Question

Two years ago a woman was 7 times as old as her daughter, but in 3 years times she be one times old as her the girl. How old are they?

Answer 1

Answer:

There present ages are

Mothers age = 9 years

Daughters age = 3 years

( Surprising but Mathematically correct )

Step-by-step explanation:

let the daughters age be X

Two years ago the age was X-2

at that time her mother's age was 7×(X - 2 )

3 years time she will be one times old

After 3 years their age will be " ages 2years ago " + 3 +2

mothers age will be 7×( X-2) +5 =7X -9

daughter's age will be ( X -2 )+5= X+3

one time old means double the age

So 7X -9 =2(x+3)

7X -9 = 2X + 6

7X - 2X = 6+9

5X = 15

X = 3 years

mothers age = two years ago mothers age +2

= 7( X-2) +2

= 7 ( 3 -2) +2

= 7 (1) +2

= 9 years