Box, Bags, and Marbles!!

[gig92]

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.

OA - Later

[anshumishra]

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.

OA - Later

n - number of bags of marbles
m - number of marbles (which are same in all but one bag)

n*m + 1 = 2001
=> n*m = 2000 = 2*2*2*2*5*5*5

n = ?

Statement 1:
13 <= n <=23

Since; n*m = 2*2*2*2*5*5*5 = ( 2*2*2*2) *(5*5*5) = 16*125; n can be 16
Also; n*m = 2*2*2*2*5*5*5 = (2*2*5) * (2*2*5*5) = 20*100 ; n can be 20 --- Insufficient

Statement 2:
There is an even number of bags, and there is an even number of marbles in the bag containing the extra marble.
=> n = even, m+1 = even => m = odd
n*m = 2000 = ( 2*2*2*2) *(5*5*5) => n = 16, m = 125
n*m = 2000 = ( 2*2*2*2*5) *(5*5) => n=80, m = 25

Insufficient

Combining 1 and 2 :
13 <= n <=23 ; n = even and m=odd

Only n = 16 is the solution, as shown above .
So, C

[Rahul@gurome]

Let the number of bags be n and the number of marbles in each bag except one be x.
So one bag has x+1 marble.
Or (n-1)*x + (x+1) = 2001.
Or nx + 1 = 2001.
Or nx = 2000.
Note that both n and x are positive integers and are factors of 2000.
First consider (1) alone .
The factors of 2000 between 13 and 23 are 16 and 20.
So n can be 16 or 20.
Or (1) alone is not sufficient.
Next consider (2) alone.
This means n is even and x+1 is even.
Or x is odd.
Now 2000 = 16*125 = 80*25
So if n = 16 and x is 125, n is even and x is odd.
Also if n = 80 and x is 25, n is even and x is odd.
Again we are not getting any unique value of n.
Or (2) alone is not sufficient.
Next combine both the statements together and check.
On combining, the only possible value of n is n = 16.
Or both statements together are sufficient.

The correct answer is (C).

[gig92]

Let the number of bags be n and the number of marbles in each bag except one be x.
So one bag has x+1 marble.
Or (n-1)*x + (x+1) = 2001.
Or nx + 1 = 2001.
Or nx = 2000.
Note that both n and x are positive integers and are factors of 2000.
First consider (1) alone .
The factors of 2000 between 13 and 23 are 16 and 20.
So n can be 16 or 20.
Or (1) alone is not sufficient.
Next consider (2) alone.
This means n is even and x+1 is even.
Or x is odd.
Now 2000 = 16*125 = 80*25
So if n = 16 and x is 125, n is even and x is odd.
Also if n = 80 and x is 25, n is even and x is odd.
Again we are not getting any unique value of n.
Or (2) alone is not sufficient.
Next combine both the statements together and check.
On combining, the only possible value of n is n = 16.
Or both statements together are sufficient.

The correct answer is (C).

OA - C