Question

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

Answer 1

Let the larger number be x and the smaller number be y.
Then, we have:
(x + y) = 16 ....(i)
And, $\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$ ....(ii)
⇒ $\frac{x+y}{xy}=\frac{1}{3}$
⇒ 3(x + y) = xy
⇒ 3 × 16 = xy [Since from (i), we have: x + y = 16]
∴ xy = 48 ....(iii)
We know:
(x − y)² = (x + y)² − 4xy
(x − y)² = (16)² − 4 × 48 = 256 − 192 = 64
∴ (x − y) = $\pm \sqrt{64}=\pm 8$
Since x is larger and y is smaller, we have:
x − y = 8 .....(iv)
On adding (i) and (iv), we get:
2x = 24
⇒ x = 12
On substituting x = 12 in (i), we get:
12 + y = 16 ⇒ y = (16 − 12) = 4
Hence, the required numbers are 12 and 4.