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Fathers and Sons (Posted on 2003-06-11) Difficulty: 3 of 5
Father A is twice the age of the difference in years of the ages of Father B and Son A, who is one and a half times older than Son B.

Father B is currently twice the age of Son A is going to be when Son B will be double the age he is now.

All of the ages are multiples of five.

How old is Father A?

See The Solution Submitted by Lewis    
Rating: 3.5000 (6 votes)

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Too easy | Comment 3 of 14 |
let A' is the age of son A
and B' is the age of son B

therefore,
A=2(B-A')
again,
A'=(1+1/2)B'=3/2 B'

since, all the ages are multiles of five

therefore,
B' as a age must be a number divisble by 5 and by 2.......also A' must be a number divisble by 5 and 3.....
i.e. B'={10,20,30,40,50,.......}
A'={15,30,45,60,75,.......}

now we must use the last data to get the truely age of A, this data is that....

B=2A' at B"=2B'

therefore,
B={30,60,90,120,......}
also B"={20,40,60,80,..}

as B=2A'=2*3/2B"=3B"

now it is obivous that the age of Father B is 60 yrs and the age of B"( B" is the double of the age of B' as to use B=2A') is 20 also the age of A' is 30 yrs old

now it is too easy to know that Father A has
A=2(B-A')=2(60-30)=60 yrs
finally, Father A has 60 yrs old



  Posted by Ahmed on 2003-06-12 03:58:40
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