Question

10 years time a father will be twice as old as his daughter. 10 years ago he was six times as old as his daughter. how old is each now?

Answer 1

Answer by mananth(15550) (Show Source): You can put this solution on YOUR website!

let fathers age now be x

daughters age now be y

10 years hence

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

x+10 = 2y+20

x-2y=10

10 years ago

(x-10)= 6(y-10)

x-10 = 6y-60

x-6y=-50

x-2y=10

x-6y=-50

subtract the equations

4y=60

y=15 daughter's age now

substitute y in x-2y=10

so x=40 present father's age

hope it's helps to you....

Answer 2

Let the age 10 years ago of daughter be (x years ) and age of father would be ( 6x years )

After 10 years, age of father = ( 6x + 10 + 10 )

=( 6x + 20 ) years

And Age of daughter will be = ( x + 10 + 10 )

= (x + 20) years

ATQ,

6x + 20 = 2(x + 20)

6x + 20 = 2x + 40

6x - 2x = 40 - 20

4x = 20

x = 5

Hence, Present age of daughter = (x + 10 )

= ( 5 + 10 )

= 15 years

Present age of father = ( 6x + 10 )

= 6( 5 ) + 10

= 30 + 10

= 40 years

this is the answer

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