Difference between revisions of "1953 AHSME Problems/Problem 2"

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== Solution ==
 
== Solution ==
  
The first discount takes off <math>20\%</math> of the price, so the cost of the refrigerator is <math>0.8\cdot250</math>. The next discount takes off <math>15\%</math>, so the cost of the refrigerator is now <math>0.8\cdot0.85\cdot250=0.68\cdot250</math>. Thus, the sale price of the refrigerator is <math>\boxed{\textbf{(D)} \text{68\% of 250.00}}</math>.
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The first discount takes off <math>20\%</math> of the price, so the cost of the refrigerator is <math>0.8\cdot250</math>. The next discount takes off <math>15\%</math>, so the cost of the refrigerator is now <math>0.8\cdot0.85\cdot250=0.68\cdot250</math>. Thus, the sale price of the refrigerator is <math>\boxed{\textbf{(D)}\ \text{68\% of 250.00}}</math>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:49, 16 January 2019

Problem

A refrigerator is offered at sale at $250.00 less successive discounts of 20% and 15%. The sale price of the refrigerator is:

$\textbf{(A) } \text{35\% less than 250.00} \qquad \textbf{(B) } \text{65\% of 250.00} \qquad \textbf{(C) } \text{77\% of 250.00} \qquad  \textbf{(D) } \text{68\% of 250.00} \qquad \textbf{(E) } \text{none of these}$

Solution

The first discount takes off $20\%$ of the price, so the cost of the refrigerator is $0.8\cdot250$. The next discount takes off $15\%$, so the cost of the refrigerator is now $0.8\cdot0.85\cdot250=0.68\cdot250$. Thus, the sale price of the refrigerator is $\boxed{\textbf{(D)}\ \text{68\% of 250.00}}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AHSME Problems and Solutions

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