Two years ago a man was six times as old as his son. In 15years he will be twice as old as his son
Determine the present ages.
Answers
Answer :
The present age of man = 27.5 years
The present age of his son = 6.25 years
Step-by-step explanation :
Given :
- Two years ago, a man was six times as old as his son.
- In 15 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 the present age of his son be y years.
Two years ago,
Man's age = (x - 2) years
His son's age = (y - 2) years
According to the given relation,
x - 2 = 6(y - 2)
x - 2 = 6y - 12
6y - x = 12 - 2
6y - x = 10 ➙ [1]
After 15 years,
Man's age = (x + 15) years
His son's age = (y + 15) years
According to the given relation,
x + 15 = 2(y + 15)
x + 15 = 2y + 30
x - 2y = 30 - 15
x - 2y = 15 ➙ [2]
Add equation [1] and equation [2],
6y - x + x - 2y = 10 + 15
6y - 2y = 25
4y = 25
y = 25/4
y = 6.25
Substitute y = 6.25 in equation [1],
x - 2y = 15
x - 2(6.25) = 15
x - 12.5 = 15
x = 15 + 12.5
x = 27.5
Therefore,
The present age of man = 27.5 years
The present age of his son = 6.25 years
Verification :
Two years ago,
man's age = 27.5 - 2 = 25.5 years
his son's age = 6.25 - 2 = 4.25 years
man's age = 6 times his son's age
25.5 = 6(4.25)
25.5 = 25.5
LHS = RHS
After 15 years,
man's age = 27.5 + 15 = 42.5
his son's age = 6.25 + 15 = 21.25
man's age = 2 times his son's age
42.5 = 2(21.25)
42.5 = 42.5
LHS = RHS
Hence verified!