Difference between revisions of "1956 AHSME Problems/Problem 2"
Megaboy6679 (talk | contribs) (→Solution) |
Megaboy6679 (talk | contribs) m (→Solution) |
||
| Line 14: | Line 14: | ||
== Solution == | == Solution == | ||
| − | For the first pipe, we are told that his profit was | + | 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>. |
| − | For the second pipe, we are told that his loss was | + | 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>. |
Latest revision as of 16:01, 14 March 2023
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
