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GMAT Club DS question

by jsasipriya » Mon Apr 04, 2011 5:15 pm
If a, b , and c are positive distinct integers, is (a/b)/c an integer?

1. c=2
2. a=b+c

OA B

Explanation provided is:

"There is no combination that would allow divisibility into an integer for distinct integers such as a ,b and c ."

Statement 2. I picked numbers. a=6; b=2; c =3. This combination yields integer value. However, when a=6; b=4; c=3 this combination yields a fraction value. Hence I decided B is insufficient. And my answer was wrong.

Can someone explain this?
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by GMATGuruNY » Mon Apr 04, 2011 5:46 pm
jsasipriya wrote:If a, b , and c are positive distinct integers, is (a/b)/c an integer?

1. c=2
2. a=b+c

OA B

Explanation provided is:

"There is no combination that would allow divisibility into an integer for distinct integers such as a ,b and c ."

Statement 2. I picked numbers. a=6; b=2; c =3. This combination yields integer value. However, when a=6; b=4; c=3 this combination yields a fraction value. Hence I decided B is insufficient. And my answer was wrong.

Can someone explain this?
Question rephrased: Is a/bc an integer?

Statement 1: c=2.
No information about a and b.
Insufficient.

Statement 2: a=b+c.
a/bc = (b+c)/bc = b/bc + c/bc = 1/c + 1/b.
If b=2 and c=2, then 1/c + 1/b = 1/2 + 1/2 = 1.
No other combination will result in an integer value for 1/c + 1/b.
Since b and c must be distinct, 1/c + 1/b -- and thus a/bc -- cannot be an integer.
Sufficient.

The correct answer is B.

The values that you plugged in do not satisfy the condition that a = b+c:

You plugged in a=6, b=2, c=3:
6 = 2+3. Doesn't work.

You plugged in a=6, b=4, c=3.
6 = 4+3. Doesn't work.

The values that you plug in must satisfy the statement that you are evaluating.
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by rohu27 » Mon Apr 04, 2011 8:46 pm
jsasipriya wrote:If a, b , and c are positive distinct integers, is (a/b)/c an integer?

1. c=2
2. a=b+c

OA B

Explanation provided is:

"There is no combination that would allow divisibility into an integer for distinct integers such as a ,b and c ."

Statement 2. I picked numbers. a=6; b=2; c =3. This combination yields integer value. However, when a=6; b=4; c=3 this combination yields a fraction value. Hence I decided B is insufficient. And my answer was wrong.

Can someone explain this?
I guess you interpreted the Q wrong, ac/b? it is a/bc as Mitch pointed out.
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by Stuart@KaplanGMAT » Tue Apr 05, 2011 11:03 am
jsasipriya wrote:If a, b , and c are positive distinct integers, is (a/b)/c an integer?

2. a=b+c

Statement 2. I picked numbers. a=6; b=2; c =3. This combination yields integer value. However, when a=6; b=4; c=3 this combination yields a fraction value. Hence I decided B is insufficient. And my answer was wrong.

Can someone explain this?
Hi,

Mitch's explanation is spot on (as usual), I just want to focus on the specific mistake you made, since it's probably the most common way that test takers go wrong in data sufficiency.

When you evaluate a statement, you must take its truth for granted. In other words, the only numbers that exist are the ones that agree with the statement.

So, when you pick numbers in DS, there are two key steps to follow:

1) pick numbers that are "legal" (i.e. they follow the rules laid down in the question stem and the statement);

and then

2) plug those numbers back into the original question.

For this particular question, we have a governing rule from the stem (a, b and c are positive distinct integers) and, when looking at statement (1), the equation provided.

Accordingly, the numbers you chose are illegal - they violate the law set out in statement (1). Illegal numbers provide neither a yes nor a no answer to the question - they merely don't exist for the purposes of the question.

This principle can be illustrated with a much simpler example:

Q: what's the value of x?

(1) x = 3

Now, when evaluating statement (1), do you think "well, what if x is 4 or 5 or 6?" Of course not - it's obvious that those numbers are inconsequential. Similarly, for the question you posted, since we KNOW that:

a = b + c

we have to ignore a = 6, b = 3 and c = 2, since those numbers don't go along with the stated law.
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by jsasipriya » Tue Apr 05, 2011 4:09 pm
Thank you very much experts! Indeed, your explanation clarifies where I went wrong. I was too much into the problem that I ignored the parent condition "a=b+c"
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