Question

mr john has a son and a daughter , the son being 4 years older than daughter.Five years ago the age of mr john was 23 years more than the sum of ages of his son and daughter. After 18 years from now the age of mr john will be 17 times the difference of ages of his son and daughter.Find their present ages?

Answer 1

First, let us define our variables.

Let

x = age of the father

y = age of the son

z = age of the daughter

 

5 years ago..

(x – 5) = age of the father

(y – 5) = age of the son

(z – 5) = age of the daughter

 

After 18 years...

(x + 18) = age of the father

(y + 18) = age of the son

(z +18) = age of the daughter

 

From the first condition

y = z + 5 à eqn 1

5 years ago

(x-5) = 23 + (y-5) + (z-5)

x –y –z = 18 à eqn 2

after 18 years

(x+18) = 17*[(y +18 –z – 18)]

x – 17y + 17z = -18 à eqn 3

 

solving the 3 equations simultaneously

x = 67

y = 27

z = 22