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七年级数学选择题一般
题目
用若干量载重量为88吨的汽车运一批货物,若每辆货车只装44吨,则剩下2020吨货物;若每辆货车装88吨,则最后一辆车装的货物不满也不空,若设有xx辆货车,则xx应满足的不等式组是( )
A.
{4x+208x04x+208(x1)0\left\{\begin{array}{l}{4x+20-8x>0}\\{4x+20-8(x-1)>0}\end{array}\right.
B.
{4x+208x04x+208(x1)0\left\{\begin{array}{l}{4x+20-8x>0}\\{4x+20-8(x-1)<0}\end{array}\right.
C.
{4x+208x04x+208(x1)0\left\{\begin{array}{l}{4x+20-8x<0}\\{4x+20-8(x-1)<0}\end{array}\right.
D.
{4x+208x04x+208(x1)0\left\{\begin{array}{l}{4x+20-8x<0}\\{4x+20-8(x-1)>0}\end{array}\right.
知识点:由实际问题抽象出一元一次不等式组章节:第11章 不等式与不等式组 / 11.3 一元一次不等式组

答案与解析

答案

D

解析

设有xx辆车,则有(4x+20)\left(4x+20\right)吨货物.
\because每辆汽车装满88吨,则最后一辆汽车不满也不空,
\therefore装满的有x1x-1辆车,
由题意,得{0(4x+20)8(x1)(4x+20)8(x1)8\left\{\begin{array}{l}{0<(4x+20)−8(x−1)}\\{(4x+20)−8(x−1)<8}\end{array}\right.
即:{4x+208x04x+208(x1)0\left\{\begin{array}{l}{4x+20-8x<0}\\{4x+20-8(x-1)>0}\end{array}\right..
故选:DD.

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