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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 ..​

Respuesta :

Let

10years ago

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

10years after

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

ATQ

[tex]\\ \tt\hookrightarrow 6x+20=2(x+20)[/tex]

[tex]\\ \tt\hookrightarrow 6x+20=2x+40[/tex]

[tex]\\ \tt\hookrightarrow 4x=20[/tex]

[tex]\\ \tt\hookrightarrow x=5[/tex]

  • Father's present age=6x+10=6(5)+10=30+10=40yrs
  • Daughter's present age=x+10=5+10=15yrs
Avcad0rable

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 ~

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