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The problem of junior one mathematics scheme
(1) There are more than 50 people in group B. Because 1392 yuan divided by 13 is not an integer, there are more than 50 people in group B.:

(2) Because 1392 yuan divided by the highest fare 13 yuan is greater than 100 people, it can be determined that the total number of people exceeds 100 people;

It can be concluded that the total number of people is: 1080/9= 120.

Set up an x person and a b person; De: x = 120-y

13x+11y =1392 instead of x =120-y.

de: 13 *( 120-y)+ 1 1y = 1392。

13 * 120- 13Y+ 1 1Y = 1392

13 * 120- 1392 = 2Y

Y=84 people

A: There are 36 people in Group A and 84 people in Group B;

If the store needs to buy X ties, it should buy them according to the first scheme, and the payable amount is 200×20+(x-20)×40=40x+3 200 (Yuan).

If purchased according to Scheme II, the payable amount is (200×20+40x)×90%=36x+3 600 (RMB).

Let y = (40x+3 200)-(36x+3600) =4x-400 (yuan).

(1) When y

(2) When y =0, 4x-400=0, that is, x = 100, so scheme 1 saves money as scheme 2;

(3) when y > 0, 4x-400 >: 0, that is, x > 100, and scheme 2 saves money than scheme 1.

To sum up, when buying more than 20 ties but less than 100, choose option 1 to save money; When buying a tie is equal to 100, choosing two schemes will also save money; Buy ties above 100 and choose the second option to save money.