Question

ten years ago a father was six times as old as his daughter .after 10years ,he will be twice as old as his daughter . determine their present ages

Answer 1

Let the daughter's age 10 years ago be 'd'
Given that father's age at the same time was 6 times daughter's age..
Father's age = 6*d
After 10 years(that is 10 years from now)
daughter's age = d+10 +10 = d+20
Father's age = 6*d + 10 + 10 = 6d+20
Given that father's age = 2 times daughter's
6d + 20 = 2*(d+20)
6d + 20 = 2d + 40
4d = 20....d= 5 years..
Hence present age of daughter = 5+10 = 15 years
Present age of father = 6d+10 = 40 years

Answer 2

Let the fathers present age be X
Let the daughters present age be Y
10 years ago their ages will be 
daughters age is 6 times as his father
so equation will be
=x-10=6{y-10}
x-10=6y-60
x-6y=-50-----equation 1


After 10 years there ages will be 
x+10=2{y+10}
x+10=2y+20
x-2y=10--------equation 2

solving 1 and 2 we get 
-4y=-60

 y=15
substituting y=15 in eq 2
x-2{15}=10
x-30=10
x=40
Fathers age is 40 and daughters age is 15years '

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