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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,
FatherтАЩs age = x тАУ 10
and sonтАЩs age = y тАУ 10
x тАУ 10 = 12(y тАУ 10)
=> x тАУ 10 = 12y тАУ 120
=> x тАУ 12y = -120 + 10 = -110
=> x тАУ 12y = -110 тАж.(i)
10 years hence,
FatherтАЩs age = x + 10
and his sonтАЩs age = y + 10
10y = 120
y = 12
From (ii), x тАУ 2y = 10
x тАУ 2 x 10 = 10
=> x тАУ 24 = 10
=> x = 10 + 24
=> x = 34
Present age of father = 34 years
and age of his son = 12 years

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