Question

A man is 4 times as old as his son. after 16 years he will only twice as old as his son find their present age​

Answer 1

Step-by-step explanation:

Let man's age be x

and son's age be y

ATQ =>

x = 4y

x -4y = 0.........(1)

ATQ =>

(x+10) = 2(y+10)

x + 10 = 2y +20

x -2y - 10 = 0

multiplying by 2=>

2x -4y = 20........(2)

from (1)&(2)=>

x = 20

put x = 20 in (1)

x -4y = 0

4y = 20

y = 5

So father's age is 20

and sons age is 5

Answer 2

Answer:

son's age=8 father's age=32

Step-by-step explanation:

present age of son=x

present age of father=4x

after 16 years father's age =4x+16

age of son =x+16

from this 4x+16=2(x+16)

4x+16=2x+32

4x-2x=32-16

2x=16

x=8

age of son =8 years

age father =8x4=32 years