Question

21. Five years ago a man was seven times as old as his son. Five years hence, the father will be
three times as old as his son. Find their present ages.
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Answer 1

Answer:

Let present age of man = x years

Let present age of son = y years

A.T.Q

$(x - 5) = 7(y - 5) \\ x - 5 = 7y - 35 \\ x - 7y + 30 = 0 \: \: \: .....(1)$

Also

$(x + 5) = 3(y - 5) \\ x + 5 = 3y + 15 \\ x - 3y = 10 \: \: \: . ...(2)$

sub eq (1) from eq (2)

$x - 3y = \: \: \: 10 \\ x - 7y = - 30 \\ - \: + \: \: \: \: \: \: \: \: \: + \\ 4y \: \: \: \: \: \: \: \: = 40 \\ y \: \: \: \: \: \: \: \: \: \: = 10$

put the value of y in eq 1 we get

x=40

Hence the present ages are 40 years and 10years

hope it helps you