# 1952 AHSME Problems/Problem 20

## Problem

If $\frac{x}{y}=\frac{3}{4}$, then the incorrect expression in the following is:

$\textbf{(A) \ }\frac{x+y}{y}=\frac{7}{4} \qquad \textbf{(B) \ }\frac{y}{y-x}=\frac{4}{1} \qquad \textbf{(C) \ }\frac{x+2y}{x}=\frac{11}{3} \qquad$

$\textbf{(D) \ }\frac{x}{2y}=\frac{3}{8} \qquad \textbf{(E) \ }\frac{x-y}{y}=\frac{1}{4}$

## Solution

$\boxed{A.} \quad \frac{x+y}{y}=\frac{x}{y}+\frac{y}{y}=\frac{3}{4}+1=\boxed{\frac{7}{4}}$

$\boxed{B.} \quad \frac{y-x}{y}=\frac{y}{y}-\frac{x}{y}=1-\frac{3}{4}=\frac{1}{4}\implies \left(\frac{y-x}{y}\right)^{-1}=\frac{y}{y-x}=\boxed{\frac{4}{1}}$

$\boxed{C.} \quad \frac{x+2y}{x}=\frac{x}{x}+\frac{2y}{x}=1+2\cdot \frac{4}{3}=\boxed{\frac{11}{3}}$

$\boxed{D.} \quad \frac{x}{2y}=\frac{1}{2}\cdot \frac{3}{4}=\boxed{\frac{3}{8}}$

$\boxed{E.} \quad \frac{x-y}{y}=\frac{x}{y}-\frac{y}{y}=\frac{3}{4}-1=\boxed{-\frac{1}{4}}\neq \frac{1}{4}$

Hence, the incorrect expression is $\boxed{\textbf{(E)}\ \frac{x-y}{y}=\frac{1}{4}}$.

Note: If you're bad at fraction manipulation, you can plug in $x=3$ and $y=4$ and find which expression is false.