Question

ten years ago father was six times as old as his daughter.After ten years ,he will be twice as old as his daughter.find their present ages.

Answer 1

Answer:

Let the age of the daughter be x.

Ten years ago

Age of father= 6(x-10) = 6x-60

Age of daughter= x-10

Ten years After

Age of father= 2 (x+10) = 2x+20

Age of daughter= x+10

Present Age

6x-60 + x-10 = 2x+20 + x+10

7x-70 = 3x+30

7x-3x= 30+70

4x = 100

x=25

Present age of daughter= 25

Present age of father = 60

Answer 2

Answer:

Step-by-step explanation:

Present age of daughter = x

6 yrs ago ,her age = x-6 yrs

6 yrs ago , fathers age = 6 { x -6 }

10 yrs hence daughter age = x + 10

10 yrs hence father age = 6 {x - 6 } + 16

btp

6 { x - 6 }+16 = 2 { x + 10 }

6x - 36 +16 = 2x = 20

6x - 2x = 20 + 20

4x = 40

x = 10

Daughter's present age = 10 yrs

father's present age = 6 *4 + 6= 24+6 =30 yrs