151k views
7 votes
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 ..​

2 Answers

7 votes

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
User Jamie Hutber
by
9.2k points
8 votes

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 ~

User Aditi Kaushal
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories