The school calligraphy interest group is going to the stationery store to buy two kinds of brushes, A and B. The sales method of the stationery store is: buy no more than 20 A brushes at one time.

Solution: (1) Let the retail prices of brushes A and B be X yuan/piece and Y yuan/piece respectively.

rule

∴ The retail prices of A and B brushes are 2 yuan/branch and 3 yuan/branch respectively;

(2) According to the meaning of the question and (1):

* * * Pay 20× 2+(15× 2-20 )× (2-0.4)+15× 3 =101(Yuan);

(3) Assume that the purchase cost according to the original sales method is y 1 yuan, and the purchase cost according to the new sales method is y 2 yuan.

Then: when 0≤a≤20, y1= 2a+15× 3+(20-15 )× (3-0.6) = 2a+57.

y ^ 2 = a×0.9×2+ 15×3+(20- 15)×(3-0.6)= 1.8a+57

∴y 1-y ^ 2 = 2a+57-( 1.8a+57)= 0.2a

∫0≤a≤20,?

∴0.2a≥0, ∴y 1 -y 2 ≥0, that is, y1≥ y2;

When a > 20, y1= 20× 2+(a-20 )× (2-0.4)+15× 3+(20-15 )× (3-0.6) =/kloc-

y ^ 2 = a×0.9×2+ 15×3+(20- 15)×(3-0.6)= 1.8a+57

∴y 1-y ^ 2 = 1.6a+65-( 1.8a+57)=-0.2a+8 =-0.2(a-40)

∴ when a < 40, y1-y 2 > 0, that is, y1> y 2;

When a=40, y1-y 2 = 0, that is, y1= y 2;

When a > 40, y1-y 2 < 0 0, that is, y1< y 2;

To sum up, when 0 < a < 40, it costs less to buy according to the new sales method;

When a=0 or a=40, the two methods cost as little as buying;

When a > 40, it costs less to buy according to the original sales method.