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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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.

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

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