mjohn has a son and a daughter ,the son being 4 years more than the daughter .5 years ago the age of

Asked by  | 20th Dec, 2009, 08:39: PM

Expert Answer:

Let x, y and z be the present ages of John, his son and his daughter respectively.

y=z+4......(1)

(x-5) = (y-5)+(z-5) + 23 ...(2)

and

(x+18)= 17((y+18)-(z+18))

(x+18)= 17(y-z)

(x+18)= 17(4)........from (1)

x = 50 yrs.

Now the (2) becomes,

y+z = 32,

Solving this with (1) gives,

y = 18 yrs and z = 14 yrs.

Hence the ages,

John = 50 yrs,

Son = 18 yrs,

Daughter = 14 yrs.

Regards,

Team,

TopperLearning.

 

 

 

Answered by  | 20th Dec, 2009, 09:07: PM

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