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Question

Ten years ago a father was four times as old as his son .After 20 years he will be twice as old as his son .Determine the present age of the son.

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Solution

Y if age of father
x is age of son
Ten years ago,
y-10=4(x-10)
y-10=4x-40
y+30=4x.....(1)

y+20=2(x+20)
y+20=2x+40
y-20=2x
y=2x+20....(2)
Solving 1 and 2

2x+20+30=4x
2x=50
x=25

Now son is 25 years old.
Father is y=2×25+20=70

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