Difference between revisions of "2014 AMC 10A Problems/Problem 11"
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==Solution 2 (Using The Answers)== | ==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. | 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. | ||
| + | |||
| + | ==Solution 3== | ||
| + | Since all the answer choices are above <math>\$100</math>, let the price be <math>\$100 + x</math>, where <math>x > 0</math>. Then, the discount from Coupon 1 is <math>\$10 + \frac{x}{10}</math>, the discount from Coupon 2 is <math>\$20</math>, and the discount from Coupon 3 is <math>\frac{9}{50}x</math>. 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 <math>x > 100</math> and <math>x < 125</math>. Thus the price is between <math>\$200</math> and <math>\$225\Rightarrow \boxed{\text{(C)}}</math>. | ||
==Video Solutions== | ==Video Solutions== | ||
Latest revision as of 07:40, 22 October 2025
- 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.
Solution 3
Since all the answer choices are above , let the price be 
, where 
. Then, the discount from Coupon 1 is 
, the discount from Coupon 2 is 
, and the discount from Coupon 3 is 
. 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 and 
. Thus the price is between 
and 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
