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>(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{ | + | 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{B}</math> and we are done. |
-Brudder | -Brudder | ||
| − | Edited by Andrew Lu | + | |
| + | Edited by Andrew Lu (originally answer was C) | ||
== See Also == | == See Also == | ||
Latest revision as of 18:46, 4 October 2023
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:
Solution
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
Edited by Andrew Lu (originally answer was C)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
