Difference between revisions of "1952 AHSME Problems/Problem 20"

(Created page with "== Problem== If <math> \frac{x}{y}=\frac{3}{4} </math>, then the incorrect expression in the following is: <math> \textbf{(A) \ }\frac{x+y}{y}=\frac{7}{4} \qquad \textbf{(B) \...")
 
 
Line 23: Line 23:
 
Hence, the incorrect expression is <math> \boxed{\textbf{(E)}\ \frac{x-y}{y}=\frac{1}{4}} </math>.
 
Hence, the incorrect expression is <math> \boxed{\textbf{(E)}\ \frac{x-y}{y}=\frac{1}{4}} </math>.
  
 +
Note: If you're bad at fraction manipulation, you can plug in <math>x=3</math> and <math>y=4</math> and find which expression is false.
 
==See also==
 
==See also==
 
{{AHSME 50p box|year=1952|num-b=19|num-a=21}}
 
{{AHSME 50p box|year=1952|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:00, 24 January 2024

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

Let's consider each answer choice.

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

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

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

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

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

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.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png