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OG#93

by tapanmittal » Sun Jul 19, 2015 8:43 am
Is the number of members of Club X > the number of members of club Y?

1)Of the members of Club X,20% are also members of club Y
2) Of the members of Club Y,30% are also members of club X

OA-C

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by Brent@GMATPrepNow » Sun Jul 19, 2015 8:55 am
Is the number of members of Club X greater than the number of members of Club Y?

1) Of the members of Club X, 20 percent are also members of Club Y.
2) Of the members of Club Y, 30 percent are also members of Club X.
One approach is to use the Double Matrix Method. This technique can be used for most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of people, and the two characteristics are:
- member of Club X or not a member of Club X
- member of Club Y or not a member of Club Y

So, we can set up our diagram as follows:
Image

Since we're not told any populations, let's assign some variables.
Let X = # of Club X members
Let Y = # of Club Y members
So, we now have a diagram that looks like this:
Image

Okay, now let's solve the question...

Target question: Is X greater than Y?

Statement 1: Of the members of Club X, 20 percent are also members of Club Y.
If X people are in Club X, then the number of THESE people whose are ALSO in Club Y = 20% of X (aka 0.2X)
So, let's add this to our diagram:
Image

Does this provide enough information to determine whether or not X is greater than Y?
No. The reason is that we have no information about the bottom-left box:
Image

Since there are no restrictions on the bottom-left box, there are many possible ways to complete the diagram so that we get CONFLICTING answers to the target question. Here are two:
Case a:
Image
In this case X = 10 and Y = 2, which means X is GREATER THAN Y

Case b:
Image
In this case X = 10 and Y = 32, which means X is LESS THAN Y

Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Of the members of Club Y, 30 percent are also members of Club X.
If Y people are in Club Y, then the number of THESE people whose are ALSO in Club X = 30% of Y (aka 0.3Y)
So, let's add this to our diagram:
Image

Using logic similar to the logic we used in statement 1, we can conclude that statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
When we combine the information we get TWO POSSIBLE expressions for the top-left corner:
Image
So, these two expressions must be equal.
In other words, 0.2X = 0.3Y
Divide both sides by 0.2 to get: X = (0.3/0.2)Y
Simplify to get: X = 1.5Y
Since X and Y must be positive integers, the expression X = 1.5Y tells us that X is 1.5 TIMES as big as Y
In other words, X is definitely greater than Y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent


----------------------------
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by GMATGuruNY » Sun Jul 19, 2015 9:07 am
Is the number of members of Club X greater than the number of members of Club Y ?

(1) Of the members of Club X, 20 percent are also members of Club Y.
(2) Of the members of Club Y, 30 percent are also members of Club X.
This is a question about OVERLAPPING groups.
Each statement offers information about the OVERLAP -- the people in BOTH X AND Y.
Let B = the members who are in both groups.

Statement 1:
20% of X is also in Y.
In other words, 20% of X is in BOTH groups:
.2X = B.
No information about Y.
INSUFFICIENT.

Statement 2:
30% of Y is also in X.
In other words, 30% of Y is in BOTH groups:
.3Y = B.
No information about X.
INSUFFICIENT.

Statements 1 and 2 together:
.2X = B.
.3Y = B.
Thus:
.2X = .3Y.
X = (3/2)Y.
Thus, X>Y.
SUFFICIENT.

The correct answer is C.
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by [email protected] » Sun Jul 19, 2015 12:28 pm
HI tapanmittal,

This DS question can be solved by TESTing VALUES.

We're asked if the number of members of Club X is greater than the number of members of Club Y? This is a YES/NO question. In these sorts of situations, it's common for some members to belong to BOTH Clubs, so we have to keep careful track of the numbers and possibilities....

Fact 1: 20% of the members of Club X are ALSO members of Club Y

IF...
Club X has 100 members, then 20 of those members ALSO belong to Club Y.
IF Club Y has 0 unique members, then the answer to the question is YES.
IF Club Y as 1,000 unique members, then the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact 2: 30% of the members of Club Y are ALSO members of Club X

This Fact offers the same general logic as Fact 1 (above). Without knowing the number of unique members in Club X, the answer to the question could be either YES or NO.
Fact 2 is INSUFFICIENT

Combined, we know...
20% of the members of Club X are ALSO members of Club Y
30% of the members of Club Y are ALSO members of Club X
These specific members are the SAME PEOPLE...

This means that .2(X) = .3(Y)

2X = 3Y
X = (3/2)(Y)

This means that X MUST be greater than Y, so the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer: C

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by nikhilgmat31 » Wed Jul 22, 2015 12:26 am
Great Explanation.
Combined, we know...
20% of the members of Club X are ALSO members of Club Y
30% of the members of Club Y are ALSO members of Club X
These specific members are the SAME PEOPLE...

This means that .2(X) = .3(Y)

2X = 3Y
X = (3/2)(Y)


However I solved it in following way :-

lets say X has 100 members
out of which 80 are X members alone
out of which 20 are both X & Y members


lets say Y has 100 members
out of which 70 are Y members alone
out of which 30 are both X & Y members

so X has total of 80 + 30 = 110 members

so Y has total of 70 + 20 = 90 members

X > Y

Answer is C

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by gmatbeater1989 » Sat Oct 10, 2015 8:56 am
Can I use Group 1 + Group 2 + Neither - Both = Total to solve this?

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by [email protected] » Sat Oct 10, 2015 9:17 am
Hi gmatbeater1989,

The Overlapping Sets Formula tends to work 'best' when you're dealing with actual values and you have 4 of the 5 'pieces' (and are trying to calculate the 5th piece). This question isn't designed in that way, so while you might be able to get to the solution with that Formula, you'd really have to think about how to make it all work. You can see from the various solutions in this thread that there are a variety of ways to go about dealing with this prompt, so you might be better served here with some basic Tactics and Algebra.

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by gmatbeater1989 » Mon Oct 12, 2015 8:58 am
[email protected] wrote:Hi gmatbeater1989,

The Overlapping Sets Formula tends to work 'best' when you're dealing with actual values and you have 4 of the 5 'pieces' (and are trying to calculate the 5th piece). This question isn't designed in that way, so while you might be able to get to the solution with that Formula, you'd really have to think about how to make it all work. You can see from the various solutions in this thread that there are a variety of ways to go about dealing with this prompt, so you might be better served here with some basic Tactics and Algebra.

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Thanks Rich that helps.
What about the double matrix method? When does that work and when doesn't it work.

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by [email protected] » Mon Oct 12, 2015 9:04 am
Hi gmatbeater1989,

The Tic-Tac-Toe Board (or Double Matrix or Grid or whatever you want to call it) 'works' on almost all Overlapping Sets questions (the primary exception being the rare 3-group prompt). A Venn Diagram (or the Formula that it's based on) only deals with 5 of the 'pieces', while the Tic-Tac-Toe Board includes those 5 pieces AND 4 others. It's a far more comprehensive approach to answering those types of prompts (and the 'work' involved is pretty straight-forward, so it's a great 'default' approach when dealing with these types of prompts).

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by gmatbeater1989 » Mon Oct 12, 2015 9:35 am
[email protected] wrote:Hi gmatbeater1989,

The Tic-Tac-Toe Board (or Double Matrix or Grid or whatever you want to call it) 'works' on almost all Overlapping Sets questions (the primary exception being the rare 3-group prompt). A Venn Diagram (or the Formula that it's based on) only deals with 5 of the 'pieces', while the Tic-Tac-Toe Board includes those 5 pieces AND 4 others. It's a far more comprehensive approach to answering those types of prompts (and the 'work' involved is pretty straight-forward, so it's a great 'default' approach when dealing with these types of prompts).

GMAT assassins aren't born, they're made,
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Thanks Rich!

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by Max@Math Revolution » Wed Oct 21, 2015 11:33 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the number of members of Club X > the number of members of club Y?

1)Of the members of Club X,20% are also members of club Y
2) Of the members of Club Y,30% are also members of club X

In the original condition, there are only club X, olny club Y, both club X and Y, so 3 variables are possible. This means we need 3 equations, whereas only 2 equations are given from the conditions. Therefore, there is high chance (E) will be our answer.
Combining the 2 conditions, if we let the number of members of Club X=x, the number of members of Club Y=y, we can get 0.2x=0.3y --> 2x=3y, so x>y, and the answer to the question always becomes 'yes', meaning this is a sufficient set of conditions and the answer becomes (C).

For cases where we need 3 more equation, such as original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

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