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Difference between revisions of "1950 AHSME Problems/Problem 36"

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==Solution==
 
==Solution==
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Since there is a 20% off sale, then the new price is <math>0.8\cdot2.50</math>. Albert buys five of them, so the amount saved is the price he had to pay with the 20% sale subtracted from the price he had to pay with no sale:
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Without loss of generality, we can set the list price equal to <math>100</math>. The merchant buys the goods for <math>100*.75=75</math>. Let <math>x</math> be the marked price.
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We then use the equation <math>0.8x-75=25</math> to solve for <math>x</math> and get a marked price of <math>\boxed{\textbf{(A)}\ 125\%}</math>.
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<math>(5\cdot2.50)-(5\cdot0.8\cdot2.50)=(5\cdot2.50)-(4\cdot2.50)</math>
 
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<math>=(5-4)(2.50)</math>
 
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<math>=2.50</math>
 
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<math>\text{Answer: }\boxed{\mathbf{(C)}}</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 11:43, 20 September 2024

Problem

A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?

$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)}\ 120\% \qquad \textbf{(D)}\ 80\% \qquad \textbf{(E)}\ 75\%$

Solution

Without loss of generality, we can set the list price equal to $100$. The merchant buys the goods for $100*.75=75$. Let $x$ be the marked price. We then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\boxed{\textbf{(A)}\ 125\%}$.

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 35 		Followed by
Problem 37
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All AHSME Problems and Solutions

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