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Difference between revisions of "1955 AHSME Problems/Problem 6"

(Created page with "Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges. That means...")
 
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A merchant buys a number of oranges at <math>3</math> for <math>10</math> cents and an equal number at <math>5</math> for <math>20</math> cents. To "break even" he must sell all at:
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<math>\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents}</math>
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Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges.
 
Since we are buying at <math>3</math> for <math>10</math> cents and <math>5</math> for <math>20</math> cents, let's assume that together, we are buying 15 oranges.
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That means that we are getting a total of <math>30</math> oranges for <math>(10*5) + (20*3)</math> cents.
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That means that we are getting a total of <math>30</math> oranges for <math>(10x5) + (20x3)</math> cents.
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That comes to a total of <math>30</math> oranges for <math>110</math> cents. <math>110/30</math> = <math>11/3</math> and we are done.
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That comes to a total of <math>30</math> oranges for <math>110</math> cents. <math>110/30</math> = <math>11/3</math>. This leads us to <math>3</math> for <math>11</math> cents which is <math>C</math> and we are done.
 
-Brudder
 
-Brudder

Revision as of 10:20, 28 April 2019

A merchant buys a number of oranges at $3$ for $10$ cents and an equal number at $5$ for $20$ cents. To "break even" he must sell all at:

$\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents}$



Since we are buying at $3$ for $10$ cents and $5$ for $20$ cents, let's assume that together, we are buying 15 oranges. That means that we are getting a total of $30$ oranges for $(10x5) + (20x3)$ cents. That comes to a total of $30$ oranges for $110$ cents. $110/30$ = $11/3$. This leads us to $3$ for $11$ cents which is $C$ and we are done. -Brudder

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