1, solution: suppose x Chinese knots are needed, and y 1 is the total purchase price. Y2 is the total price made by ourselves, depending on the problem: y 1 = 9x2 = 5x+ 160 when y 1 > y2 when y 1=y2 when y 1 5x+1. 9x = 5x+1609x < 5x+1604x >1604x =1604x <160x > 40x = 40x < 40 ∴ when China needs it. The same is true when the number is 40. When the quantity is less than 40, buy it cheaply. 2, 1) Solution: Let's rent X cars. Rent a car B (8-x) {40x+30 (8-x) ≥ 290① {10x+20 (8-x) ≥100② Solve the inequality 140x+240-30x ≥ 29010x+240. 60-20x ≥ 6010x+160 ≥100-10x ≥-60x ≤ 6 ∴ 5 ≤ x ≤ 6 and ∵x is an integer ∴x is 5. 2) Solution: ∵ A car is more expensive than B car ∴ You should rent as many B cars as possible ∴ You should rent 5 A cars. 3 B car 5× 1000+3×800=7400 (yuan) 3. Solution: Let the retail price be X yuan1≤ X-14 ≤14×15%15 ≤ X.1) Solution: Let's make a general snack X box. Delicate snack (50-x) box {0.3x+0.1(50-x) ≤10.2① {0.1x+0.3 (50-x) ≤10.2② 0.3x+ Kloc-0/0.2x ≤-4.8x ≥ 24 ∴ 24 ≤ x ≤ 26 ∴x is an integer ∴ X is 24, 25, 26 50-x is 26, 25, 242) Solution: ∵ The price of exquisite snacks is higher than that of ordinary snacks.