Question

ten years ago ,the father was six times as old as his daughter .After 10 years ,he will be twice as old as his daughter. find their present ages​

Answer 1

Answer:

bro it is very easy late father's age be X and daughter b y 10 years after father's age is 2 x y use by substituting X is equals to 2 y in the age of daughter you can get your require answer

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

let the present age of father and daughter be x years and y years respectively.

ten years ago

father's age=(x-10) years

and daughter's age =(y-10) years

according to question

⇒  (x-10) =6(y-10)

⇒ x-10=6 y-60

⇒ x-6 y= -50⇒A

ten years later

father's age= (x+10)years

daughter 's age=(y+10)years

according to given condition

⇒  father's age=2( daughter's age)

⇒   (x+10)=2(y+10)

⇒ x+10=2 y+20

⇒ x-2 y=10

⇒ x=10+2 y  ⇒⇒B

using in A

x-6 y= -50

⇒ (10+ 2 y) -6 y= -50

⇒ 10+2 y-6 y=-50

⇒    -4 y= -50-10

⇒    -4 y= -60

⇒      y= 60/4

⇒  y=15

     using B

x=10+2 y

⇒x=10+2(15)

⇒ x=10+30

⇒x=40

∴ father's age=x=40 years

and daughter's age=y= 15 years