Question

The differences of the present ages of a father and his daughter is 30 years. Four years ago father was seven times as old as his daughter was. Find their present ages.​

Answer 1

Step-by-step explanation:

let the present age of daughter = x

and the present age of Father =x+30

4 years ago,

father's age =(x+30)-4

and daughter's age=(x-4)

Now according to the question

(x+30)-4=7(x-4)

x+30-4 = 7x-28

6x = 30+28-4

6x =54

x = 9

Hence ...... Father's present age is 30+9 = 39 years

and Daughter's present age is = 9 years

Answer 2

Answer:

Step-by-step explanation:

Let the present age of father be x and the present age of daughter be y.

According to the question.

x - y = 30

4 years ago the age of father and daughter

i.e = (30 + y) - 4........(1)

.x =7(y-4).......(2)

and

according to the question,

30 + y - 4 = 7(y-4)

y + 26 = 7y -28

26 + 28 = 7y - y.

54 = 6y

; . y = 54\6

y= 9

again the value of y put in (1) we get,

x = 30+9 [:. Y = 9]

x = 39.

so,, the present age of father be 39 years old.

the present age of daughter be 9 years old.