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There are 51 people in a certain community who belong to the local

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Veritas Prep GMAT Instructor
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New post Updated on: 24 Aug 2016, 04:20
2
15
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

63% (02:55) correct 37% (03:03) wrong based on 235 sessions

HideShow timer Statistics

Responding to a pm:

There are 51 people in a certain community who belong to the local golf club, tennis club or both clubs. If twice as many people belong to the tennis club as belong to the golf club, what could be the number of people who belong to both clubs?

1) 4
2) 7
3) 17
4) 21
5) 27

Total 51 people who belong to one or both clubs.
"if twice as many people belong to the tennis club as belong to the golf club" - this means if there are x people who belong to golf club, 2x belong to tennis club.
Say anyone who is in one club is not in the other i.e. there is no overlap.

Total = n(G) + n(T) - Both
51 = x + 2x - Both
Both = 3x - 51 = 3*(x - 17)
'Both' has to be a multiple of 3.

Only option (D) satisfies this condition.

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Originally posted by VeritasKarishma on 12 Aug 2016, 02:40.
Last edited by Bunuel on 24 Aug 2016, 04:20, edited 1 time in total.
Added the OA.
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New post 12 Aug 2016, 04:26
x -> Golf Club
2x -> Tennis Club
y -> both

x+2x-y = 51
3x - y =51

multiples of 3 greater than 51
54
57
60
63
66
69
72 - 21 = 51

IMO (D)
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New post 23 Aug 2016, 13:33
Why not E i.e. 27 which is also a multiple of 3 ?
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New post 24 Aug 2016, 02:15
VeritasPrepKarishma wrote:
Responding to a pm:

There are 51 people in a certain community who belong to the local golf club, tennis club or both clubs. If twice as many people belong to the tennis club as belong to the golf club, what could be the number of people who belong to both clubs?

1) 4
2) 7
3) 17
4) 21
5) 27

Total 51 people who belong to one or both clubs.
"if twice as many people belong to the tennis club as belong to the golf club" - this means if there are x people who belong to golf club, 2x belong to tennis club.
Say anyone who is in one club is not in the other i.e. there is no overlap.

Total = n(G) + n(T) - Both
51 = x + 2x - Both
Both = 3x - 51 = 3*(x - 17)
'Both' has to be a multiple of 3.

Only option (D) satisfies this condition.


VeritasPrepKarishma : Why not its E..? 27 is also a multiple of 3..

As Both = 3x - 51

if i assume both = 27 = 3(26) - 51 ? Am i doing something wrong ?
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Re: There are 51 people in a certain community who belong to the local  [#permalink]

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New post 24 Aug 2016, 04:09
VeritasPrepKarishma wrote:
Responding to a pm:

There are 51 people in a certain community who belong to the local golf club, tennis club or both clubs. If twice as many people belong to the tennis club as belong to the golf club, what could be the number of people who belong to both clubs?

1) 4
2) 7
3) 17
4) 21
5) 27

Total 51 people who belong to one or both clubs.
"if twice as many people belong to the tennis club as belong to the golf club" - this means if there are x people who belong to golf club, 2x belong to tennis club.
Say anyone who is in one club is not in the other i.e. there is no overlap.

Total = n(G) + n(T) - Both
51 = x + 2x - Both
Both = 3x - 51 = 3*(x - 17)
'Both' has to be a multiple of 3.

Only option (D) satisfies this condition.



The key point in solving this question is first of all, the number of people belonging must be a non negative integer. This in turn means that if we substitute the answer choice in the equation, the resulting number must be an integer. Only Choice D suffices this condition.

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New post 24 Aug 2016, 04:38
thapliya wrote:
Why not E i.e. 27 which is also a multiple of 3 ?






27 also can be the answer.
16-8-27
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New post 25 Sep 2016, 09:12
3
azamaka wrote:
thapliya wrote:
Why not E i.e. 27 which is also a multiple of 3 ?






27 also can be the answer.
16-8-27


It is given that "twice as many people belong to the tennis club as belong to the golf club".
If # of people in golf club = x
Then # of people in tennis club = 2x.
2x<51 => x <26.5

Since the number of people in both clubs can not be greater then number of people in either club hence:
Answer less then 26.5
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New post 16 Nov 2016, 11:01
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1
thapliya wrote:
Why not E i.e. 27 which is also a multiple of 3 ?


Responding to a pm:

Let's find the range of Both and hence the range of x.

Minimum value of Both: Say there is no overlap in the two sets.

51 = x + 2x
x = 17
In this case Both = 0

Maximum value of Both: Say there is maximum overlap between the two sets. One set has x so it could be a subset of the set 2x.
51 = 2x
But x cannot be a decimal so (x-1) could be the overlap (= Both) and there could be 1 in only the Golf club.
51 = 2x + x - (x - 1)
x = 25
In this case Both = 25 - 1 = 24

So Both lies between 0 and 24.
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New post 13 Aug 2017, 23:07
VeritasPrepKarishma wrote:

There are 51 people in a certain community who belong to the local golf club, tennis club or both clubs. If twice as many people belong to the tennis club as belong to the golf club, what could be the number of people who belong to both clubs?

1) 4
2) 7
3) 17
4) 21
5) 27



Let x be the number of people who belongs to Golf club. So, 2x is the number of people who belongs to Tennis club.
Let y be the people who belongs to both clubs.

So, (x-y)+(2x-y) +y = 51
y = 3x-51......................(i)

So, y must be a multiple of 3.
We can eliminate option A,B,C

Also, Since total number of people in a club can't be greater than 51
So, 2x <51
x<26

Now we check y = 21 and 27 in eq (i)
At y = 21, x = (21+51)/3 = 24

At y = 27, x = (27+51)/3 = 26 (Not possible)

So, y = 21 is correct ..

Answer D
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New post 18 Aug 2017, 09:06
Quote:

There are 51 people in a certain community who belong to the local golf club, tennis club or both clubs. If twice as many people belong to the tennis club as belong to the golf club, what could be the number of people who belong to both clubs?

1) 4
2) 7
3) 17
4) 21
5) 27


We can let the number of people who belong to the golf club = x, making people who belong to the tennis club = 2x, and the number who belong to both = b. Thus:

51 = x + 2x - b

Note that we must subtract b from the right side of the equation because we have double-counted the “b” individuals (those who belong to both clubs) as members of both the golf club and the tennis club.

51 = 3x - b

51 + b = 3x

(51 + b)/3 = x

So (51 + b) must be a multiple of 3. Since 51 is a multiple of 3, b must also be a multiple of 3. Of our answer choices, both D and E are multiples of 3, so let’s test each one.

D) If b = 21, then:

(51 + 21)/3 = x

24 = x

So 2x = 48.

E) If b = 27, then:

(51 + 27)/3 = x

26 = x

So 2x = 52.

However, we see that answer choice E can’t be correct, since that would mean the number of people belonging to the tennis club, 52, was greater than the given total of 51 people. Thus, the correct answer choice is D.

Answer: D
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New post 18 Aug 2017, 11:05
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If anyone was curious about the source of this problem, it's from one of my high-level problem sets (#55 in my 2nd problem set). I'd definitely consider it a very hard problem, even if at a glance it might appear simple.

The TC has twice as many members as the GC. When we count the number of people in the TC, or in the GC, we're also counting people in both clubs. So if the GC has x members, then we know the TC has 2x members. If we have b people in both clubs, a Venn diagram would look like this:

only in TC: 2x - b
in both TC and GC: b
only in GC: x - b

These add to 51, so

3x - b = 51

We care about b, so we should isolate b:

b = 51 - 3x

This equation means: "b" and "51 - 3x" are the exact same number. So, since 3 is a factor of "51 - 3x", 3 must be a factor of b, so b must be divisible by 3. Only two answer choices are candidates: 21 and 27. You don't need to use any algebra to rule out 27. If 27 people were in both clubs, all 27 of those people are in the GC, so there would naturally be at least 27 people in the GC. Since the TC has twice as many members as the GC, there would then need to be at least 54 people in the TC. But we don't have 54 people (there are only 51 people in total), so 27 is impossible, and the only possible answer is 21.
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