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Help on a Data Sufficiency Problem

This topic has 3 expert replies and 3 member replies
wlvoh Newbie | Next Rank: 10 Posts Default Avatar
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Help on a Data Sufficiency Problem

Post Wed Feb 07, 2007 6:37 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    I got the following question on a practice test, and cannot figure it out. The correct answer, according to PR, is A. I thought initially it was D. If someone can explain to me the rationale, most importantly why B is not sufficient, I would greatly appreciate it.

    " A bowl is filled with consecutively numbered tiles from 1 to x. Joe pulls out a tile
    and uses it to contruct sequence Q, which consists of 10 consecutive integers starting
    with the number drawn. If Joe then selects one number from Sequence Q, what is the
    probability that the selected number is a multiple of 3?

    1.) The last number in sequence Q is a prime number that is less than 20.

    2.) x (is less than or equal to) 10

    "

    Thanks.

    wlvoh

    Thanked by: shantanu290
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    Post Wed Feb 07, 2007 11:07 pm
    Wow - tough one. The wording is pretty funky - I had to read it twice to make sure I understood what was going on! Note that he's not pulling 10 tiles out of the bowl - he's pulling 1 tile, and then whatever that number is, he's just using the next 9 consecutive numbers after it to fill out the sequence. (Eg, he pulls a "3" from the bowl and that makes the sequence 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.)

    Every 3rd number in a set of consecutive numbers is divisible by 3. We have 10 consecutive numbers in sequence Q (so 10 will be the bottom of my probability fraction).

    I have two possibilities for how many of those 10 numbers are divisible by 3. If the first number in the sequence is divisible by 3, then so are the 4th, 7th and 10th, for a total of 4. If the 2nd number is div. by 3, then so are the 5th and 8th, for a total of 3. If the 3rd number is div. by 3, then so are the 6th and 9th, for a total of 3. I don't have to check "if the 4th number is div by 3" b/c that would mean the 1st number is too, so I've already covered all my options.

    So, basically, I need to know whether my starting or ending number is divisible by 3. If so, then my probability will be 4/10. If not, my probability will be 3/10.

    Statement 1 tells me the last number in the sequence is prime. Prime numbers aren't divisible by 3. So my probability is 3/10. Sufficient.

    Statement 2 tells me x is less than or equal to 10, which means my possible starting numbers start at 1 and can go as high as 10. If it's equal to 10, which is not div. by 3, then my prob. is 3/10. If it's equal to, say, 9, which is div. by 3, my prob. is 4/10. Insufficient.

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    Post Thu Feb 08, 2007 12:21 am
    Stacey did a wonderful job in answering this convoluted question. The problem I have with questions like these is they don't reflect the actual GMAT. They are unnecessarily complicated. Unless the question was not copied verbatim, I think it lacks precision and is poorly worded, which is never the case with ETS/Pearson Vue. The questions on the test are far simpler, however, there is always some special constraint or a little twist. I personally would not spend too much time going over really hard probability/combinations problems.

    Just my two cents.
    Mark

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    Stacey Koprince GMAT Instructor
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    Post Thu Feb 08, 2007 12:43 am
    What is the source on this one?

    Totally agree that this one is more convoluted than is typical on the real thing. We (ManhattanGMAT) have a special set of questions we call Challenge Questions - half of them are harder than anything you'd actually see on the test. (And we say they're really just for "fun" or challenging yourself, but they're not part of the main curriculum.) Maybe other companies also have something like this and this question came from that?

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    wlvoh Newbie | Next Rank: 10 Posts Default Avatar
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    Post Thu Feb 08, 2007 5:06 am
    The question as I typed it was verbatim from the practice test. A huge thanks to Stacey on the explanation. I took the wording to mean he actually drew 10 tiles for his sequence, thus he would have had 1 through 10. The source of this is a practice test from Princeton Review.

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    Post Thu Feb 08, 2007 10:54 am
    Yeah - I assumed that too as I first read it, but then I realized that didn't make sense when I looked at the statements, so I read it again to figure out what was really going on. As Mark said, the language isn't "clean" enough to be an official question - they do try to trap you on the test, but not through deliberately convoluted language.

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    Cybermusings Legendary Member Default Avatar
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    Post Wed Mar 28, 2007 5:53 am
    It's A...Very much in the way explained...If the sequence starts with multiples of 3 then it will have 4 elements which are divisible by 3. Then the probabiltiy becomes 4/10

    If the sequence starts with any other number (numbers which are not multiples of 3) then the probability becomes 3/10

    Hence B is insufficient alone

    A is sufficient

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