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

(→‎Solution 2: Using the Answer Choices)
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==Solution 1==
 
==Solution 1==
−
Let the listed price be <math>x</math>. Since all the answer choices are above <math>\textdollar100</math>, we can assume <math>x > 100</math>. Thus the prices after coupons will be as follows:
+
Let the listed price be <math>x</math>. Since all the answer choices are above <math>\textdollar100</math>, we can assume <math>x > 100</math>. Thus the discounts after the coupons are used will be as follows:
  
−
Coupon 1: <math>x-10\%\cdot x=90\%\cdot x</math>
+
Coupon 1: <math>x\times10\%=.1x</math>
  
−
Coupon 2: <math>x-20</math>
+
Coupon 2: <math>20</math>
  
−
Coupon 3: <math>x-18\%\cdot(x-100)=82\%\cdot x+18</math>
+
Coupon 3: <math>18\%\times(x-100)=.18x-18</math>
  
−
For coupon <math>1</math> to give a better price reduction than the other coupons, we must have <math>90\%\cdot x < x-20</math> and <math>90\%\cdot x < 82\%\cdot x+18</math>.
 
  
−
From the first inequality, <math>90\%\cdot x+(-90\%\cdot x) +(20)< x-20+(-90\%\cdot x)+(20)\Rightarrow 20 < 10\%\cdot x\Rightarrow 200 < x</math>.
+
For coupon <math>1</math> to give a greater price reduction than the other coupons, we must have <math>.1x>20\implies x>200</math> and <math>.1x>.18x-18\implies.08x<18\implies x<225</math>.
  
−
From the second inequality, <math>90\%\cdot x +(-82\%\cdot x)< 82\%\cdot x+18+(-82\%\cdot x)\Rightarrow 8\%\cdot x < 18\Rightarrow x < 225</math>.
+
From the first inequality, the listed price must be greater than <math>\textdollar200</math>, so answer choices <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> are eliminated.
  
−
The only answer choice that satisfies these constraints is <math>\boxed{\textbf{(C) }\textdollar219.95}</math>
+
From the second inequality, the listed price must be less than <math>\textdollar225</math>, so answer choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math> are eliminated.
 +
 
 +
The only answer choice that remains is <math>\boxed{\textbf{(C) }\textdollar219.95}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:46, 1 February 2015

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 $.1x>.18x-18\implies.08x<18\implies x<225$.

From the first inequality, the listed price must be greater than $\textdollar200$, so answer choices $\textbf{(A)}$ and $\textbf{(B)}$ are eliminated.

From the second inequality, the listed price must be less than $\textdollar225$, so answer choices $\textbf{(D)}$ and $\textbf{(E)}$ are eliminated.

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

See Also

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. AMC Logo.png

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