Difference between revisions of "1982 AHSME Problems/Problem 5"
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<math>\text{(A)} \ \frac{ac}{b} \qquad \text{(B)} \ \frac{bc-ac}{b} \qquad \text{(C)} \ \frac{ac}{a+b} \qquad \text{(D)}\ \frac{bc}{a+b}\qquad \text{(E)}\ \frac{ac}{b-a}</math> | <math>\text{(A)} \ \frac{ac}{b} \qquad \text{(B)} \ \frac{bc-ac}{b} \qquad \text{(C)} \ \frac{ac}{a+b} \qquad \text{(D)}\ \frac{bc}{a+b}\qquad \text{(E)}\ \frac{ac}{b-a}</math> | ||
+ | |||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | We can write 2 equations. | ||
+ | |||
+ | <math>\frac{x}{y}=\frac{a}{b}</math> | ||
+ | |||
+ | and | ||
+ | |||
+ | <math>x+y=c</math> | ||
+ | |||
+ | Solving for <math>x</math> and <math>y</math> in terms of <math>a, b, c</math> we get : | ||
+ | |||
+ | <math>x=\frac{ac}{a+b}</math> and <math>y=\frac{bc}{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.