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Question -

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages. (C.B.S.E. 1994)



Answer -

Let present age of father = x years
and age of his son = y years
10 years ago,
Father’s age = x – 10
and son’s age = y – 10
x – 10 = 12(y – 10)
=> x – 10 = 12y – 120
=> x – 12y = -120 + 10 = -110
=> x – 12y = -110 ….(i)
10 years hence,
Father’s age = x + 10
and his son’s age = y + 10
10y = 120
y = 12
From (ii), x – 2y = 10
x – 2 x 10 = 10
=> x – 24 = 10
=> x = 10 + 24
=> x = 34
Present age of father = 34 years
and age of his son = 12 years

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