Question

A father's age is twice his daughter age but 16 years ago the father age was 4 times his daughter age. Calculate their ages???? Can anyone explain this question?

Answer 1

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

Given that Father's age is twice his daughter.

y = 2x  ----- (1)

Given that 16 years ago the father age was 4 times his daughter age.

(y - 16) = 4(x - 16)

(2x - 16) = 4x - 64

2x = 4x - 48

2x = 48

x = 24.

Then the age of father y = 2 * 24

                                          = 48


Therefore the age of father = 48 years.

The age of daughter = 24 years.


Hope this helps!

Answer 2

For these type of questions always let the present ages.

So,

Here we go,

Let the present age of daughter be x years
then,
Father's age = 2x

16 years ago,

Daughter's age = (x-16)
Father's age = (2x-16)-------------(i)

A/q

Father's age = 4(x-16)-------------(ii)

From (i) & (ii)


2x-16=4x-64
64-16=2x
2x=48
x=24

Daughter's age= 24years
Father's age = 2x=48 years