Question

five years ago a man was seven times as old as his son. five years hence the gather would be three times as old as his son. find their present ages

Answer 1

father = x
son = y
x-5 = 7(y-5)
x +5 = 3(y+5) solve the system

x-5 = 7y-35
x+5 = 3y +15

x-5 = 7y-35
-x-5 = -3y -15
-10 = 4y -50
+50 +50
40 = 4y
10 = y

Solve for X
x+5 = 3(10) + 15
x+5 = 45
x = 40

The father is 40yrs old and the son is 10yrs old.

Answer 2

Explanation -:

Given :

• Five years ago a man was seven times as old as his son.
• Five years hence the man would be three times as old as his son

Find :

• Their present ages

Solution :

Five years ago a man was seven times as old as his son.

Let us assume

• Son's age as 'x'
• So, man's age as 7x

son's age = (x - 5) years

man's age = (7x - 5) years

After five years

Son's age = (x + 5) years

Man's age = 3(x + 5) years

Now we will calculate son's present age

According to question

3(x + 5) = 7(x - 5) + 10

→ 3x + 15 = 7x - 35 + 10

→ 7x - 3x = -35 + 10 + 15

→ 4x = 40

→ x = $\frac{40}{4}$

x = 10

Son's present age = 10 years

Calculating man's present age

Putting the value of x = 10 years

3(10 + 5) - 5

→ (30 + 15) - 5

→ 45 - 5

→ 40 years

Man's present age = 40 years

Final Answer :

• Man's present age is 40 years
• Son's present age is 10 years.