Question

Ten years ago the father was six times as old as his daughter. After ten years, he will be twice as old as his father . Determine their present age ..​

Answer 1

Let

10years ago

• Daughter's age=x
• Father's age=6x

10years after

• Daughter's age=x+20
• Father's age=6x+20

ATQ

$\\ \tt\hookrightarrow 6x+20=2(x+20)$

$\\ \tt\hookrightarrow 6x+20=2x+40$

$\\ \tt\hookrightarrow 4x=20$

$\\ \tt\hookrightarrow x=5$

• Father's present age=6x+10=6(5)+10=30+10=40yrs
• Daughter's present age=x+10=5+10=15yrs

Answer 2

Answer :-

• 15 years

Solution :-

▪︎If ten years ago , the age of daughter was x years , then , ten years ago the age of her father was 6x years ▪︎

• Present age of father = ( 6x + 10 ) years
• Present age of the daughter = ( x + 10 ) years

Ten years after ,

• Father's age = ( 6x + 10 + 10 ) years
• => ( 6x + 20 ) years

• Daughter's age = ( x + 10 + 10 ) years
• => ( x + 20 ) years

According to the question,

▪︎6x + 20 = 2 ( x + 20 )

• 6x + 20 =! 2x + 40
• 6x - 2x = 40 - 40
• 4x = 20
• x = 5

Therefore , Present age of the Father = 6x + 10 = 6 × 5 + 10 = 40 years and present age of the daughter,

• x + 10 = 5 + 10 = 15 years

hence , the age of his daughter is 15 years

Hope it helps ~