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

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==Solution 1==
 
==Solution 1==
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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:
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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:
  
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Coupon 1: <math>x-10\%\cdot x=90\%\cdot x</math>
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Coupon 1: <math>x\times10\%=.1x</math>
  
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Coupon 2: <math>x-20</math>
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Coupon 2: <math>20</math>
  
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Coupon 3: <math>x-18\%\cdot(x-100)=82\%\cdot x+18</math>
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Coupon 3: <math>18\%\times(x-100)=.18x-18</math>
  
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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>.
 
  
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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>.
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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>.
  
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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>.
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The only choice that satisfies such conditions is <math>\boxed{\textbf{(C)}\ \textdollar219.95}</math>
  
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The only answer choice that satisfies these constraints is <math>\boxed{\textbf{(C) }\textdollar219.95}</math>
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==Solution 2 (Using The Answers)==
 +
For coupon <math>1</math> 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 <math>\boxed{\textbf{(C) }\textdollar219.95}</math> is the smallest number over 200.
 +
 
 +
==Video Solutions==
 +
===Video Solution 1===
 +
https://youtu.be/aGEB3ykW1BM
 +
 
 +
~savannahsolver
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 +
===Video Solution 2===
 +
https://youtu.be/rJytKoJzNBY
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2014|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2014|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Introductory Algebra Problems]]

Revision as of 10:44, 7 September 2021

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 number over 200.

Video Solutions

Video Solution 1

https://youtu.be/aGEB3ykW1BM

~savannahsolver

Video Solution 2

https://youtu.be/rJytKoJzNBY

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

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

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