Difference between revisions of "2014 AMC 10A Problems/Problem 11"

Revision as of 14:41, 8 February 2014

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

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

Coupon 1: $x-10\%\cdot x=90\%\cdot x$

Coupon 2: $x-20$

Coupon 3: $x-18\%\cdot(x-100)=82\%\cdot x+18$

For coupon $1$ to give a better price reduction than the other coupons, we must have $90\%\cdot x < x-20$ and $90\%\cdot x < 82\%\cdot x+18$ .

From the first inequality, $90\%\cdot x+(-90\%\cdot x) +(20)< x-20+(-90\%\cdot x)+(20)\Rightarrow 20 < 10\%\cdot x\Rightarrow 200 < x$ .

From the second inequality, $90\%\cdot x +(-82\%\cdot x)< 82\%\cdot x+18+(-82\%\cdot x)\Rightarrow 8\%\cdot x < 18\Rightarrow x < 225$ .

The only answer choice that satisfies these constraints is $\boxed{\textbf{(C) }\textdollar219.95}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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-{{AMC12 box|year=2014|ab=A|num-b=6|num-a=8}}
+{{AMC12 box|year=2014|ab=A|num-b=7|num-a=9}}
 {{MAA Notice}}

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Difference between revisions of "2014 AMC 10A Problems/Problem 11"

Revision as of 14:41, 8 February 2014

Problem

Solution

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