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og-ds 125

by bblast » Tue Jun 28, 2011 10:40 am
IF # REPRESENTS EITHER +,- OR *.
is k#(l+m) = (k#l)+(k#m) for all numbers k,l and m.

1>k#1 is not = 1#k for some numbers k.
2># is subtraction.


[spoiler]obviously both statements mean we are talking about subtraction. So considering k,l and m, all 3 equal to 0 and other cases answer should be E. But OG says- D. ??[/spoiler]
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by dabral » Tue Jun 28, 2011 11:04 am
Here is a video solution to this Official Guide Data Sufficiency problem:

https://www.gmatquantum.com/list-of-vide ... ds125.html

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by krishnasty » Tue Jun 28, 2011 7:53 pm
bblast wrote:IF # REPRESENTS EITHER +,- OR *.
is k#(l+m) = (k#l)+(k#m) for all numbers k,l and m.

1>k#1 is not = 1#k for some numbers k.
2># is subtraction.


[spoiler]obviously both statements mean we are talking about subtraction. So considering k,l and m, all 3 equal to 0 and other cases answer should be E. But OG says- D. ??[/spoiler]
I agree...the ans should be E.
why the ans is D? both the statements cleary talk about subtraction.
Please explain..
P.S. - cant see the video. Hence, someone explaining on the forum would be really helpful
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Appreciation in thanks please!!
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by bblast » Tue Jun 28, 2011 9:17 pm
dabral wrote:Here is a video solution to this Official Guide Data Sufficiency problem:

https://www.gmatquantum.com/list-of-vide ... ds125.html

Dabral

Hi Dabral, the explanation in the video was really good. But again you forgot to consider the case when all 3 numbers are 0. So my doubt still stands !!!
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by Frankenstein » Tue Jun 28, 2011 9:25 pm
bblast wrote:
Hi Dabral, the explanation in the video was really good. But again you forgot to consider the case when all 3 numbers are 0. So my doubt still stands !!!
Hi,
As you are convinced that each of the statements conclude that subtraction is the operations, I will continue from this.
Is k-(l+m) = (k-l)+(k-m)?
i.e. Is k = 2k for all k?
The answer is simply NO.
So, isn't sufficient?
If you are not convinced, consider an analogy. Is every prime number odd? What will be your answer? Is it 'NO' or Can't say?
Cheers!

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by dabral » Tue Jun 28, 2011 9:58 pm
It is a confusing question so I understand your difficulty. Let me try another way.

Let's look at statement 2 that says the operation stands for subtraction. The question is asking if k*(l+m)= (k*m) + (k*l) for all values of k,l,m? Meaning, if you selected k=0, l=0, and m=0 then yes the above relationship holds true, but the question is asking if that relationship is true for all values of k, l, and m, not just one particular value of k, l, and m. The answer is a definite NO to that question.

Contrast this with this modified question.

Is k*(l+m)= (k*m) + (k*l) where *=+, -, or x, and k, l, and m are integers?

2)* = subtraction

In this case, the statement is insufficient because when k=l=m=0, the answer to the question is Yes, and if k=3, l=2, and m=1, then the answer to the question is a NO.

I hope this makes sense.

And yes all the video solutions to the official guide problems can be accessed for free.

Cheers,
Dabral

p.s: Frankenstein gives a good analogy, let me try to give a similar example.

Q#1: Is n^2 = n ?
1) n is an integer.

Here the statement is insufficient because if n=0 or n=1, then the answer to the question is YES, but if n=2, 3, or any other number the answer is NO.
We don't have a definite outcome to the answer of the question.

If instead the question was:
Q#1: Is n^2 = n for all values of n?
1) n is an integer.

Here the statement is sufficient, because the relationship n^2=n does not hold true for all values of n, even though it does hold true for some values, namely n=0 or n=1, and we have a definite NO as an answer to the question.
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by Ian Stewart » Tue Jun 28, 2011 11:13 pm
dabral wrote:
If instead the question was:
Q#1: Is n^2 = n for all values of n?
1) n is an integer.
That question doesn't make sense as a DS question. Clearly n^2 is not equal to n "for all values of n", so you wouldn't need any statements to answer the question. Statement 1 then tells us we're not concerned with "all values of n", but rather only with integer values of n, so the statement contradicts the stem. I think you mean to ask something like:

Is n^2 = n for all values of n in set S?
1. Set S contains all of the integers from 0 to 20 inclusive.

in which case, Statement 1 is sufficient, since we know n^2 = n is *not* true for *all* values of n in our set, only for some values.

In case there's still any confusion about the question in the original post:

IF # REPRESENTS EITHER +,- OR *. , is k#(l+m) = (k#l)+(k#m) for all numbers k,l and m.

1>k#1 is not = 1#k for some numbers k.
2># is subtraction.

I think everyone has agreed that each statement tells you that # is subtraction. The question then becomes whether it is true that

k-(l+m) = (k-l) + (k-m)
k = 2k

for *all* numbers k, l and m. Clearly k=2k is *not* true for *all* values of k, which is what the question asks, so we know the answer to the question is 'no', and each statement is sufficient. It's irrelevant that k = 2k is true for some value of k (and it is, for k=0), since we're asked whether it's true for *every* possible value of k.

This is the only DS question in the OG, incidentally, where you have sufficient information to get a 'no' answer to the question.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by dabral » Tue Jun 28, 2011 11:24 pm
Hey Ian,

Thanks for making the correction, the GMAT will never ask a question where the statement is superfluous. I like your modification to the question, that is a much better way to clarify the confusion that the original poster had.

Cheers,
Dabral
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by bblast » Tue Jun 28, 2011 11:43 pm
Thanks Frank, Ian and Dabral. As the last line in Ian's post mentions. There is a big takeaway in this question which no other question in OG-DS section has.

Is "XYZ" true for all numbers k,l and m ? We are able to answer here- "yes, its true for k,l,m(0,0,0) but its not true for any other value of k,l,m"

So even when we have multiple solutions still the statements are sufficient, with the answer- a big NO - "XYZ is not true in all cases".


Lemme know if I still missed something.
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