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
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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