Question

in 10 years time a father Will be twice as old as his daughter.10 years ago he was six time as old as his daughter.How old is each now
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Answer 1

Answer:

Step-by-step explanation:

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

pls mark brainliest

Answer 2

Answer:

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

Step-by-step explanation:

your answer Miss.