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Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 8"

(Created page with "== Problem == <math>\frac{a^(-1) + b^(-1)}{(a+b)^(-1)}</math> is equal to <math>\text{(A)} \quad K-10 \quad \text{(B)} \quad 5K-20 \quad \text{(C)} \quad \frac{K-4}{5} \quad...")
 
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
<math>\frac{a^(-1) + b^(-1)}{(a+b)^(-1)}</math> is equal to
+
<math>\frac{a^{-1} + b^{-1}}{(a+b)^{-1}}</math> is equal to
  
−
<math>\text{(A)} \quad K-10 \quad \text{(B)} \quad 5K-20 \quad \text{(C)} \quad \frac{K-4}{5} \quad \text{(D)} \quad \frac{K}{5}-4 \quad \text{(E) NOTA}</math>
+
<math>\text{(A)} \quad \frac{(a+b)^2}{ab} \quad \text{(B)} \quad \frac{1}{ab} \quad \text{(C)} \quad ab \quad \text{(D)} \quad \frac{ab}{(a+b)^2} \quad \text{(E)} a+b</math>
  
 
== Solution ==
 
== Solution ==
−
Note that there are <math>5</math> dimes in one half dollar, so there are
+
<math>\frac{\frac{1}{a} + \frac{1}{b}}{\frac{1}{a+b}} = \frac{\frac{a+b}{ab}}{\frac{1}{a+b}} = \frac{a+b}{ab} \cdot (a+b) = \boxed{\text{(A)} \quad \frac{(a+b)^2}{ab}}</math>
−
<math>(K-4)(5) = 5K-20</math> dimes in <math>K-4</math> half dollars. The answer is <math>\boxed{\text{(B)}}.</math>
 
−
 
 
 
== See Also ==
 
== See Also ==
 
{{Succession box
 
{{Succession box

Latest revision as of 22:55, 1 February 2021

Problem

$\frac{a^{-1} + b^{-1}}{(a+b)^{-1}}$ is equal to

$\text{(A)} \quad \frac{(a+b)^2}{ab} \quad \text{(B)} \quad \frac{1}{ab} \quad \text{(C)} \quad ab \quad \text{(D)} \quad \frac{ab}{(a+b)^2} \quad \text{(E)} a+b$

Solution

$\frac{\frac{1}{a} + \frac{1}{b}}{\frac{1}{a+b}} = \frac{\frac{a+b}{ab}}{\frac{1}{a+b}} = \frac{a+b}{ab} \cdot (a+b) = \boxed{\text{(A)} \quad \frac{(a+b)^2}{ab}}$

See Also

1963 TMTA High School Mathematics Contests (Problems)
Preceded by
Problem 7 		TMTA High School Mathematics Contest Past Problems/Solutions Followed by
Problem 9
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