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GMAT Club World Cup 2026 (Day 1): Six carsA, B, C, D, E, and Fare

1 2 3 4 5 6 7

zerioi

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Using the First staement, there will be a total of 6 arrangements possible when A and B are placed together and B and C are in even numbered spots.

Arrangement 1: 1-A, 2-B, 4-C, rest empty (3,5,6)
Arrangement 2: 1-A, 2-B, 6-C, rest empty (3,4,5)
Arrangement 3: 2-C, 3-A, 4-B, rest empty (1,5,6)
Arrangement 4: 3-A, 4-B, 6-C, rest empty (1,2,5)
Arrangement 5: 4-C, 5-A, 6-B, rest empty (1,2,3)
Arrangement 6: 2-C, 5-A, 6-B, rest empty (1,3,4)

Now, using statement 1 alone, there are other spots than spot 1 where F can be placed, eg: spot 3 in arrangement 1 or 4 in arrangement 4.

If we further use statement 2, there will still be cases remaning where F can be put apart from spot 1, therefore even if both of the statements are taken together then also it is not sufficient to answer where F is in spot 1 or not.

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06 Jul 2026, 09:36

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Its given A and B has to be together.

Considering Statement 1:
B and C are in even numbered spaces. So it can be in 2, 4 and 6.
It doesn't clarify if F is in Space 1 or not.

Considering Statement 2:
D and E Occupy the odd spaces. So it can be 1, 3 or 5.
If its 1 then F cannot be 1 and if 3 or 5 then F may be 1. So no clarity yet.

Combining both the statements:
For Even Spaces: B and C occupies 2 places.
For odd spaces: D and E occupies 2 spaces.
Since A has to be attached to B then A occupies odd spaces.
Since we have 3 odd spaces guaranteed F cannot go in 1. Hence Sufficent.

Ans is C IMO.

Bunuel

Six cars—A, B, C, D, E, and F—are parked in six adjacent spaces numbered 1 through 6, from left to right. Cars A and B are parked next to each other. Is Car F parked in space 1?

(1) Car B and Car C are both parked in even-numbered spaces.
(2) Car D and Car E are both parked in odd-numbered spaces.

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06 Jul 2026, 09:41

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Constraints: A and B need to be parked together
Ask: Is F in spot 1?

1. B and C in even spaces -> Not sufficient. F could be in 1, B2, A3, C4.... or A1 B2 F3 C4....
2. D and E in odd -> If D3 and E5, then only way AB are together is in spots 1 and 2, so F not in 1
D or E in 1 means F can't be in 1. Sufficient.

B.

Bunuel

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Answer: B) Statement two alone is sufficient.

Using statement one alone, we can find arrangements where F is in spot 1 but also arrangements where it isn't, so we do not have sufficient info.
Using the second statement, we cannot find an arrangement where F is in spot 1 because it forces D and E into spots 3 and 5 (the only two odd-numbered spots remaining) and doesn't allow for B and A to be next to each other (which was given in the prompt).

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Question: is F in first place?
Given: parking spaces numbered 1 - 6 from left to right and AB always together.

Lets start with (1)
There are 3 possibilities
_ B _ C _ _ ( F in this case could be in position 1 or 2 we cannot be sure)
_ _ _ B _ C (In this case If A and B occupy the first two spaces F will not be in 1st position else F might be in position 1 or 5)
_ B _ _ _ C (Similarly we can't say for sure where F would be)

Statement 1 is insufficient. Option A and D can be ruled out.

Statement (2)
Again 3 possibilities
D _ E _ _ _ (F will not be in position 1 for sure)
D _ _ _ E _ (Again F will not be in position 1 for sure)
_ _ D _ E _ ( F will not be in position 1 since AB have to be together and the only way for that is if they occupy the first two spots )

Statement B is sufficient.

Note: In the exam during a time crunch. For statement two the first two cases can be ruled out mentally without the need to draw and only the third case may be drawn to be tested out. Start with option 2 since it talks about occupying odd spaces and the question asks for whether F occupies position 1 which is odd. Might save you time.

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06 Jul 2026, 09:59

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According to me - It is important to visualise the question.
It says there are 6 spaces

1 2 3 4 5 6 ( Numbered left to right)
_ _ _ _ _ _

Now Car A and B are always together is given in the question.

Question asks - Whether Car F is getting parked at Space 1 ?

Option A - Car B and Car C are parked at even numbered spaces. So If Car B is parked at Space 2, Car A will be Parked at Space 1 and hence Car F won’t be at Space 1. But if Car B is parked at Space 4, Car A will be in Space 3 and hence Car F can be at Space 1. Therefore this option is insufficient.

Option B - If Car D and E are both parked at odd numbered spaces - then Let’s say D is at 3 and E is at 5,( or vice versa) in such case - Car A and B mandatorily have to be at Space 1 and 2 since they are together. Therefore Car F can’t be at space 1 in such a case. Even if D is at space 1 and E at space 3, then Car F can’t be at space 1. Therefore this option is sufficient.

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06 Jul 2026, 10:02

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This is a good question to trap you if you do not carefully solve each statement alone:
Remember AB is together - AB or BA would be the case - JUST CHECK IF F WOULD BE AT 1

S1: B & C even numbered places
cases: _ B _ C _ _
FBACDE : F would be at 1
ABFDEC: F not at 1
CANNOT SAY ANYTHING

S2: D and E both parked in odd-numbered spaces
D_E_AB or ABD_E_
In any case F cannot be placed at 1st position so we would definitely say that F would not be at 1.

OA: B

PS - I fell for the trap as I rushed to check both the statements together and while writing the solution I realized my mistake. Be careful for this in DS guys!
Good Luck

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Always helps writing out A-F.

Looking at option 1: can Car F be in space 1? Lets see, Even numbers are 2,4,6. We are told B and C are even numbered, so lets say B-2, C-4. Then A can be 1, which means F is not 1.

However, what if B-4, C-2, A-5 and so F can be 1.

Option 1 not sufficient

Option 2: Odd numbers are 1,3,5
Which means, D could be 3 and E could be 5, A-B must be 1,2. So F not in 1.
However, lets say D is 1 and E is 3, which again means F cannot be in 1. So sufficient hence B

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BOTH STATEMENTS TOGETHER NOT SUFFICIENT

Bunuel

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Bunuel

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we know A and B has to be together, either like AB or BA.
considering statement (1) knowing B and C occupy even spaces leaves too many options open. _B_C_ _, _B_ _ _ C, _C_B_ _, _C_ _ _ B...., so it isnt sufficient.

considering statement (2) D and E occupy odd spaces. so we have odd spaces at 1, 3 and 5. Now A and B must be together, which means A and B will acquire positions even and odd not particularly in that order. If D and E will take up odd positions and A or B will take the final 3rd odd position then all odd positions will be full. example : D_E_ _ _, _ _D_E_ etc, AB has to take adjacent positions.
ultimately F cannot be in position 1.
This statement alone is sufficient to answer whether F occupies position 1 or not.

Therefore, option B, statement (2) alone is sufficient, but statement (1) alone is not.

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06 Jul 2026, 10:13

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Given in Q stem A, B are next to each other. We have to find if F is in 1 (Yes/No)
1.Statemnet 1- B & C in even which means For example if B,C= 2,4 (A=1,3 or 3,5) or 2,6 or 4,6 . Placing A next to B, F, D, E can be in any remaining locations - Not sufficient
2. Statement 2- D & E in odd position ---> D,E = 1,3 or 3,5 or 1,5. A, B, C & F can be in any remaining location. Cant say if F in 1 or any other location - Not sufficient
3. Combining st 1 & st 2: Since B & C are in even position & A will always be next to B, A will take a odd number. The remaining 2 odd nos will be taken by D&E. Which means F can never be in 1 (odd) since all odd positions are taken by D, E & A
Hence Sufficient. Ans C

Bunuel

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06 Jul 2026, 10:23

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Drawing the sports from one to six in the writing notepad helps a lot. On top of it the critical information here is that A and B are parked next to each other.

Bunuel

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Statement one: there are two situations when B and C are parked in two and four or in four and six. Now in either case we don't know that A is on the left or on the right of B. Here in both the situations the answer for the question asked is yes and no. Therefore this statement is not sufficient.

Statement 2: Now here also we come across two situations:
1. D,E parked at 1,3. Here we are sure that the answer to the question of car F is no.
2. D,E parked at 3,5. In this situation as well we know that the only adjacent empty spaces are 1,2, which will be filled by A and B. Therefore the answer to the question that car F is parked in space 1 is still no and hence statement 2 is sufficient to answer the question.

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06 Jul 2026, 10:31

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Answer(B)
123456 spaces
ABCDEF cars
Question wants to know if we can certainly answer that F is in space 1.
SO assuming F is in space 1.
For Statement 1:
123456
FDABEC or
ABFCDE, therefore, not certain, so not option(A)
For statement 2:
123456
F
As there are only 2 odd places left, D and E will have to go there.
123456
F_D _E_, but A and B should be adjacent, therefore certainly not possible.
Therefore 2 is sufficient.
Answer(B)

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06 Jul 2026, 10:56

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What we know from statement:
6 cars
A and B adjacent to each other
Yes/No DS
F in 1 - Yes
Any other car in 1 - No

Statement 1 B and C on even place
Working with constrains

B - 2nd
C - 4th

A can be on 1st place - DS NO

Now B - 4th Place
C - 6th Place

A can be on 3rd or 5th place. F can be on 1st place - DS Yes

So Statement Not sufficient we get both Yes and No

Statement 2 D and E on Odd place
Working with Constrain

D - 1st place
E - 3rd place

A and B can be accommodated on 4th 5th 6th place adjacent to each other

This will give DS - No because F can’t be on 1st place.

D - 1st Place
E- 5th Place

A and B can be accommodated on 2nd 3rd 4th place adjacent to each other.

This will give DS - No because F can’t be on 1st place

Last case

D - 3rd Place
E - 5th Place

A and B has only 1st and 2nd place left to accommodate to stay adjacent.

Hence, this will give DS - No because F can’t be on 1st place.

From statement 2 we get definite No. Hence B is the answer.

Considering statement Constraint first to create cases, then in each cases try 1st position with F and Try 1st position without F. Statement 1 both cases possible there for NS and Statement 2 only 1 scenario possible that is 1st position without F for all cases Hence Sufficient.

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Statement 1: Given A & B are parked adjacent and B,C are parked at even-numbered spaces, so A is at odd -numbered space
Now, 1 Even & 2 Odd spaces left , so F can be in either space. So, not sufficient

Statement 2: D & E are both at odd-numbered space and A-B are adjacent so one of them would be parked at Odd-Numbered space
Thus F has to occupy even-numbered space and cannot park in space 1. Sufficient

Option B

Bunuel

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06 Jul 2026, 11:07

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We have A and B next to each other

(1) B and C are in even places.
Possible: F can be in space 1 but does not have to be
NOT SUFFICIENT

(2) D and E are both in odd places.

There are three odd places 1,3,5
If we put F at 1, the two remaining odd places are occupied by D and E => there is no place for A and B to be next to each other

So, the answer is NO, F can't be in space 1, so statement (2) provides sufficient information to answer the question

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06 Jul 2026, 11:12

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made cases so
for option 1
put AB in first 2 so dcef need to be arranged but we dont know where from 1 so 1 not sufficient
now de is in odd spaces but no info on abcf so 2 not sufficienct
combining bc can be in 246 and de in 135 that leaves a and f to fill remaining spaces
we get that f is not in the first space so both together help us answer this answer C

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06 Jul 2026, 11:14

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To find out if F is parked at position 1.

Analysing statements

A) B and C are parked in even numbered spaces.
With AB is together, we can form multiple arrangements such as
A B D C E F
F B A C E D
Since F may or may not take first position, This is not sufficient

B) D and E are parked in odd numbered spaces
with AB together,
A or B needs to take one odd numbered position.
So we have D , E and (A or B) occupying the odd positions.

This means F cannot occupy position 1.
This is sufficient.

Bunuel

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06 Jul 2026, 11:28

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Statement 2 alone is sufficient; Statement 1 alone is not.

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06 Jul 2026, 11:32

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No F is not in space1

Explanation

A and B are next to each other in an adjacent parking where B is even numbered , so A is odd

Total spaces is 6
D , E and A are odd numbers now
B , C and F should be even then with this logic so F is not a odd number so it cannot be 1

Bunuel

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