# Two years ago father was three times as old as his son and two years hence twice his age will be equal to five times that of his son.find their present age.(don't use x and y two variable,use only one variable)

2
by raviverma02486

2014-11-21T00:29:44+05:30

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Using one variable.
Let age of Son two years ago be = Son.
Then father's age two years ago was = 3 * Son

Son's age two years hence (after the present ) = Son + 4
Father's age after two years hence = 3 * Son + 4

hence,   2 * (3 * Son + 4) = 5 ( Son + 4)
Son = 12
Son was 12 years old , two years ago.  Hence father was 36 yrs old.
They are now  14 years and 38 years old.

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using two variables:

two years ago:

Father - 2 years  = 3 * ( son - 2 years)

=>   Father = 3 * son - 4    ---  equation 1

after two years from now , in future

2 * (father + 2 years ) = 5 * (son + 2 years)

=>   2 * father = 5 * son + 6      ---  equation 2

Multiply equation 1 by 2 and subtract equation 2 from it:

0 = son - 14       =>  son is 14 years old now

Father = 3 * 14 - 4 = 38 years now
thanks
2014-11-21T06:33:04+05:30
Let age of Son two years ago be = Son.
Then father's age two years ago was = 3  × Son

Son's age two years hence (after the present ) = Son + 4
Father's age after two years hence = 3 × Son + 4

Hence,   2 × (3 × Son + 4) = 5 ( Son + 4)
6Son + 8 = 5Son + 20
Son = 20 - 8
Son = 12 years
Son was 12 years old , two years ago.  Hence father was 36 years old.

Present age of Son = 14 years . Present age of father = 38 years.