Question

Eight years ago, the ratio of the ages of Samir and Varun was 4:3. Eight years hence, the ratio of their ag
6:5. What is the sum of their present ages?​

Answer 1

GIVEN :-

• Eight years ago, the ratio of the ages of Samir and Varun was 4:3.
• Eight years hence, the ratio of their age 6:5.

TO FIND :-

• Sum of their present ages.

SOLUTION :-

Let present age of Samir be 'x'yrs and Varun be 'y'yrs.

CASE 1 —

Eight years ago, the ratio of the ages of Samir and Varun was 4:3.

Eight years ago,

• Age of Samir → 'x - 8' yrs
• Age of Varun → 'y - 8' yrs
• Ratio was 4:3.

$\\ \sf \: \dfrac{x - 8}{y - 8} = \dfrac{4}{3} \\ \\ \sf \: 3(x - 8) = 4(y - 8) \\ \\ \sf \: 3x - 24 = 4y - 32 \\ \\ \sf \: 3x - 4y = - 32 + 24 \\ \\ \sf \: 3x - 4y = - 8$

Multiplying whole equation by 5,

$\\ \sf \: 15x - 20y = - 40 \: \: \: \: \: \: \: - - - (1)$

CASE 2—

Eight years hence, the ratio of their age 6:5.

Eight years hence ,

• Age of Samir → 'x + 8' yrs
• Age of Varun → 'y + 8' yrs
• Ratio will be 6:5.

$\\ \sf \: \dfrac{x + 8}{y + 8} = \dfrac{6}{5} \\ \\ \sf \: 5(x + 8) = 6(y + 8) \\ \\ \sf \: 5x + 40 = 6y + 48 \\ \\ \sf \: 5x - 6y = 48 - 40 \\ \\ \sf \: 5x - 6y = 8$

Multiplying whole equation by 3,

$\\ \sf \: 15x - 18y = 24 \: \: \: \: \: \: \: - - - (3)$

Subtracting equation (2) by equation (1) ,

→ 15x - 20y - (15x - 18y) = -40 - (24)

→ 15x - 20y - 15x + 18y = -40 - 24

→ -2y = -64

→ y = -64/-2

→ y = 32

Putting y=32 in equation (3) ,

→ 15x - 18(32) = 24

→ 15x - 576 = 24

→ 15x = 24 + 576

→ 15x = 600

→ x = 600/15

→ x = 40

Hence , age of Samir is 40 years and age of Varun is 32years.

Sum of their ages = 40 + 32 = 72years.

Hence , sum of the ages of Samir and Varun is 72 years.