The sum of two numbers is 16 and the sum of their reciprocals is 1/3 . Find the numbers.

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1 Answers

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) = ±√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

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