239637239582721","questionId":"796239636337807361","content":" 初始值、线性迭代式、收敛条件","answer":0,"chooseValue":"A"},{"id":"796239637260554241","questionId":"796239636337807361","content":" 线性目标函数、线性进度计划、资源分配、可能的问题与应对措施","answer":0,"chooseValue":"B"},{"id":"796239637285720065","questionId":"796239636337807361","content":" 线性目标函数、线性约束条件、变量非负条件","answer":1,"chooseValue":"C"},{"id":"796239637302497281","questionId":"796239636337807361","content":" 网络计划图、资源分配","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239626003042305","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>线性规划问题就是求出一组变量，在一组线性约束条件下，使某个线性目标函数达到极大（小）值。满足线性约束条件的变量区域称为可行解区。由于可行解区的边界均是线性的（平直的），属于单纯形，所以线性目标函数的极值只要存在，就一定会在可行解区边界的某个顶点达到。因此，在求解线性规划问题时，如果容易求出可行解区的所有顶点，那么只要在这些顶点处比较目标函数的值就可以了。<br>例如，线性规划问题：maxS=x+y (求S=x+y的最大值)：2x+y<=7,x+2y<=8,x>=0, y>=0的可行解区是由四条直线2x+y=7;x+2y=8，x=0,y=0, 围成的，共有四个顶点。除了原点外，其他三个顶点是（53）。因此，该线性规划问题的解为(54)","analyze":"本题考查应用数学（线性规划）基础知识。<br>本题中的可行解区是由4条直线2x+y=7, x+2y=8, x=0, y=0围成的，可行解区的每 个顶点都是由两条直线相交得到的。<br>2x+y=7与x=0的交点（0,7)不符合条件x+2y<=8，因此（0,7)不是可行解区的顶点（落在可行解区外)》<br>x+2y=8与y=0的交点（8,0)不符合条件2x+y<=7，因此（8,0)不是可行解区的顶点（落在可行解区外）。<br>2x+y=7与x+2y=8的交点（2，3)，2x+y=7与y=0的交点（3.5，0)，x+2y=8与x=0 的交点（0, 4), x=0与y=0的交点（0, 0)都属于可行解区的顶点。在这4个顶点中， ；x=2,y=3可使目标函数S达到极大值5。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239627043229697"],"itemList":[{"id":"796239627043229697","questionId":"796239626003042305","content":" x=2, y=3","answer":1,"chooseValue":"A"},{"id":"796239627080978433","questionId":"796239626003042305","content":" x=0, y=7","answer":0,"chooseValue":"B"},{"id":"796239627114532865","questionId":"796239626003042305","content":" x=0, y=4","answer":0,"chooseValue":"C"},{"id":"796239627152281601","questionId":"796239626003042305","content":" x=8, y=0","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239558931927041","title":"某企业开发了一种新产品，拟定的价格方案有三种：较高价、中等价、较低价。估计这种产品的销售状态也有三种：销路较好、销路一般、销路较差。根据以往的销售经验，他们算出，这三种价格方案在三种销路状态下的收益值如下表：<br><img alt=\"\" width=\"552\" height=\"87\" src=\"https://image.chaiding.com/ruankao/eb9e5954e1b8c242ad59d3f079908b4f.jpg?x-oss-process=style/ruankaodaren\"><br>企业一旦选择了某种决策方案，在同样的销路状态下，可能会产生后悔值（即所选决策方案产生的收益与最佳决策收益值的差值)。例如，如果选择较低价决策，在销路较好时，后悔值就为8万元。因此，可以根据上述收益值表制作后悔值表如下（空缺部分有待计算):<br><img alt=\"\" width=\"535\" height=\"85\" src=\"https://image.chaiding.com/ruankao/38c0f47cb05d643fa460f37d8268f04f.jpg?x-oss-process=style/ruankaodaren\"><br>企业做定价决策前，首先需要选择决策标准。该企业决定采用最小-最大后悔值决策标准（坏中求好的保守策略)，为此，该企业应选择决策方案（58）。","analyze":"本题考查应用数学基础知识。<br>首先算出各种方案在各种销路状态下的后悔值，填写后悔值表中的空缺部分，并算出每种方案的最大后悔值。<br><img alt=\"\" width=\"535\" height=\"88\" src=\"https://image.chaiding.com/ruankao/08f27cc1062625e69b5b994cd2724ad3.jpg?x-oss-process=style/ruankaodaren\"><br>按照最小最大后悔值决策标准（坏中求好的保守策略），应根据最大后悔值中的最小值来选择对应的决策方案。上表中，最大后悔值中的最小值为4万元（对应中等价），所以决定采用中等价方案。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239559833702401"],"itemList":[{"id":"796239559816925185","questionId":"796239558931927041","content":" 较高价","answer":0,"chooseValue":"A"},{"id":"796239559833702401","questionId":"796239558931927041","content":" 中等价","answer":1,"chooseValue":"B"},{"id":"796239559850479617","questionId":"796239558931927041","content":" 较低价","answer":0,"chooseValue":"C"},{"id":"796239559867256833","questionId":"796239558931927041","content":" 中等价或较低价","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239594046640129","title":"某项目包括A、B、C、D、E五个作业，各个作业的紧前作业、所需时间和所需人数如下表：<br><img alt=\"\" width=\"546\" height=\"92\" src=\"https://image.chaiding.com/ruankao/183e9460599ae9edee2342ca4d245886.jpg?x-oss-process=style/ruankaodaren\"><br>假设该项目的起始时间为0 (单位：周），为使该项目各作业的进度和人力资源安排更合理，各作业的起始时间应分别为（57）。","analyze":"本题考查应用数学基础知识。<br>根据题意，该项目的网络计划图如下：<br><img alt=\"\" width=\"382\" height=\"142\" src=\"https://image.chaiding.com/ruankao/40797b012665a4a6d28883bdbd97fadd.jpg?x-oss-process=style/ruankaodaren\"><br>该项目的关键路径是ACE，共需要4周。作业A应安排在第0周，作业C应安排在第1、2周，作业E应安排在第3周。作业B可以安排在0~2周的某一周，作业D可以安排在1~3周的某一周。现在需要再考虑人力资源的合理安排。<br>先做出作业初步安排的表如下：<br><img alt=\"\" width=\"541\" height=\"108\" src=\"https://image.chaiding.com/ruankao/35e306efcae49def993b5ff5c008cc21.jpg?x-oss-process=style/ruankaodaren\"><br>显然，将作业B和D分别安排在第1、2周可使总人数需求最少（最多需要15人）。如果将作业B安排在第1周，将作业D安排在第2周，则各周需要的人数为10、15、13、15。<br>如果将作业D安排在第1周，将作业B安排在第2周，则各周需要的人数为10、13、15、15。<br>后一种情况人数是逐渐增加的。前一种情况人数是波动的，人员的调度安排常会有些难度。<br>因此，本题较为合理（人力资源均衡分配）的安排如下:<br><img alt=\"\" width=\"554\" height=\"87\" src=\"https://image.chaiding.com/ruankao/8682c2ef25c634779990b778628d0345.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796239595015524353"],"itemList":[{"id":"796239594940026881","questionId":"796239594046640129","content":" 0，0，1，1，3","answer":0,"chooseValue":"A"},{"id":"796239594965192705","questionId":"796239594046640129","content":" 0，2，1，2，3","answer":0,"chooseValue":"B"},{"id":"796239594990358529","questionId":"796239594046640129","content":" 0，1，2，4，5","answer":0,"chooseValue":"C"},{"id":"796239595015524353","questionId":"796239594046640129","content":" 0，2，1，1，3","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239586077462529","title":"某工程项目包括8个作业A～H。各作业的紧前作业、所需天数、所需人数见下表：<br><img alt=\"\" width=\"572\" height=\"103\" src=\"https://image.chaiding.com/ruankao/8aa317bf8128d0c616bcd0ad070e5ea5.jpg?x-oss-process=style/ruankaodaren\"><br><br type=\"_moz\">该项目共有10人，各作业必须连续进行，至少需要（55）天才能完成。","analyze":"本题考查数学应用（进度计划网络图）能力。<br>该项目的进度计划网络图如下，各作业上标注了“作业名（天数，人数)\" ,<br><img alt=\"\" width=\"337\" height=\"161\" src=\"https://image.chaiding.com/ruankao/aa405c85535a91503f4b6491b58fc06b.jpg?x-oss-process=style/ruankaodaren\"><br>如果不考虑人数的限制，关键路径为C-F-G-H，该项目共需2+2+3+4=11天。<br>作业D必须在作业G前完成，但D不能与F并行（DF人数超过10)，所以只能CD 并行2天，F推迟1天开始，导致该项目总天数延长1天。<br>作业B可以在作业C完成后立即开始，并与F并行2天。作业E将与作业G并行。<br>作业A可以与作业H并行。考虑到人数分配的平衡性，可以让AH同时开始。这样，整个项目最后1天只需1人，其他人可以转做别的项目。<br>各作业进度以及人数安排如下表：<br><img alt=\"\" width=\"559\" height=\"217\" src=\"https://image.chaiding.com/ruankao/50752cb13269fdd35f721c0b16bc0999.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"B","chooseItem":["796239586979237889"],"itemList":[{"id":"796239586962460673","questionId":"796239586077462529","content":" 11","answer":0,"chooseValue":"A"},{"id":"796239586979237889","questionId":"796239586077462529","content":" 12","answer":1,"chooseValue":"B"},{"id":"796239587000209409","questionId":"796239586077462529","content":" 13","answer":0,"chooseValue":"C"},{"id":"796239587021180929","questionId":"796239586077462529","content":" 14","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239564707483649","title":"某厂准备生产甲、乙、丙三种产品，生产每件产品所需的A、B两种原料数量，能获得的利润，以及工厂拥有的原料数量如下表：<br><img alt=\"\" width=\"537\" height=\"110\" src=\"https://image.chaiding.com/ruankao/3a3f9046e93ff2a11523fc183926a667.jpg?x-oss-process=style/ruankaodaren\"><br>根据该表，只要安排好生产计划，就能获得最大利润（54）万元。","analyze":"本题考查数学应用（线性规划）能力。<br>设该厂计划生产甲x件，乙y件，丙z件，则有线性规划模型：<br>Max S=3x+4y+z<br>6x+5y+3z≤45<br>3x+5y+4z≤30<br>xyz≤0<br>线性规划问题的最优解必然在可行解区的顶点处达到。 <br>由于产品丙对利润的贡献最低，不妨先假设z=0。<br>此时，容易解得，在x=5，y=3时能获得最大利润27万元。 <br>当z=△＞0时，<br>Max S=3x+4y+△<br>6x+5y≤45-3△<br>3x+5y≤30-4△<br>xy≥0<br>可以得到最优解：x=5+△/3,y=3-△，s=27-2△。<br>即z增加某个增量时，总利润将减少2倍的这些增量。 <br>因此，在x=5, y=3, z=0时能获得最大利润27万元。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239565672173569"],"itemList":[{"id":"796239565626036225","questionId":"796239564707483649","content":" 25","answer":0,"chooseValue":"A"},{"id":"796239565651202049","questionId":"796239564707483649","content":" 26","answer":0,"chooseValue":"B"},{"id":"796239565672173569","questionId":"796239564707483649","content":" 27","answer":1,"chooseValue":"C"},{"id":"796239565697339393","questionId":"796239564707483649","content":" 28","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239550094528513","title":"山区某乡的6个村之间有山路如下图所示，其中的数字标明了各条山路的长度（公里）。<br><img alt=\"\" width=\"250\" height=\"110\" src=\"https://image.chaiding.com/ruankao/9dcdc51e3580a872940219f3a2f30248.jpg?x-oss-process=style/ruankaodaren\"><br>乡政府决定沿山路架设电话线。为实现村村通电话，电话线总长至少为（59）公里。","analyze":" 本题需要在给定的图上寻找最小支撑树。<br> 图由若干个结点以及结点之间的连线组成，每条连线上标记了权数（本题为长度)。<br> 最小支撑树实际上是其中的一个子图，它包括所有的结点以及部分连线，这些连线需要连接所有的结点，但其总权数（长度）最小。<br> 从本题应用看，就是要在上述山路图中确定部分山路，使其能连接6个村，又能使总长度最短。<br> 最小支撑树的求解方法：先选择最短的一条线（如有多条，可以任选一条)，它已经连接了2个点。从这2点出发，再找出能连接其他一个点的最短线（如有多条，可以任选一条)。这样，就已经用2条线连接了3个点。依此类推，逐步做下去，连线也逐步增多，连接的点也逐步增多，直到所有的点都连上为止。这样求出的若千条连线以及所有结点就组成了最小支撑树。<br> 本题求出的一种最小支撑树如下：<br><img alt=\"\" width=\"279\" height=\"130\" src=\"https://image.chaiding.com/ruankao/0b80eb517a8d3b0c440ed1755c56282e.jpg?x-oss-process=style/ruankaodaren\"><br> 其连线的总长度等于14公里，连接了6个村。<br> 在同一个图中，最小支撑树的方案可能有多个，但其连线的总长度是相等的。<br> 这是运筹学求解最优问题的普遍原则：最优值如果有，则必是唯一的，但达到最优值的方案可能不止一个。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239551218601985"],"itemList":[{"id":"796239551193436161","questionId":"796239550094528513","content":" 11","answer":0,"chooseValue":"A"},{"id":"796239551218601985","questionId":"796239550094528513","content":" 14","answer":1,"chooseValue":"B"},{"id":"796239551243767809","questionId":"796239550094528513","content":" 18","answer":0,"chooseValue":"C"},{"id":"796239551273127937","questionId":"796239550094528513","content":" 33","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239623159304193","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>线性规划问题就是求出一组变量，在一组线性约束条件下，使某个线性目标函数达到极大（小）值。满足线性约束条件的变量区域称为可行解区。由于可行解区的边界均是线性的（平直的），属于单纯形，所以线性目标函数的极值只要存在，就一定会在可行解区边界的某个顶点达到。因此，在求解线性规划问题时，如果容易求出可行解区的所有顶点，那么只要在这些顶点处比较目标函数的值就可以了。<br>例如，线性规划问题：maxS=x+y (求S=x+y的最大值)：2x+y<=7,x+2y<=8,x>=0, y>=0的可行解区是由四条直线2x+y=7;x+2y=8，x=0,y=0, 围成的，共有四个顶点。除了原点外，其他三个顶点是（53）。因此，该线性规划问题的解为(54)","analyze":"本题考查应用数学（线性规划）基础知识。<br>本题中的可行解区是由4条直线2x+y=7, x+2y=8, x=0, y=0围成的，可行解区的每个顶点都是由两条直线相交得到的。<br>2x+y=7与x=0的交点（0,7)不符合条件x+2y<=8，因此（0,7)不是可行解区的顶点（落在可行解区外)》<br>x+2y=8与y=0的交点（8,0)不符合条件2x+y<=7，因此（8,0)不是可行解区的顶点（落在可行解区外）。<br>2x+y=7与x+2y=8的交点（2，3)，2x+y=7与y=0的交点（3.5，0)，x+2y=8与x=0的交点（0, 4), x=0与y=0的交点（0, 0)都属于可行解区的顶点。在这4个顶点中 ；x=2,y=3可使目标函数S达到极大值5。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239624082051073"],"itemList":[{"id":"796239624044302337","questionId":"796239623159304193","content":" (2,3), (0,7), (3.5,0)","answer":0,"chooseValue":"A"},{"id":"796239624056885249","questionId":"796239623159304193","content":" (2,3). (0,4), (8,0)","answer":0,"chooseValue":"B"},{"id":"796239624069468161","questionId":"796239623159304193","content":" (2,3)，(0,7), (8,0)","answer":0,"chooseValue":"C"},{"id":"796239624082051073","questionId":"796239623159304193","content":" (2,3), (0,4), (3.5,0)","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239588971532289","title":"某批发站准备向甲、乙、丙、丁四家小商店供应5箱商品。批发站能取得的利润（单位：百元）与分配的箱数有关（见下表）。<br><img alt=\"\" width=\"560\" height=\"161\" src=\"https://image.chaiding.com/ruankao/515111756d4dee2ac8e6183b267e990e.jpg?x-oss-process=style/ruankaodaren\"><br>批发站为取得最大总利润，应分配__(57)__。","analyze":"本题考查数学应用（最优分配）能力。<br>该批发站如将5箱都分配给1家，则最大总利润为9百元（给乙5箱)；<br>如分配给2家（1-4箱或2-3箱)，则最大总利润分别为12或13百元；<br>如分配给3家（1-1-3箱)，则最大总利润为15百元；<br>如分配给3家（1-2-2箱），则最大总利润为16百元（给甲、丙各2箱，给丁1箱); <br>如分配给4家（1-1-1-2箱)，则最大总利润为16百元（给甲、乙、丁各1箱，给丙2箱)。<br>因此，该批发站有两种最优分配方案能取得最大利润16百元。这两种方案中，都需要给丙分配2箱。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239589944610817"],"itemList":[{"id":"796239589898473473","questionId":"796239588971532289","content":" 给甲、丙各1箱","answer":0,"chooseValue":"A"},{"id":"796239589923639297","questionId":"796239588971532289","content":" 给乙2箱","answer":0,"chooseValue":"B"},{"id":"796239589944610817","questionId":"796239588971532289","content":" 给丙2箱","answer":1,"chooseValue":"C"},{"id":"796239589965582337","questionId":"796239588971532289","content":" 给丁2箱","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239619808055297","title":"线性规划问题不可能（）。","analyze":"本题考査应用数学基础知识。<br>线性规划问题的可行解区是一个凸集。如果线性规划问题存在两个最优解，则连接这两个解点的线段上所有的点都必然是可行解。<br>设该线性规划的目标函数为f(X)=C<sub>1</sub>X<sub>1</sub>+C<sub>2</sub>X<sub>2</sub>+…+C<sub>n</sub>X<sub>n</sub>=XC’，其中向量C=(C<sub>1</sub>，C<sub>2</sub>，…，C<sub>n</sub>)，X=(X<sub>1</sub>，X<sub>2</sub>，…，X<sub>n</sub>)。如果f(Y<sub>1</sub>)=f(Y<sub>2</sub>)=M，则连接Y<sub>1</sub>与Y<sub>2</sub>的线段内的任一点λY<sub>1</sub>+μY<sub>2</sub>(λ,μ≥0，λ+μ=1)，也有f(λY<sub>1</sub>+μY<sub>2</sub>)=λf(Y<sub>1</sub>)+μf(Y<sub>2</sub>)=M.也就是说，如果有两个不同的最优解(达到极值M)，则连接这两个点的线段内所有的点也都是最优解(达到同样的极值M)，即必然有无穷多个最优解。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239620823076865"],"itemList":[{"id":"796239620793716737","questionId":"796239619808055297","content":" 没有最优解","answer":0,"chooseValue":"A"},{"id":"796239620810493953","questionId":"796239619808055297","content":" 只有一个最优解","answer":0,"chooseValue":"B"},{"id":"796239620823076865","questionId":"796239619808055297","content":" 只有2个最优解","answer":1,"chooseValue":"C"},{"id":"796239620844048385","questionId":"796239619808055297","content":" 有无穷多个最优解","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239646047621121","title":"载重量限24吨的某架货运飞机准备选装若干箱金属原料运往某地。供选择的各箱原料的重量、运输利润如下表所示。<br><img alt=\"\" width=\"618\" height=\"81\" src=\"https://image.chaiding.com/ruankao/45c4549972040d6191d7035a4a2d7781.jpg?x-oss-process=style/ruankaodaren\"><br>经优化安排，该飞机本次运输可以获得的最大利润为（58）千元。","analyze":"在重量有限制的条件下，为取得最大的利润，显然应优先选择装载“利润重量比”大的货物。先列出每箱货物的利润/重量比如下：<br><img alt=\"\" width=\"614\" height=\"132\" src=\"https://image.chaiding.com/ruankao/92bc31bab67b4881f808ca3477837fc6.jpg?x-oss-process=style/ruankaodaren\"><br>根据利润重量比优先原则，应先装第4箱、第6箱货物。重量己达到16吨，离最大载重量还差8吨，只能再装第1箱，或第3箱，或第5箱。为取得最大利润，再装第1箱更好。<br>所以最优方案是装运箱号为1、4、6的三箱，总利润为3+4+3=10千元。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239647108780033"],"itemList":[{"id":"796239647096197121","questionId":"796239646047621121","content":" 11","answer":0,"chooseValue":"A"},{"id":"796239647108780033","questionId":"796239646047621121","content":" 10","answer":1,"chooseValue":"B"},{"id":"796239647125557249","questionId":"796239646047621121","content":" 9","answer":0,"chooseValue":"C"},{"id":"796239647138140161","questionId":"796239646047621121","content":" 8","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239629069078529","title":"线性规划问题就是面向实际应用，求解一组非负变量，使其满足给定的一组线性约束条件，并使某个线性目标函数达到极值。满足这些约束条件的非负变量组的集合称为可行解域。可行解域中使目标函数达到极值的解称为最优解。以下关于求解线性规划问题的叙述中，不正确的是（56）。","analyze":"线性规划的可行解域是由一组线性约束条件形成的，从几何意义来说，就是由一些线性解面围割形成的区域。由于线性规划的目标函数也是线性的，因此，目标函数的等值域是线性区域。如果在可行解域中的某内点处目标函数达到最优值，则通过该内点的目标函数等值域与可行解域边界的交点也能达到最优解。所以，第一步的结论是：最优解必然会在可行解域的边界处达到。由于目标函数的各个等值域是平行的，而且目标函数的值将随着该等值域向某个方向平行移动而增加或减少（或不变)。如果最优解在可行解域边界某个非顶点处达到，则陣着等值域向某个方向移动，目标函数的值会增加或减少（与最优解矛盾）或没有变化（在此段边界上都达到最优解），从而仍会在可行解域的某个顶点处达到最优解。<br>既然可行解域是由一组线性约束条件所对的线性区域围成的，那么再增加一个约束条件时，要么缩小可行解域（新的约束条件分割了原来的可行解域)，要么可行解域不变（新的约束条件与原来的可行解域不相交)。<br>如果可行解域是无界的，那么目标函数的等值域向某个方向平移（目标函数的值线性变化）时，可能出现无限增加或无限减少的情况，因此有可能没有最优解。当然，有时，即使可行解域是无界的，但仍然有最优解，但确实会有不存在最优解的情况。<br>由于线性规划的可行解域是凸域，区域内任取两点，则这两点的连线上所有的点都属于可行解域（线性函数围割而成的区域必是凸域)。如果线性规划问题在可行解域的某两个点上达到最优解（等值），则在这两点的连线上都能达到最优解（如果目标函数的等值域包括某两个点，则也会包括这两点连线上的所有点）。因此，线性规划问题的最优解要么是0个（没有)，要么是唯一的（1个)，要么有无穷个（只要有2个，就会有无穷个）。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239630008602625"],"itemList":[{"id":"796239629962465281","questionId":"796239629069078529","content":" 线性规划问题如果有最优解，则一定会在可行解域的某个顶点处达到","answer":0,"chooseValue":"A"},{"id":"796239629987631105","questionId":"796239629069078529","content":" 线性规划问题中如果再增加一个约束条件，则可行解域将缩小或不变","answer":0,"chooseValue":"B"},{"id":"796239630008602625","questionId":"796239629069078529","content":" 线性规划问题如果存在可行解，则一定有最优解","answer":1,"chooseValue":"C"},{"id":"796239630029574145","questionId":"796239629069078529","content":" 线性规划问题的最优解只可能是0个、1个或无穷多个","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239556088188929","title":"某书店准备向出版社订购一批本地旅游新版书，书的定价为每本30元，订购价为每本15元。如果该书在年底前尚未售出，则不得不以每本5元的价格退回给出版社。根据以往经验，按定价售出150本、160本、170本、180本的概率分别为0.1、0.2、0.4、0.3。为获取最大期望利润，该书店应订购此书（57）本。","analyze":"本题考查应用数学（决策论）知识。<br>根据题意，我们先对订购150本、151~159本、160本、161~169本、170本、171~ 179、180本的多种情况，以及按书定价售出150本、160本、170本、180本的多种可能，分别计算其利润值，填入下表（设0<x<10):<br><img alt=\"\" width=\"450\" height=\"151\" src=\"https://image.chaiding.com/ruankao/5f163b2cfd39cb30d33a4e2b7b1f3570.jpg?x-oss-process=style/ruankaodaren\"><br>再根据各种销售情况的概率计算出期望利润。从表中看出，在订购170本时能获得最大利润2450元。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239556994158593"],"itemList":[{"id":"796239556968992769","questionId":"796239556088188929","content":" 160","answer":0,"chooseValue":"A"},{"id":"796239556981575681","questionId":"796239556088188929","content":" 161〜169","answer":0,"chooseValue":"B"},{"id":"796239556994158593","questionId":"796239556088188929","content":" 170","answer":1,"chooseValue":"C"},{"id":"796239557002547201","questionId":"796239556088188929","content":" 171 〜180","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239561792442369","title":"某公司计划开发一种新产品，其开发前景有成功、较成功与失败三种可能情况。根据该公司的技术水平与市场分析，估计出现这三种情况的概率分别为40%、40%和20%。现有三种开发方案可供选择，每种方案在不同开发前景下估计获得的利润（单位：万元）如下表：<br><img alt=\"\" width=\"550\" height=\"111\" src=\"https://image.chaiding.com/ruankao/2f4c8922677b6d3a00d54a1f30cbd163.jpg?x-oss-process=style/ruankaodaren\"><br><br>为获得最大的期望利润，该公司应选择（60）。","analyze":"本题考查应用数学的基础知识。<br>根据题意，通过计算可以得到：<br>方案1的期望利润为20*40%+5*40%-10*20%=8 (万元）<br>方案2的期望利润为16*40%+8*40%-5*20%=8.6 (万元）<br>方案3的期望利润为12*40%+5*40%-2*20%=6.4 (万元）<br>为获得最大的期望利润，该公司应选择方案2。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239562715189249"],"itemList":[{"id":"796239562698412033","questionId":"796239561792442369","content":" 方案1","answer":0,"chooseValue":"A"},{"id":"796239562715189249","questionId":"796239561792442369","content":" 方案2","answer":1,"chooseValue":"B"},{"id":"796239562727772161","questionId":"796239561792442369","content":" 方案3","answer":0,"chooseValue":"C"},{"id":"796239562740355073","questionId":"796239561792442369","content":" 方案1或方案2","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239553219284993","title":"某IT企业计划对一批新招聘的技术人员进行岗前脱产培训，培训内容包括编程和测试两个专业，每个专业要求在基础知识、应用技术和实际训练三个方面都得到提高。根据培训大纲，每周的编程培训可同时获得基础知识3学分、应用技术7学分以及实际训练10学分；每周的测试培训可同时获得基础知识5学分、应用技术2学分以及实际训练7学分。企业要求这次岗前培训至少能完成基础知识70学分，应用技术86学分，实际训练185学分。以上说明如下表所示：<br><img alt=\"\" width=\"620\" height=\"104\" src=\"https://image.chaiding.com/ruankao/f947e32b83ee6f3e71e2c62fb78630b1.jpg?x-oss-process=style/ruankaodaren\"><br>那么这样的岗前培训至少需要（56）周时间才能满足企业的要求。","analyze":"设安排编程培训x周，测试培训y周，则可以建立本题的线性规划模型如下:<br>目标函数：x+y，求最小值<br>约束条件：3x+5y≥70<br> 7x+2y≥86<br> 10x+7y≥185<br>非负条件：x,y≥0<br>该线性规划问题的图解法如下：<br>在坐标系第一象限内（因为要求x,y≥0)：<br>画直线L1: 3x+5y=7≤0(一定通过点（0，14)与（70/3,0))<br>所以，3x+5y≥70表示在直线L1之上的区域。<br>画直线L2: 7x+2y=86(―定通过点（0,43)与（86/7,0))<br>所以，7x+2y≥86表示在直线L2之上的区域。<br>画直线 L3: 10x+7y=185(―定通过点（0，185/7)与（20,18.5))<br>所以，10x+7y≥185表示在直线L3之上的区域。<br>上述三个约束条件以及变量非负条件组成的可行解区域见下图。<br><img alt=\"\" width=\"268\" height=\"252\" src=\"https://image.chaiding.com/ruankao/747f2a4671b14a5e53a99a59dc12f5ee.jpg?x-oss-process=style/ruankaodaren\"><br>根据线性规划方法，目标函数的最小值一定会在可行解区的顶点处到达。<br>因此，只要考察直线L1与L3的交点以及直线L2与L3的交点处目标函数的值。<br>L1与L3的交点满足：<br>3x+5y=70 <br>10x+7y=185<br>可以求出可行解区的一个顶点为（15,5),因此，x+y=20。<br>L2与L3的交点满足：<br>7x+2y=86 <br>10x+7y=185<br>可以求出可行解区的另一个顶点为（8,15),因此，x+y=23。<br>比较这两个顶点处的x+y值，就能知道本题的最优解就是：<br>x=15 (周），y=5 (周），x+y的最小值为20(周）。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239554133643265"],"itemList":[{"id":"796239554108477441","questionId":"796239553219284993","content":" 15","answer":0,"chooseValue":"A"},{"id":"796239554121060353","questionId":"796239553219284993","content":" 18","answer":0,"chooseValue":"B"},{"id":"796239554133643265","questionId":"796239553219284993","content":" 20","answer":1,"chooseValue":"C"},{"id":"796239554150420481","questionId":"796239553219284993","content":" 23","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239541659783169","title":"下面的网络图表示从城市A到城市B运煤的各种路线。各线段上的数字表示该线段运煤所需的费用（百元/车）。城市A有三个装货点，城市B有三个卸货点，各点旁标注的数字表示装/卸煤所需的费用（百元/车）。根据该图，从城市A的一个装卸点经过一条路线到城市B的一个卸货点所需的装、运、卸总费用至少为（56）（百元/车）。<br><img alt=\"\" width=\"443\" height=\"257\" src=\"https://image.chaiding.com/ruankao/3718f0df010090e83eea7ef6482ff95b.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考査数学应用（最优路径）能力。<br>从A线出发经过中间5点可以到达B线。首先，很容易计算并标注各条路线从第5点到达B线并卸货的最少费用，可将其标注在相应的点旁。据此就容易计算并标注从第4点到达B线并卸货的最少费用，并将其标注在相应的点旁，依次类推。<br><img alt=\"\" width=\"391\" height=\"199\" src=\"https://image.chaiding.com/ruankao/e4b94541e7d483579e633d6a4c5b4e29.jpg?x-oss-process=style/ruankaodaren\"><br>从A的下端出发，向上、上、下、上、上、下到达B的中间点，总费用=3+（2+2+3+2+2+3) + 2=19 (百元/车）最少。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239542553169921"],"itemList":[{"id":"796239542553169921","questionId":"796239541659783169","content":" 19","answer":1,"chooseValue":"A"},{"id":"796239542578335745","questionId":"796239541659783169","content":" 20","answer":0,"chooseValue":"B"},{"id":"796239542599307265","questionId":"796239541659783169","content":" 21","answer":0,"chooseValue":"C"},{"id":"796239542628667393","questionId":"796239541659783169","content":" 22","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235170289045505","title":"加工某种零件需要依次经过毛坯、机加工、热处理和检验四道工序。各道工序有多种方案可选，对应不同的费用。下图表明了四道工序各种可选方案(连线)的衔接关系，线旁的数字表示该工序加工一个零件所需的费用(单位:元)。从该图可以推算出，加工一个零件的总费用至少需要（ ）元。<br><img width=\"386\" height=\"280\" alt=\"\" src=\"https://image.chaiding.com/ruankao/3e48161b0e6bbc8cd56d1275d11b25d6.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学（运筹学—最短路径）基础知识。<br>用倒推方法计算如下：<br>G-I需要20元，H-I需要10元。<br>D-I最少需要60元，E-I最少需要40元（EGI)，F-I最少需要60元。<br>B-I最少需要80元(BEGI), C-I最少需要100元。<br>A-I最少需要130元（ABEGI)。","multi":0,"questionType":1,"answer":"B","chooseItem":["796235171337621505"],"itemList":[{"id":"796235171312455681","questionId":"796235170289045505","content":" 120","answer":0,"chooseValue":"A"},{"id":"796235171337621505","questionId":"796235170289045505","content":" 130","answer":1,"chooseValue":"B"},{"id":"796235171362787329","questionId":"796235170289045505","content":" 140","answer":0,"chooseValue":"C"},{"id":"796235171387953153","questionId":"796235170289045505","content":" 150","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233958546231297","title":"各种线性规划模型都可以将其标准化。线性规划模型标准形式的特点不包括（）。","analyze":"本题考查线性规划相关知识。<br> 线性规划的标准型（standard form of linearprogramrmng）是线性规划模型的标准形式。其主要特征为：<br> （1）目标函数为极大化类型；<br> （2）所有的约束条件都是等式；<br> （3）所有约束方程右端的常数都是非负的，C选项描述不够准确；<br> （4）所有决策变量都是非负的。","multi":0,"questionType":1,"answer":"C","chooseItem":["796233959582224385"],"itemList":[{"id":"796233959548669953","questionId":"796233958546231297","content":" 目标函数达到最大化（或最小化）","answer":0,"chooseValue":"A"},{"id":"796233959561252865","questionId":"796233958546231297","content":" 约束条件都是线性等式","answer":0,"chooseValue":"B"},{"id":"796233959582224385","questionId":"796233958546231297","content":" 约束条件中的常数系数均为非负","answer":1,"chooseValue":"C"},{"id":"796233959599001601","questionId":"796233958546231297","content":" 所有的决策变量均为非负","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234921835253761","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某厂拥有三种资源A、B、C，生产甲、乙两种产品。生产每吨产品需要消耗的资源、可以获得的利润见下表。目前，该厂拥有资源A、资源B和资源C分別为12吨、7吨和12吨。根据上述说明，适当安排甲、乙两种产品的生产量，就能获得最大总利润（53）。如果生产计划只受资源A和C的约束，资源B很容易从市场上以每吨0.5百万元购得，则该厂宜再购买（54）资源B，以获得最大的总利润。<br><img alt=\"\" src=\"https://image.chaiding.com/ruankao/39b86dfae6ac2a20d7e7756d375fa17c.jpg?x-oss-process=style/ruankaodaren\" height=\"169\" width=\"499\">","analyze":"本题考查应用数学（运筹学-线性规划）基础知识。<br>设产品甲生产x吨，产品乙生产y吨，则线性规划模型为：<br><img width=\"405\" height=\"170\" alt=\"\" src=\"https://image.chaiding.com/ruankao/3109888c80594311c99d3b169d59619c.jpg?x-oss-process=style/ruankaodaren\"><br>可行解区为左下的五边形，其顶点为：（0,0)，（6,0)，（5,2)，（2,5)，(0,6)。 <br>显然，在顶点（5,2)处目标函数S达到最大值19。即产品甲生产5吨，产品乙生生产2吨，可以取得最大总利润19百万元。<br>如果资源B没有约束，可以外购，设新购n吨，则需要多花费0.5n百万元，则线性规划模型修改为：<br>求maxS=3x+2y-0.5n <br>约束条件：2x+y≤12 <br>x+y≤7+n <br>x+2y≤12 <br>x,y,n≥0<br>从上图看出，当n≥1时，直线x+y=7+n,对形成可行解区不起作用。<br>此时，可行解区四边形顶点为（0,0)，（6,0)，（4,4)，（0,6)。<br>只有当x=y=4时S取得最大值，maxS=max20-0.5n}。只有当n=1时取得最大值19.5。 <br>当0≤n≤1时，可行解区五边形的顶点为：<br>(0,0)，（6,0),(5-n，2+2n），(2+2n,5-n),(0,6)。<br>maxS=max{-0.5n,18-0.5n，19+0.5n，16+3.5n，12-0.5n}(0≤n≤1),只有当n=1时S取得最大值19.5。<br>因此，在资源B无约束条件下，为取得最大总利润，应增购1吨资源B。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234922812526593"],"itemList":[{"id":"796234922762194945","questionId":"796234921835253761","content":" 16百万元","answer":0,"chooseValue":"A"},{"id":"796234922791555073","questionId":"796234921835253761","content":" 18百万元","answer":0,"chooseValue":"B"},{"id":"796234922812526593","questionId":"796234921835253761","content":" 19百万元","answer":1,"chooseValue":"C"},{"id":"796234922837692417","questionId":"796234921835253761","content":" 20百万元","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234924809015297","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某厂拥有三种资源A、B、C，生产甲、乙两种产品。生产每吨产品需要消耗的资源、可以获得的利润见下表。目前，该厂拥有资源A、资源B和资源C分別为12吨、7吨和12吨。根据上述说明，适当安排甲、乙两种产品的生产量，就能获得最大总利润（53）。如果生产计划只受资源A和C的约束，资源B很容易从市场上以每吨0.5百万元购得，则该厂宜再购买（54）资源B，以获得最大的总利润。<br><img alt=\"\" src=\"https://image.chaiding.com/ruankao/191a2f23196a27e77c637c4c90fadca1.jpg?x-oss-process=style/ruankaodaren\" height=\"169\" width=\"499\">","analyze":"本题考查应用数学（运筹学-线性规划）基础知识。<br>设产品甲生产x吨，产品乙生产y吨，则线性规划模型为：<br><img width=\"405\" height=\"170\" alt=\"\" src=\"https://image.chaiding.com/ruankao/2e6149b1fc7d7676083c60b56cd57440.jpg?x-oss-process=style/ruankaodaren\"><br>可行解区为左下的五边形，其顶点为：（0,0)，（6,0)，（5,2)，（2,5)，(0,6)。 <br>显然，在顶点（5,2)处目标函数S达到最大值19。即产品甲生产5吨，产品乙生生产2吨，可以取得最大总利润19百万元。<br>如果资源B没有约束，可以外购，设新购n吨，则需要多花费0.5n百万元，则线性规划模型修改为：<br>求maxS=3x+2y-0.5n <br>约束条件：2x+y≤12 <br>x+y≤7+n <br>x+2y≤12 <br>x,y,n≥0<br>从上图看出，当n≥1时，直线x+y=7+n,对形成可行解区不起作用。<br>此时，可行解区四边形顶点为（0,0)，（6,0)，（4,4)，（0,6)。<br>只有当x=y=4时S取得最大值，maxS=max20-0.5n}。只有当n=1时取得最大值19.5。 <br>当0≤n≤1时，可行解区五边形的顶点为：<br>(0,0)，（6,0),(5-n，2+2n），(2+2n,5-n),(0,6)。<br>maxS=max{-0.5n,18-0.5n，19+0.5n，16+3.5n，12-0.5n}(0≤n≤1),只有当n=1时S取得最大值19.5。<br>因此，在资源B无约束条件下，为取得最大总利润，应增购1吨资源B。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234925853396993"],"itemList":[{"id":"796234925853396993","questionId":"796234924809015297","content":" 1吨","answer":1,"chooseValue":"A"},{"id":"796234925878562817","questionId":"796234924809015297","content":" 2吨","answer":0,"chooseValue":"B"},{"id":"796234925899534337","questionId":"796234924809015297","content":" 3吨","answer":0,"chooseValue":"C"},{"id":"796234925924700161","questionId":"796234924809015297","content":" 4吨","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235411222450177","title":"某地区仅有甲、乙两个企业为销售同种电子产品竞争市场份额。甲企业有三种策略A、B、C，乙企业也有三种策略Ⅰ、Ⅱ、Ⅲ。两企业分别独立地选择各种策略时，预计甲企业将增加的市场份额（百分点）见下表（负值表示乙企业将增加的市场份额）。若两企业都采纳稳妥的保守思想（从最坏处着想，争取最好的结果），则（57）。<br><img alt=\"\" width=\"500\" height=\"96\" src=\"https://image.chaiding.com/ruankao/b798a617f386d2e0ff3ac260b811bf4d.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学基础知识（运筹一对策）。<br>甲企业若选择策略A，则最差情况会失去市场1个百分点；<br>甲企业若选择策略B，则最差情况会失去市场5个百分点；<br>甲企业若选择策略C，则最差情况市场份额没有变化，<br>因此甲企业决定选择策略C。<br>乙企业若选择策略Ⅰ，则最差情况会失去市场12个百分点；<br>乙企业若选择策略Ⅱ，则最差情况会失去市场10个百分点；<br>乙企业若选择策略Ⅲ，则最差情况会失去市场5个百分点，<br>因此乙企业决定选择策略Ⅲ。","multi":0,"questionType":1,"answer":"D","chooseItem":["796235412245860353"],"itemList":[{"id":"796235412149391361","questionId":"796235411222450177","content":" 甲选择策略B，乙选择策略Ⅲ","answer":0,"chooseValue":"A"},{"id":"796235412182945793","questionId":"796235411222450177","content":" 甲选择策略A，乙选择策略Ⅱ","answer":0,"chooseValue":"B"},{"id":"796235412216500225","questionId":"796235411222450177","content":" 甲选择策略B，乙选择策略Ⅱ","answer":0,"chooseValue":"C"},{"id":"796235412245860353","questionId":"796235411222450177","content":" 甲选择策略C，乙选择策略Ⅲ","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235414380761089","title":"某工厂每年需要铁矿原料100万吨，切假设全年对这种原料的消耗是均匀的。为了减少库存费用，准备平均分多批进货。库存费按平均年库存量（每次进货量的一半）以每万吨500元计算。由于每次进货需要额外支出订单费1000元，所以每次进货次数也不能太多。为节省库存费和订货费总支出，最经济的办法是（58）。","analyze":"本题考查应用数学基础知识（运筹-库存）。<br>设每次进货x万吨，则平均库存量为x/2万吨，年库存费=500x/2=250x元，<br>年订货次数=100/x，年订货费=1000*100/x=100000/x元。<br>总支出y=250x+100000/x元。<br>通过求导数分析极值知，当x=20时，Y取得最小值。","multi":0,"questionType":1,"answer":"C","chooseItem":["796235415328673793"],"itemList":[{"id":"796235415303507969","questionId":"796235414380761089","content":" 每年进货2次，每次进货50万吨","answer":0,"chooseValue":"A"},{"id":"796235415316090881","questionId":"796235414380761089","content":" 每年进货4次，每次进货25万吨","answer":0,"chooseValue":"B"},{"id":"796235415328673793","questionId":"796235414380761089","content":" 每年进货5次，每次进货20万吨","answer":1,"chooseValue":"C"},{"id":"796235415341256705","questionId":"796235414380761089","content":" 每年进货10次，每次进货10万吨","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235404670947329","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某工程包括A、B、C、D、E、F六个作业，分别需要5、7、3、4、15、12天。A必须在C、D开始之前完成，B、D必须在E开始之前完成，C必须在F开始之前完成，F不能在B、D完成之前开始。该工程的工期至少需要（55）天。若作业E缩短4天，则整个工期可以缩短（56）天。","analyze":"本题考查应用数学基础知识（运筹一网络计划图）。<br>根据题意画出网络计划图如下:<br><img alt=\"\" width=\"489\" height=\"165\" src=\"https://image.chaiding.com/ruankao/b19abf0b1fc869f2f40057e6c13538ee.jpg?x-oss-process=style/ruankaodaren\"><br>关键路径（最长工期路径）为ADE，工期为5+4+15=24天。时间安排如下图：<br><img alt=\"\" width=\"463\" height=\"121\" src=\"https://image.chaiding.com/ruankao/b6ae4c748f989288bd4e6603b5725c35.jpg?x-oss-process=style/ruankaodaren\"><br>作业B可以在前9天内安排7天进行；作业C和F可以在第5到24天内依次安排，但作业F必须安排在第9天以后。如果作业E缩短4天，变成E11，则关键路径变成ADF，工期变成21天，缩短3天。","multi":0,"questionType":1,"answer":"C","chooseItem":["796235405690163201"],"itemList":[{"id":"796235405618860033","questionId":"796235404670947329","content":" 21","answer":0,"chooseValue":"A"},{"id":"796235405652414465","questionId":"796235404670947329","content":" 22","answer":0,"chooseValue":"B"},{"id":"796235405690163201","questionId":"796235404670947329","content":" 24","answer":1,"chooseValue":"C"},{"id":"796235405723717633","questionId":"796235404670947329","content":" 46","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235408156413953","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某工程包括A、B、C、D、E、F六个作业，分别需要5、7、3、4、15、12天。A必须在C、D开始之前完成，B、D必须在E开始之前完成，C必须在F开始之前完成，F不能在B、D完成之前开始。该工程的工期至少需要（55）天。若作业E缩短4天，则整个工期可以缩短（56）天。","analyze":"本题考查应用数学基础知识（运筹一网络计划图）。<br>根据题意画出网络计划图如下:<br><img alt=\"\" width=\"489\" height=\"165\" src=\"https://image.chaiding.com/ruankao/aad0867cd1ceb754aa1a231d0b438ac2.jpg?x-oss-process=style/ruankaodaren\"><br>关键路径（最长工期路径）为ADE，工期为5+4+15=24天。时间安排如下图：<br><img alt=\"\" width=\"463\" height=\"121\" src=\"https://image.chaiding.com/ruankao/1b425113c020dad5c229d2389e00a4a0.jpg?x-oss-process=style/ruankaodaren\"><br>作业B可以在前9天内安排7天进行；作业C和F可以在第5到24天内依次安排，但作业F必须安排在第9天以后。如果作业E缩短4天，变成E11，则关键路径变成ADF，工期变成21天，缩短3天。","multi":0,"questionType":1,"answer":"C","chooseItem":["796235409225961473"],"itemList":[{"id":"796235409167241217","questionId":"796235408156413953","content":" 1","answer":0,"chooseValue":"A"},{"id":"796235409196601345","questionId":"796235408156413953","content":" 2","answer":0,"chooseValue":"B"},{"id":"796235409225961473","questionId":"796235408156413953","content":" 3","answer":1,"chooseValue":"C"},{"id":"796235409259515905","questionId":"796235408156413953","content":" 4","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235163813040129","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某工程有七个作业A~G，按计划，完成各作业所需的时间以及作业之间的衔接关系见下表：<br><img width=\"618\" height=\"76\" alt=\"\" src=\"https://image.chaiding.com/ruankao/d3b25e31a14a1be8170e4c5a0ab566e1.jpg?x-oss-process=style/ruankaodaren\"><br>按照上述计划，该工程的总工期预计为（54）周。<br>在工程实施了10周后，经理对进度进行了检查，结果是：作业A和B已经完成，作业D完成了30% ，作业E完成了25% ，其他作业都还没有开始。<br>如果随后完全按原计划实施，则总工期将（55）完成。","analyze":"本题考查应用数学（运筹学-网络计划图）基础知识。<br>根据题意，绘制该工程的网络计划图如下：<br><img width=\"543\" height=\"138\" alt=\"\" src=\"https://image.chaiding.com/ruankao/9d0ac00b373cefd256e81a93c23b50cf.jpg?x-oss-process=style/ruankaodaren\"><br>因此，关键路径为B-D-G，预计总工期=6+10+4=20周。 <br>还可以画出甘特图如下：<br><img width=\"449\" height=\"177\" alt=\"\" src=\"https://image.chaiding.com/ruankao/3f78e1e4eea4f0889ad438911eaab3fa.jpg?x-oss-process=style/ruankaodaren\"><br>若按照原计划，该工程开始10周后，作业A和B必须完成，作业C将完成4/5=80% (4周工作量），作业D必须完成4/10=40% (4周工作量)，作业E将完成4/8=50% (4周工作量）。<br>在工程开始10周后实际进行检查时，作业D只完成了30% （3周的工作量）因此作业D不得不推迟1周完成。B-D-G路径需要21周完成。<br>由于检查时，作业C尚未开始，所以它推迟了4周。不过，作业C和F还可以在前18周内完成，没有影响总工期。<br>由于检查时，作业E只完成了25% (2周工作量)，相当于它推迟了2周。不过，它还可以在前16周完成，并不影响总工期。<br>综合看，如果随后的时间内完全按原计划实施，则该工程将推迟1周完成。","multi":0,"questionType":1,"answer":"A","chooseItem":["796235164735787009"],"itemList":[{"id":"796235164735787009","questionId":"796235163813040129","content":" 20","answer":1,"chooseValue":"A"},{"id":"796235164756758529","questionId":"796235163813040129","content":" 25","answer":0,"chooseValue":"B"},{"id":"796235164773535745","questionId":"796235163813040129","content":" 33","answer":0,"chooseValue":"C"},{"id":"796235164790312961","questionId":"796235163813040129","content":" 41","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235167030071297","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某工程有七个作业A~G，按计划，完成各作业所需的时间以及作业之间的衔接关系见下表：<br><img width=\"618\" height=\"76\" alt=\"\" src=\"https://image.chaiding.com/ruankao/a716c18a6411cc0ac29b33a5fd4ae72a.jpg?x-oss-process=style/ruankaodaren\"><br>按照上述计划，该工程的总工期预计为（54）周。<br>在工程实施了10周后，经理对进度进行了检查，结果是：作业A和B已经完成，作业D完成了30% ，作业E完成了25% ，其他作业都还没有开始。<br>如果随后完全按原计划实施，则总工期将（55）完成。","analyze":"本题考查应用数学（运筹学-网络计划图）基础知识。<br>根据题意，绘制该工程的网络计划图如下：<br><img width=\"545\" height=\"138\" alt=\"\" src=\"https://image.chaiding.com/ruankao/6921d27ac0bb758d14afb0a912e4eee4.jpg?x-oss-process=style/ruankaodaren\"><br>因此，关键路径为B-D-G，预计总工期=6+10+4=20周。 <br>还可以画出甘特图如下：<br><img width=\"450\" height=\"178\" alt=\"\" src=\"https://image.chaiding.com/ruankao/3ff8310e7f690eb791dac8176c921c54.jpg?x-oss-process=style/ruankaodaren\"><br>若按照原计划，该工程开始10周后，作业A和B必须完成，作业C将完成4/5=80% (4周工作量），作业D必须完成4/10=40% (4周工作量)，作业E将完成4/8=50% (4周工作量）。<br>在工程开始10周后实际进行检查时，作业D只完成了30% （3周的工作量）因此作业D不得不推迟1周完成。B-D-G路径需要21周完成。<br>由于检查时，作业C尚未开始，所以它推迟了4周。不过，作业C和F还可以在前18周内完成，没有影响总工期。<br>由于检查时，作业E只完成了25% (2周工作量)，相当于它推迟了2周。不过，它还可以在前16周完成，并不影响总工期。<br>综合看，如果随后的时间内完全按原计划实施，则该工程将推迟1周完成。","multi":0,"questionType":1,"answer":"B","chooseItem":["796235168032509953"],"itemList":[{"id":"796235168011538433","questionId":"796235167030071297","content":" 提前1周","answer":0,"chooseValue":"A"},{"id":"796235168032509953","questionId":"796235167030071297","content":" 推迟1周","answer":1,"chooseValue":"B"},{"id":"796235168049287169","questionId":"796235167030071297","content":" 推迟2周","answer":0,"chooseValue":"C"},{"id":"796235168070258689","questionId":"796235167030071297","content":" 推迟3周","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234451674746881","title":"某运输网络图（见下图）有A~E五个结点，结点之间标有运输方向箭线，每条箭线旁标有两个数字，前一个是单位流量的运输费用，后一个是该箭线所允许的单位时间内的流量上限。从结点A到E可以有多种分配运输量的方案。如果每次都选择最小费用的路径来分配最大流量，则可以用最小总费用获得最大总流量的最优运输方案。该最优运输方案中，所需总费用和达到的总流量分别为(56)。<br> <img src=\"https://image.chaiding.com/ruankao/14d2bddfbf60f6e6e4d7abfd550fecd2.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"430\" height=\"114\" title=\"\" align=\"\">","analyze":"本题考查应用数学-运筹学-网络图的基础知识。<br> 从原图中的运输费用来看，从A到E的路径ACBE上单位流量的总费用最低，为1+2+1=4,最多可以分配流量min{8，5，7}=5。除去流量5后得到如下图：<br> <img width=\"578\" height=\"113\" src=\"https://image.chaiding.com/ruankao/3a4ae0c3f1bab650aa1919b6eddf18fd.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 从该图中的运输费用来看，从A到E的路径ABE上单位流量的总费用最低，为4+1=5, 最多可以分配流量min{10，2}=2。除去流量2后得到如下图：<br> <img width=\"576\" height=\"101\" src=\"https://image.chaiding.com/ruankao/bff489961a229afa0754abf83ccf8f51.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 从该图中的运输费用来看，从A到E的路径ACDE上单位流量的总费用最低，为1+3+2=6,最多可以分配流量min{3，10，4}=3。除去流量3后得到如下图：<br> <img width=\"581\" height=\"107\" src=\"https://image.chaiding.com/ruankao/74fa580b07e4fa4501a0e2a10a92252d.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 从该图看，从A到E只有路径ABDE，单位流量的总费用=4+6+2=12,最多可以分配流量min{8，2，1}=1。<br> 上述运输方案，总流量=5+2+3+1=11，总费用=5X4+2X5+3X6+1X12=60。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234452752683009"],"itemList":[{"id":"796234452681379841","questionId":"796234451674746881","content":" 4，5","answer":0,"chooseValue":"A"},{"id":"796234452714934273","questionId":"796234451674746881","content":" 12，16","answer":0,"chooseValue":"B"},{"id":"796234452752683009","questionId":"796234451674746881","content":" 60，11","answer":1,"chooseValue":"C"},{"id":"796234452790431745","questionId":"796234451674746881","content":" 71，11","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234442308866049","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某项目有8个作业A~H，每个作业的紧前作业、所需天数和所需人数见下表。由于整个项目团队总共只有9人，各个作业都必须连续进行，中途不能停止，因此需要适当安排施工方案，使该项目能尽快在（53）内完工。在该方案中,作业A应安排在（54）内进行。<br> <img src=\"https://image.chaiding.com/ruankao/9004017545afea5e2e2e950c4b9b8749.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"650\" height=\"109\" title=\"\" align=\"\">","analyze":"本题考查应用数学-运筹学-网络计划图的基础知识。<br> 根据题中各作业的紧前作业和所需天数，可绘制网络计划图如下：<br> <img src=\"https://image.chaiding.com/ruankao/a440e9f0430bfc1762350db0078b6528.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"350\" height=\"110\" title=\"\" align=\"\"><br> 根据该图，如果不考虑人数限制，该项目的关键路径为C-E-F-H，需要2+2+3+3=10天。 <br> 先考虑安排关键路径上这几个作业顺序进行：第1~2天安排5人做作业C，第3~4天安排1人做作业E，第5~7天安排1人做作业F，第8~10天安排6人做作业H，图示如下:<br> <img width=\"766\" height=\"42\" src=\"https://image.chaiding.com/ruankao/699e6d6b091ce2ead075def5c9c31344.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 由于作业B必须在F之前进行，需要8人做1天，只能安排在第3或第4天进行。<br> 由于作业D和G必须在H之前进行，作业D需要4人做2天，安排在第1~2天为好。而作业G需要7人做2天。将作业B安排在第3天，将作业G安排在第4~5天为好。 <br> 作业A虽然可以在全程安排，但由于需要7人，所以安排在第6~8天为好。然而第8天作业H已暂时安排6人，这样会引发第8天人数（6+7)超出9人的限制。最好的解决办法是将作业H推迟1天。从而，在每天人数限制9人的条件下，项目最快能在11天完成， 实施方案图示如下：<br> <img width=\"768\" height=\"122\" src=\"https://image.chaiding.com/ruankao/2b4d0051b7bf89f97394a040f0b74897.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br></br>\n","multi":0,"questionType":1,"answer":"B","chooseItem":["796234443365830657"],"itemList":[{"id":"796234443319693313","questionId":"796234442308866049","content":" 10天","answer":0,"chooseValue":"A"},{"id":"796234443365830657","questionId":"796234442308866049","content":" 11天","answer":1,"chooseValue":"B"},{"id":"796234443399385089","questionId":"796234442308866049","content":" 12天","answer":0,"chooseValue":"C"},{"id":"796234443420356609","questionId":"796234442308866049","content":" 13天","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234445496537089","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某项目有8个作业A~H，每个作业的紧前作业、所需天数和所需人数见下表。由于整个项目团队总共只有9人，各个作业都必须连续进行，中途不能停止，因此需要适当安排施工方案，使该项目能尽快在（53）内完工。在该方案中,作业A应安排在（54）内进行。<br> <img src=\"https://image.chaiding.com/ruankao/c684e42d5cf6677de83f911a794787d6.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"650\" height=\"109\" title=\"\" align=\"\">","analyze":"本题考查应用数学-运筹学-网络计划图的基础知识。<br> 根据题中各作业的紧前作业和所需天数，可绘制网络计划图如下：<br> <img src=\"https://image.chaiding.com/ruankao/a440e9f0430bfc1762350db0078b6528.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"350\" height=\"110\" title=\"\" align=\"\"><br> 根据该图，如果不考虑人数限制，该项目的关键路径为C-E-F-H，需要2+2+3+3=10天。 <br> 先考虑安排关键路径上这几个作业顺序进行：第1~2天安排5人做作业C，第3~4天安排1人做作业E，第5~7天安排1人做作业F，第8~10天安排6人做作业H，图示如下:<br> <img width=\"766\" height=\"42\" src=\"https://image.chaiding.com/ruankao/699e6d6b091ce2ead075def5c9c31344.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br> 由于作业B必须在F之前进行，需要8人做1天，只能安排在第3或第4天进行。<br> 由于作业D和G必须在H之前进行，作业D需要4人做2天，安排在第1~2天为好。而作业G需要7人做2天。将作业B安排在第3天，将作业G安排在第4~5天为好。 <br> 作业A虽然可以在全程安排，但由于需要7人，所以安排在第6~8天为好。然而第8天作业H已暂时安排6人，这样会引发第8天人数（6+7)超出9人的限制。最好的解决办法是将作业H推迟1天。从而，在每天人数限制9人的条件下，项目最快能在11天完成， 实施方案图示如下：<br> <img width=\"768\" height=\"122\" src=\"https://image.chaiding.com/ruankao/2b4d0051b7bf89f97394a040f0b74897.jpg?x-oss-process=style/ruankaodaren\" alt=\"\"><br></br>\nps:由于A和G任务的顺序是可以换的，此题有两个答案C/D","multi":0,"questionType":1,"answer":"D","chooseItem":["796234446591250433"],"itemList":[{"id":"796234446503170049","questionId":"796234445496537089","content":" 第3-5天","answer":0,"chooseValue":"A"},{"id":"796234446532530177","questionId":"796234445496537089","content":" 第4-6天","answer":0,"chooseValue":"B"},{"id":"796234446561890305","questionId":"796234445496537089","content":" 第5-7天","answer":0,"chooseValue":"C"},{"id":"796234446591250433","questionId":"796234445496537089","content":" 第6-8天","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234689009438721","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某项目有A~H八个作业，各作业所需时间（单位：周）以及紧前作业如下表:<br> <img alt=\"\" src=\"https://image.chaiding.com/ruankao/31ee6863e8dc7a5bfc6ad1a4885cfb8a.jpg?x-oss-process=style/ruankaodaren\" width=\"646\" height=\"82\"><br> 该项目的工期为（54）周。如果作业C拖延3周完成，则该项目的工期（55）。","analyze":"<br> <img alt=\"\" src=\"https://image.chaiding.com/ruankao/c1350bf7cffd0e33392d7537a8c89438.jpg?x-oss-process=style/ruankaodaren\" width=\"646\" height=\"200\"><br>通过绘图找最长路径可知，关键路径为：ADFH，长度为13，所以项目的工期为13周。 当C拖延3周之后，关键路径变为：ACEH，长度为15，所以工期拖延2周。","multi":0,"questionType":1,"answer":"B","chooseItem":["796234689944768513"],"itemList":[{"id":"796234689932185601","questionId":"796234689009438721","content":" 12","answer":0,"chooseValue":"A"},{"id":"796234689944768513","questionId":"796234689009438721","content":" 13","answer":1,"chooseValue":"B"},{"id":"796234689957351425","questionId":"796234689009438721","content":" 14","answer":0,"chooseValue":"C"},{"id":"796234689965740033","questionId":"796234689009438721","content":" 15","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234691987394561","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某项目有A~H八个作业，各作业所需时间（单位：周）以及紧前作业如下表:<br> <img alt=\"\" src=\"https://image.chaiding.com/ruankao/5b4dab2f971ac99d45b260dc70fdeaa6.jpg?x-oss-process=style/ruankaodaren\" width=\"646\" height=\"82\"><br> 该项目的工期为（54）周。如果作业C拖延3周完成，则该项目的工期（55）。","analyze":"<br> <img alt=\"\" src=\"https://image.chaiding.com/ruankao/c1350bf7cffd0e33392d7537a8c89438.jpg?x-oss-process=style/ruankaodaren\" width=\"646\" height=\"200\"><br>\n通过绘图找最长路径可知，关键路径为：ADFH，长度为13，所以项目的工期为13周。 当C拖延3周之后，关键路径变为：ACEH，长度为15，所以工期拖延2周。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234692947890177"],"itemList":[{"id":"796234692897558529","questionId":"796234691987394561","content":" 不变","answer":0,"chooseValue":"A"},{"id":"796234692922724353","questionId":"796234691987394561","content":" 拖延1周","answer":0,"chooseValue":"B"},{"id":"796234692947890177","questionId":"796234691987394561","content":" 拖延2周","answer":1,"chooseValue":"C"},{"id":"796234692973056001","questionId":"796234691987394561","content":" 拖延3周","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234934103592961","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>某项目有A~H八个作业，各作业所需时间（单位：周）以及紧前作业如下表：<br><img alt=\"\" src=\"https://image.chaiding.com/ruankao/a305dfa7dae24ba1cbc3e96a36873439.jpg?x-oss-process=style/ruankaodaren\" height=\"97\" width=\"535\"><br>该项目的工期为（56）周。如果作业C拖延3周完成，则该项目的工期（57）。","analyze":"本题考查应用数学（运筹学-网络计划图）基础知识。<br>先根据题中给出的表绘制如下的网络计划图：<br><br><img width=\"441\" height=\"115\" alt=\"\" src=\"https://image.chaiding.com/ruankao/6713bad46a1070966751f94c2d6e5547.jpg?x-oss-process=style/ruankaodaren\"><br>关键路径为从起点到终点所需时间最长的路径：A-D-F-H,工期为1+5+6+1=13周。 <br>若作业C拖延3周完成，则关键路径为A-C-E-H，工期为1+6+7+1=15周，拖延2周。","multi":0,"questionType":1,"answer":"C","chooseItem":["796234935227666433"],"itemList":[{"id":"796234935135391745","questionId":"796234934103592961","content":" 不变","answer":0,"chooseValue":"A"},{"id":"796234935181529089","questionId":"796234934103592961","content":" 拖延1周","answer":0,"chooseValue":"B"},{"id":"796234935227666433","questionId":"796234934103592961","content":" 拖延2周","answer":1,"chooseValue":"C"},{"id":"796234935261220865","questionId":"796234934103592961","content":" 拖延3周","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234930991419393","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>某项目有A~H八个作业，各作业所需时间（单位：周）以及紧前作业如下表：<br><img src=\"https://image.chaiding.com/ruankao/42b818ddf08c7576741688fc0593fc7d.jpg?x-oss-process=style/ruankaodaren\" height=\"97\" width=\"100%\"><br>该项目的工期为（56）周。如果作业C拖延3周完成，则该项目的工期（57）。","analyze":"本题考查应用数学（运筹学-网络计划图）基础知识。<br>先根据题中给出的表绘制如下的网络计划图：<br><img width=\"441\" height=\"115\" alt=\"\" src=\"https://image.chaiding.com/ruankao/3def670731a6058e014d4677d1053161.jpg?x-oss-process=style/ruankaodaren\"><br>关键路径为从起点到终点所需时间最长的路径：A-D-F-H,工期为1+5+6+1=13周。 <br>若作业C拖延3周完成，则关键路径为A-C-E-H，工期为1+6+7+1=15周，拖延2周。","multi":0,"questionType":1,"answer":"B","chooseItem":["796234932010635265"],"itemList":[{"id":"796234931981275137","questionId":"796234930991419393","content":" 12","answer":0,"chooseValue":"A"},{"id":"796234932010635265","questionId":"796234930991419393","content":" 13","answer":1,"chooseValue":"B"},{"id":"796234932039995393","questionId":"796234930991419393","content":" 14","answer":0,"chooseValue":"C"},{"id":"796234932073549825","questionId":"796234930991419393","content":" 15","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234458217861121","title":"根据历史数据和理论推导可知，某随机变量x的分布密度函数为f(x)=2x，(0<x<1)。这意味着，当Δx充分小时，随机变量x落在区间(x，x+ΔX)内的概率约等于f(x)Δx。为此，在电脑上可采用（58）来模拟该随机变量，其中，r1和r2为计算机产生的、均匀分布在(0，1)区间的两个伪随机数，且互相独立。","analyze":"本题考查应用数学-运筹学-随机模拟的基础知识。<br> 用计算机来模拟随机系统往往需要模拟实际的随机变量。根据历史数据或理论推导可以得到随机变量的分布密度函数，而根据分布密度函数设计计算机抽样方法，可用于模拟随机<br> 变量。<br> 本题中，若ΔX充分小，随机变量max(r1，r2)落在区间(x，x+ΔX)内的事件A，是事件A1、 A2和A3的并集。事件A1为r1落在区间(x，x+ΔX)内，而r2<x;事件A2为r1<x, 而r2落在区间(x，x+ΔX)内；事件A3为r1和r2都落在区间(X，X+ΔX)内。这三个事件互相没有交集。因此概率P(A)=P(A1)+P(A2)+P(A3)=ΔX*x+x*ΔX+ΔX*ΔX≈2xΔX=f(x)ΔX。因此，max(r1，r2)可以<br> 用来模拟随机变量X。<br> 定性地选择该题的正确答案也不难:（0,1)区间内的分布密度函数2x，意味着随着x的增大出现的概率也线性地增大。显然，对于min(r1，r2)，出现较小的数值的概率更大些；r1*r2 (两个小于1的数相乘会变得更小）也会这样。对于随机变量(r1+r2)/2,出现中等大小数值的概率更大一些，出现较大的或较小值的概率会小一些，其分布密度函数会呈凸型。只有max(r1 ,r2)，出现较大数值的概率更大些。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234459161579521"],"itemList":[{"id":"796234459161579521","questionId":"796234458217861121","content":" max(r1,r2)","answer":1,"chooseValue":"A"},{"id":"796234459199328257","questionId":"796234458217861121","content":" min(r1，r2)","answer":0,"chooseValue":"B"},{"id":"796234459228688385","questionId":"796234458217861121","content":" r1*r2","answer":0,"chooseValue":"C"},{"id":"796234459249659905","questionId":"796234458217861121","content":" (r1+r2)/2","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235173321527297","title":"根据历史统计情况，某超市某种面包的日销量为100、110、120、130、140个的概率相同，每个面包的进价为4元，销售价为5元，但如果当天没有卖完，剩余的面包次日将以每个3元处理。为取得最大利润，该超市每天应进货这种面包（ ）个。","analyze":"本题考查应用数学（运筹学-决策）基础知识。<br>这种面包各种进货情况和销售情况下，所得利润如下表：<br><img width=\"614\" height=\"183\" alt=\"\" src=\"https://image.chaiding.com/ruankao/c0c9a61a6759418645c878784c7ec871.jpg?x-oss-process=style/ruankaodaren\"><br>因此，每天进货120个面包时，能得到最大利润108元。","multi":0,"questionType":1,"answer":"B","chooseItem":["796235174470766593"],"itemList":[{"id":"796235174441406465","questionId":"796235173321527297","content":" 110","answer":0,"chooseValue":"A"},{"id":"796235174470766593","questionId":"796235173321527297","content":" 120","answer":1,"chooseValue":"B"},{"id":"796235174504321025","questionId":"796235173321527297","content":" 130","answer":0,"chooseValue":"C"},{"id":"796235174533681153","questionId":"796235173321527297","content":" 140","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235401097400321","title":"用一辆载重量为10吨的卡车装运某仓库中的货物（不用考虑装车时货物的大小），这些货物单件的重量和运输利润如下表。适当选择装运一些货物各若干件，就能获得最大总利润（54）元。<br><img alt=\"\" width=\"514\" height=\"66\" src=\"https://image.chaiding.com/ruankao/a3eb3c4c4fc188fcfb6d38eb7a7bc3e2.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考查应用数学（运筹一最优化分配）。<br>先计算各类货物的单位运输利润如下：<br><img alt=\"\" width=\"554\" height=\"91\" src=\"https://image.chaiding.com/ruankao/0e438b95ca43c637b89c06e423d1c6ee.jpg?x-oss-process=style/ruankaodaren\"><br>货物A重量为1吨，利润53元，用它可以代替所有利润/重量之比不超过53元的货物B、C、E、F。首先选择装运利润/重量之比最大的货物D，可以装2件，10吨卡车占了8吨，再选择货物A，可装2件。总共可获得利润2*216+2*53=538元。","multi":0,"questionType":1,"answer":"D","chooseItem":["796235402154364929"],"itemList":[{"id":"796235402049507329","questionId":"796235401097400321","content":" 530","answer":0,"chooseValue":"A"},{"id":"796235402083061761","questionId":"796235401097400321","content":" 534","answer":0,"chooseValue":"B"},{"id":"796235402120810497","questionId":"796235401097400321","content":" 536","answer":0,"chooseValue":"C"},{"id":"796235402154364929","questionId":"796235401097400321","content":" 538","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234439347687425","title":"线性规划问题由线性的目标函数和线性的约束条件(包括变量非负条件）组成。约束条件的所有解的集合称为可行解区。既满足约束条件，又使目标函数达到极值的解称为最优解。以下关于可行解区和最优解的叙述中，正确的是 (52)。","analyze":"本题考查应用数学-运筹学-线性规划的基础知识。<br> 线性规划问题的可行解区可能无界；如果增加一个线性约束条件，则可行解区可能缩小也可能不变；如果存在两个最优解，则连接这两点的线段内所有的点都是最优解，而线段两端延长线上可能会超出可行解区；如果最优解存在且唯一，则目标函数的极值一定会在某个顶点处达到，这就为方便计算开辟了道路。","multi":0,"questionType":1,"answer":"D","chooseItem":["796234440354320385"],"itemList":[{"id":"796234440262045697","questionId":"796234439347687425","content":" 可行解区一定是封闭的多边形或多面体","answer":0,"chooseValue":"A"},{"id":"796234440291405825","questionId":"796234439347687425","content":" 若增加一个线性约束条件，则可行解区可能会扩大","answer":0,"chooseValue":"B"},{"id":"796234440320765953","questionId":"796234439347687425","content":" 若存在两个最优解，则它们的所有线性组合都是最优解","answer":0,"chooseValue":"C"},{"id":"796234440354320385","questionId":"796234439347687425","content":" 若最优解存在且唯一，则可以从可行解区顶点处比较目标函数值来求解","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235157911654401","title":"线性规划问题由线性的目标函数和线性的约束条件（包括变量非负条件）组成。满足约束条件的所有解的集合称为可行解区。既满足约束条件，又使目标函数达到极值的解称为最优解。以下关于可行解区和最优解的叙述中，正确的是（ ）。","analyze":"本题考查应用数学（运筹学-线性规划）基础知识。<br>线性规划问题的可行解区可能不存在。例如：两个约束条件（不等式）矛盾，没有交集。可行解区可能无界。例如，X+Y>1，X≥0，Y≥0。当可行解区无界时，可能仍存在最优解。例如：min S=X+2Y;X+Y>1，X≥0，Y≥0。如果最优解存在，并且在可行解区的内点或边界（非顶点）内点达到，则目标函数的等值线（面、体）要么还可以在可行解区内移动，扩大和缩小目标函数的值；要么已经包含了某些顶点。","multi":0,"questionType":1,"answer":"D","chooseItem":["796235158863761409"],"itemList":[{"id":"796235158813429761","questionId":"796235157911654401","content":" 线性规划问题的可行解区一定存在","answer":0,"chooseValue":"A"},{"id":"796235158826012673","questionId":"796235157911654401","content":" 如果可行解区存在，则一定有界","answer":0,"chooseValue":"B"},{"id":"796235158842789889","questionId":"796235157911654401","content":" 如果可行解区存在但无界，则一定不存在最优解","answer":0,"chooseValue":"C"},{"id":"796235158863761409","questionId":"796235157911654401","content":" 如果最优解存在，则一定会在可行解区的某个顶点处达到","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235398144610305","title":"设三个煤场A1、A2、A3分别能供应煤7、12、11万吨，三个工厂B1、B2、B3分别需要煤10、10、10万吨，从各煤场到各工厂运煤的单价（百元／吨）见下表方框内的数字。只要选择最优的运输方案，总的运输成本就能降到（53）百万元。<br><img alt=\"\" width=\"547\" height=\"113\" src=\"https://image.chaiding.com/ruankao/44790c493c749180dac532ecc866b23e.jpg?x-oss-process=style/ruankaodaren\">","analyze":"本题考査应用数学基础知识（运筹一运输问题）。<br>先做出初始方案（第1、2列按最便宜运输，第3列再配齐，总运费61百万元）。<br><img alt=\"\" width=\"548\" height=\"168\" src=\"https://image.chaiding.com/ruankao/e464f7a843d2b678cc374d24931a359b.jpg?x-oss-process=style/ruankaodaren\"><br>再改进此方案（按第1行最便宜运输，调整其他项，总运费40百万元）。<br><img alt=\"\" width=\"551\" height=\"161\" src=\"https://image.chaiding.com/ruankao/b9ac058b2e4c74ca6a8394ebc56a325e.jpg?x-oss-process=style/ruankaodaren\"><br>各空格处若再增加运量，都不能再减少运费，因此最低总运费为40百万元。<br>初始方案可以不同，最优方案也可以不同，但最低运费一定相同。关键是对改进的方案经过各种试验已不能再调整来降低总运费了。","multi":0,"questionType":1,"answer":"B","chooseItem":["796235399079940097"],"itemList":[{"id":"796235399063162881","questionId":"796235398144610305","content":" 30","answer":0,"chooseValue":"A"},{"id":"796235399079940097","questionId":"796235398144610305","content":" 40","answer":1,"chooseValue":"B"},{"id":"796235399096717313","questionId":"796235398144610305","content":" 50","answer":0,"chooseValue":"C"},{"id":"796235399113494529","questionId":"796235398144610305","content":" 60","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234686014705665","title":"设三个煤场A、B、C分别能供应煤12、14、10万吨，三个工厂X、Y、Z分别需要煤11、12、13万吨，从各煤场到各工厂运煤的单价（百元/吨）见下表方框内的数字。只要选择最优的运输方案，总的运输成本就能降到（ ）百万元。<br> <img width=\"550\" height=\"131\" alt=\"\" src=\"https://image.chaiding.com/ruankao/c28cf5f0845a94d0e75f42e0173e3b31.jpg?x-oss-process=style/ruankaodaren\" title=\"\" align=\"\">","analyze":"本题考查应用数学（运筹学一运输问题）基础知识。<br> 先按最低运费单价1和2(百元/吨）尽量多运，做出如下初始方案，总运费11X1+10X2+1X4+3X3+10X8=124百万元。<br> <img src=\"https://image.chaiding.com/ruankao/eeb2e1c4bc8d2c48387b5a2a538cd74f.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"550\" height=\"169\" title=\"\" align=\"\"><br> 再改进此方案。按最高运费单价8百元/吨尽量少运，再调整其他项，得到如下方案，总运费11X1+1X4+13X3+10X3=84百万元。<br> <img src=\"https://image.chaiding.com/ruankao/6ac0eec98ef6fcc5eaaf960781239bd5.jpg?x-oss-process=style/ruankaodaren\" alt=\"\" width=\"550\" height=\"169\" title=\"\" align=\"\"><br> 现在，每个未运格若再增加运量，都将增加运费。<br> 例如，若AX格增加1吨运输（运费增加5百元)，则其他格的运量需要做相应调整。<br> 可以有两种情况：(1)AX,AY,CY,CX分别增、减、增、减1吨运量，则运费变化为+5-1+6-3=+7(增加7百万元）；(2)AX,AY,BY,BZ、CZ、CX分别增、减、增、减、增、减1吨运量，则运费变化为+5-1+4-3+8-3=+10(增加10百万元)。<br> 全是增加运费的。其余类推。因此最低总运费为84百万元。（实际解答时，许多明显不合理的途径不用计算就可以舍去）<br> 运输问题的初始方案可以不同，最优方案也可以不同，但最低运费一定相同。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234686929063937"],"itemList":[{"id":"796234686929063937","questionId":"796234686014705665","content":" 83","answer":1,"chooseValue":"A"},{"id":"796234686954229761","questionId":"796234686014705665","content":" 91","answer":0,"chooseValue":"B"},{"id":"796234686983589889","questionId":"796234686014705665","content":" 113","answer":0,"chooseValue":"C"},{"id":"796234687012950017","questionId":"796234686014705665","content":" 153","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796234928009269249","title":"设三个煤场A、B、C分别能供应煤12、14、10万吨，三个工厂X、Y、Z分别需要煤11、12、13万吨，从各煤场到各工厂运煤的单价（百元/吨）见下表方框内的数字。只要选择最优的运输方案，总的运输成本就能降到（ ）百万元。<br><img alt=\"\" src=\"https://image.chaiding.com/ruankao/81a8668ec4a84e1c0f58e7664eab4209.jpg?x-oss-process=style/ruankaodaren\" height=\"157\" width=\"573\">","analyze":"本题考查应用数学（运筹学-运输问题）基础知识。<br>先按最低运费单价1和2(百元/吨）尽量多运，做出如下初始方案，总运费12×1+11×2+3×3+10×7=113 百万元。<br><img width=\"535\" height=\"169\" alt=\"\" src=\"https://image.chaiding.com/ruankao/a3f43b11240091e9bec80370af1ca62d.jpg?x-oss-process=style/ruankaodaren\"><br>再改进此方案。按最高运费单价7百元/吨尽量少运，再调整其他项，得到如下方案， 总运费12×1+1×2+13×3+10×3=83 百万元。<br><img width=\"535\" height=\"181\" alt=\"\" src=\"https://image.chaiding.com/ruankao/48ea9d6eae5004a08f7a927aad005c2f.jpg?x-oss-process=style/ruankaodaren\"><br>现在，每个未运格若再增加运量，都将增加运费。<br>例如，若AX格增加1吨运输（运费增加5百万元)，则其他格的运量需要做相应调整。可以有三种情况:(1)AX，AY，BY，BX分别增、减、增、减1吨运量，则运费变化为+5-1+4-2=+6(增加6百万元);(2)AX，AY，CY，CX分別增、减、增、减1吨运量，则运费变化为+5-1+6-3=+7(增加7百万元);(3)AX，AY，BY，BZ、CZ、CX分别增、减、增、减、增、减1吨运量，则运费变化为+5-1+4-3+7-3=+10(增加10百万元)。全部都是增加运费的。其余类推。因此最低总运费为83百万元。（实际解答时，许多明显不合理的途径不用计算就可以舍去。)<br>运输问题的初始方案可以不同，最优方案也可以不同，但最低运费一定相同。","multi":0,"questionType":1,"answer":"A","chooseItem":["796234928978153473"],"itemList":[{"id":"796234928978153473","questionId":"796234928009269249","content":" 83","answer":1,"chooseValue":"A"},{"id":"796234929007513601","questionId":"796234928009269249","content":" 91","answer":0,"chooseValue":"B"},{"id":"796234929028485121","questionId":"796234928009269249","content":" 113","answer":0,"chooseValue":"C"},{"id":"796234929053650945","questionId":"796234928009269249","content":" 153","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239654683693057","title":"人们需要用观测或测量得到的原始数据建立数学模型来解决实际问题，这种方法称为数据建模法。在建模过程中，下面关于原始数据作用的叙述，不正确的是（55）。","analyze":"从实际问题中观察或测量得到的原始数据，通常是不太精确的，也难以完整。需要透过现象看本质，去伪存真，建立比较合理的模型，并求解。建模的过程通常是个渐进的过程。<br>首先，要根据原始数据初步判断应架构什么样的模型。例如，将一批二维数据画在平面坐标系内，观察它们的分布趋势，初步判断采用什么样的曲线进行拟合比较合适。写出大致的曲线函数表达式，其中必然带有待定的参数。<br>然后，通过原始数据来估计模型中的参数。算出了参数后，初步的模型就已经建立。但是，该模型是否符合实际，还需要用原始数据来检验。如果发现有些偏差，则需要调整模型或调整参数。<br>一般的建模过程往往要反复多次经历上述过程，逐步优化得到比较合理、适用的模型，然后再选用适当的数值方法进行求解。<br>针对不太精确、不大完整的原始数据建立起比较合理的数学模型，并获得满意的（不一定最优的）解答，是应用数学工作者能力、水平和经验的体现。","multi":0,"questionType":1,"answer":"C","chooseItem":["796239655660965889"],"itemList":[{"id":"796239655610634241","questionId":"796239654683693057","content":" 原始数据能够对构建什么样的模型给予提示","answer":0,"chooseValue":"A"},{"id":"796239655631605761","questionId":"796239654683693057","content":" 原始数据可以帮助对模型的参数给出估计","answer":0,"chooseValue":"B"},{"id":"796239655660965889","questionId":"796239654683693057","content":" 模型的合理性取决于原始数据的精确性和完整性","answer":1,"chooseValue":"C"},{"id":"796239655686131713","questionId":"796239654683693057","content":" 原始数据可以帮助检验模型、优化模型","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239660706713601","title":"面对复杂的实际问题，常需要建立数学模型来求解，但根据数学模型求出的解答可能不符合实际情况，故还需分析模型参数和输入数据的微小变化是否会引起输出结果的很大变化。这种分析常称为（54）。","analyze":"本题考查应用数学基础知识。<br>面对复杂的实际问题，常需要建立近似的数学模型来求解，但根据数学模型求出的解答可能不符合实际情况。有时模型参数和输入数据的微小变化会引起输出结果的很大变化，也就是说，模型的计算结果对模型参数和输入数据非常敏感，这种计算结果就很不可靠。因为模型参数和输入数据都是近似的，它的误差可能严重影响计算结果，此时就需要修正这种数学模型。因此，在建立数学模型并求解后，还需要分析计算结果对模型参数和输入数据的敏感程度。这种分析常称为敏感度分析（或灵敏度分析)。这一步骤 在实际应用中非常重要。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239661642043393"],"itemList":[{"id":"796239661625266177","questionId":"796239660706713601","content":" 准确度分析","answer":0,"chooseValue":"A"},{"id":"796239661642043393","questionId":"796239660706713601","content":" 敏感度分析","answer":1,"chooseValue":"B"},{"id":"796239661654626305","questionId":"796239660706713601","content":" 可靠性分析","answer":0,"chooseValue":"C"},{"id":"796239661671403521","questionId":"796239660706713601","content":" 风险分析","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239657724563457","title":"采用数学模型求解实际问题常会有误差，产生的原因不包括（58）。","analyze":"本题考查应用数学的基础知识。<br>数学研究的对象包括数、形和模型三大类。求解实际问题通常需要先建立数学模型。由于实际问题大多是很复杂的，所以只能考虑主要因素，建立近似的模型。因此，模型的假设总是会产生一定的误差。其次，模型的参数常需要测量得到。而测量也会发生误差。还有，多数情况很难精确求解模型，只能采用近似解法，而且求解的计算过程也会产生误差。手工计算会产生误差，计算机计算也会产生误差（局限的字长位数也使实数的表示以及计算产生误差）。由于以上原因，计算的结果当然是有误差的，但这不是求解模型产生误差的原因。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239658710224897"],"itemList":[{"id":"796239658630533121","questionId":"796239657724563457","content":" 模型假设的误差","answer":0,"chooseValue":"A"},{"id":"796239658655698945","questionId":"796239657724563457","content":" 数据测量的误差","answer":0,"chooseValue":"B"},{"id":"796239658680864769","questionId":"796239657724563457","content":" 近似解法和计算过程的误差","answer":0,"chooseValue":"C"},{"id":"796239658710224897","questionId":"796239657724563457","content":" 描述输出结果","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239651785428993","title":"两学生分别在笔直的高速公路A、B两处对车流进行记录。设A和B相距d米，车1和车2先后匀速行驶依次经过了A、B处，车1经过A、B处的时间分别为T1A和T1B车2经过A、B处的时间分别为T2A和T2B，则当车2经过B处时，与车1的距离为（59）米。","analyze":"本题考察应用数学（数学建模）知识。<br>根据题意，车1的速度为<img alt=\"\" width=\"60\" height=\"15\" src=\"https://image.chaiding.com/ruankao/02762db0bfcc258ea58a865b8bd0760f.jpg?x-oss-process=style/ruankaodaren\">，车2的速度为<img alt=\"\" width=\"60\" height=\"14\" src=\"https://image.chaiding.com/ruankao/05621d205130c3ee39cfa8b0226531e2.jpg?x-oss-process=style/ruankaodaren\">。两车的行驶过程如下图所示：<br><br><br><img alt=\"\" width=\"300\" height=\"204\" src=\"https://image.chaiding.com/ruankao/0ea3d92d2b3c689670eb320f0d91f3dc.jpg?x-oss-process=style/ruankaodaren\"><br>当车2在B处时（时间为T2B),车2距离A处d米，而车1已从A处出发行驶了时间（T2B-T1A)，因此距离A 处<img alt=\"\" width=\"150\" height=\"18\" src=\"https://image.chaiding.com/ruankao/715b497c07c8cdbbeb39116761a3a5f7.jpg?x-oss-process=style/ruankaodaren\">米。<br>因此两车相距<img alt=\"\" src=\"https://image.chaiding.com/ruankao/a6a0c1ae15c9e4e5ecc26cac9a03aac4.jpg?x-oss-process=style/ruankaodaren\" height=\"22\" width=\"291\">米。<br>对于<img alt=\"\" src=\"https://image.chaiding.com/ruankao/0c47190494b2895b7570fa1ea36b13d6.jpg?x-oss-process=style/ruankaodaren\" height=\"19\" width=\"328\">多种情形，计算结果（注意绝对值）相同（有可能车2比车1先到达B处)。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239652678815745"],"itemList":[{"id":"796239652678815745","questionId":"796239651785428993","content":" <img alt=\"\" width=\"115\" height=\"16\" src=\"https://image.chaiding.com/ruankao/9c40041ae38be3aa5769b4c77cf71b8e.jpg?x-oss-process=style/ruankaodaren\">","answer":1,"chooseValue":"A"},{"id":"796239652708175873","questionId":"796239651785428993","content":" <img alt=\"\" width=\"135\" height=\"14\" src=\"https://image.chaiding.com/ruankao/c0131c7d637007282dc415c97a27d396.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"B"},{"id":"796239652737536001","questionId":"796239651785428993","content":" <img alt=\"\" width=\"0\" height=\"0\" src=\"https://image.chaiding.com/ruankao/a5feb0cd871d91526ce474d750941f4f.jpg?x-oss-process=style/ruankaodaren\"><img alt=\"\" width=\"159\" height=\"24\" src=\"https://image.chaiding.com/ruankao/c91bc61e1f0c5edd295da2483fdf10fd.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"C"},{"id":"796239652766896129","questionId":"796239651785428993","content":" <img alt=\"\" width=\"130\" height=\"13\" src=\"https://image.chaiding.com/ruankao/2fe1096773db53d04b4ee8dd70221b22.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239419156746241","title":"评价信息系统经济效益的方法不包括（59）。 ","analyze":"本题考查应用数学（量化管理）的基础知识。<br>评价信息系统经济效益常用的方法主要有成本效益分析法、投入产出分析法和价值工程方法。盈亏平衡法常用于商品的销售定价。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239420083687425"],"itemList":[{"id":"796239420083687425","questionId":"796239419156746241","content":" 盈亏平衡法","answer":1,"chooseValue":"A"},{"id":"796239420129824769","questionId":"796239419156746241","content":" 成本效益分析法","answer":0,"chooseValue":"B"},{"id":"796239420175962113","questionId":"796239419156746241","content":" 投入产出分析法","answer":0,"chooseValue":"C"},{"id":"796239420192739329","questionId":"796239419156746241","content":" 价值工程方法","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239427390164993","title":"设某信息系统明年初建成后预计在第i (i=l、2、...n)年将能获得经济效益Ci元，则该系统总的经济效益可以估计为（53）元，其中r是贴现率（利率)。","analyze":"本题考查应用数学（系统效益评估）知识。<br>某信息系统明年初建成。第1年能获得经济效益C1元，按现值计算就是C1/(l+i)元，因为，现值*(l+i)= C。第2年获得经济效益C2元，按现值计算就是C2/(1+i)<sup>2</sup>元。依次类推，n年后，获得的总经济效益应等于<img alt=\"\" src=\"https://image.chaiding.com/ruankao/00247ba7acc8510a6c3fe5a218da4b4b.jpg?x-oss-process=style/ruankaodaren\" height=\"48\" width=\"96\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796239428367437825"],"itemList":[{"id":"796239428291940353","questionId":"796239427390164993","content":" <img alt=\"\" width=\"30\" height=\"37\" src=\"https://image.chaiding.com/ruankao/6f14b7b33325f6968c292c5e88655cc4.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"A"},{"id":"796239428317106177","questionId":"796239427390164993","content":" <img alt=\"\" width=\"68\" height=\"34\" src=\"https://image.chaiding.com/ruankao/e7edf3757a496ca194abb9c1556ba9b0.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"B"},{"id":"796239428342272001","questionId":"796239427390164993","content":" <img alt=\"\" width=\"80\" height=\"41\" src=\"https://image.chaiding.com/ruankao/d60d741126fc289c94f9c7ade353d6ed.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"C"},{"id":"796239428367437825","questionId":"796239427390164993","content":" <img alt=\"\" width=\"80\" height=\"40\" src=\"https://image.chaiding.com/ruankao/7ce68aa9b2255a47be4f8feb14cc94fd.jpg?x-oss-process=style/ruankaodaren\">","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235318436057089","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>成本是信息系统生命周期内各阶段的所有投入之和，按照成本性态分类，可以分为固定成本、变动成本和混合成本。其中（28）属于固定成本，（29）属于变动成本。","analyze":"本题考查软件工程中成本管理的相关基础知识。<br>成本是信息系统生命周期内各阶段的所有投入之和，按照成本性态分类，可以分为固定成本、变动成本和混合成本。<br>①固定成本。固定成本是指其总额在一定期间和一定业务量范围内，不受业务量变动的影响而保持固定不变的成本。例如，管理人员的工资、办公费、固定资产折旧费、员工培训费等。固定成本又可分为酌量性固定成本和约束性固定成本。酌量性固定成本是指管理层的决策可以影响其数额的固定成本，例如，广告费、员工培训费、技术开发经费等；约束性固定成本是指管理层无法决定其数额的固定成本，即必须开支的成本，例如，办公场地及机器设备的折旧费、房屋及设备租金、管理人员的工资等。<br>②变动成本。变动成本也称为可变成本，是指在一定时期和一定业务量范围内其总额随着业务量的变动而成正比例变动的成本。例如，直接材料费、产品包装费、外包費用、开发奖金等。变动成本也可以分为酌量性变动成本和约束性变动成本。开发奖金、外包费用等可看作是酌量性变动成本；约束性变动成本通常表现为系统建设的直接物耗成本，以直接材料成本最为典型。<br>③混合成本。混合成本就是混合了固定成本和变动成本性质的成本。例如，水电费、电话费等。这些成本通常有一个基数，超过这个基数就会随业务量的增大而增大。例如，质量保证人员的工资、设备动力费等成本在一定业务量内是不变的，超过了这个量便会随业务量的增加而增加。有时，员工的工资也可以归结为混合成本，因为员工平常的工资一般是固定的，但如果需要加班，则加班工资与时间的长短便存在着正比例关系。","multi":0,"questionType":1,"answer":"A","chooseItem":["796235319388164097"],"itemList":[{"id":"796235319388164097","questionId":"796235318436057089","content":" 固定资产折旧费","answer":1,"chooseValue":"A"},{"id":"796235319409135617","questionId":"796235318436057089","content":" 直接材料费","answer":0,"chooseValue":"B"},{"id":"796235319430107137","questionId":"796235318436057089","content":" 产品包装费","answer":0,"chooseValue":"C"},{"id":"796235319451078657","questionId":"796235318436057089","content":" 开发奖金","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235321430790145","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>成本是信息系统生命周期内各阶段的所有投入之和，按照成本性态分类，可以分为固定成本、变动成本和混合成本。其中（28）属于固定成本，（29）属于变动成本。","analyze":"本题考查软件工程中成本管理的相关基础知识。<br>成本是信息系统生命周期内各阶段的所有投入之和，按照成本性态分类，可以分为固定成本、变动成本和混合成本。<br>①固定成本。固定成本是指其总额在一定期间和一定业务量范围内，不受业务量变动的影响而保持固定不变的成本。例如，管理人员的工资、办公费、固定资产折旧费、员工培训费等。固定成本又可分为酌量性固定成本和约束性固定成本。酌量性固定成本是指管理层的决策可以影响其数额的固定成本，例如，广告费、员工培训费、技术开发经费等；约束性固定成本是指管理层无法决定其数额的固定成本，即必须开支的成本，例如，办公场地及机器设备的折旧费、房屋及设备租金、管理人员的工资等。<br>②变动成本。变动成本也称为可变成本，是指在一定时期和一定业务量范围内其总额随着业务量的变动而成正比例变动的成本。例如，直接材料费、产品包装费、外包費用、开发奖金等。变动成本也可以分为酌量性变动成本和约束性变动成本。开发奖金、外包费用等可看作是酌量性变动成本；约束性变动成本通常表现为系统建设的直接物耗成本，以直接材料成本最为典型。<br>③混合成本。混合成本就是混合了固定成本和变动成本性质的成本。例如，水电费、电话费等。这些成本通常有一个基数，超过这个基数就会随业务量的增大而增大。例如，质量保证人员的工资、设备动力费等成本在一定业务量内是不变的，超过了这个量便会随业务量的增加而增加。有时，员工的工资也可以归结为混合成本，因为员工平常的工资一般是固定的，但如果需要加班，则加班工资与时间的长短便存在着正比例关系。","multi":0,"questionType":1,"answer":"D","chooseItem":["796235322462588929"],"itemList":[{"id":"796235322382897153","questionId":"796235321430790145","content":" 员工培训费","answer":0,"chooseValue":"A"},{"id":"796235322416451585","questionId":"796235321430790145","content":" 房屋租金","answer":0,"chooseValue":"B"},{"id":"796235322441617409","questionId":"796235321430790145","content":" 技术开发经费","answer":0,"chooseValue":"C"},{"id":"796235322462588929","questionId":"796235321430790145","content":" 外包费用","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239299015102465","title":"<p><strong>请作答第<span style=\"color: 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软件需求说明书：也称软件规格说明书。其中对所开发软件的功能、性能、用户界面机运行环境等作出详细的说明。它是用户与开发人员双方对软件需求取得共同理解基础上达成的协议，也是实施开发工作的基础。<br>(3) 数据要求说明书：该说明书应当给出数据逻辑描述和数据采集的各项要求，为生成和维护系统的数据文件做好准备。<br>(4) 概要设计说明书：该说明书是概要设计工作阶段的成果。它应当说明系统的功能分配、模块划分、程序的总体结构、输入输出及接口设计、运行设计、数据结构设计和出错处理设计等，为详细设计奠定基础。<br>(5) 详细设计说明书：着重描述每一个模块是如何实现的，包括实现算法、逻辑流程等。<br>(6) 用户手册：详细描述软件的功能、性能和用户界面，使用户了解如何使用该软件。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239300134981633"],"itemList":[{"id":"796239300076261377","questionId":"796239299015102465","content":" 项目开发计划","answer":0,"chooseValue":"A"},{"id":"796239300097232897","questionId":"796239299015102465","content":" 软件需求说明书","answer":0,"chooseValue":"B"},{"id":"796239300114010113","questionId":"796239299015102465","content":" 数据需求说明书","answer":0,"chooseValue":"C"},{"id":"796239300134981633","questionId":"796239299015102465","content":" 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软件需求说明书：也称软件规格说明书。其中对所开发软件的功能、性能、用户界面机运行环境等作出详细的说明。它是用户与开发人员双方对软件需求取得共同理解基础上达成的协议，也是实施开发工作的基础。<br>(3) 数据要求说明书：该说明书应当给出数据逻辑描述和数据采集的各项要求，为生成和维护系统的数据文件做好准备。<br>(4) 概要设计说明书：该说明书是概要设计工作阶段的成果。它应当说明系统的功能分配、模块划分、程序的总体结构、输入输出及接口设计、运行设计、数据结构设计和出错处理设计等，为详细设计奠定基础。<br>(5) 详细设计说明书：着重描述每一个模块是如何实现的，包括实现算法、逻辑流程等。<br>(6) 用户手册：详细描述软件的功能、性能和用户界面，使用户了解如何使用该软件。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239303008079873"],"itemList":[{"id":"796239303008079873","questionId":"796239302093721601","content":" 项目开发计划","answer":1,"chooseValue":"A"},{"id":"796239303020662785","questionId":"796239302093721601","content":" 详细设计说明书","answer":0,"chooseValue":"B"},{"id":"796239303029051393","questionId":"796239302093721601","content":" 用户手册","answer":0,"chooseValue":"C"},{"id":"796239303041634305","questionId":"796239302093721601","content":" 概要设计说明书","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239419156746241","title":"评价信息系统经济效益的方法不包括（59）。 ","analyze":"本题考查应用数学（量化管理）的基础知识。<br>评价信息系统经济效益常用的方法主要有成本效益分析法、投入产出分析法和价值工程方法。盈亏平衡法常用于商品的销售定价。","multi":0,"questionType":1,"answer":"A","chooseItem":["796239420083687425"],"itemList":[{"id":"796239420083687425","questionId":"796239419156746241","content":" 盈亏平衡法","answer":1,"chooseValue":"A"},{"id":"796239420129824769","questionId":"796239419156746241","content":" 成本效益分析法","answer":0,"chooseValue":"B"},{"id":"796239420175962113","questionId":"796239419156746241","content":" 投入产出分析法","answer":0,"chooseValue":"C"},{"id":"796239420192739329","questionId":"796239419156746241","content":" 价值工程方法","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239427390164993","title":"设某信息系统明年初建成后预计在第i (i=l、2、...n)年将能获得经济效益Ci元，则该系统总的经济效益可以估计为（53）元，其中r是贴现率（利率)。","analyze":"本题考查应用数学（系统效益评估）知识。<br>某信息系统明年初建成。第1年能获得经济效益C1元，按现值计算就是C1/(l+i)元，因为，现值*(l+i)= C。第2年获得经济效益C2元，按现值计算就是C2/(1+i)<sup>2</sup>元。依次类推，n年后，获得的总经济效益应等于<img alt=\"\" src=\"https://image.chaiding.com/ruankao/00247ba7acc8510a6c3fe5a218da4b4b.jpg?x-oss-process=style/ruankaodaren\" height=\"48\" width=\"96\">","multi":0,"questionType":1,"answer":"D","chooseItem":["796239428367437825"],"itemList":[{"id":"796239428291940353","questionId":"796239427390164993","content":" <img alt=\"\" width=\"30\" height=\"37\" src=\"https://image.chaiding.com/ruankao/6f14b7b33325f6968c292c5e88655cc4.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"A"},{"id":"796239428317106177","questionId":"796239427390164993","content":" <img alt=\"\" width=\"68\" height=\"34\" src=\"https://image.chaiding.com/ruankao/e7edf3757a496ca194abb9c1556ba9b0.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"B"},{"id":"796239428342272001","questionId":"796239427390164993","content":" <img alt=\"\" width=\"80\" height=\"41\" src=\"https://image.chaiding.com/ruankao/d60d741126fc289c94f9c7ade353d6ed.jpg?x-oss-process=style/ruankaodaren\">","answer":0,"chooseValue":"C"},{"id":"796239428367437825","questionId":"796239427390164993","content":" <img alt=\"\" width=\"80\" height=\"40\" src=\"https://image.chaiding.com/ruankao/7ce68aa9b2255a47be4f8feb14cc94fd.jpg?x-oss-process=style/ruankaodaren\">","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235318436057089","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>成本是信息系统生命周期内各阶段的所有投入之和，按照成本性态分类，可以分为固定成本、变动成本和混合成本。其中（28）属于固定成本，（29）属于变动成本。","analyze":"本题考查软件工程中成本管理的相关基础知识。<br>成本是信息系统生命周期内各阶段的所有投入之和，按照成本性态分类，可以分为固定成本、变动成本和混合成本。<br>①固定成本。固定成本是指其总额在一定期间和一定业务量范围内，不受业务量变动的影响而保持固定不变的成本。例如，管理人员的工资、办公费、固定资产折旧费、员工培训费等。固定成本又可分为酌量性固定成本和约束性固定成本。酌量性固定成本是指管理层的决策可以影响其数额的固定成本，例如，广告费、员工培训费、技术开发经费等；约束性固定成本是指管理层无法决定其数额的固定成本，即必须开支的成本，例如，办公场地及机器设备的折旧费、房屋及设备租金、管理人员的工资等。<br>②变动成本。变动成本也称为可变成本，是指在一定时期和一定业务量范围内其总额随着业务量的变动而成正比例变动的成本。例如，直接材料费、产品包装费、外包費用、开发奖金等。变动成本也可以分为酌量性变动成本和约束性变动成本。开发奖金、外包费用等可看作是酌量性变动成本；约束性变动成本通常表现为系统建设的直接物耗成本，以直接材料成本最为典型。<br>③混合成本。混合成本就是混合了固定成本和变动成本性质的成本。例如，水电费、电话费等。这些成本通常有一个基数，超过这个基数就会随业务量的增大而增大。例如，质量保证人员的工资、设备动力费等成本在一定业务量内是不变的，超过了这个量便会随业务量的增加而增加。有时，员工的工资也可以归结为混合成本，因为员工平常的工资一般是固定的，但如果需要加班，则加班工资与时间的长短便存在着正比例关系。","multi":0,"questionType":1,"answer":"A","chooseItem":["796235319388164097"],"itemList":[{"id":"796235319388164097","questionId":"796235318436057089","content":" 固定资产折旧费","answer":1,"chooseValue":"A"},{"id":"796235319409135617","questionId":"796235318436057089","content":" 直接材料费","answer":0,"chooseValue":"B"},{"id":"796235319430107137","questionId":"796235318436057089","content":" 产品包装费","answer":0,"chooseValue":"C"},{"id":"796235319451078657","questionId":"796235318436057089","content":" 开发奖金","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235321430790145","title":"<p><strong>请作答第<span style=\"color: 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员工培训费","answer":0,"chooseValue":"A"},{"id":"796235322416451585","questionId":"796235321430790145","content":" 房屋租金","answer":0,"chooseValue":"B"},{"id":"796235322441617409","questionId":"796235321430790145","content":" 技术开发经费","answer":0,"chooseValue":"C"},{"id":"796235322462588929","questionId":"796235321430790145","content":" 外包费用","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239431223758849","title":"某软件公司分为研发、人力资源管理、市场营销等部门，通过部门之间互相协作完成项目。该公司的组织结构属于（19）。","analyze":"软件项目的开发必须建立在设计良好的组织结构之上，组织结构是指对实施与一个项目相关的各项活动所需资源的确认与组织。常见的组织结构包括矩阵型组织结构、职能型组织结构和产品型组织结构三种。根据题干的描述，说明该软件公司分为几个部门，每个部门负责特定的职能，且项目的总部通常位于特定的职能部门中。这是典型的职能型组织结构的特征。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239432154894337"],"itemList":[{"id":"796239432125534209","questionId":"796239431223758849","content":" 矩阵型组织结构","answer":0,"chooseValue":"A"},{"id":"796239432154894337","questionId":"796239431223758849","content":" 职能型组织结构","answer":1,"chooseValue":"B"},{"id":"796239432184254465","questionId":"796239431223758849","content":" 产品型组织结构","answer":0,"chooseValue":"C"},{"id":"796239432209420289","questionId":"796239431223758849","content":" 协作型组织结构","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239434197520385","title":"某软件公司分为研发、人力资源管理、市场营销等部门，通过部门之间互相协作完成项目。该公司的组织结构属于（）。","analyze":"本题考查企业组织结构的基础知识。<br>根据题干描述，该软件公司的组织是按照职能进行划分的，即分为研发、人力资源管理、市场营销等业务部门，部门之间通过协作完成工作，因此该公司属于职能型组织结构。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239435145433089"],"itemList":[{"id":"796239435116072961","questionId":"796239434197520385","content":" 矩阵型组织结构","answer":0,"chooseValue":"A"},{"id":"796239435145433089","questionId":"796239434197520385","content":" 职能型组织结构","answer":1,"chooseValue":"B"},{"id":"796239435174793217","questionId":"796239434197520385","content":" 产品型组织结构","answer":0,"chooseValue":"C"},{"id":"796239435204153345","questionId":"796239434197520385","content":" 协作型组织结构","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null}]}}
