Difference between revisions of "1955 AHSME Problems/Problem 6"
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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. | ||
| − | That means that we are getting a total of <math>30</math> oranges for <math>( | + | 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>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 | ||
Revision as of 22:07, 18 August 2019
A merchant buys a number of oranges at 
for 
cents and an equal number at 
for 
cents. To "break even" he must sell all at:
Since we are buying at for cents and for cents, let's assume that together, we are buying 15 oranges.
That means that we are getting a total of 
oranges for 
cents.
That comes to a total of oranges for 
cents. 
= 
. This leads us to for 
cents which is 
and we are done.
-Brudder
