Difference between revisions of "1995 AJHSME Problems/Problem 10"

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==Problem==
 
==Problem==
  
A jacket and a shirt originally sold for <math></math>80<math> and <dollar/>40, respectively.  During a sale Chris bought the <dollar/>80 jacket at a </math>40\%<math> discount and the <dollar/>40 shirt at a </math>55\%<math> discount.  The total amount saved was what percent of the total of the original prices?
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A jacket and a shirt originally sold for <math>80</math> dollars and <math>40</math> dollars, respectively.  During a sale Chris bought the <math>80</math> dollar jacket at a <math>40\%</math> discount and the <math>40</math> dollar shirt at a <math>55\%</math> discount.  The total amount saved was what percent of the total of the original prices?
  
</math>\text{(A)}\ 45\% \qquad \text{(B)}\ 47\dfrac{1}{2}\% \qquad \text{(C)}\ 50\% \qquad \text{(D)}\ 79\dfrac{1}{6}\% \qquad \text{(E)}\ 95\%$.
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<math>\text{(A)}\ 45\% \qquad \text{(B)}\ 47\dfrac{1}{2}\% \qquad \text{(C)}\ 50\% \qquad \text{(D)}\ 79\dfrac{1}{6}\% \qquad \text{(E)}\ 95\%</math>.
  
 
==Solution==
 
==Solution==

Latest revision as of 16:02, 12 November 2019

Problem

A jacket and a shirt originally sold for $80$ dollars and $40$ dollars, respectively. During a sale Chris bought the $80$ dollar jacket at a $40\%$ discount and the $40$ dollar shirt at a $55\%$ discount. The total amount saved was what percent of the total of the original prices?

$\text{(A)}\ 45\% \qquad \text{(B)}\ 47\dfrac{1}{2}\% \qquad \text{(C)}\ 50\% \qquad \text{(D)}\ 79\dfrac{1}{6}\% \qquad \text{(E)}\ 95\%$.

Solution

Chris would have spent $40 + 80 = 120$ dollars if there were no sale.

The sale means that he saved $80(0.40) + 40(0.55) = 32 + 22 = 54$ dollars.

He saved $\frac{54}{120}\cdot 100\% = 0.45 \cdot 100\% = 45\%$ off the total of the original prices, and the answer is $\boxed{A}$.

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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

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