Difference between revisions of "1953 AHSME Problems/Problem 1"
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| + | ==Solution 2== | ||
| + | The boy buys <math>3</math> oranges for <math>10</math> cents. He sells them at <math>5</math> for <math>20</math>. So, he buys <math>15</math> for <math>50</math> cents and sells them a <math>15</math> for <math>60</math> cents, so he makes <math>10</math> cents of profit on every <math>5</math> oranges. To make <math>100</math> cents of profit, he needs to sell <math>10\cdot 5</math>, or <math>50</math> oranges. | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1953|before=First Question|num-a=2}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:12, 1 April 2017
A boy buys oranges at 
for 
cents. He will sell them at 
for 
cents. In order to make a profit of 
, he must sell:
Solution
The boy buys oranges for cents or 
orange for 
cents. He sells them at 
cents each.
That means for every orange he sells, he makes a profit of 
cents.
To make a profit of 
cents, he needs to sell 
~mathsolver101
Solution 2
The boy buys oranges for cents. He sells them at for . So, he buys 
for 
cents and sells them a for 
cents, so he makes cents of profit on every oranges. To make cents of profit, he needs to sell 
, or oranges.
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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. 
