Difference between revisions of "2014 AMC 10A Problems/Problem 11"
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| + | {{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #8]] and [[2014 AMC 10A Problems|2014 AMC 10A #11]]}} | ||
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==Problem== | ==Problem== | ||
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\textbf{(E) }\textdollar259.95\qquad</math> | \textbf{(E) }\textdollar259.95\qquad</math> | ||
| − | ==Solution== | + | ==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 discounts after the coupons are used will be as follows: | ||
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| + | Coupon 1: <math>x\times10\%=.1x</math> | ||
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| + | Coupon 2: <math>20</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 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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| + | The only choice that satisfies such conditions 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 answer choice over 200. | ||
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| + | ==Video Solutions== | ||
| + | ===Video Solution 1=== | ||
| + | https://youtu.be/aGEB3ykW1BM | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ===Video Solution 2=== | ||
| + | https://youtu.be/rJytKoJzNBY | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2014|ab=A|num-b=10|num-a=12}} | ||
| + | {{AMC12 box|year=2014|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
Latest revision as of 13:12, 3 November 2024
- The following problem is from both the 2014 AMC 12A #8 and 2014 AMC 10A #11, so both problems redirect to this page.
Contents
Problem
A customer who intends to purchase an appliance has three coupons, only one of which may be used:
Coupon 1: 
off the listed price if the listed price is at least 
Coupon 2: 
off the listed price if the listed price is at least 
Coupon 3: 
off the amount by which the listed price exceeds
For which of the following listed prices will coupon 
offer a greater price reduction than either coupon 
or coupon 
?
Solution 1
Let the listed price be 
. Since all the answer choices are above , we can assume 
. Thus the discounts after the coupons are used will be as follows:
Coupon 1: 
Coupon 2: 
Coupon 3: 
For coupon to give a greater price reduction than the other coupons, we must have 
and 
.
The only choice that satisfies such conditions is 
Solution 2 (Using The Answers)
For coupon 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 
is the smallest answer choice over 200.
Video Solutions
Video Solution 1
~savannahsolver
Video Solution 2
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. 
