Two years ago ,a woman was 7 times as old as her daughter, but in 3 years time ,she would be 2 times as old as the girl, how old are they now?.

Let present age of women and her daughter be x and y respectively.

According to the question,

Case 1 :

Two years ago,

Woman age = ( x - 2 ) years

Her daughter age = ( y - 2 ) years

Woman was 7 times old as her daughter. [ Given ]

x - 2 = 7 ( y - 2 )

=> x - 2 = 7y - 14

=> x - 2 + 14 = 7y

=> x + 12 = 7y ( i )

Case 2 :

After Three years ,

Woman age = x + 3

Her daughter age = y + 3

she would be 4 times old as the girl. [ Given ]

x + 3 = 4 ( y + 3 )

=> x + 3 = 4y + 12

=> x = 4y + 12 - 3

=> x = 4y + 9( ii )

Now,

★ Substituting the value of x = 4y + 9 from equation ( ii ) in equation ( i ),we get

x + 12 = 7y

=> 4y + 9 + 12 = 7y

=> 21 = 7y - 4y

=> 21 = 3y

=> 3y = 21

=> y = 21/3

=> y = 7

And,

x = 4y + 9

★ Substituting the value of y in equation ( ii ), we get

x = 4 × 7 + 9

x = 28 + 9

x = 37

Hence, the present age of women is 37 years and her daughter age is 7.