Difference between revisions of "1982 AHSME Problems/Problem 5"
(→Solution) |
J314andrews (talk | contribs) (added navbox) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 21: | Line 21: | ||
Since we know <math>a</math> is less than <math>b</math> and <math>\frac{x}{y}=\frac{bc}{a+b}</math>, the smaller of <math>x</math> and <math>y</math> must be <math>x</math>. Therefore the answer is <math>\boxed{\textbf{(C) }\frac{ac}{a+b}}</math>. | Since we know <math>a</math> is less than <math>b</math> and <math>\frac{x}{y}=\frac{bc}{a+b}</math>, the smaller of <math>x</math> and <math>y</math> must be <math>x</math>. Therefore the answer is <math>\boxed{\textbf{(C) }\frac{ac}{a+b}}</math>. | ||
| + | |||
| + | |||
| + | |||
| + | ~superagh | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1982|num-b=4|num-a=6}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 22:08, 29 June 2025
Problem
Two positive numbers 
and 
are in the ratio 
where 
. If 
, then the smaller of and is
Solution
We can write 2 equations.
and
Solving for and in terms of 
we get :
and 
Since we know 
is less than 
and 
, the smaller of and must be . Therefore the answer is 
.
~superagh
See Also
| 1982 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
