Question

Two positive integers differ by 4 and sum of their reciprocals is $\frac{10}{21}$. Then one of the numbers is: (CAT 1995)

• A. 3
• B. 1
• C. 5
• D. 21

Answer 1

The correct option is A 3

Let one number be x, then the second number will be (x + 4).
Thus $:\frac{1}{x}+\left(\frac{1}{x+4}\right)=\frac{10}{21}$
$⟹\frac{(x+x+4)}{x(x+4)}=\frac{10}{21}$
$⟹\frac{(2x+4)}{x(x+4)}=\frac{10}{21}$
$⟹x=3$

Elimination strategy:

By looking at the answer options, the sum of reciprocal of the numbers will have to give $\frac{10}{21}$.
The number has to be a factor or multiple of 21. So directly option (c) can be eliminated.
Now the numbers that we will have to check will be 3 & 7, 1 & 5, 17 & 21, 21 & 25.
The only possibility from here is 3 & 7. Hence option (a).