Question

Ten years ago father was six times old as his daughter. After 10 years, he will be twice as old as his daughter. Determine their present ages ?

Answer 1

Answer:

15 years and 40 years

Step-by-step explanation:

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