Riddle me this!

[mayerling]

PSers, I need help!

I'm studying for an exam I have to take soon (you can call it IQ, aptitude, logic, whatever really), and it involves figuring out brain teasers. I'm going over past papers to try and get myself ready and I'm having trouble with one particular question. Could anyone help? Here goes:

Four couples (Mr and Mrs A, Mr and Mrs B, Mr and Mrs C, and Mr and Mrs D) are seated around a round table, and each person is equidistant from the next person. They are seated according to the following rules:

a. Mr and Mrs B are not seated next to Mr and Mrs C
b. Mr and Mrs C are seated next to each other
c. Mr A is seated directly across from Mr D

Question:
If Mr B is seated equidistantly from Mr A and Mr D, then which of the following people could be seated directly across from him (Mr B)?
a. Mrs B
b. Mrs A
c. Mrs D
d. Mrs C

What's the answer and WHY?

 

[kefira]

I love brain teasers!

The answer is Mrs. C. If you draw the scenario as a pie, it's pretty easy. Think of the slices as 1-8, with Mr. A in slice 1, and Mr. D is in slice 5 (directly across the pie). Mr. B then has to be right between them, in slice 3 or 7 (doesn't matter which), and you are solving for the one you didn't choose. I put Mr. B in slice 7, so I was solving for slice 3, the slice directly across from him.

Since Mr. & Mrs. C are sitting beside each other, AND aren't next to Mr. & Mrs. B, they have to be in the slices 2, 3, or 4, with one of them having to occupy slice 3. The quick route here is that none of the answer choices are Mr. C, so the answer can only be Mrs. C.

(Let's hope I'm right so I don't look like a dummy! )

 

[monarch64]

Out of the choices given for answers, it has to be Mrs. C. Drawing it out a bit, following the rules (esp Mr. A being directly across from Mr. D.) logic'd me into that conclusion.

 

[mayerling]

Thanks, Kefira and Monarch. That's actually the answer DH and I got, but unfortunately it's the wrong one. We have the answer sheet which gives the correct answer but doesn't explain why. So I'm hoping somebody will get it right and explain it to me!

Spoiler alert! I'll give the correct answer below for anyone who's interested but please help me understand why it's the correct answer.

c. Mrs D

 

[Loves Vintage]

Mrs. C here as well . . . even though you said it's incorrect already! Is it possible that your answer sheet is wrong? I'll be interested to see if anyone comes up with another answer!!

 

[mayerling]

Mrs. C here as well . . . even though you said it's incorrect already! Is it possible that your answer sheet is wrong? I'll be interested to see if anyone comes up with another answer!!

I suppose it is possible but I have no way of finding out if so. Does anyone think that the correct answer could even conceivably be true? I really don't get it...

 

[kefira]

I don't think the answer you were given can be right. If every other space is taken by Mr. A, Mr.B, Mr. D, and Mrs. D (highlight), it leaves nowhere for Mr. & Mrs. C to sit together.

 

[VapidLapid]

Im jumping ahead to avoid the spoiler. Is it safe to assume that they are all seated as couples and boy girl? I get Mrs A

 

[VapidLapid]

 

[fleur-de-lis]

Alas, Vapid, Mr B is not seated equidistantly from A and D in your illustration.

 

[fleur-de-lis]

If the couples do not have to sit together, across from Mr B *could* sit Mrs C as well; if you want I can explain/draw.

Even when we assume couples do not have to sit with each other, there are two seating options which fulfill all requirements: one has Mr B across from Mrs C, the other has Mr B across from Mr C. I really, really suspect the answer in your book will be revealed to be a publishing error by your instructor. (BTW, is this for the LSAT?)

 

[davi_el_mejor]

It has to be C because the only spaces for the C's to sit next to each other is opposite Mr B.

 

[Lady_Disdain]

Sorry, but the book is wrong. If Mrs D is placed opposite Mr B, then Mr and Mrs C have to be split up, since they can only fit on the opposite side of the table.

 

[mayerling]

Im jumping ahead to avoid the spoiler. Is it safe to assume that they are all seated as couples and boy girl? I get Mrs A

No, only Mr and Mrs have to sit together and it doesn't have to be boy-girl.

 

[mayerling]

If the couples do not have to sit together, across from Mr B *could* sit Mrs C as well; if you want I can explain/draw.

Even when we assume couples do not have to sit with each other, there are two seating options which fulfill all requirements: one has Mr B across from Mrs C, the other has Mr B across from Mr C. I really, really suspect the answer in your book will be revealed to be a publishing error by your instructor. (BTW, is this for the LSAT?)

No, it's not for the LSAT.

 

[VapidLapid]

It is all that Mrs. C's fault. I swear I am never going to dinner with them again.

 

[amc80]

It is all that Mrs. C's fault. I swear I am never going to dinner with them again.

Hey, I'm Mrs. C!

 

[partgypsy]

I get Mrs (or Mr) C as the answer. It can be either of those two. Basically starting from the top going clockwise have the two C's, Mrs D, Mr D, Mrs B, Mr B, Mrs A, Mr. A, then back to the two C's.

Alternative could be CC, AA, BB, DD. But still doesn't change who is across from the C's, who stick together.
could the booklet be wrong?

 

[rubybeth]

I tried to figure it out, and Mrs. D can't be correct because Mr. and Mrs. C need to be seated together, leaving Mrs. D, Mrs A., and Mrs. B to fill in the other seats. The answer has to be Mrs. C *or* Mr. C, because they have to be seated together at that side of the table.

I am guessing it is a typo. I wish they would have used names like Mr. Adams, Mrs. Brown, Mr. Connelly, and Mrs. Daniels, because it would be easy to make a typo for the answer in this case if it's just a matter of hitting "D" or "C" as the letters are close together on the keyboard.

 

[mayerling]

Okay, I'll assume it's a typo and stop torturing myself trying to make it work.

For those who said they enjoy riddles, I have another one for you:

There's a mayoral election going on. The five candidates, A, B, C, D, and E, have agreed to hold three rallies at which they will give speeches following a specific order. The order they will speak in must satisfy the following conditions:
a. Each candidate must go either first or second in at least one of the three rallies.
b. Whoever goes fifth at any one of the rallies, must go first in at least one of the other rallies.
c. No candidate can go fourth at more than one rallies.

Which of the following statements could be true?
A. D goes first in all three rallies.
B. D goes after B in all three rallies.
C. D goes after C and E in all three rallies.
D. D goes before B and E in all three rallies.

Please explain your thought process when you answer this so I can understand.

 

[VapidLapid]

B. d goes after b in all three

 

[kefira]

Spoilering my solution below for those who want to puzzle it without influence.

(highlight)

By process of elimination, I get statement B for my answer.

It cannot be answer A because if D goes first each time, there are only 3 spots where others can go second, with 4 remaining candidates needing to go first or second. This goes against rule 1.

It cannot be answer C because if D goes after at least 2 others at each rally, D will never go first of second. Breaks the same rule as above.

It cannot be answer D because D must go first at 2 rallies, and third at one rally, in order for the first rule to be fulfilled, and to leave room for B & E to both go after. This means that either B or E will go fifth at one rally, but never first, breaking rule 2.

My minimap I used to show this:

Rally 1: D, B, x, x, x
Rally 2: D, E, x, x, x
Rally 3: A<->C, D, B<->E

 

[VapidLapid]

there are a total of six first and second positions.
If D goes first in all three that leaves three positions for four candidates - "A" is false

If D goes after C and E in all three, then the highest position D will ever occupy is third

If D always goes before B and E then the highest position E will ever occupy is third

That leaves B.

But since B always goes before D, B can never be fourth because that would put D in fifth which would require D to be first in another and if D were first then D could not be after B.

whoever goes fifth must go first also, (but that does not mean that one being first also had to be fifth once)
And fifth place can be occupied more than once, so

 

[kefira]

VL, although we got the same answer, I noticed that you discounted answer D by faulty logic. D going before B and E could be D, B, E, or D, E, B, (with possibly others in between, of course) since nothing forces the order of E and B in the statement. Just wanted to point that out for mayerling's notes.

 

[mayerling]

Thanks, Kefira and VL.

Last one, I promise (unless you want me to post more ) I got an answer for this one and was proud of myself until I checked the booklet which gave a different answer

A, B, C, and D need to cross a bridge made of rope during a dark night. The bridge can withstand two people at a time. In order to cross, they need to carry a flashlight, and they have only one flashlight at their disposal. A can cross the bridge in two minutes, B can cross it in three, C can cross it in five, and D can cross it in ten. What's the shortest amount of time needed for all of them to cross?
a. 20 mins
b. 21 mins
c. 22 mins
d. 23 mins
e. 24 mins

 

[kefira]

No, keep them coming! I love trying them. Again, spoilering.

I got answer B. But I'm not 100% sure if I'm right. Here's my solution:

A+B --> 3min
<-- B 3min
C+D --> 10min
<-- A 2min
A+B --> 3min

total = 21 min

What does the answer sheet say?

 

[davi_el_mejor]

Thanks, Kefira and VL.

Last one, I promise (unless you want me to post more ) I got an answer for this one and was proud of myself until I checked the booklet which gave a different answer

A, B, C, and D need to cross a bridge made of rope during a dark night. The bridge can withstand two people at a time. In order to cross, they need to carry a flashlight, and they have only one flashlight at their disposal. A can cross the bridge in two minutes, B can cross it in three, C can cross it in five, and D can cross it in ten. What's the shortest amount of time needed for all of them to cross?
a. 20 mins
b. 21 mins
c. 22 mins
d. 23 mins
e. 24 mins

Can "A" stand in the middle of the bridge and shine light from the middle while the rest cross?

 

[mayerling]

Perhaps. I don't know...

 

[VapidLapid]

VL, although we got the same answer, I noticed that you discounted answer D by faulty logic. D going before B and E could be D, B, E, or D, E, B, (with possibly others in between, of course) since nothing forces the order of E and B in the statement. Just wanted to point that out for mayerling's notes.

Maybe faulty logic, maybe different. having eliminated the three and deduced that B was true, I incorporated the rule of B into the set-up. Since the other three were patently wrong I proceeded with the idea that this one was true and looked for an instance that showed that, not a rule that proved it.

 

[VapidLapid]

I get B 21 minutes, but I fear there may be a trick to all these questions now