Difference between revisions of "1980 AHSME Problems/Problem 3"

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Cross multiplying gets  
 
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<math>6x-3y=2x+2y\\4x=5y\\ \dfrac{x}{y}=\dfrac{5}{4}\\ \boxed{(E)}</math>
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<math>6x-3y=2x+2y\\4x=5y\\ \dfrac{x}{y}=\dfrac{5}{4}\\ \boxed{(E)}</math>  
 
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== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=2|num-a=4}}
 
{{AHSME box|year=1980|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:06, 2 September 2024

Problem

If the ratio of $2x-y$ to $x+y$ is $\frac{2}{3}$, what is the ratio of $x$ to $y$?

$\text{(A)} \ \frac{1}{5} \qquad \text{(B)} \ \frac{4}{5} \qquad \text{(C)} \ 1 \qquad \text{(D)} \ \frac{6}{5} \qquad \text{(E)} \ \frac{5}{4}$

Solution

Cross multiplying gets

$6x-3y=2x+2y\\4x=5y\\ \dfrac{x}{y}=\dfrac{5}{4}\\ \boxed{(E)}$

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See also

1980 AHSME (ProblemsAnswer KeyResources)
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
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