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1955 AHSME Problems/Problem 6: Difference between revisions

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== Solution ==




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That means that we are getting a total of <math>30</math> oranges for <math>(10\times5) + (20\times3)</math> cents.
That means that we are getting a total of <math>30</math> oranges for <math>(10\times5) + (20\times3)</math> cents.
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>\boxed{C}</math> and we are done.
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>\boxed{C}</math> and we are done.
-Brudder
-Brudder
== See Also ==
{{AHSME 50p box|year=1955|num-b=5|num-a=7}}
{{MAA Notice}}

Revision as of 09:39, 15 February 2021

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:

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


Solution

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 $(10\times5) + (20\times3)$ 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 $\boxed{C}$ and we are done.

-Brudder

See Also

1955 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AHSME Problems and Solutions


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