2010 AMC 10B Problems/Problem 11

Revision as of 13:33, 4 February 2012 by Juluqu (talk | contribs) (Made ending a bit clearer, x and y aren't the amount by which the price exceeds $100 (p), they're the actual price)

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

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions
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