Difference between revisions of "1980 AHSME Problems/Problem 3"
Mrdavid445 (talk | contribs) (Created page with "==Problem== If the ratio of <math>2x-y</math> to <math>x+y</math> is <math>\frac{2}{3}</math>, what is the ratio of <math>x</math> to <math>y</math>? <math>\text{(A)} \ \frac{1...") |
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<math>\text{(A)} \ \frac{1}{5} \qquad \text{(B)} \ \frac{4}{5} \qquad \text{(C)} \ 1 \qquad \text{(D)} \ \frac{6}{5} \qquad \text{(E)} \ \frac{5}{4}</math> | <math>\text{(A)} \ \frac{1}{5} \qquad \text{(B)} \ \frac{4}{5} \qquad \text{(C)} \ 1 \qquad \text{(D)} \ \frac{6}{5} \qquad \text{(E)} \ \frac{5}{4}</math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | <math>\frac{2x-y}{x+y} = \frac{2}{3}</math> | ||
+ | |||
+ | Cross multiplying yields | ||
+ | |||
+ | <math>6x-3y=2x+2y</math> | ||
+ | |||
+ | <math>4x=5y</math> | ||
+ | |||
+ | <math>\dfrac{x}{y}= \boxed{(\textbf{E}) \dfrac{5}{4}}</math> | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1980|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:20, 14 July 2025
Problem
If the ratio of to
is
, what is the ratio of
to
?
Solution
Cross multiplying yields
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.