Question

Two years ago,the age of a father was three and half times the age of his daughter then.Six years hence ,the age of father will be ten years more than twice the age of his daughter then.Find there present ages ​

Answer 1

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

2 yrs ago, age of daughter =y-2
age of father. = x-2

ATQ,
x-2 = 3whole 1 by 2 (y-2)
x-2 = 7/2 (y-2)
2(x-2)=7y-14
2x-4=7y -14
2x= 7y-14...........(I)

6 years hence,
daughter's age =y+6
father's age. =x+6
ATQ,
x+6 = (y+6)+10
x = y+10...... ii

substituting value of x from equation I in equation ii.
2(y+10) =7y-10
2y+20=7y-10
5y=30
y= 6

therefore, daughter present age is 6 yrs and father age is 16.