Question

The difference of the present age of a and b is 10 years. if 8 years back their ages were in the ratio 4:3, what is the average age of a and b at present?

Answer 1

» The difference in ages of a and b is 10.

→ a - b = 10

→ a = 10 + b ____ (eq 1)

» 8 years back, their ages were in the ratio 4:3.

• Age of a = (a - 8) years
• Age of b = (b - 8) years

According to question,

→ $\dfrac{a\:-\:8}{b\:-\:8}\:=\:\dfrac{4}{3}$

Cross multiply them

→ 3(a - 8) = 4(b - 8)

→ 3a - 24 = 4b - 32

→ 3a - 4b = - 32 + 24

→ 3(10 + b) - 4b = - 8

→ 30 + 3b - 4b = - 8

→ - b = - 8 - 30

→ b = 38

Put value of b in (eq 1)

→ a = 10 + 38

→ a = 48

• Present age of a = 48 years.
• Present age of b = 38 years

We have to find the average age of a and b at present.

Means, we have to find the sum of a and b and then divide them by 2.

→ $\dfrac{38\:+\:48}{2}$

→ $\dfrac{86}{2}$

→ $43$

∴ Average age of a and b is 43 years.

Answer 2

a - b = 10

a = 10 + b

Now

(a-8)/(b-8) = 4/3

After solving we get

a = 48 and b = 38 years

Their average = (48+38)/2

= 43 years