Question

The sum of two numbers is 1/6 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) = ±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.