Question

Father is twice as his son .20 yrs ago the father was 6 times as old his son .find their present ages​

Answer 1

Answer:

Let the age of the son be = x

let the age of the father be = 2x

20 years ago ,

the age of the son is = x-20

and the age of the father is = 2x-20

( 2x-20 ) = 12 × ( x-20 )

( 2x-20 ) = ( 12x -240 )

2x = 12x - 240 + 20

2x = 12x-220

-10x= -220

-220 ÷ (-10 ) = 22

➡ SO THE PRESENT AGE OF THE SON IS = 22 years.

➡ THE PRESENT AGE OF THE FATHER IS = 2×22 = 44 years

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HOPE IT HELPS YOU.✌

THANKS.❤

Answer 2

Step-by-step explanation:

suppose, father's age be "x" &

son's age be "y"

1st condition,

x=2y

x-2y=0_____[१]

20 years ago,

x-20

y-20

2nd condition,

x-20=6(y-20)

x-20=6y-120

x-6y= -120+20

x-6y= -100____[२]

subtracting equation (१)&(२)

x-2y=0

x-6y= -100

- +. = +

________

4y=100

y=25

put in equation (1)

x-2y=0

x-2(25)=0

x-50=0

x=50

father's present age is 50

and son's age is 25