Elena, Marco, and Sofia each had one folder to copy to an external

Question

Elena, Marco, and Sofia each had one folder to copy to an external drive that was nearly full. From the sizes of the three folders, they knew that before any space was cleared, Sofia’s folder and exactly one of the other two folders could be copied to the drive, while the remaining folder would have to wait. Which folder could be copied along with Sofia’s folder before any space was cleared?

(1) Marco’s folder was smaller than Elena’s folder.
(2) After Sofia’s folder was copied, the remaining free space would be less than the size of Elena’s folder.

M47-107

Bunuel · Accepted Answer

GMAT Club Official Solution:

Elena, Marco, and Sofia each had one folder to copy to an external drive that was nearly full. From the sizes of the three folders, they knew that before any space was cleared, Sofia’s folder and exactly one of the other two folders could be copied to the drive, while the remaining folder would have to wait. Which folder could be copied along with Sofia’s folder before any space was cleared?

Only one of these two possibilities can work:

Sofia + Elena
Sofia + Marco

We need to determine which one.

(1) Marco’s folder was smaller than Elena’s folder.

Suppose Sofia’s folder could be copied with Elena’s folder. Then Sofia’s folder could also be copied with Marco’s folder, because Marco’s folder is smaller. But the stem says only one of those two pairings is possible. So Sofia’s folder cannot be copied with Elena’s folder. Therefore, Sofia’s folder must be copied with Marco’s folder. Sufficient.

(2) After Sofia’s folder was copied, the remaining free space would be less than the size of Elena’s folder.

That means Elena’s folder could not be copied after Sofia’s folder. Since one of the two folders can be copied with Sofia’s folder, it must be Marco’s folder. Sufficient.

Answer: D.

Adit_ · Answer

Statement 1:
This explicitly states that Marco<Elena and since we know one of them is for sure going to copy to the drive before waiting for clearing, Marco is the one to join with Sofia.
SUFFICIENT.

Statement 2:
Again it indirectly says Marco is going to join with Sofia as Elena doesn't have enough space remaining after Sofia copies her folder.
SUFFICIENT.

Answer:  Option D

thisisakr · Answer

Statement 1: it makes it very clear that marco folder is smaller so obviously that would be copied considering we are using smaller files due to space constraints. ——  sufficient

Statement 2: This says the space left is less than Elena so again Marco becomes obvious answer ——  sufficient

D

dp1234 · Answer

IMO  B.

Statement 1: M<E . But questions says either M or either E could fit in the drive. So either M or E can fit.  Not sufficient .
Statement 2: After copying S to drive, free space on drive is less than E. So E can't be copied. But questions says one can be copied - so its M.  Sufficient.

Adit_ · Answer

Statement 1 is sufficient IMO.
Think of it like this:

There are 3 people.
One person is already in (Sofia in this case).
ONLY one more WILL join for sur but you dont know who.
The statement says Marco<Elena.
Think the reverse way, can Elena actually join our list?

Let's assume Elena does get in.
Sofia and Elena is in our list now. 
If Elena is in then how did Marco not make it? It is logically not possible that Marco with lesser space couldn't get in but Elena with more space got in.
This contradicts with the statement we are provided with, this Marco indeed joins with Sofia and Elena is out.

truplayer256 · Answer

This is not necessarily true regarding statement 1 being sufficient. What if the amount of remaining space was X and Elena had E amount of space where E < X, and X - E < M, then Marco will not make it. On the other hand, what if X - M < E, then Elena won't make it.

Adit_ · Answer

If you notice you actually took two different cases to make your point.

If X-E<M then Marcos won't make it but Elena also will NOT make it then.
Is that possible? No.
Because the stem says that one of them WILL make it.
So your case actually contradicts the stem itself.

BongBideshini · Answer

1. Marco’s folder was smaller than Elena’s folder.

Therefore, only the smaller folder (Marco's) can be the one that fits.
 Sufficient.

2. After Sofia’s folder was copied, the remaining free space would be less than the size of Elena’s folder.

This statement directly tells us Sofia + Elena  cannot  fit. So, it has to be Marco.
 Sufficient.

Correct Answer: D

NidhiYad · Answer

I think the answer should be B because:

Statement 1: M<E,
 Say the total space is 20Mb S occupied 10 Mb. 10 MB remaining. M is 9Mb and E is 10 Mb. In this case exactly one of M and E can occupy the space the other has to wait, but which one? we cannot say, because any one can. Insufficient!

Statement 2: Total - S <E
 So in this case we know that E can for sure not fit. Question stem says S and exactly 1 other can, so we know it must be M. Sufficient!

Adit_ · Answer

You actually contradicted with the stem while stating it yourself. Exactly one of them CAN occupy the space meaning only either one of them can, each case you take. You can't take a case where you can have either one of them get in the queue as the stem states exactly one of them only can enter the space.
I understand how you probably interpreted it like, one of them enters and the other waits, which is what your case ideally does, except for the fact that it contradicts "exactly one" of them getting in.

Hope it helps!

BPranav7 · Answer

Bunuel  ; 
Is my reasoning correct?

Question is asking which of the 2 possibilities  COULD BE TRUE  before deleting any files?
1. Sofia + Marco(S+M)
2. Sofia + Elena(S+E)
And Key Point is : Either (1) or either (2) only one combination is possible.     Both can't be possible    
Which folder could be copied along with Sofia’s folder before any space was cleared?

(1) Marco’s folder was smaller than Elena’s folder.
M's file is small than E's =>S+E or S+M both Could be Possible,
but Since already given that only one is possible => We Rule out the possibility of S+E 
=> Sufficient

(2) After Sofia’s folder was copied, the remaining free space would be less than the size of Elena’s folder.
Space Remaining < E's Size=> S+E not possible 
But since already given that only one is possible & S+E is out=> S+M is possible
=> Sufficient

prragya · Answer

Bunuel 
I selected B (Statement 2 alone is sufficient), here is my reasoning:

Let the remaining space after Sofia's folder is copied be R.

Statement 1: Marco's folder is smaller than Elena's (M<E).

I don't think this is sufficient. For example, suppose R=50.

Case 1: M=40, E=60 → Marco can be copied, but Elena cannot.
Case 2: M=49, E=50 → either Marco or Elena could be copied since both satisfy the remaining space, but the question only asks which folder could be copied along with Sofia. We do not know for certain whose folder gets copies

Both cases satisfy M<E, but they don't give the same answer. Therefore, Statement 1 is  insufficient.

Statement 2: After Sofia's folder is copied, the remaining free space is less than Elena's folder. Therefore Elena cannot be copied. Marco must be the one that can be copied.  Sufficient .

Bunuel · Answer

Your second case is  not allowed by the stem . The stem already tells us that   exactly  one  of Marco and Elena can be copied along with Sofia.

So with R = 50, M = 49, and E = 50, both Marco and Elena would fit, which contradicts the given information.

Given M < E, if Elena could fit, Marco would necessarily fit as well. Since the stem says   exactly   one can fit, Elena cannot be the one. Therefore Marco must be the folder that can be copied with Sofia.

So Statement (1) is  sufficient .

OddManatee · Answer

Bunuel - Using your interpretation of the stem, I understand why the answer is D. But can you please explain how you know that is the correct interpretation?

I feel that the stem line : " Sofia’s folder and exactly one of the other two folders could be copied to the drive" can be interpreted in two ways:

"exactly one of the other two folders" - meaning "exactly a quantity of one folder can fit at a time. Not zero, not two, but exactly one at a time" 
This means that perhaps each E and M could fit separately, but they cannot both fit together. (This is how prragya seemed to interpret it as well, since their explanation was exactly what I was thinking.)As another 'example situation": If I had two tickets to a game (one for me, one for a friend), and I have two friends named A and B, then I can bring exactly one of A and B to the game... (but obviously I could take A OR B. It's not like B is secretly banned from the stadium). This is how I interpreted the stem. 
  "exactly one of the other two folders" - meaning "looking at them independently, only one of M or E could ever fit (alongside S), regardless of whether the other one was already there" 
This is how you interpreted it to get the correct answer.

As I said at the start, can you explain why my interpretation is wrong and/or how you knew the second one is right?  Basically how do we know if 'exactly one' means 'exactly a quantity of one' or 'one specific person/thing'?  Maybe this is always what GMAT means by this sort of language or something? I just want to know for future questions.

Thanks!

Bunuel · Answer

The key is the complete wording:  “Sofia’s folder and  exactly one of the other two folders  could be copied ..., while  the remaining folder would have to wait .”  This tells us that, of Elena’s and Marco’s folders,  one can be copied with Sofia’s and the other cannot .

If both Elena’s and Marco’s folders could each fit individually with Sofia’s, then it would not be true that exactly one of them  could be copied along with Sofia’s folder . Both could.

The phrase “while the remaining folder would have to wait” reinforces this interpretation: there is one particular folder that can be copied and one particular folder that cannot.

Arpit_Manchanda · Answer

Another way of dismissing case 2 (M 49, E 50) is: Given total space is S + R (S has to be the biggest size for this ques to hold, i.e. S>E and S>M) Then in case 2 we can paste E + M together without the need to delete additional file which shatters the ques stem (S still waits). Eg. Total space is 101 GB - S 51 GB, M = 49 GB , E = 50 GB Then we can paste folder M+E and still be under the space constraint. Also, ques is stating that the total space is S + exactly one other (one of the two, not either of two). I did the same mistake while solving. All the best

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