Difference between revisions of "1953 AHSME Problems/Problem 1"
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==Solution 2== | ==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 | + | The boy buys <math>3</math> oranges for <math>10</math> cents. He sells them at <math>5</math> for <math>20</math> cents. So, he buys <math>15</math> for <math>50</math> cents and sells them at <math>15</math> for <math>60</math> cents, so he makes <math>10</math> cents of profit on every <math>15</math> oranges. To make <math>100</math> cents of profit, he needs to sell <math>15 \cdot \frac{100}{10} = \boxed{150}</math> oranges. |
==See Also== | ==See Also== | ||
Latest revision as of 03:01, 14 June 2022
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 cents. So, he buys 
for 
cents and sells them at for 
cents, so he makes cents of profit on every oranges. To make cents of profit, he needs to sell 
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 | ||
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