Difference between revisions of "1969 AHSME Problems/Problem 2"

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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
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Let <math>c</math> be the cost. Selling the item for <math>x</math> dollars equates to losing <math>15\%</math> of <math>c</math>, so <math>x=.85c</math>. Selling the item for <math>y</math> dollars equates to profiting by <math>15\%</math> of <math>c</math>, so <math>y=1.15c</math>. Therefore <math>\frac{y}{x}=\frac{1.15c}{.85c}=\frac{23}{17}</math>. The answer is <math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:27, 10 July 2015

Problem

If an item is sold for $x$ dollars, there is a loss of $15\%$ based on the cost. If, however, the same item is sold for $y$ dollars, there is a profit of $15\%$ based on the cost. The ratio of $y:x$ is:

$\text{(A) } 23:17\quad \text{(B) } 17y:23\quad \text{(C) } 23x:17\quad \\ \text{(D) dependent upon the cost} \quad \text{(E) none of these.}$

Solution

Let $c$ be the cost. Selling the item for $x$ dollars equates to losing $15\%$ of $c$, so $x=.85c$. Selling the item for $y$ dollars equates to profiting by $15\%$ of $c$, so $y=1.15c$. Therefore $\frac{y}{x}=\frac{1.15c}{.85c}=\frac{23}{17}$. The answer is $\fbox{A}$.

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 26 27 28 29 30 31 32 33 34 35
All AHSME Problems and Solutions

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