Difference between revisions of "1964 AHSME Problems/Problem 23"
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<math>\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96</math> | <math>\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96</math> | ||
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Set the two numbers as <math>x</math> and <math>y</math>. Therefore, | Set the two numbers as <math>x</math> and <math>y</math>. Therefore, | ||
Latest revision as of 02:00, 19 June 2019
Problem
Two numbers are such that their difference, their sum, and their product are to one another as 
. The product of the two numbers is:
Solution
Set the two numbers as 
and 
. Therefore,
, and
. Simplifying the first equation gives 
. Substituting for in the second equation gives 
Solving yields 
or 
. Substituting back into the first equation yields 
which is false, so is not valid and . Substituting into gives 
and 
.
See More
| 1964 AHSME (Problems • Answer Key • Resources) | ||
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
