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

(Problem)
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
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>?
+
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>
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
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>.
+
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>.
  
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>
+
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==

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png