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?
Answers
Answered by
56
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!
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!
siddhartharao77:
:-))
Answered by
19
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
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
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