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

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==Problem==
 
==Problem==
  
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><dollar/>5</math> and sells them at a price of <math>3</math> for <math><dollar/>5</math>.  How many computer disks must she sell in order to make a profit of <math><dollar/>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>

Revision as of 23:29, 17 January 2016

Problem

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

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

Solution

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

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

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
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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