Question

2 years ago a father was 5 times as old as his son 2 years later his age will be 8 more than three times the age of the son find the present age of father and son

Answer 1

Let son's age = x
Let father's age = y

5(x-2) = y-2
5x-10 = y-2
5x-y = -2+10
5x-y = 8 -¡

3(x+2)+8 = y+2
3x+6+8 = y+2
3x-y =2-6-8
3x-y =-12 -¡¡

Subtracting ¡¡ from ¡

5x-y=8
-3x+y=+12
________
2x = 20
x = 10
________

5x-y=8
5*10 -y= 8
-y = 8-50
-y = -42
y = 42

Answer 2

Answer:

Present age of son is 11 years and the present age of father is 47 years.

Step-by-step explanation:

Let,

• Father's present age = x years
• Son's present age = y years

2 years ago,

• Father's age = (x-2) years
• Son's age = (y-2) years

A/q, ( 1st condition)

x-2 = 5(y-2)

→ x-2 = 5y - 10

→ x = 5y-10+2

→ x= 5y-8...................eq(1)

2 years later,

• Father's age = (x+2) years
• Son's age = (y+2) years

A/q, (2nd condition)

x+2 = 3(y+2)+8

→ 5y-8+2 = 3y+6+8 [ Put x = 5y-8 from eq (I)]

→ 5y - 3y = 20

→ 2y = 20

→ y = 10

• Present age of son = 10 years
• Present age of father = 5×10 -8 = 42 years.