The sum of two numbers is 16 and the sum of their reciprocals is \(\fr

1.	The sum of two numbers is 16 and the sum of their reciprocals is \(\frac{1}3\) . Find the numbers.
Answer» 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.

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

Answer»

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.