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Difference between revisions of "1956 AHSME Problems/Problem 2"

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== Solution ==
 
== Solution ==
  
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For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or <math>\frac{6}{5}</math> of its original value. This tells us that the original price was <math>\frac{5}{6} * 1.20 = \$1.00</math>.
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For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or <math>\frac{6}{5}</math> of its original value. This tells us that the original price was <math>\frac{5}{6}\cdot1.20 = \$1.00</math>.
  
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For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or <math>\frac{4}{5}</math> of its original value. This tells us that the original price was <math>\frac{5}{4} * 1.20 = \$1.50</math>.
+
For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or <math>\frac{4}{5}</math> of its original value. This tells us that the original price was <math>\frac{5}{4}\cdot1.20 = \$1.50</math>.
  
  
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Therefore, he <math>\fbox{(D) lost 10 cents}</math>.
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Therefore, he <math>\boxed{\textbf{(D)}\ \text{lost }10\text{ cents}}</math>.
  
 
== See Also ==
 
== See Also ==
 +
 +
{{AHSME 50p box|year=1956|num-b=1|num-a=3}}
 +
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[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 16:01, 14 March 2023

Problem #2

Mr. Jones sold two pipes at $\textdollar{ 1.20}$ each. Based on the cost, his profit on one was

$20$% and his loss on the other was $20$%. On the sale of the pipes, he:

$\textbf{(A)}\ \text{broke even}\qquad \textbf{(B)}\ \text{lost }4\text{ cents} \qquad\textbf{(C)}\ \text{gained }4\text{ cents}\qquad \\ \textbf{(D)}\ \text{lost }10\text{ cents}\qquad \textbf{(E)}\ \text{gained }10\text{ cents}$

Solution

For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or $\frac{6}{5}$ of its original value. This tells us that the original price was $\frac{5}{6}\cdot1.20 = $1.00$.

For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or $\frac{4}{5}$ of its original value. This tells us that the original price was $\frac{5}{4}\cdot1.20 = $1.50$.


Thus, his total cost was $$2.50$ and his total revenue was $$2.40$.


Therefore, he $\boxed{\textbf{(D)}\ \text{lost }10\text{ cents}}$.

See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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