2014 AMC 10A Problems/Problem 11

The following problem is from both the 2014 AMC 12A #8 and 2014 AMC 10A #11, so both problems redirect to this page.

Problem

A customer who intends to purchase an appliance has three coupons, only one of which may be used:

Coupon 1: $10\%$ off the listed price if the listed price is at least $\textdollar50$

Coupon 2: $\textdollar 20$ off the listed price if the listed price is at least $\textdollar100$

Coupon 3: $18\%$ off the amount by which the listed price exceeds $\textdollar100$

For which of the following listed prices will coupon $1$ offer a greater price reduction than either coupon $2$ or coupon $3$ ?

$\textbf{(A) }\textdollar179.95\qquad \textbf{(B) }\textdollar199.95\qquad \textbf{(C) }\textdollar219.95\qquad \textbf{(D) }\textdollar239.95\qquad \textbf{(E) }\textdollar259.95\qquad$

Solution 1

Let the listed price be $x$ . Since all the answer choices are above $\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:

Coupon 1: $x\times10\%=.1x$

Coupon 2: $20$

Coupon 3: $18\%\times(x-100)=.18x-18$

For coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\implies x>200$ and $.1x>.18x-18\implies.08x<18\implies x<225$ .

The only choice that satisfies such conditions is $\boxed{\textbf{(C)}\ \textdollar219.95}$

Solution 2 (Using The Answers)

For coupon $1$ to be the most effective, we want 10% of the price to be greater than 20. This clearly occurs if the value is over 200. For coupon 1 to be more effective than coupon 3, we want to minimize the value over 200, so $\boxed{\textbf{(C) }\textdollar219.95}$ is the smallest answer choice over 200.

Solution 3

Since all the answer choices are above $\$100$ , let the price be $\$100 + x$ , where $x > 0$ . Then, the discount from Coupon 1 is $\$10 + \frac{x}{10}$ , the discount from Coupon 2 is $\$20$ , and the discount from Coupon 3 is $\frac{9}{50}x$ . Thus we obtain the following inequalities: \begin{align*} &10 + \frac{x}{10} > 20 \\ &10 + \frac{x}{10} > \frac{9}{50}x \end{align*} Solving them, we obtain $x > 100$ and $x < 125$ . Thus the price is between $\$200$ and $\$225\Rightarrow \boxed{\text{(C)}}$ .

Video Solutions

Video Solution 1

https://youtu.be/aGEB3ykW1BM

~savannahsolver

Video Solution 2

https://youtu.be/rJytKoJzNBY

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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