Question

the age of father 8years ago was 5 times that of his son. 8 years hence his age will be 8 years more than twice the age of son. find the present age of father and son?. ​

Answer 1

Step-by-step explanation:

8 years ago the son = X

his father's age = 5x

present age of his son= X+8

" " " of his father = 5x+8

so,

8 years after,

his son age will be= X+8+8

his father age will be= 5x +8+8

a/q,

5x+8+8= 2(X+8+8)+8

=> 5x+16 = 2x+32+8

=> 5x+16 = 2x+40

=> 5x-2x = 40-16

=> 3x= 24

=> X= 8

present age of son is= 8+8=16

". ". " his father is = 5×8+8= 40+8= 48

hope this answer is clear to you ..

Answer 2

Given

1. The age of father 8 years ago was 5 times that of his son.

2. 8 years hence his age will be 8 years more than twice the age of son

To Find

Present age of father and son

Solution

Let the present age of father be x

Present age of his son be y

A/c to condition 1 ,

⇒ ( x - 8 ) = 5 ( y - 8 )

⇒ x - 8 = 5y - 40

⇒ x = 5y - 32 ... (1)

A/c to 2nd condition ,

⇒ ( x + 8 ) = 2 ( y + 8 ) + 8

⇒ x + 8 = 2y + 16 + 8

⇒ x - 2y = 16

Sub. (1) here ,

⇒ ( 5y - 32 ) - 2y = 16

⇒ 3y = 48

⇒ y = 16  $\pink{\bigstar}$

Sub. y value in (1) ,

⇒ x = 5(16) - 32

⇒ x = 80 - 32

⇒ x = 48  $\green{\bigstar}$

So ,

→ Present age of father = 48 years  $\green{\bigstar}$

→ Present age of son = 16 years  $\pink{\bigstar}$