Mathematics
One year ago, a man was 8 times as old as his son. Now his age is equal to square of his son's age. Find their present ages.
Answer
Let one year ago,
son's age = x years
So, man's age = 8x years
Present age of,
son = x + 1
man = (x + 1)2
We can write,
⇒ (x + 1)2 - 1 = 8x
⇒ x2 + 1 + 2x - 1 = 8x
⇒ x2 + 2x - 8x = 0
⇒ x2 - 6x = 0
⇒ x(x - 6) = 0
⇒ x = 0 or x - 6 = 0
Since, age cannot be zero,
∴ x = 6
∴ x + 1 = 7 and (x + 1)2 = 72 = 49.
Hence, present age of son = 7 years and man = 49 years.
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