Five years ago a man was seven times as old as his son. Five years hence, the father will be
three times as old as his son. Find their present ages.
Answers
To Find :
- we need to find the present age of son and his father.
Solution :
Five years ago a man was seven times as old as his son.
Five years ago : --
- Let the age of son be x years
- Age of father = 7x
So ,
- present age of son = x + 5
- present age of father = 7x + 5
Five years hence, the father will be three times as old as his son.
After 5 year's : --
- Age of son = x + 5 + 5 = x + 10
- Age of father = 7x + 10
But age of father is 3 times age of son .
So,
↛ 7x + 10 = 3(x + 10)
↛ 7x + 10 = 3x + 30
↛ 7x - 3x = 30 - 10
↛ 4x = 20
- ›› x = 5
So,
- Present age of father : -
= 7x + 5
= 7 × 5 + 5
= 35 + 5
= 40 years.
- present age of son
= x + 5
= 5 + 5
= 10 years.
Hence,
━━━━━━━━━━━━━━━━━━━━━━━━━
Answer:
Present age of man = 40
Present age of son = 10
Step-by-step explanation:
Let the present age of the man = x
let the present age of son = y
man's age five years ago = x-5
son's age five years ago = y-5
five years ago man's age was seven times his son's age. therefore the equation will be
x-5 = 7(y-5)
x-5 = 7y - 35
x-7y= -35+5
x-7y = -30 eq. 1
man's age 5 years from now = x+5
son's age 5 years from now = y+5
after 5 years the man's age will be 3 times his son's age. Therefore the equation will be
x+5 = 3(y+5)
x+5 = 3y+15
x-3y = 15-5
x-3y = 10 eq. 2
subtract eq 1 and eq 2
(x-7y) - (x-3y) = -30-10
x-7y-x+3y = -40
-4y = 40
y = -40 ÷ -4
y = 10
since son's present age = y
and y = 10
therefore son's present age= 10
Now substitute the value of y= 10 in eq 1
x-7y = -30
x - 7×10 = -30
x - 70= -30
x = -30 + 70
x = 40
Therefore man's present age= 40