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. ⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

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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

amansoni5

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1) B and C are in even spaces. Since A must be adjacent to B, F can be in space 1 in some arrangement and not in others - NS
2) D and E are in odd spaces. F can still be in space 1 depending on the arrangement - NS
3) Together, the three odd spaces (1,3,5) must be occupied by A, D and E (A must be odd because B is even)
4) Therefore, F must be in an even space, so F cannot be in 1.
Hence, C is the answer.

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B- St (2) is correct because as D & E can only have odd places and A&B have to be together.
Case:1- If AB is in 1 &2- D&E have to be 3 and 5, F can't be in 1. Not possible
Case-2: If D&E in 1 &3, A&B have to be 4/5 or 5/6 but F can't be in 1. Not possible

Even if we put F in 1st place to check if this option is false- we will find that D &E can only take places 3 and 5. and if it happens only 2,4,6 spaces will be left and A&B would never be together (violating main condition)- hence F can never be in 1st place

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The answer is B because in every possible case F is never in space 1. Since this gives one consistent answer, "No," in every single scenario, statement (2) alone is sufficient.
Bunuel

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A , B, C ,D , E , F have a total of 6 spots available . Question stem says AB are adjacent that means A is at all odd places (1,3,5)and B is at all even places (2,4,6). We need to find if F is in 1 ?

S-1 B ,C are in even places (2,4,6) that means A has to be at odd places (1,3,5). If A is at 1 then F cannot be at 1 - NO. If A is at 3 or 5 then F can be 1

S1 is not sufficient

S - 2 D and E are at odd places (1,3,5) so if D and E is 3,5 then 1 can be taken by A, B , C,F so not sure if F can be 1. If D and E are 1, 3 then for sure F cannot be 1

S2 not sufficient

Combining 1 and 2 we know D and E (1,3,5) and B,C can be even (2,4,6) then A has to be odd . If D,E is 1,3 then A is 5 , F wont get 1 . If D,E is 3,5 then A =1 f again wont get 1 . Sufficient - C.

Sorry for the excessive highlighting i am unable to unhighlight some glitch

Bunuel

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(1)
cars FBACDE answer is yes
cars ABFCDE answer is no

Insufficient

(2)
If D or E are parked in space 1 the answer is no.
If D or E are not parked in space 1 then cars are parked XXDXEX or XXEXDX. In boths cases it is impossible that F is parked in space 1 because the only two adjacent spaces (where A and B
must be parked) are space 1 and space 2, so F is in space 4 or space 6 and the answer is no.

Sufficient

IMO B

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spaces 1 to 6
cars A B C D E F

(1)
multiple solutions with F parked in space 1 [F E A B D C] or in some other place [A B D C E F].

Condition insufficient

(2)
If F could be parked in space 1 [F X D X E X], A and B couldn't be next to each other -> F is nor parked in space 1.

Condition sufficient

Answer B

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Given: A and B are parked next to each other
(1) B and C are in even places
It is not sufficient to answer as we can create both cases where F is at 1 place and where it’s not - given these conditions

(2) D and E are in odd places
In this case F cannot be at 1 because if it is at 1 place, B and A cannot be together
Hence the answer is (B)

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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.
B&C are 2,4,6 parking spaces. No info about DE.
Insufficient

(2) Car D and Car E are both parked in odd-numbered spaces.
D&E are in 1,3,5 parking spaces. No info about others.
Insufficient

1&2 together
B is parked in 2.
ABD_E_
(D/E)BA_(E/D)_
B is parked in 4.
(D/E)CAB(E/D)F
(D/E)C(E/D)BAF
B is parked in 6.
(D/E)C(E/D)_AB
In every order F has to be in 2,4,6 space. Car F is not in space 1.
Sufficient

C

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Car A & B are adjacent.
Statement 1: B & C in even spaces, i.e., 2,4,6. If B is in 2 and C is in 4, A can be in space 1. Therefore, F cannot go to 1st space. If A is in space 3, F can go to the 1st space. Insufficient, as there could be Yes/No situation.
Statement 2: D & E in odd spaces, i.e., 1,3,5. If D & E are in 3 and 5, there are 4 spots left (1,2,4,6). Since A & B needs to be adjacent, we can give them in spot 1 and 2. Therefore, F cannot be in the first spot. Sufficient.

B is the answer, as Statement 2 survives in the test.

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The answer is no as Car B and Car C are in even and Car E and Car D are in odd and also Car A is beside Car B thus Car A is also in odd so all positions filled for odd. Thus both statements together are sufficient.

But if we look at statement 2 alone then according to it Car D and Car E are placed in odd space. Thus only 1 spot left for odd and if Car A and Car B must be parked next to each other, 1 of them will be parked in that empty odd space.
So the final answer is Statement 2 alone is sufficient.

Bunuel

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I couldnt get a better way to do this. So did this via a lot of trial and error. Placed the cars as per the statements in various slots. Ultimately got B as an answer. Can anyone please help with a faster method on this?

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Statement (1) tells us that B and C could be in any of the spaces 2, 4 or 6. Not sufficient, as it leaves 1,3, 5 and one even numbered space free. Statement (2) tells us that D and E can be in 1,3 or 5. Not sufficient, as it leaves 2,4,6 and one odd numbered space free. Taken together, there still leaves one even and one odd numbered space free, telling us nothing about which car occupies space 1. So, E - 1&2 together are not sufficient.

Bunuel

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Bunuel

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Assuming 6 spots to access

_ _ _ _ _ _
We know that A and B are parked next to each other so, any space from 1-5 can be taken by A/B and 2-6 by B/A

Now,
Statement 1: Car-B & C are both parked in even-numbered spaces. Basically, both can be at 2nd, 4th and 6th spot. Keeping in mind if B is 2nd A is 1/3 and D is 4/6 so on. However, this gives us no info for F.

Statement 2: D & E both parked in odd spaces: again gives no info for F

Let's combine and try to form a structure with F at 1st. What you realise is that its not possible hence C is sufficient to answer.

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Since Both either of individual statements doesn't completely give idea of all the cars placement option A and B alone cant be alone sufficient. Both combined cannot be sufficient since if F is in first place and A and B should be adjacent, it wont be logically correct. So answer is E

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Looking at statement (1) places the (AB) or (BA) block in 1, 2 or 2,3 or 3,4 or 4,5 or 5,6 and C either in 2,4 or 6, so it's not helpful and A,C & D are eliminated.

Statement (2) Places D and E in places 1, 3, 5. If F is in 1, then D and E fall into 3 and 5 and the (AB)/(BA) has no room, so F cannot be in 1, so the answer is B) statement (2) alone is sufficient, but statement (1) alone is not sufficient.

Bunuel

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Analyzing Statement 1:
B can be at 2 but then A can be on 1 or 3 cant say , therefore this statement is not sufficient

Analyzing Statement 2:
D and E are at 3 and 5 position ,this doest tell about how can we derive who is at postion 1. Hence not sufficient

Analyzing 1+2:
If D and E are at odd position , AB are together and B can be only at even position
then the only choice of B is to be at 2 because it cannot be at 4 as D and E has already occupied adjacent positions
B cannot be at 6 as then A cannot be placed adjacent to it.
Therefore only place left is 2 where 3 is already occupied by D/E so only position left for A is 1. Hence sufficient

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(1)
F D E C A B -> answer is yes
D F E C A B -> answer is no

Condition (1) is insufficient

(2)
D or E are in space 1 -> answer is no
D and E aren't in space 1 -> F isn't in space 1 because space 1 and space 2 must be AB or BA -> answer is no

Condition (2) is sufficient

The answer is B

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(1)
F is parked in space 1 in FBACDE and F isn't parked in space 1 in ABFCDE

Insufficient

(2)
If F is parked in spaace 1: F _ D _ E _ or F _ E _ D _
It is impossible because then A and B cannot be parked next to each other.

Sufficient

The correct answer is B

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And- B

Solution in PIC attached.

Attachments

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Bunuel

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1) Knowing that B and C are in even-numbered spaces does not determine whether F is in space 1. You can create valid arrangements where F is in space 1 and others where it is not. Hence, S1 alone is not sufficient.

2) The odd-numbered spaces are 1, 3, and 5. Since D and E take two of those three odd spaces, only one odd space is left for A, B, C, or F. But A and B must be parked next to each other. Any adjacent pair uses one odd space and one even space.

Therefore, the remaining odd space must belong to A or B. That means all three odd spaces are already taken by D, E, and one of either A or B.

So F cannot be in an odd-numbered space, and F cannot be in space 1.

Therefore, the answer to the question "Is F parked in space 1?" is No, and Statement (2) alone is sufficient.

Option B