A man is 20 year elder to his son. After 10 years, his age becomes twice the age of his son. Find their present ages.
Answers
Let son's present age be x years.
Let father's present age be 20+x years.
Son's age after 10 years = 10+x years
Father's age after 10 years = 10+(20+x) years.
Equation:
10+(20+x) = 2(10+x)
10+20+x = 20+2x
30+x = 20+2x
30-20 = 2x-x
10 years = x ⇒ Present age of son
Father's present age = 20+x
= 20+10
= 30 years
Answer: Present ages of son and father are 10 years and 30 years respectively.
Check:
Putting the value of x as 10 in the equation,
10+(20+x) = 2(10+x)
10+(20+10) = 2(10+10)
10+30 = 2×20
40 = 40,
LHS = RHS
Hence Verified
Step-by-step explanation:
Let son's age be = x
Father's age = x+20
After 10 years
Son's age = x+10
Father's age = x+20+10 = x+30
2(x+10) = x+30
2x+20 = x+30
2x-x = 30 -20
x= 10
Final Answer =
Present age of son = 10 years
Present age of father = 30 years
Hope it Helps