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Math Fractions in Grade Two of Junior High School
(1) Suppose the price of computer A is X yuan in March this year, then100000/(x+1000) = 80000/x, and the solution is X=4000. Test.

⑵ If it is set to computer Y, it is set to computer B (15-Y) to get the inequality group.

3800Y+3000( 15-Y)≥48000

3800Y+3000( 15-Y)≤50000

The solution is Y≥3.75 and Y≤6.25, because y is a positive integer, ∴ y = 4,5,6.

∴ * * There are three purchase schemes:

A enters 4 sets, and B enters 1 1 set;

A into 5 sets, B into 10 sets;

A into 6 units, B into 9 units;

(3) A earns 500 yuan for each set, and B earns 800 yuan for each set.

Scheme 1: A enters 4 sets and B enters 1 1 set, earning 4× 500+11× 800 =10800.

Scheme 2: A enters 5 sets and B enters 10 sets, earning 5× 500+10× 800 =10500.

Scheme 3: A enters 6 sets and B enters 9 sets, earning 6×500+9×800= 10200.

According to the topic:10800-11a =10500-10a =10200-9a,

A=300, that is to say, each B computer only earns money from one computer.

The title requires that in order to open up the market, the more computers B sell, the better, so the first scheme is adopted.