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This is a badly-worded exercise, as well as logically incorrect in its phrasing. I had to skip this exercise during the diagnostic test.

Quote:

GMAT Diagnostic Test Question 24

Among 60 members of a club, 6p players play soccer, 11p players play tennis, 8p players play golf and 2p players play none of the games. If p players play all of the games, how many players play only one game?

(1) The number of players who play soccer and golf is half of the players who play each of the rest two games (2) p = 3

The OA is fine, but I had to read it to actually understand the question.

All the problems are in this statement "The number of players who play soccer and golf is half of the players who play each of the rest two games " which is completely meaningless.

1) Reading the OA, what you meant to say is this: "The number of players who only play soccer and golf..." Why? For the same reason that "6p players play soccer" means "6p players play soccer, or soccer and another sport, or all 3 sports". So the meaning of the word "only" is absolutely critical if you want the OA to actually be correct.

2) As other posters pointed out, "players who play each of the rest two games" has no meaning in English (or even if translated in any other language).

3) Grammar parallelism rules need to be followed. So compare a "number of a player" with another "number of players" not with just "players".

I would write statement as (1):

"The number of players who play soccer and golf but not tennis is half the number of players who play any other combination of two sports."

GMAT Diagnostic Test Question 24 Field: word problems (overlapping sets) Difficulty: 750

Rating:

Among 60 members of a club, 6p players play soccer, 11p players play tennis, 8p players play golf and 2p players play none of the games. If p players play all of the games, how many players play only one game?

(1) The number of players who play soccer and golf but not tennis is half the number of players who play any other combination of two sports (2) p = 3

A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient D. EACH statement ALONE is sufficient E. Statements (1) and (2) TOGETHER are NOT sufficient

I have a question. The statement didn't tell us about the people who play two types of sport here. Thus, can we still apply the equation here? Thank you!

Q)Among 60 members of a club, 6p players play soccer, 11p players play tennis, 8p players play golf and 2p players play none of the games. If p players play all of the games, how many players play only one game? (1) The number of players who play soccer and golf but not tennis is half the number of players who play any other combination of two sports (2) p = 3

I have read through this and am EXTREMELY confused and believe it to be wrong, so someone please straighten my logic out if it is incorrect! Obviously the answer is B, but the actual mathematical answer is bugging many number crunchers such as myself. And we can't seem to get the story straight.

Statement 1 says: The number of players who play soccer AND golf but not tennis is half the number of players who play any other combination of two sports.

Translation of statement 1: 1/2(SG) = GT = ST. In words that is the people who play BOTH soccer AND golf are half the number of people who play any other combo of TWO sports. Thus, the number of people who play only two sports (not 3 and not 1) is 5*SG. You don't need this, but it is as far as you can take it...thus statement 1 is insufficient.

So, I come up with the following:

# of individuals = 6p+11p+8p - (people playing 2 sports) + (people playing all 3 sports) [forget for a second the players who play none as they are irrelevant to this count]. This logic right here is where the first source of confusion seems.

6p+11p+8p is the total # of players in each sport INCLUDING those who play 2 sports or 3 sports (thus accounting for your # being over 60). Now, if you take away the extra player counts for individuals playing 2 sports and individuals playing all 3 sports...then that is logically how you get that this is the # of individuals playing at least one sport.

Next you find out that p = 3.

So, 75 - D - 3 = 54 (or you can add back on the 6 for those who do not play any sport but are a member...in this case 60-6 = 54). Here "D" is for doubles i.e. those playing 2 sports.

Next you have: 72 - D = 54

Which yields: -D = -18 or D = 18

So, it can be solved, but you do not have a whole #. You don't have to...this isn't real data! But, you do have a solution possible with the second statement, thus the answer is B.

I agree that statement 2) alone, is sufficient. However, for reasons which (as far as I've noticed) have been overlooked, statement 1) should also be sufficient:

Consider that simply by the givens of the problem, we can narrow down possible values of p, which is necessarily a positive integer. Since 11p players play tennis, we know that 11p is less than or equal to 60. We immediately know that p is a natural number no greater than 5.

Now, we can also build upon our work with statement 2). Supposing we use the following labels, based on the work of flyingbunny:

suppose there are: s: players only play soccer t: only tennis g: only golf x: only play soccer and golf y: only play soccer and tennis z: only play tennis and golf p: play all the three.

we can have: a) s+t+g+x+y+z+p+2p=60 ==> s+t+g+x+y+z=60-3p ==>> a1) 2*(s+t+g)+2*(x+y+z)=120-6p b) s+x+y+p=6p c) t+y+z+p=11p d) g+x+z+p=8p

from b) c) d), we have bcd) s+t+g+2*(x+y+z)=22p

From this work, we can double the bcd) and arrive at 2s+2t+2g+4*(x+y+z)=44p. We can subtract the a1) equation from this, to get 2*(x+y+z)=44p-(120-6p)=50p-120, which simplifies to a2) x+y+z=25p-60.

However, statement 1) tells us that half the players who only play soccer and golf but not tennis is half the number of players who play any other combination of two sports. In other words x=(1/2)*(z+y), or 2x=y+z. Plugging this into a2) gives us 3x=25p-60. From the beginning, we narrowed p down to 5 options, (1, 2, 3, 4, or 5). We can immediately eliminate 1 and 2 as solutions for p. Note also, that 4 and 5 are not possible solutions for p, since 125-60 and 100-60 are both not divisible by 3. In fact, the only solution for p is 3. Essentially statement 1) implies statement 2), and is therefore, also sufficient by itself.

The fact which I believe was overlooked is that when we are told that one group is half the sum of the other two groups, we can deduce that the sum of all 3 groups is of a size divisible by 3. This, coupled with basic eliminations regarding possible solutions for p, limits our solution to one possible answer.

Please comment if there is something I have missed, or erroneously concluded. I'd like to see this post addressed, whether or not I am correct.

The below explanation is very nice. Though most others got the answer B (by considering 6p as those who play ONLY soccer), it would have been incorrect if its a problem solving question.

Venn diagrams make it easy. +1 kudos

whiplash2411 wrote:

Since I love Venn Diagrams, I am going to use one to solve this. (Hussain15 - Bear with me :D)

Here's a Venn Diagram of the situation where I've represented unknown quantities with variables a, b, c, x, y, and z respectively.

Attachment:

Club Sports.jpg

To find: \(a+b+c\)

Given Information:

Total\(= a+b+c+x+y+z+p+2p = 60\)

\(a+b+c+x+y+z = 60 - 3p\) (1)

Total soccer = a + x + y + p = 6p

\(x+y = 5p - a\) (2)

Total tennis\(= b + x + z + p = 11 p\)

\(x+z = 10p - b\) (3)

Total golf \(= c + y + z + p = 8p\)

\(y+z = 7p - c\) (4)

Adding (2), (3) and (4) and substituting into (1) we get:

\(22p - 2(x+y+z) = (a+b+c)\)

\(22p - 2(x+y+z) + (x+y+z) = 60 - 3p\)

\((x+y+z) = 25p - 60\)

\((a+b+c) = 22p - 25p + 60 = 60 - 3p\)

Statement 1:

\(y = \frac{(x+z)}{2}\)

Insufficient, because this doesn't tell us anything.

Statement 2

p = 3

Substituting p = 3 in the equation obtained for (a+b+c) we get a numeric answer. Hence sufficient. So, the answer is B.

The below explanation is very nice. Though most others got the answer B (by considering 6p as those who play ONLY soccer), it would have been incorrect if its a problem solving question.

Venn diagrams make it easy. +1 kudos

whiplash2411 wrote:

Since I love Venn Diagrams, I am going to use one to solve this. (Hussain15 - Bear with me :D)

Here's a Venn Diagram of the situation where I've represented unknown quantities with variables a, b, c, x, y, and z respectively.

Attachment:

Club Sports.jpg

To find: \(a+b+c\)

Given Information:

Total\(= a+b+c+x+y+z+p+2p = 60\)

\(a+b+c+x+y+z = 60 - 3p\) (1)

Total soccer = a + x + y + p = 6p

\(x+y = 5p - a\) (2)

Total tennis\(= b + x + z + p = 11 p\)

\(x+z = 10p - b\) (3)

Total golf \(= c + y + z + p = 8p\)

\(y+z = 7p - c\) (4)

Adding (2), (3) and (4) and substituting into (1) we get:

\(22p - 2(x+y+z) = (a+b+c)\)

\(22p - 2(x+y+z) + (x+y+z) = 60 - 3p\)

\((x+y+z) = 25p - 60\)

\((a+b+c) = 22p - 25p + 60 = 60 - 3p\)

Statement 1:

\(y = \frac{(x+z)}{2}\)

Insufficient, because this doesn't tell us anything.

Statement 2

p = 3

Substituting p = 3 in the equation obtained for (a+b+c) we get a numeric answer. Hence sufficient. So, the answer is B.

Hope this helps.

Again, the work showing statement 2 to be sufficient is correct. But I take exception to the following:

\(y = \frac{(x+z)}{2}\) Insufficient, because this doesn't tell us anything.

This does, in fact, tell us something. It tells us that 2y=x+z, which, in turn, tells us that 3y=25p-60. Therefore, 25p=60+3y, from which we can deduce that 25p is a multiple of 3. We also know, since 11p players play tennis, yet there are only 60 people in the club, that p, a natural number, cannot be greater than 5. There is only one solution for p which is not greater than 5 and for which 25p is a multiple of 3 - namely, p=3. Therefore, statement 1) is also sufficient

Please check my logic, and if possible, provide a counterexample where p is not 3, yet all the conditions in statement 1) still hold.

Didn't like the explanations above. So here is an alternative.

From the Venn Diagram we create the following table None: 2p Only 1 sport: (S+T+G) = X Two Sports: (A+B+C) Three Sports: p Total = 60 Then equation one is: 60 = 2p + (S+T+G) + (A+B+C) + p

Now, we know that (11p+6p+8p)... 25p = (S+T+G) + 2(A+B+C) + 3p Simplyfing we obtain equation two: 22p = (S+T+G) + 2(A+B+C)

We need to find X = (S+T+G) using those two equations.

(1) 2A=B+C Therefore we have 60 = 2p + X + (3A) + p and 22p = X+ 6A

NOT SUFFICIENT

(2) p=3 Using equation 1 we have 60 = 6 + (S+T+G) + (A+B+C) + 3 51 = X + (A+B+C)

Using equation 3 we have 66 = X + 2(A+B+C)

Two equations with two unknowns: X and (A+B+C). Can be solved. Therefore SUFFICIENT. Answer B is the correct answer. Just for completeness let's find the number substracting equation 1 to equation 2. 15 = (A+B+C) Therefore X = 36

leandrobrufman (or anyone else who might be reading this):

I have yet to see my specific query addressed regarding the first statement. Everyone on this thread seems to be in agreement that the statement p=3 is sufficient by itself.

However, the other statement, that "(1) The number of players who play soccer and golf but not tennis is half the number of players who play any other combination of two sports" should also be sufficient.

In this case, I will cite the work of leandrobrufman. The conclusion that most of us come to (some of us use different variables, but the idea is identical every time) is:

" 1) 2A=B+C Therefore we have 60 = 2p + X + (3A) + p and 22p = X+ 6A

NOT SUFFICIENT"

I disagree that that information is insufficient. Here is why: let's take the two equations within the quotation marks above, and simplify as much as possible.

We end up with 60=3p+X+3A=3p+(X+6A)-3A=25p-3A. In other words, 60+3A=25p. Therefore, p MUST BE A MULTIPLE OF 3.

As I mentioned in an earlier post, p has to be less than 6, because there are 60 people in total, and 11p of them play tennis. So p MUST BE 3.

What we have here are two statements, the first of which, if true, necessarily implies the that the second is true.

Since we all agree that the second statement is sufficient, we now can conclude that the first one is.

Statement 1) is actually meaningful because it tells us that p is a multiple of 3, which narrows down the value of p to one possibility.

Will somebody please address this, perhaps by finding a counterexample?

leandrobrufman (or anyone else who might be reading this):

I have yet to see my specific query addressed regarding the first statement. Everyone on this thread seems to be in agreement that the statement p=3 is sufficient by itself.

However, the other statement, that "(1) The number of players who play soccer and golf but not tennis is half the number of players who play any other combination of two sports" should also be sufficient.

In this case, I will cite the work of leandrobrufman. The conclusion that most of us come to (some of us use different variables, but the idea is identical every time) is:

" 1) 2A=B+C Therefore we have 60 = 2p + X + (3A) + p and 22p = X+ 6A

NOT SUFFICIENT"

I disagree that that information is insufficient. Here is why: let's take the two equations within the quotation marks above, and simplify as much as possible.

We end up with 60=3p+X+3A=3p+(X+6A)-3A=25p-3A. In other words, 60+3A=25p. Therefore, p MUST BE A MULTIPLE OF 3.

As I mentioned in an earlier post, p has to be less than 6, because there are 60 people in total, and 11p of them play tennis. So p MUST BE 3.

What we have here are two statements, the first of which, if true, necessarily implies the that the second is true.

Since we all agree that the second statement is sufficient, we now can conclude that the first one is.

Statement 1) is actually meaningful because it tells us that p is a multiple of 3, which narrows down the value of p to one possibility.

Will somebody please address this, perhaps by finding a counterexample?

You are right. Well done! OA should be changed to reflect that. Both statements are sufficient. Indeed, even if statement 1 provides one equation with two unknowns, we still have other information in the useful in the question stem (p is an integer, 0<p<6), and as you showed, statement 1 requires p to be a multiple of 3.

leandrobrufman (or anyone else who might be reading this):

I have yet to see my specific query addressed regarding the first statement. Everyone on this thread seems to be in agreement that the statement p=3 is sufficient by itself.

However, the other statement, that "(1) The number of players who play soccer and golf but not tennis is half the number of players who play any other combination of two sports" should also be sufficient.

In this case, I will cite the work of leandrobrufman. The conclusion that most of us come to (some of us use different variables, but the idea is identical every time) is:

" 1) 2A=B+C Therefore we have 60 = 2p + X + (3A) + p and 22p = X+ 6A

NOT SUFFICIENT"

I disagree that that information is insufficient. Here is why: let's take the two equations within the quotation marks above, and simplify as much as possible.

We end up with 60=3p+X+3A=3p+(X+6A)-3A=25p-3A. In other words, 60+3A=25p. Therefore, p MUST BE A MULTIPLE OF 3.

As I mentioned in an earlier post, p has to be less than 6, because there are 60 people in total, and 11p of them play tennis. So p MUST BE 3.

What we have here are two statements, the first of which, if true, necessarily implies the that the second is true.

Since we all agree that the second statement is sufficient, we now can conclude that the first one is.

Statement 1) is actually meaningful because it tells us that p is a multiple of 3, which narrows down the value of p to one possibility.

Will somebody please address this, perhaps by finding a counterexample?

You are right. Well done! OA should be changed to reflect that. Both statements are sufficient. Indeed, even if statement 1 provides one equation with two unknowns, we still have other information in the useful in the question stem (p is an integer, 0<p<6), and as you showed, statement 1 requires p to be a multiple of 3.

No need for counterexample. I came to the same conclusion, solved similarly. We should bare in mind that equations involving variables who can take on only integer values, sometimes can have uniques solution, as in the above case (here, due to multiple constraints).
_________________

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