Exam pack 2 DS - What is going on ?Scratching my head!

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My approach :-

1-->now there are 9 people who must have chosen some number.-IS
2--->okay so people have chosen '5' more than '4'--IS

Well combining I have 9 people so there can be many situations:-
I- 5 5 4 3 3 2 2 2 2 Answer NO
II-5 5 5 4 4 3 2 1 1 Answer Yes

Hence,E! It can be C only if we are not allowed to repeat the integers. But if thats the case should it be mentioned?
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by [email protected] » Wed May 25, 2016 10:12 am
Hi Neilsheth2,

Your approach to this question is spot-on. However, all of your examples use 9 people when the prompt clearly states that there are 12 people. Try redoing your work with 12 people and you'll get the correct answer.

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by Neilsheth2 » Wed May 25, 2016 9:45 pm
Still not clear Rich . Pls help.


5 5 4 No no no no 3 3 3 3 3 - Atleast 1-->NO
5 5 5 4 4 No no no no 1 1 1-Altleast 1 --> yes
Still E

What am i doing wrong?

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by [email protected] » Thu May 26, 2016 9:29 am
Hi Neilsheth2,

We're told that there are 12 people who each choose a number from 1 to 5 (inclusive). By definition, this means that there will be at least one 'repeat number' in the group. We're asked if at least one of those twelve people chose the number 1. This is a YES/NO question.

1) No number was chosen by more than 3 people.

With this Fact, we can certainly avoid choosing the number 1 IF we choose each of the other numbers three times each (but it's not a given)...

IF.... the numbers chosen are
1 1 1 2 2 2 3 3 3 4 4 4, then the answer to the question is YES.
2 2 2 3 3 3 4 4 4 5 5 5, then the answer to the question is NO.
Fact 1 is INSUFFICIENT

2) More people chosen the number 5 than the number 4

With this Fact, we have no idea how many times each number was chosen.

IF.... the numbers chosen are
1 1 1 2 2 2 3 3 3 4 5 5, then the answer to the question is YES.
2 2 2 3 3 3 4 4 5 5 5 5, then the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, since we know there are MORE 5s than 4s, then we could have up to three 5s, but that means that there could be at most TWO 4s. With the restriction that we cannot choose more than three of any number, we could have eleven of the numbers as....

2 2 2 3 3 3 4 4 5 5 5

BUT that's only eleven numbers. We have to pick 12 numbers, so that last number would have to be a 1.

1 2 2 2 3 3 3 4 4 5 5 5

Combined, SUFFICIENT

Final Answer: C

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by Matt@VeritasPrep » Thu May 26, 2016 3:02 pm
Neilsheth2 wrote:Still not clear Rich . Pls help.


5 5 4 No no no no 3 3 3 3 3 - Atleast 1-->NO
5 5 5 4 4 No no no no 1 1 1-Altleast 1 --> yes
Still E

What am i doing wrong?
We're told that everybody chose a number, and "no" isn't a number. (It is a response, but it violates what we're told by the prompt.)