Question

The sum of two numbers is $16$ and the sum of their reciprocal is $\frac{1}{3}$. Find the numbers.

• A. $4,-12$
• B. $4,12$
• C. $8,8$
• D. None

Answer 1

Answer: (B)

$4,12$

Let the first number be $x$ then the second number will be $16-x$ so, that the sum of the two numbers remains $16.$

Sum of their reciprocal will be,

$\frac{1}{x}+\frac{1}{16-x}=\frac{1}{3}$

$\Rightarrow \frac{16-x+x}{x(16-x)}=\frac{1}{3}$

$\Rightarrow 16\times 3=16x-x^2$

$\Rightarrow x^2-16x+48=0$

$\Rightarrow x-12x-4x+48=0$

$\Rightarrow (x-12)(x-4)=0$

$\Rightarrow x=12,4$

If we choose the first number to be equal to $12$ then the second will be $4$ and if we choose the first number to be equal to $4$ then the second number will be $12.$

So, we can say that the two numbers are $4,12.$