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Difference between revisions of "1988 AJHSME Problems/Problem 23"

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==Problem==
 
==Problem==
  
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Maria buys computer disks at a price of <math>4</math> for <dollar/><math>5</math> and sells them at a price of <math>3</math> for <dollar/><math>5</math>.  How many computer disks must she sell in order to make a profit of <dollar/><math>100</math>?
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Maria buys computer disks at a price of <math>4</math> for <math>\$5</math> and sells them at a price of <math>3</math> for <math>\$5</math>.  How many computer disks must she sell in order to make a profit of <math>\$100</math>?
  
 
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 1200</math>
 
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 1200</math>
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==Solution==
 
==Solution==
  
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This is the equivalent of saying she buys <math>12</math> for <dollar/><math>15</math> and sells <math>12</math> for <dollar/><math>20</math>, so for every dozen disks she sells, she profits <dollar/><math>5</math>.
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This is the equivalent of saying she buys <math>12</math> for <math>\$15</math> and sells <math>12</math> for <math>\$20</math>, so for every dozen disks she sells, she profits <math>\$5</math>.
  
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She needs to profit <dollar/><math>100</math>, so she needs to sell <math>\frac{100}{5}=20</math> dozen disks, which is <math>240\rightarrow \boxed{\text{D}}</math>
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She needs to profit <math>\$100</math>, so she needs to sell <math>\frac{100}{5}=20</math> dozen disks, which is <math>240\rightarrow \boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1988|num-b=22|num-a=24}}
 
{{AJHSME box|year=1988|num-b=22|num-a=24}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:46, 24 June 2016

Problem

Maria buys computer disks at a price of $4$ for $$5$ and sells them at a price of $3$ for $$5$. How many computer disks must she sell in order to make a profit of $$100$?

$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 1200$

Solution

This is the equivalent of saying she buys $12$ for $$15$ and sells $12$ for $$20$, so for every dozen disks she sells, she profits $$5$.

She needs to profit $$100$, so she needs to sell $\frac{100}{5}=20$ dozen disks, which is $240\rightarrow \boxed{\text{D}}$

See Also

1988 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
All AJHSME/AMC 8 Problems and Solutions

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