Difference between revisions of "1952 AHSME Problems/Problem 7"

(Created page with "== Problem== When simplified, <math> (x^{-1}+y^{-1})^{-1} </math> is equal to: <math> \textbf{(A) \ }x+y \qquad \textbf{(B) \ }\frac{xy}{x+y} \qquad \textbf{(C) \ }xy \qquad \...")
 
m (See also)
 
Line 10: Line 10:
  
 
==See also==
 
==See also==
{{AHSME 50p box|year=1952|num-b=4|num-a=6}}
+
{{AHSME 50p box|year=1952|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:10, 1 January 2014

Problem

When simplified, $(x^{-1}+y^{-1})^{-1}$ is equal to:

$\textbf{(A) \ }x+y  \qquad \textbf{(B) \ }\frac{xy}{x+y} \qquad \textbf{(C) \ }xy \qquad \textbf{(D) \ }\frac{1}{xy} \qquad \textbf{(E) \ }\frac{x+y}{xy}$

Solution

$(x^{-1}+y^{-1})^{-1}=\left(\frac{1}{x}+\frac{1}{y}\right)^{-1}=\left(\frac{x+y}{xy}\right)^{-1}=\boxed{\textbf{(B)}\ \frac{xy}{x+y}}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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