2010 AMC 10B Problems/Problem 11

Problem

A shopper plans to purchase an item that has a listed price greater than $\textdollar 100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\textdollar 30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\textdollar 100$ .
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 C. What is $y - x$ ?

$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ 80 \qquad \textbf{(E)}\ 100$

Solution

Let the listed price be $(100 + p)$ , where $p > 0$

Coupon A saves us: $0.15(100+p) = (0.15p + 15)$

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 gives us the following inequalities:

$A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100$

$A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150$

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 $y - x = 250 - 200 = \boxed{\textbf{(A)}\ 50}$

Video Solution

https://youtu.be/vYXz4wStBUU

~IceMatrix

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2010 AMC 10B Problems/Problem 11

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Video Solution

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