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Difference between revisions of "2013 AMC 8 Problems/Problem 12"

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To find the percent saved, we have <math>100 \% -70 \%= \boxed{\textbf{(B)}\ 30 \%}</math>
 
To find the percent saved, we have <math>100 \% -70 \%= \boxed{\textbf{(B)}\ 30 \%}</math>
 +
 +
==Solution 2==
 +
Step 1: Find the total price
 +
The first sandal is 50<math>. The second sandal is (1-0.4)50</math>+1/2*50<math> = </math>105
 +
Step 2: Divide by 150
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<math>105/</math>150 = $21/30 = 7/10 = 70%.
 +
Step 3: Subtract from 100% to find the amount he saved.
 +
100%-70% = B)30%
  
 
==Video Solution==
 
==Video Solution==

Revision as of 14:39, 24 August 2025

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?

$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 40 \qquad \textbf{(E)}\ 45$

Solution

First, find the amount of money one will pay for three sandals without the discount. We have $\textdollar 50\times 3 \text{ sandals} = \textdollar 150$.

Then, find the amount of money using the discount: $50 + 0.6 \times 50 + \frac{1}{2} \times 50 = \textdollar 105$.

Finding the percentage yields $\frac{105}{150} = 70 \%$.

To find the percent saved, we have $100 \% -70 \%= \boxed{\textbf{(B)}\ 30 \%}$

Solution 2

Step 1: Find the total price The first sandal is 50$. The second sandal is (1-0.4)50$+1/2*50$=$105 Step 2: Divide by 150 $105/$150 = $21/30 = 7/10 = 70%. Step 3: Subtract from 100% to find the amount he saved. 100%-70% = B)30%

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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