Question

人教版七年级下册数学复习题八7、8、9、10、11题的答案(要有过程)

Answer 1

7.
解：需要A型钢板x 块，需要B型钢板y块。
则 2x+y=15
X+2y=18
解得 x=4,y=7
答：需要A型钢板4块，需要B型钢板7块.

8.
解：设1个大桶可以盛酒X壶,1个小桶可以盛酒Y壶,列方程式:
5x+y=3 ①
x+5y=2 ②
②式两边各乘5得5x＋25y＝10
②－①得 24y＝7
∴ y=7/24
将 y=7/24代入①得　 5x＋7/24=3
5x=3-7/24
5x=65/24
x=13/24
答：1个桶可以盛酒13/24壶，1个小桶可以酒7/24

9.
解：设弹簧原长为xcm ，进度系数为kN/m，根据题意，得：
k(16.4-x)=2 ①
k(17.9-x)=5 ②

由①式得：k=2/(16.4-x) ③
将③代入②，得：
3x=46.2
则x=15.4
答：弹簧应取15.4cm。

10.解：设取1角、5角、1元各X、Y、Z枚
由题意得：
X+5Y+10Z=70 ①
X+Y+Z=15 ②
消去x得 4y+9z=55.
z=(55-4y)/9=6+(1-4y)/9
y=7.
或 x=(55-9z)/4=13+(3-9z）/4
z=3.
∴x=5,y=7,z=3.
答：取1角、5角、1元各5枚、7枚、3枚 。

11.
解：设上坡路程为xkm,下坡路程为ykm，则平路有（3.3-x-y）km，可得方程
x/3+y/5+(3.3-x-y)/4=51/60
x/5+y/3+(3.3-x-y)/4=53.4/60
解得x=1.2 y=1.5
∴平坡路程为3.3-1.2-1.5=0.6（km)
答：从甲地到乙地时上坡，平路，下坡的路程分别是1.2km、0.6km、1.5km

Answer 2

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