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A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{{13}}{{12}}$, then the numbers are
$\frac{1}{4},\;\frac{4}{1}$
$\frac{3}{4},\;\frac{4}{3}$
$\frac{2}{5},\;\frac{5}{2}$
$\frac{3}{2},\;\frac{2}{3}$
Solution
(d) 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 $ $a + \frac{1}{a} = \frac{{13}}{6} \Rightarrow 6{a^2} – 13a + 6 = 0$
$ \Rightarrow $ $\left( {a – \frac{3}{2}} \right)\,\left( {a – \frac{2}{3}} \right) = 0$
$ \Rightarrow $$a = \frac{3}{2}$ and $b = \frac{2}{3}$
or $a = \frac{2}{3}$ and $b = \frac{3}{2}$.
Trick : Find the $A.M.$ of option $(a), (b), (c), (d)$ one by one.
