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Mathematics

Forty years hence, Mr. Pratap's age will be the square of what it was 32 years ago. Find his present age.

Quadratic Equations

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Answer

Let the present age of Mr. Pratap be x years.

After 40 years his age will be (x + 40) years

32 years before his age was (x - 32) years

According to question ,

⇒(x−32)2=(x+40)⇒x2+1024−64x=x+40⇒x2−64x−x+1024−40=0⇒x2−65x+984=0⇒x2−24x−41x+984=0⇒x(x−24)−41(x−24)=0⇒(x−41)(x−24)=0⇒x−41=0 or x−24=0x=41 or x=24\Rightarrow (x - 32)^2 = (x + 40) \\[1em] \Rightarrow x^2 + 1024 - 64x = x + 40 \\[1em] \Rightarrow x^2 - 64x - x + 1024 - 40 = 0 \\[1em] \Rightarrow x^2 - 65x + 984 = 0 \\[1em] \Rightarrow x^2 - 24x - 41x + 984 = 0 \\[1em] \Rightarrow x(x - 24) - 41(x - 24) = 0 \\[1em] \Rightarrow (x - 41)(x - 24) = 0 \\[1em] \Rightarrow x - 41 = 0 \text{ or } x - 24 = 0 \\[1em] x = 41 \text{ or } x = 24

But x ≠ 24 as it is less than 32.

∴ x = 41.

The present age of Mr. Pratap is 41 years.

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