Difference between revisions of "2014 AMC 10A Problems/Problem 11"
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==Solution== | ==Solution== | ||
+ | Let the listed price be <math>x</math>. Since all the answer choices are above <math>$100</math>, we can assume <math>x > 100</math>. Thus the prices after coupons will be as follows: | ||
+ | |||
+ | Coupon 1: <math>x-10\%\cdot x=90\%\cdot x</math> | ||
+ | |||
+ | Coupon 2: <math>x-20</math> | ||
+ | |||
+ | Coupon 3: <math>x-18\%\cdot(x-100)=82\%\cdot x+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>. | ||
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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 answer choice that satisfies these constraints is <math>\boxed{\textbf{(C) }\textdollar219.95}</math> | ||
==See Also== | ==See Also== |
Revision as of 23:15, 6 February 2014
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
Let the listed price be . Since all the answer choices are above , we can assume . Thus the prices after coupons will be as follows:
Coupon 1:
Coupon 2:
Coupon 3:
For coupon to give a better price reduction than the other coupons, we must have and .
From the first inequality, .
From the second inequality, .
The only answer choice that satisfies these constraints is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.