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The most difficult application problem in junior one mathematics
Use (1) X trucks: each truck loads (320-20)/x=300/x (boxes).

The same number of B trucks: each truck carries (320+30)/x=350/x (boxes).

When fully loaded, each Type A car is less loaded than Type B car 10 case: 350/x-300/x= 10.

The solution is x=5 (cars), so each car of a truck can hold 300/5=60 cases.

Each B-type truck can hold 350/5=70 cases.

(2) According to the meaning of the question: w=320a+350b and 60a+70b≥320,

When positive integers A and B are a= 1 and b=4, the solution is w=320a+350b= 1720 yuan.

When a=2 and b=3, the solution is w=320a+350b= 1690 yuan.

When a=3 and b=2, the solution is w=320a+350b= 1660 yuan.

When a=4 and b=2, the solution is w=320a+350b= 1980 yuan.

When a=5, b= 1, and the solution is w=320a+350b= 1950 yuan.

That is, when a=3 and b=2 are adopted, the lowest cost is 1660 yuan.