Question

The sum of two numbers is 16 and the sum of their reciprocals is `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, `1/x + 1/y = 1/3`   ...(ii)

⇒ 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) = ± sqrt(64)`

∴ (x – y) = ±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)

⇒ y = 4

Hence, the required numbers are 12 and 4.