Difference between revisions of "1953 AHSME Problems/Problem 1"
(Created page with "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...") |
|||
| (5 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
| + | 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: | ||
| + | |||
| + | <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. | ||
| Line 5: | Line 11: | ||
~mathsolver101 | ~mathsolver101 | ||
| + | |||
| + | ==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. | ||
| + | |||
| + | ==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 | |
| 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. 
