Difference between revisions of "2013 AMC 8 Problems/Problem 12"
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==Problem== | ==Problem== | ||
+ | At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of \$50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the \$150 regular price did he save? | ||
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+ | <math>\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 40 \qquad \textbf{(E)}\ 45</math> | ||
==Solution== | ==Solution== | ||
+ | First, find the amount of money one will pay for three sandals without the discount. We have <math>\textdollar 50\times 3 \text{ sandals} = \textdollar 150</math>. | ||
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+ | Then, find the amount of money using the discount: <math>50 + 0.6 \times 50 + \frac{1}{2} \times 50 = \textdollar 105</math>. | ||
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+ | Finding the percentage yields <math>\frac{105}{150} = 70 \%</math>. | ||
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+ | To find the percent saved, we have <math>100 \% -70 \%= \boxed{\textbf{(B)}\ 30 \%}</math> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/VQPKh5hc3xY ~savannahsolver | ||
==See Also== | ==See Also== | ||
− | {{AMC8 box|year=2013| | + | {{AMC8 box|year=2013|num-b=11|num-a=13}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:20, 16 November 2024
Contents
Problem
At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?
Solution
First, find the amount of money one will pay for three sandals without the discount. We have .
Then, find the amount of money using the discount: .
Finding the percentage yields .
To find the percent saved, we have
Video Solution
https://youtu.be/VQPKh5hc3xY ~savannahsolver
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.