MGMAT CAT 1: PS Problem 2

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 72
Joined: Sun Nov 03, 2013 9:29 am
Thanked: 1 times

MGMAT CAT 1: PS Problem 2

by josh80 » Mon Nov 11, 2013 11:57 am
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?

- 2
- 5
- 6
- 8
- 9

Ans. C
Source: — Problem Solving |

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Nov 11, 2013 4:15 pm
josh80 wrote:Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?

- 2
- 5
- 6
- 8
- 9
Here is a helpful formula for 3 overlapping groups:

T = A + B + C - (AB + AC + BC) - 2(ABC)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in A, everyone in B, and everyone in C:
Those in exactly 2 of the groups (AB+AC+BC) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (ABC) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
T = 59.
Poetry = 22.
History = 27.
Writing = 28.
Exactly 2 of the groups = 6.
Let x = the number in all 3 groups.

Plugging these values into the formula, we get:
59 = 22 + 27 + 28 - 6 - 2x
59 = 71 - 2x
x=6.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

User avatar
Master | Next Rank: 500 Posts
Posts: 490
Joined: Thu Jul 04, 2013 7:30 am
Location: Chennai, India
Thanked: 83 times
Followed by:5 members

by Uva@90 » Mon Nov 11, 2013 7:30 pm
josh80 wrote:Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?

- 2
- 5
- 6
- 8
- 9

Ans. C
Hi Josh80,
In question if they mention as EXACTLY 2- group Overlaps then you can use the below formula,
Total = A+B+C -(Sum of EXACTLY 2-group overlaps) -2*(All three) + Neither

From Question we know,
Total = 59, A= 22,B= 27, C = 28 and they said 6 students sign up of Exactly 2- groups and Neither will be O as they mentioned they should sign up for minimum one group.

Hence,
59 = 22+27+28 - 6 -2(All-Three) + 0
Solve it, you will get All-Three = 6

Answer is C

Hope it helps you.

Regards,
Uva.
Known is a drop Unknown is an Ocean

Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Tue Nov 12, 2013 1:58 am
Here's a reasoning that seems valid, but does not produce the right answer.
Can anyone understand firstly the method, then see where the error is, and why?

Consider these 3 lines (each represents an arbitrary club) and their relative positions:
(Ignore the dots; they are purely for spacing purposes)

A: ......_________
B: ______
C: ...____

Lines existing at the same time denote multiple membership.

Not drawn to scale, the length of each line is proportional to it's number of club members.

If not overlapping, we get::

A: .........._________
B: ______
C: ..........................____

Total length (A+B+C) = 77

Now break up club C and overlap 6 people between B and C:

A: ........._________
B: ______
C: ......__................__
Total length = A+B+C-6 = 77-6 = 71

To reduce this to 59, we need to overlap more:

A: ......_________
B: ______
C: ...____

71-59 = 12

Therefore answer = 12 people.

However, this would mean:
Club 22 = 6 + 12 + 4
Club 28 = 6 + 12 + 0 + 10
Club 27 = 0 + 12 + 0 + 0 + 15
6 + 12 + 4 + 10 + 15 = 47, not 59.

So why and where is the method flawed?