The age of a student becomes 5 times after 24 years, and then his age after 3 years will be:
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Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.
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Let father age be x, his son age be y.
Before 2 years,
⇒(x−2)=5(y−2)−−−−−−−(1)
After 2 years
⇒(x+2)=8+3(y+2)−−−−−−−−(2)
From (1) x−5y+8=0
From (2)
⇒x+2−3y−6−8=0
⇒x−3y−12=0
Solving them
⇒−2y+20=0
⇒y=10
When y=10,x=42
So the present age of the father is 42 and the son age is 10.
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