Question

The age of a father 8 years back was 5 times that of his son. After 8 years, his age will be 8 years more than double the age of hia son. Find their prwsent ages

Answer 1

Step-by-step explanation:

let the present age of father and son be X and Y.

Before of 8 years age of father =(x-8)

Before 8 years age of son =(y-8)

A.T.Q

(x-8)= 5 (y-8)

x-8= 5y-40

x-5y= -40+8

x-5y= -32...............1

After 8 years age of father =(x+8) year

After 8 years age of son =( y+8)

A.T.Q

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

x+8=2y+16+8

x+8=2y+24

x-2y=24-8

x-2y=16........(2)

from equation (1) we get

x-5y= -32

x= -32+5y........(3)

putting the value of x in equation (2)

x-2y=16

-32+ 5y-2y=16

3y=16+32

3y=48

y=48/3

y=16 yrs

putting the value of Y in equation (3)

x=-32+5y=>-32 +5+16=> -32+80

x=48 years

So,

Age of father x= 48 years

and

Age of son is = y=16 years.

hope you understand

plz mark brainliest.

Answer 2

Answer:

age of father ♒4⃣8⃣

age of son♒1⃣7⃣