Two years ago a man age was six times as old as his son. In 18 years he will be twice as old as his son determine their present age
Answers
Answer:
The present age of man = 32 years
The present age of his son = 7 years
Step-by-step explanation:
Given :
- Two years ago a man's age was six times as old as his son.
- In 18 years he will be twice as old as his son.
To find :
their present ages
Solution :
Let the present age of man be x years and his son be y years.
Two years ago,
man's age = (x - 2) years
his son's age = (y - 2) years
As given, man's age = 6 × his son's age
x - 2 = 6(y - 2)
x - 2 = 6y - 12
x = 6y - 12 + 2
x = 6y - 10 → [1]
After 18 years,
man's age = (x + 18) years
his son's age = (y + 18) years
As given, man's age = 2 × his son's age
x + 18 = 2(y + 18)
x + 18 = 2y + 36
x - 2y = 36 - 18
x - 2y = 18
Put x = 6y - 10, [ ∵ eqn. 1 ]
6y - 10 - 2y = 18
4y - 10 = 18
4y = 18 + 10
4y = 28
y = 28/4
y = 7
Substitute in eqn. 1,
x = 6y - 10
x = 6(7) - 10
x = 42 - 10
x = 32
Therefore,
the present age of man is 32 years and that of his son is 7 years.