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Sum of two numbers is 16. The sum of their reciprocals is 1/3. Find the numbers.
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Given that the sum of two numbers is 16.
Let the two numbers be x and 16 - x
By the given hypothesis, we have
`rArr1/x+1/(16-x)=1/3`
`rArr(16-x+x)/(x(16-x))=1/3`
⇒ 48 = 16x - x2
⇒ ЁЭСе2 - 16ЁЭСе + 48 = 0
⇒ ЁЭСе2 - 12ЁЭСе - 4ЁЭСе + 48 = 0
⇒ x(x - 12) - 4(x - 12) = 0
⇒ (x - 12) (x - 4) = 0
⇒ x = 12 or x = 4
∴ The two numbers are 4 and 12.
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