Question

5 years ago man was 7 times as old as his son . 5 years hence the father's age will be three times as his son.find their present ages

Answer 1

Let fathers present age =x       Sons present age =y Five years ago: Fathers age=x-5 Sons age =y-5 Equation =x-5=7(y-5)                 =x-5=7y-35                 =x-7y=-35+5                 =x-7y=-30 -----(1) Five years hence: Fathers age =x+5 Sons age=y+5 Equation =x+5=3(y+5)                 =x+5=3y+15                 =x-3y=15-5                 =x-3y=10 -----(2) x-7y=-30 x-3y=10 -  +    - _______    -4y=-40 y=10 x-3y=10 =x-3(10)=10 =x-30=10 =x=30+10 =x=40 Fathers present age=40 years Sons present age =10 years

Answer 2

✬ Man' age = 40 Years ✬

✬ Son's age = 10 Years ✬

Step-by-step explanation:

Given:

• 5 years ago a man was seven times as old as his son.
• 5 years hence fater's age will be three times as his son.

To Find:

• What is the present ages of man's and his son?

Solution: Let the man's age be x years and son's age be y years.

• 5 years ago their ages was ➪

• Man's age = x – 5 years.
• Son's age = y – 5 years.

∴ Equation will be → x – 5 = 7 ( y – 5 )

$\implies{\rm }$ x – 5 = 7y – 35

$\implies{\rm }$ x – 7y = – 35 + 5

$\implies{\rm }$ x – 7y = – 30.......(1)

• 5 years hence their ages will be ➪

• Man's age = x + 5 years.
• Son's age = y + 5 years.

∴ Equation → x + 5 = 3 ( y + 5 )

$\implies{\rm }$ x + 5 = 3y + 15

$\implies{\rm }$ x – 3y = 15 – 5

$\implies{\rm }$ x – 3y = 10........(2)

Subtract equation (1) from (2) ,

x – 7y = – 30

x – 3y = 10

– + = –

__________

–4y = – 40

➼ 4y = 40

➼ y = 10

So, Son's present age = y = 10 years

Man's present age :- x – 3y = 10

➭ x – 3(10) = 10

➭ x = 10 + 30

➭ x = 40 Years.