Question

A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{13}{12},$ then the numbers are

• A. $\frac{1}{4},\frac{4}{1}$
• B. $\frac{3}{4},\frac{4}{3}$
• C. $\frac{2}{5},\frac{5}{2}$
• D. $\frac{3}{2},\frac{2}{3}$

Answer 1

The correct option is D $\frac{3}{2},\frac{2}{3}$

Suppose that required numbers a and b. Therefore according to the conditions $a=\frac{1}{b}$ and $\frac{a+b}{2}=\frac{13}{12}\Rightarrow a+b=\frac{13}{6}$
$\Rightarrow a+\frac{1}{a}=\frac{13}{6}\Rightarrow 6a^2-13a+6=0$

$\Rightarrow (a-\frac{3}{2})(a-\frac{2}{3})=0\Rightarrow a=\frac{3}{2}andb=\frac{2}{3}ora=\frac{2}{3}andb=\frac{3}{2}.$
Trick : Find the A.M. of option (a), (b), (c), (d) one by one.