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DS- Walter Zachary age problem

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Junior | Next Rank: 30 Posts
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by anticipation » Wed Aug 25, 2010 3:20 am
What is WalterÂ’s age now?

(1) WalterÂ’s brother Zachary is now twice as old as Walter was
exactly fiÂ…ve years ago.

(2) The positive difference between WalterÂ’s and ZacharyÂ’s
ages is less than three.

[spoiler]OA:E

Some how not convinced with OA as it can be solved using both the equations.
1)z=2(w-5)
also known w>z

2) |w-z|<3
i.e w-z=1 or w-z=2 or w-z=0 (only taking 3 possibilities since we already know w>z)
by taking w-z=1, we know w=9 , z=8 which satisfies equation 1

So I feel it should be C. Any inputs to correct me if I'm wrong would be great.
[/spoiler]
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by kvcpk » Wed Aug 25, 2010 3:41 am
anticipation wrote:What is WalterÂ’s age now?

(1) WalterÂ’s brother Zachary is now twice as old as Walter was
exactly fiÂ…ve years ago.

(2) The positive difference between WalterÂ’s and ZacharyÂ’s
ages is less than three.

[spoiler]OA:E

Some how not convinced with OA as it can be solved using both the equations.
1)z=2(w-5)
also known w>z

2) |w-z|<3
i.e w-z=1 or w-z=2 or w-z=0 (only taking 3 possibilities since we already know w>z)
by taking w-z=1, we know w=9 , z=8 which satisfies equation 1

So I feel it should be C. Any inputs to correct me if I'm wrong would be great.
[/spoiler]
YOu were right till the end. Why did u leave out w-z=2??
w-z cannot be 0 if w>z
Hence w-z can be either 1 or 2. [Assuming only integer ages are possible. I am not sure of it]

Even in case of integer ages, w-z=1 or w-z=2
when w-z =1, w=9,z=8 is possible
when w-z=2, w=8, z=6 is possible.
2 idfferent values for W.

hence INSUFF

pick E.

Hope this helps!!
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by sk818020 » Wed Aug 25, 2010 9:29 am
Just in case you wanted to see it alebraically; z=zachary w=walter

1) z=2(w-5)
We don't know z or w, so (1) is insufficient.

2) w-z<3
We don't know z or w, so (2) is insufficient.

(1) and (2), tell us;

w-[2(5-w)]<3
w-(10-2w)<3
w-10+2w<3
3w<13
w<(13/3)

Thus, 0<w<13/3, we cannot say for sure hold old Walter is. Thus E, none together or alone is sufficient.


Hope this helps.

THanks,

Jared
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