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Difference between revisions of "1955 AHSME Problems/Problem 6"
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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.
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-Brudder
 
-Brudder
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== See Also ==
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{{AHSME 50p box|year=1955|num-b=5|num-a=7}}
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{{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
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


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