Difference between revisions of "1992 AJHSME Problems/Problem 8"

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== Solution ==
 
== Solution ==
<math> 1500\times 0.1=150 </math>, so the store owner is <dollar/>150 below profit. Therefore he needs to sell <math>150+100= 250</math> dollars worth of pencils. Selling them at <dollar/>0.25 each gives <math>250/0.25= \boxed{\textbf{(C)}\ 1000}</math>.
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<math> 1500\times 0.1=150 </math>, so the store owner is <math>\$150</math> below profit. Therefore he needs to sell <math>150+100= 250</math> dollars worth of pencils. Selling them at <math>\$0.25</math> each gives <math>250/0.25= \boxed{\textbf{(C)}\ 1000}</math>.
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1992|num-b=7|num-a=9}}
 
{{AJHSME box|year=1992|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:12, 2 July 2017

Problem

A store owner bought $1500$ pencils at $$ 0.10$ each. If he sells them for $$ 0.25$ each, how many of them must he sell to make a profit of exactly $$ 100.00$?

$\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900$

Solution

$1500\times 0.1=150$, so the store owner is $$150$ below profit. Therefore he needs to sell $150+100= 250$ dollars worth of pencils. Selling them at $$0.25$ each gives $250/0.25= \boxed{\textbf{(C)}\ 1000}$.

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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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