2010年AMC10竞赛B第11题
$$2010-AMC10B-11$$  A shopper plans to purchase an item that has a listed price greater than ＄$$100$$ and can use any one of the three coupons. Coupon $$A$$ gives $$15$$\% off the listed price, Coupon $$B$$ gives ＄$$30$$ off the listed price, and Coupon $$C$$ gives $$25$$\% off the amount by which the listed price exceeds ＄$$100$$. Let $$x$$ and $$y$$ be the smallest and largest prices, respectively, for which Coupon $$A$$ saves at least as many dollars as Coupon $$B$$ or Coupon $$C$$. What is $$y- x$$?  购物者计划购买标价大于 ＄100 的商品，并且可以使用三种优惠券中的任何一种。 优惠券$$A$$ 优惠$$15$$ 元，优惠券$$B$$ 优惠$$30$$ 元，优惠券$$C$$ 优惠25\%，超过$$100$$ 元的优惠券。 设 X 和 Y 分别是最小和最大的价格，对于该价格，优惠券 A 至少节省与优惠券 B 或优惠券 C 一样多的美元。y-x 是多少？
课内体系->知识点->函数的应用->函数的实际应用->一次函数模型 美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems
Let the listed price be $$(100+p)$$, where $$p\textgreater0$$.  Coupon $$A$$ saves us: $$0.15(100+p)=(0.15 p+15)$$.  Coupon $$B$$ saves us: $$30$$.  Coupon $$C$$ saves us: $$0.25p$$.  Now, the condition is that $$A$$ has to be greater than or equal to either $$B$$ or $$C$$ which give us the following inequalities: $$A≥B⇒0.15 p+15≥30⇒p≥100$$, $$A≥C⇒0.15 p+15≥0.25p⇒p≤150$$. We see here that the greatest possible value for $$p$$ is $$150$$,~ thus $$y= 100+150=250$$ and the smallest value for $$p$$ is $$100$$ so $$x= 100+100=200$$. The difference between $$y$$ and $$x$$ is $$y-x=250-200=(\rm A) 50$$. 