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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> </math>1.00<math>, he must sell:
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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>
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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>
  
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Solution:
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==Solution==
  
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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.   
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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.   
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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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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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To make a profit of </math>100<math> cents, he needs to sell </math>\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}$
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To make a profit of <math>100</math> cents, he needs to sell <math>\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}</math>
  
 
~mathsolver101
 
~mathsolver101
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==Solution 2==
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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.
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==See Also==
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{{AHSME 50p box|year=1953|before=First Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 20:12, 1 April 2017

A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$, he must sell:

$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$

Solution

The boy buys $3$ oranges for $10$ cents or $1$ orange for $\frac{10}{3}$ cents. He sells them at $\frac{20}{5}=4$ cents each. That means for every orange he sells, he makes a profit of $4-\frac{10}{3}=\frac{2}{3}$ cents.

To make a profit of $100$ cents, he needs to sell $\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}$

~mathsolver101

Solution 2

The boy buys $3$ oranges for $10$ cents. He sells them at $5$ for $20$. So, he buys $15$ for $50$ cents and sells them a $15$ for $60$ cents, so he makes $10$ cents of profit on every $5$ oranges. To make $100$ cents of profit, he needs to sell $10\cdot 5$, or $50$ 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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