1956 AHSME Problems/Problem 2: Difference between revisions
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== Solution == | == Solution == | ||
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> | 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>. | ||
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>. | |||
Thus, his total cost was <math>\$2.50</math> and his total revenue was <math>\$2.40</math>. | |||
Therefore, he <math>\fbox{(D) lost 10 cents}</math>. | |||
== See Also == | == See Also == | ||
Revision as of 16:09, 31 December 2015
Problem #2
Mr. Jones sold two pipes at 
each. Based on the cost, his profit on one was
% and his loss on the other was %.
On the sale of the pipes, he:
Solution
For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or 
of its original value. This tells us that the original price was 
.
For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or 
of its original value. This tells us that the original price was 
.
Thus, his total cost was 
and his total revenue was 
.
Therefore, he 
.
