Question

19. Five years ago, a man was five times as old as his son. Eight years hence, the
man's age will be two years less than three times his son age. Find their present
ages.​

Answer 1

Answer:

Man's present age = 65 years

Son's present age = 17 years

Step-by-step explanation:

Let the man's present age be x, and son's present age be y.

Now,

5 years ago, their ages will be,

Man's age = (x - 5)

Son's age = (y - 5)

Then, according to the Question,

Man's age = 5 × Son's age

(x - 5) = 5(y - 5)

x - 5 = 5y - 25

x = 5y - 25 + 5

x = 5y - 20 ---- 1

Again,

8 years hence, their ages will be,

Man's age = (x + 8)

Son's age = (y + 8)

Then, according to the Question,

Man's age = 3(Son's age) - 2

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

x + 8 = 3y + 24 - 2

x + 8 = 3y + 22

x - 3y = 22 - 8

x - 3y = 14 ---- 2

Putting eq.1 in eq.2,

(5y - 20) - 3y = 14

5y - 20 - 3y = 14

2y = 20 + 14

2y = 34

y = 34/2

y = 17 years

Putting y = 17 in eq.1,

x = 5(17) - 20

x = 85 - 20

x = 65 years

Hence,

Man's present age = 65 years

Son's present age = 17 years

Hope it helped and believing you understood it........All the best

Answer 2

Given :

• Five years ago, a man was five times as old as his son.
• Eight years hence, the man's age will be two years less than three times his son age.

To find :

• The present ages of the man and his son.

Solution :

★ Let:-

• Man's present age = x
• Son's present age = y

★ Five years ago:-

• Man's age = x - 5
• Son's age = y - 5

★ By the problem:-

• (x - 5) = 5(y - 5)

→ x - 5 = 5y - 25

→ x = 5y - 25 + 5

→ x = 5y - 20 ......... equ. 1

★ Eight years hence:-

• Son's age = x + 8
• Man's age = y + 8

★ By the problem:-

• y + 8 = 3(x + 8) - 2

→ y + 8 = 3x + 24 - 2

→ y = 3x + 22 + 8

→ y = 3x + 30 ......... equ. 2

★ Putting equ. 1 in equ. 2:-

→ (5y - 20) - 3y = 14

→ 5y - 20 - 3y = 14

→ 2y - 20 = 14

→ 2y = 14 + 20

→ y = 34/2

→ y = 17

★ Putting y = 17 in equ. 1:-

→ x = 5(17) - 20

→ x = 85 - 20

→ x = 65

Answer :

• The man's age = 65 years
• The son's age = 17 years