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. |
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+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1953|before=First Question|num-a=2}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
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 | |
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