Difference between revisions of "1953 AHSME Problems/Problem 1"
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+ | A boy buys oranges at <math>3</math> for <math>10</math> cents. He will sell them at <math>5</math> for <math>20</math> cents. In order to make a profit of <math> $1.00</math>, he must sell: | ||
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+ | <math>\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
The boy buys <math>3</math> oranges for <math>10</math> cents or <math>1</math> orange for <math>\frac{10}{3}</math> cents. He sells them at <math>\frac{20}{5}=4</math> cents each. | The boy buys <math>3</math> oranges for <math>10</math> cents or <math>1</math> orange for <math>\frac{10}{3}</math> cents. He sells them at <math>\frac{20}{5}=4</math> cents each. | ||
That means for every orange he sells, he makes a profit of <math>4-\frac{10}{3}=\frac{2}{3}</math> cents. | That means for every orange he sells, he makes a profit of <math>4-\frac{10}{3}=\frac{2}{3}</math> cents. | ||
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~mathsolver101 | ~mathsolver101 | ||
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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> 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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All AHSME Problems and Solutions |
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