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A box of bags containing marbles

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A box of bags containing marbles

by Brent@GMATPrepNow » Thu Jan 29, 2009 2:38 pm
A box contains several bags of marbles. All of the bags contain the same number of marbles except one bag, which contains one marble more than the other bags contain. If the box contains a total of 2001 marbles, then the number of bags is

(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.

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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by gmatguy16 » Thu Jan 29, 2009 4:20 pm
answer is e,number of bags could either be 16 or 20.
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by dmateer25 » Thu Jan 29, 2009 5:28 pm
A box contains several bags of marbles. All of the bags contain the same number of marbles except one bag, which contains one marble more than the other bags contain. If the box contains a total of 2001 marbles, then the number of bags is

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

The number of bags can be 16 in which case there will be 125 marbles in 15 bags and 126 in 1.

The number of bags can be 20 in which case there will be 100 marbles in 19 bags and 101 in 1.

INSUFF

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


The number of bags can be 16 in which case there will be 125 marbles in 15 bags and 126 in 1.

The number of bags can be 80 in which case there will be 25 marbles in 79 bags and 26 in 1.

INSUFF

1 and 2 togther

The answer would b 16.

I will go with C.
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by Brent@GMATPrepNow » Thu Jan 29, 2009 5:33 pm
Nice work, the answer is C

Here's my solution as well:

Assume that we have b bags and almost all bags contain m marbles (except the one bag that has m+1 marbles)

We know that bm + 1 = 2001
Or we can say that bm = 2000
It will help with this question to find the prime factorization of 2000.
We get 2000 = 2x2x2x2x5x5x5

Now we can examine the statements
(1) We need to write bm = 2000 such that b is between 13 and 23. We can see that b can equal 16 (as in b=16 and m=125), or b can equal 20 (as in b=20 and m=100) INSUFF

(2) Another way to say that there is an even number of marbles in the bag containing the extra marble is to say that there is an odd number of marbles in all of the other bags. In other words, m is odd.
How many ways are there to split 2x2x2x2x5x5x5 into two parts such that one part is even and the other part is odd? Several. We can have b=16 and m=125, and we can have b=80 and m=25). INSUFF

(1) and (2): Statement (1) tells us that either b=16 (and m=125) or b=20 (and m=100). Statement (2) tells us that m must be odd. So, the only conclusion is that b=16 and m=125. SUFF
Last edited by Brent@GMATPrepNow on Wed May 28, 2014 6:51 pm, edited 1 time in total.
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by gmatguy16 » Thu Jan 29, 2009 5:38 pm
of course c ,sorry silly mistake at the end...
15*125 +126 or 19*100+101
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by hypik21 » Thu Jan 29, 2009 5:38 pm
gmatguy16 wrote:answer is e,number of bags could either be 16 or 20.
I believe it is C

an equation for this..b=bags, m=marbles

1 bag has 1 more marble than the rest so basically

bm=2000

bm can be 200,10, 40,50, etc

1. insuf..can be 16,125 or 20,100
2. insuf, 16 or 80, etc can both leave an even number of marbles in the extra bag...2000=(2^4)(5^3), 2000/16=125, extra bag will have 126; 2000/80=25, extra bag will have 26

C suff....only 16,125 fits our criteria as 20,100 will give us an odd number in the extra bag
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