中考数学应用题典型题解析

中考数学应用题典型题解析

1．（2009南宁）南宁市狮山公园计划在健身区铺设广场砖．现有甲、乙两个工程队参加竞标，甲工程队铺设广场砖的造价y甲（元）与铺设面积xm2的函数关系如图12所示；乙工程队铺设广场砖的造价y乙（元）与铺设面积xm2满足函数关系式：y乙kx． （1）根据图12写出甲工程队铺设广场砖的造价y甲（元）与铺设面积xm2的函数关系式； （2）如果狮山公园铺设广场砖的面积为1600m2，那么公园应选择哪个工程队施工更合算？

解：（1）当0≤x≤500时，设y甲k1x，把500，28000代入上式得：

28000500k1，k12800050056

y甲56x ·················································································································· 2分

28000、1000，48000代入上式得： 当x≥500时，设y甲k2xb，把500，500k2b28000··································································································· 3分 1000kb480002k240解得： ········································································································· 4分

b8000y甲40x8000

56x0≤x500···················································································· 5分 40x8000x≥500y甲（2）当x1600时，y甲401600800072000················································· 6分

y乙1600k ·············································································· 7分

①当y甲y乙时，即：720001600k

得：k45 ·················································································································· 8分

②当y甲y乙时，即：720001600k

得：0k45············································································································· 9分

③当y甲y乙时，即720001600k，k45

答：当k45时，选择甲工程队更合算，当0k45时，选择乙工程队更合算，当k45时，选择两个工程队的花费一样． ···············································································10分

y3

卫 2（09钦州）小王购买了一套经济适用房，他准备将地面铺上地

生 2砖，地面结构如图所示．根据图中的数据（单位：m），解答下列间 卧室 问题：

厨房 2（1）写出用含x、y的代数式表示的地面总面积；

（2）已知客厅面积比卫生间面积多21m2，且地面总面积是

2

卫生间面积的15倍，铺1m地砖的平均费用为80元，求铺地砖

x客厅 的总费用为多少元？

6

解：（1）地面总面积为：（6x＋2y＋18）m2； ······························································· 4分

（2）由题意，得6x2y21,6x2y18152y.······················································· 6分

x4,解之，得····················································································· 8分 3

y.2∴地面总面积为：6x＋2y＋18＝6×4＋2×∵铺1m地砖的平均费用为80元，

2

32＋18＝45（m2）． ················ 9分

∴铺地砖的总费用为：45×80＝3600（元）． ············································10分

3（09湖南）为迎接“建国60周年”国庆，我市准备用灯饰美化红旗路，需采用A、B两

2种不同类型的灯笼200个，且B灯笼的个数是A灯笼的。

3（1）求A、B两种灯笼各需多少个？ （2）已知A、B两种灯笼的单价分别为40元、60元，则这次美化工程购置灯笼需多少费用？

（1）设需A种灯笼x个，B种灯笼y个，根据题意得：

xy200， ·············································································································· 4分 2yx，3解得x120，y80；·············································································································· 6分

（2）120×40+80×60=9600（元）． ·············································································· 8分 4（09定西）图（1）是一扇半开着的办公室门的照片，门框镶嵌在墙体中间，门是向室内开的．图（2）画的是它的一个横断面．虚线表示门完全关好和开到最大限度（由于受到墙

角的阻碍，再也开不动了）时的两种情形，这时二者的夹角为120°，从室内看门框露在外面部分的宽为4cm，求室内露出的墙的厚度a的值．（假设该门无论开到什么角度，门和门框之间基本都是无缝的．精确到0.1cm，3≈1.73）

图1） 图（2）

解从图中可以看出，在室内厚为acm的墙面、宽

为4cm的门框及开成120°的门之间构成了一 个直角三角形，且其中有一个角为60°． ·········· 3分 从而 a=4×tan60° ·············································· 6分

=4×3≈6.9(cm)．··································· 8分 即室内露出的墙的厚度约为6.9cm．

5（09河池）铭润超市用5000元购进一批新品种的苹果进行试销，由于销售状况良好，超市又调拨11000元资金购进该品种苹果，但这次的进货价比试销时每千克多了0.5元，购进苹果数量是试销时的2倍．

（1）试销时该品种苹果的进货价是每千克多少元?

（2）如果超市将该品种苹果按每千克7元的定价出售，当大部分苹果售出后，余下的400千克按定价的七折（“七折”即定价的70﹪）售完，那么超市在这两次苹果销售中共盈利多少元？

解：(1)设试销时这种苹果的进货价是每千克x元，依题意，得 ······························ （1分）

11000x0.55000x2 ·············································································· （5分）

解之，得 x5 ··················································································· （6分） 经检验，x5是原方程的解．······························································· （7分） (2)试销时进苹果的数量为：

500051000 (千克)

第二次进苹果的数量为：2×10002000(千克) ········································· （8分） 盈利为： 2600×7＋400×7×0.7－5000－110004160(元) ······················ （9分） 答：试销时苹果的进货价是每千克5元，商场在两次苹果销售中共盈利4160元．

··············································（10分）

6（09南宁）如图，要设计一个等腰梯形的花坛，花坛上底长120米，下底长180米，上下底相距80米，在两腰中点连线（虚线）处有一条横向甬道，上下底之间有两条纵向甬道，各甬道的宽度相等．设甬道的宽为x米．

（1）用含x的式子表示横向甬道的面积；

（2）当三条甬道的面积是梯形面积的八分之一时，求甬道的宽；

（3）根据设计的要求，甬道的宽不能超过6米.如果修建甬道的总费用（万元）与甬道的宽度成正比例关系，比例系数是5.7，花坛其余部分的绿化费用为每平方米0.02万元，那么当甬道的宽度为多少米时，所建花坛的总费用最少？最少费用是多少万元？ ．解：（1）横向甬道的面积为：

120180218x150xm12018022··········································· 2分  ·

（2）依题意：280x150x2x2整理得：x2155x7500

············································ 4分 80 ·

······································································ 6分 x15，x2150（不符合题意，舍去） ·

甬道的宽为5米．

（3）设建设花坛的总费用为y万元．

1201802··········································· 7分 y0.0280160x150x2x5.7x·

20.04x0.5x240

2当xb2a0.520.046.25时，y的值最小．··························································· 8分

因为根据设计的要求，甬道的宽不能超过6米，

当x6米时，总费用最少． ····················································································· 9分

最少费用为：0.04620.56240238.44万元···················································10分 7（09本溪）为奖励在演讲比赛中获奖的同学，班主任派学习委员小明为获奖同学买奖品，

要求每人一件．小明到文具店看了商品后，决定奖品在钢笔和笔记本中选择．如果买4个笔记本和2支钢笔，则需86元；如果买3个笔记本和1支钢笔，则需57元．

（1）求购买每个笔记本和钢笔分别为多少元？

（2）售货员提示，买钢笔有优惠，具体方法是：如果买钢笔超过10支，那么超出部分可以享受8折优惠，若买x(x0)支钢笔需要花y元，请你求出y与x的函数关系式； （3）在（2）的条件下，小明决定买同一种奖品，数量超过10个，请帮小明判断买哪种奖品省钱．

解（1）解：设每个笔记本x元，每支钢笔y元． ························································· 1分

4x2y86，············································································································· 2分 3xy57.x14，解得

y15. ····································································································3分

····································································································4分

答：每个笔记本14元，每支钢笔15元．······································································ 5分

（2）y15x(0x≤10)12x30(x10)

·······························································6分 ·······························································7分

（3）当14x12x30时，x15；

当14x12x30时，x15；

当14x12x30时，x15． ···················································································· 8分 综上，当买超过10件但少于15件商品时，买笔记本省钱；

当买15件奖品时，买笔记本和钢笔一样； 当买奖品超过15件时，买钢笔省钱． ··········································································10分 8（09泉州）如图，等腰梯形花圃ABCD的底边AD靠

墙，另三边用长为40米的铁栏杆围成，设该花

圃的腰AB的长为x米. (1)请求出底边BC的长（用含x的代数式表示）； （2）若∠BAD=60°, 该花圃的面积为S米2.

①求S与x之间的函数关系式（要指出自变

量x的取值范围），并求当S=933时x的值；

②如果墙长为24米，试问S有最大值还是最小值？这个值是多少？

解：（1）∵AB=CD=x米，∴BC=40-AB-CD=（40-2x）

„„„„„„„„„„„„„„„„„„„„（3分） (2)①如图，过点B、C分别作BE⊥AD于E，CF⊥AD于F，在Rt△ABE中，AB=x,∠BAE=60°

∴AE=

12x,BE=

32x.同理DF=

12x,CF=

32x

又EF=BC=40-2x ∴

AD=AE+EF+DF=

12x+40-2x+

12x=40-x„„„„„„„„„„„（4分）

12∴S==

(40-2x+40-x)·

343x20232x=

34x(80-3x)

＜

x

＜

3 (0

20)„„„„„„„„„„„„„（6分）

当S=933时，解得：x1=6,x2=2023343x2023=933

（舍去）.∴x=6„„„„„„„„„„„„（8分）

②由题意，得40-x≤24,解得x≥16，

结合①得16≤x＜20„„„„„„„„„„„„„„„„„„„„„„„„（9分）

由①，S=343x2023=343(x403)240033

∵a=334＜0

∴函数图象为开口向下的抛物线的一段（附函数图象草图如左）. 其对称轴为x=

403，∵16＞

403,由左图可知，

当16≤x＜20时，S随x的增大而减小„„„„„„„„„„„（11分） ∴当x=16时，S取得最大值，„„„„„„„„„„„„„„„（12分） 此时S最大值=343162203161283.„„„„„„„（13分）

得 分

评卷人

9（09衢州）水产公司有一种海产品共2 104千克，为寻求合适的销

售价格，进行了8天试销，试销情况如下：

售价x(元/千克) 销售量y(千克)

第1天 400 30

第2天

40

第3天 250 48

第4天 240

第5天 200 60

第6天 150 80

第7天 125 96

第8天 120 100

观察表中数据，发现可以用反比例函数刻画这种海产品的每天销售量y(千克)与销售价格x(元/千克)之间的关系．现假定在这批海产品的销售中，每天的销售量y(千克)与销售价格x(元/千克)之间都满足这一关系．

(1) 写出这个反比例函数的解析式，并补全表格；

(2) 在试销8天后，公司决定将这种海产品的销售价格定为150元/千克，并且每天都

按这个价格销售，那么余下的这些海产品预计再用多少天可以全部售出？ 解：(1) 函数解析式为y填表如下：

售价x(元/千克) 销售量y(千克)

第1天 400 30

第2天

第3天 250 48

第4天 240

第5天 200 60

第6天 150 80

第7天 125 96

第8天 120 100

12000x． „„2分

300

40

50

„„2分

(2) 2 104-(30+40+48+50+60+80+96+100)=1 600， 即8天试销后，余下的海产品还有1 600千克． 分

当x=150时，y12000150 „„1

=80． „„2分

„„1评卷人

1 600÷80=20，所以余下的这些海产品预计再用20天可以全部售出． 分

10（09衢州）2009年5月17日至21日，甲型H1N1流感在日本迅

速蔓延，每天的新增病例和累计确诊病例人

数如图所示．

(1) 在5月17日至5月21日这5天中，日

本新增甲型H1N1流感病例最多的是哪一天？该天增加了多少人？

(2) 在5月17日至5月21日这5天中，日

得 分

日本2009年5月16日至5月21日

人数(人) 甲型H1N1流感疫情数据统计图 300 250 200 150 100 50 0 4 0 16

21 17 18

19

96 75 67 30 20

21 日期 74 新增病例人数 累计确诊病例人数 267 163 193 17

本平均每天新增加甲型H1N1流感确诊病例多少人？如果接下来的5天中，继续按这个平均数增加，那么到5月26日，日本甲型H1N1流感累计确诊病例将会达到多少人？

(3) 甲型H1N1流感病毒的传染性极强，某地因1人患了甲型H1N1流感没有及时隔

离治疗，经过两天传染后共有9人患了甲型H1N1流感，每天传染中平均一个人传．．．．

染了几个人？如果按照这个传染速度，再经过5天的传染后，这个地区一共将会有多少人患甲型H1N1流感？

解：(1) 18日新增甲型H1N1流感病例最多，增加了75人；

(2) 平均每天新增加

2674552.6人，

继续按这个平均数增加，到5月26日可达52.6×5+267=530人；

(3) 设每天传染中平均一个人传染了x个人，则

1xx(x1)9，(x1)29，

解得x2(x = -4舍去)． 再经过5天的传染后，这个地区患甲型H1N1流感的人数为 (1+2)7

=2 187（或1+2+6+18+54+162+486+1 458=2 187）， 即一共将会有2 187人患甲型H1N1流感．

„„4分 „„2分 „„2分

„„2分

„„2分