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Marbles in bags

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Marbles in bags

by nish9622 » Thu Nov 18, 2010 9:35 pm
A box contains bags of marbles. All of the bags hold the same number of marbles except one bag, which holds one marble more than each of the other bags hold. If the box contains a total of 2001 marbles, how many bags are in the box?

(1) The number of bags is between 13 and 23 inclusive

(2) There is an even number of bags, and there is an even number of marbles in the bag containing the extra marble.

I understand how to solve the Q. I want to know of faster ways to think of the examples that will satisfy these conditions, especially for Statement 2- since its so open ended.

OA is C
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Source: — Data Sufficiency |

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by Rahul@gurome » Thu Nov 18, 2010 9:58 pm
nish9622 wrote:A box contains bags of marbles. All of the bags hold the same number of marbles except one bag, which holds one marble more than each of the other bags hold. If the box contains a total of 2001 marbles, how many bags are in the box?

(1) The number of bags is between 13 and 23 inclusive

(2) There is an even number of bags, and there is an even number of marbles in the bag containing the extra marble.

I understand how to solve the Q. I want to know of faster ways to think of the examples that will satisfy these conditions, especially for Statement 2- since its so open ended.

OA is C
Given: Say, there are m bags and n marbles in each except one bag has one more marble than n.
Thus, total number of marbles = (mn + 1) = 2001 => mn = 2000
Therefore, m and n are factors of 2000.

Statement 1: 13 ≤ m ≤ 23
The factors of 2000 between 13 and 23 are 16 and 20. Thus m can be 16 or 20!

Not sufficient.

Statement 2: m is even and (n + 1) is even => n is odd.
Thus, we have to look for an even and an odd factor of 2000 such that their product is 2000. Now, 2000 = 5*400 = 16*125 = 25*80 etc..

The set (m, n) can be (16, 125) or (80, 25) etc...

Not sufficient.

1 & 2 Together: m is an even factor of 2000 between 13 and 23. Only two possible values of m are 16 and 20. But for m = 20, n = 100, which is not possible as n has to be odd. For m = 16, n = 125 (odd!)

Sufficient.

The correct answer is C.
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by nish9622 » Thu Nov 18, 2010 10:05 pm
Thanks, So the answer is basically that I have to prime factorize/factorize and get the pairs-- no other option :(
I was hoping for some miracle! :D
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by Rahul@gurome » Thu Nov 18, 2010 10:23 pm
nish9622 wrote:Thanks, So the answer is basically that I have to prime factorize/factorize and get the pairs-- no other option :(
I was hoping for some miracle! :D
You don't have to go for complete factorization of 2000.
A partial factorization is needed only for the analysis of statement 2. Once you've found two such set, it is sufficient to conclude that statement 2 is insufficient.
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