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

Answer: (D)

$\frac { 3 } { 2 } , \frac { 2 } { 3 }$

Answer 2

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

$\Rightarrow$ $a + b = \frac { 13 } { 6 }$

$\Rightarrow$ $a + \frac { 1 } { a } = \frac { 13 } { 6 } \Rightarrow 6 a ^ { 2 } - 13 a + 6 = 0$

$\Rightarrow$ $\left( a - \frac { 3 } { 2 } \right) \left( a - \frac { 2 } { 3 } \right) = 0$ $\Rightarrow$ $a = \frac { 3 } { 2 }$ and $b = \frac { 2 } { 3 }$

or $a = \frac { 2 } { 3 }$ and $b = \frac { 3 } { 2 }$ .

Trick : Find the A.M. of option (1), (2), (3), (4) one by one.