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06 Jul 2026, 08:50
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Given that,
Six Cars ( A, B, C, D, E, F) and Six Places (1, 2, 3, 4, 5, 6) and cars (A, B) are parked together.
To find: Is Car F at 1?
Statement 1: B and C both parked at even-numbered spaces.
Even number spaces available: 2, 4, 6
Let (B, C) be at (2 & 4) in any order, then since (A, B) are together, the possible places for A can be at (1, 3, 5)
Let (B, C ) be at (4 & 6) in any order, then since (A, B) are together, the possible places for A can be (3,5)
Let (B, C) be at (2 & 6) in any order, then the possible places for A can be (1, 3, 5)
But in all the cases, we don't know anything about the positions of D, E & F.
Hence, this statement alone is Insufficient.
Statement 2: D and E both parked at odd-numbered spaces.
Odd-number spaces available: 1,3,5
Let (D, E) be at (1, 3) or (1,5) in any order, then since place 1 is occupied by either D or E, hence car F can't be parked there anymore.
Let (D, E) be at (3,5) in any order, also (A, B) need to be parked together, hence the only possible options for A, B together are (1,2), thus in this case as well, F can't be parked at 1.
Thus, in all the cases F can't be parked at 1.
Hence this Statement alone is Sufficient.
Thus, the final answer is B. Statement 2 alone is sufficient, but Statment 1 is not
Six Cars ( A, B, C, D, E, F) and Six Places (1, 2, 3, 4, 5, 6) and cars (A, B) are parked together.
To find: Is Car F at 1?
Statement 1: B and C both parked at even-numbered spaces.
Even number spaces available: 2, 4, 6
Let (B, C) be at (2 & 4) in any order, then since (A, B) are together, the possible places for A can be at (1, 3, 5)
Let (B, C ) be at (4 & 6) in any order, then since (A, B) are together, the possible places for A can be (3,5)
Let (B, C) be at (2 & 6) in any order, then the possible places for A can be (1, 3, 5)
But in all the cases, we don't know anything about the positions of D, E & F.
Hence, this statement alone is Insufficient.
Statement 2: D and E both parked at odd-numbered spaces.
Odd-number spaces available: 1,3,5
Let (D, E) be at (1, 3) or (1,5) in any order, then since place 1 is occupied by either D or E, hence car F can't be parked there anymore.
Let (D, E) be at (3,5) in any order, also (A, B) need to be parked together, hence the only possible options for A, B together are (1,2), thus in this case as well, F can't be parked at 1.
Thus, in all the cases F can't be parked at 1.
Hence this Statement alone is Sufficient.
Thus, the final answer is B. Statement 2 alone is sufficient, but Statment 1 is not
06 Jul 2026, 08:50
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I think option B is the answer .
- 1 2 3 4 5 6 so ,, six spaces now if we saw each statement
-- statement 1 - if b and c both parked even number of places if b,c = 2,4 then a b 3 c 5 6 here , we got one case where k is not 1 ,,
now if b,c = 4,6 ,, then 1 2 3 b a c or 1 2 a b 5 c here k can be 1 so we are getting two cases that's why statement 1 It's not
possible ......... now we can exclude option no A , D
-- Statement 2 - If D and E both on odd numbers then if then d / e if 1 then the ans k is not 1
and if d and e respectively 3 , 5 then 1 2 d 4 e 6 or 1 2 e 4 d 6 for each only 1,2 left where we can put a , b as
Adjacent .
so now d is not 1 ,, So for each of the cases we are getting D is not equals to 1 ,, We can conclude that statement 2
alone is sufficient .
** I think people will tend to Separate statement one and statement two and do it together and get a wrong answer as option C **
That's my approach if anyone has any better Ways to do it feel free to share : )
- 1 2 3 4 5 6 so ,, six spaces now if we saw each statement
-- statement 1 - if b and c both parked even number of places if b,c = 2,4 then a b 3 c 5 6 here , we got one case where k is not 1 ,,
now if b,c = 4,6 ,, then 1 2 3 b a c or 1 2 a b 5 c here k can be 1 so we are getting two cases that's why statement 1 It's not
possible ......... now we can exclude option no A , D
-- Statement 2 - If D and E both on odd numbers then if then d / e if 1 then the ans k is not 1
and if d and e respectively 3 , 5 then 1 2 d 4 e 6 or 1 2 e 4 d 6 for each only 1,2 left where we can put a , b as
Adjacent .
so now d is not 1 ,, So for each of the cases we are getting D is not equals to 1 ,, We can conclude that statement 2
alone is sufficient .
** I think people will tend to Separate statement one and statement two and do it together and get a wrong answer as option C **
That's my approach if anyone has any better Ways to do it feel free to share : )
Maesha
Joined: 11 Jun 2026
Last visit: 31 Jul 2026
Posts: 29
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Concentration: Leadership, Economics
Schools: HBS '26 (S)
06 Jul 2026, 08:55
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If , A and B are together and B as well as C is in even position - the location of A,B,C can be 124 / 234/346/456 , F's position cant be determined by this as it can fall into 1 to 6 , any slot . so option one is not sufficient to solve this .
If A,B are together and D as well as E in odd slot then it is still not sufficient to solve this problem as F call occupy any slot among 2,4 ,6 , as if D and E occupies 1 and 3 , A B occupies 4,5 the remaining space are 2 and 6 . Also if D and E occupies 3 5 , A B must occupy 1 2 , so only possible space for B are 4,6 . so , option two is also not sufficient .
If we use consider both statement , when D and E sit in 3 , 5 space , AB must occupy 1 ,2 . So only left space for F will be 4 and 6 . If D and E sit is 1 , 3 space A, B, C are destined to occupy 4,5,6 as A and B are always together . In conclusion , the only possible sit for F is 2 or 4 or 6 .
Hence, both statement together are sufficient to solve this problem. As they answered the question - does F sit in space 1 . Ans is No.
If A,B are together and D as well as E in odd slot then it is still not sufficient to solve this problem as F call occupy any slot among 2,4 ,6 , as if D and E occupies 1 and 3 , A B occupies 4,5 the remaining space are 2 and 6 . Also if D and E occupies 3 5 , A B must occupy 1 2 , so only possible space for B are 4,6 . so , option two is also not sufficient .
If we use consider both statement , when D and E sit in 3 , 5 space , AB must occupy 1 ,2 . So only left space for F will be 4 and 6 . If D and E sit is 1 , 3 space A, B, C are destined to occupy 4,5,6 as A and B are always together . In conclusion , the only possible sit for F is 2 or 4 or 6 .
Hence, both statement together are sufficient to solve this problem. As they answered the question - does F sit in space 1 . Ans is No.
