Last visit was: 21 Dec 2024, 23:18 It is currently 21 Dec 2024, 23:18
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

The 20 people at a party are divided into n mutually exclusi

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

avatar
HarveyKlaus
Intern
Intern
Joined: 22 Aug 2016
Posts: 33
Own Kudos [?]: 80 [19]
Given Kudos: 0
Send PM
The 20 people at a party are divided into n mutually exclusi [#permalink] New post   Updated on: 14 Nov 2018, 13:18
3
1
15
Bookmarks
00:00

Question Stats:

51% (01:44) correct 48% (01:48) wrong based on 310 sessions

Hide Show timer Statistics

The 20 people at a party are divided into n mutually exclusive groups in such a way that the number of people in any group does not exceed the number in any other group by more than 1.

Quantity A
Quantity B
The value of n if at least one of the groups consists of 3 people
6


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
ShowHide Answer
Official Answer
D

Originally posted by HarveyKlaus on 24 Mar 2017, 14:48.
Last edited by Carcass on 14 Nov 2018, 13:18, edited 1 time in total.
Edited by Carcass
Most Helpful Expert Reply
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 471 [6]
Given Kudos: 0
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 18:38
5
1
Expert Reply
HarveyKlaus wrote:
The 20 people at a party are divided into n mutually exclusive groups in such a way that the number of people in any group does not exceed the number in any other group by more than 1.

Quantity A
Quantity B
The value of n if at least one of the groups consists of 3 people
6


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



Question is basically asking us to distribute 20 into groups with each having a difference of max 1 with another.
What does it tell us that all the groups will have the strength within two consecutive numbers for example x and x+1.

Now they have given a value as 3, so different ways it can have is - a combinations of 2 and 3, or a combinations of 3 and 4.

1) Max possible value of n
when most of them have 2. since one of them has 3 => so 20-3=17 left.
Now we cannot have all of them as 2 since we will be left with 17-2*8=1 in the end and we cannot have 1..It has to be only 2 and 3
so another group will have 3, 17-3=14..
the remaining 14 can be divided in 14/2 = 7 group
so MAX is 1+1+7=9.....Here A>B

2) Min possible value of n
when most of them have 4. since one of them has 3 => so 20-3=17 left.
Now we cannot have all of them as 4 since we will be left with 17-4*4=1 in the end and we cannot have 1..It has to be only 4 and 3
so another group will have 3, 17-3=14 but again 14 is not divisible by 4. Hence two more group will have 3 members => 4 group of 3 each = 4*3=12..
the remaining 20-12=8 can be divided in 8/4 = 2 groups
so MIN is 4+2=6..... here A=B

therefore we can have different answers

D
Most Helpful Community Reply
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12234 [7]
Given Kudos: 136
Send PM

Top Member of the Month
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   24 Mar 2017, 15:32
5
2
Bookmarks
HarveyKlaus wrote:
The 20 people at a party are divided into n mutually exclusive groups in such a way that the number of people in any group does not exceed the number in any other group by more than 1.

Quantity A: The value of n if at least one of the groups consists of 3 people
Quantity B: 6


Let's first see what happens when n = 6
We could divide the 20 people into 6 groups as follows: 3, 3, 3, 3, 4, 4
So, n COULD equal 6.
This means the correct answer is either C or D


Are there other possible values of n that satisfy the given conditions?
For example, can n = 7?
Well, we could divide the 20 people into 7 groups as follows: 3, 3, 3, 3, 3, 3, 2
So, n COULD equal 7.

This means the correct answer is
Show: ::
D


Cheers,
Brent
General Discussion
avatar
AE
Director
Director
Joined: 09 Nov 2018
Posts: 505
Own Kudos [?]: 133 [0]
Given Kudos: 0
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 09:44
Is there any easy way?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [0]
Given Kudos: 26096
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 11:52
Expert Reply
Brent showed you two approaches.

I do not think there is one more.

Regards
User avatar
sandy
Retired Moderator
Joined: 07 Jun 2014
Posts: 4815
Own Kudos [?]: 11267 [2]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 13:21
2
Expert Reply
AE wrote:
Is there any easy way?


It is the same way as brent

Take the number of groups as 2. Then you can have 10,10 split. So 2 is a possible solution.

Take number of groups as 3. You can have 7, 7, 6 split also a possible solution.

Take number 4 You can have 5, 5, 5, 5

Clearly all numbers below 6 are possible.

Now if you look at 6 Split is 3, 3, 3, 3, 4, 4.

Clearly the quanity A is less than quantity B is true and quantity A = quantity B is also true. Hence D
avatar
Fatemeh
Manager
Manager
Joined: 22 Feb 2018
Posts: 163
Own Kudos [?]: 215 [0]
Given Kudos: 0
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 21:33
answer: D
At least number of people in a group is 3, considering the constraint in the question that no group number can be more than 1 different from any other group number, the group members can be just in groups of 3,4 or 3,2. We write the possible combination of 2and3 & 3and4 which equal 20:
3*2 + 2*7 = 20 (9 groups)
3*4 + 2*4 = 20. (8 groups)
3*6 +2 = 20 (7 groups)
(There can’t be odd number of 3, because then we can’t have 20)

3*4 + 4*2. (6 groups)
So n can be 6,7,8 or 9, the answer is D.
avatar
Salsanousi
Manager
Manager
Joined: 07 Aug 2016
Posts: 59
Own Kudos [?]: 68 [0]
Given Kudos: 0
GRE 1: Q166 V156
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   20 Nov 2018, 08:03
HarveyKlaus wrote:
The 20 people at a party are divided into n mutually exclusive groups in such a way that the number of people in any group does not exceed the number in any other group by more than 1.

Quantity A
Quantity B
The value of n if at least one of the groups consists of 3 people
6


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


It doesn't say that the group number is distinct. So we can safely assume it could be repetitive.

One option is 2,3,4,5,6 the sum will be 20. Number of groups 5.

If we have 3,3,3,3,4,4 = 12 + 8 = 20 we have 6 groups.

We could also have it 3,3,3,3,3,3,2 = 3*6 + 2 = 20. 7 groups.

Since we have different answers it is D.
avatar
arvindpr95
Intern
Intern
Joined: 18 Jul 2022
Posts: 5
Own Kudos [?]: 10 [2]
Given Kudos: 4
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   04 Aug 2022, 18:13
2
The problem can be summed as two equations:-
4a+3b=20; where a is number of 4 member group & b "3" member group
2a+3b=20; where a is the number of 2 member group and b "3" member group.
given b>=1. so in first case (a,b)= (4,2) ; 4+2=6.
in second case (a,b)= (2,7)=9 (4,4)= 8, (1,6)=7

therefor solution is D.
avatar
B
KC15
Intern
Intern
Joined: 08 Aug 2022
Posts: 49
Own Kudos [?]: 38 [1]
Given Kudos: 98
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   10 Aug 2022, 18:54
1
Since there are 3 people in at least one group, and the prompt also indicates no groups can have beyond 1 more or less than that, we have 2 options:

Option 1: groups of 3 and 4
Option 2: groups of 3 and 2

Let's look at option 1:
The only way I can think to make this add up to 20 people is:
4 groups x 3 people each = 12
2 groups x 4 people each = 8
Total number of groups = 6 groups, which equals our quantity B above.

What about option 2:
One possible solution is:
2 groups x 3 people each = 6
7 groups x 2 people each = 14
Total number of groups = 9 groups, which does not equal our quantity B above.

Since we have already found one arrangement that equals 6 and one that does not, we do not need to keep thinking of all possible arrangements. The answer is D, cannot be determined.
avatar
B
aurora07
Intern
Intern
Joined: 15 Sep 2024
Posts: 14
Own Kudos [?]: 3 [0]
Given Kudos: 15
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   15 Nov 2024, 07:16
Hi, why can't the number of people in groups be 2,3,4,5,6? That would make 5 groups and satisfy condtions: atleast 3 in one group as well as not exceeding more than 1 in any other group?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [0]
Given Kudos: 26096
Send PM
Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   15 Nov 2024, 08:27
Expert Reply
The information given by Quantity A says that there is at least one group with 3 people, so all possible combinations are either 2 and 3 or 3 and 4.

(1) For the combination of 2 and 3, find the minimum and maximum values ​​of n. The minimum value is obtained when the combination of 3 is the most, and the maximum value is obtained when the combination of 2 is the most.
-minimum: 3 people can form 6 groups at most, in which case 2 people can form 1 group, n=7
-maximum: 2 people can form 7 groups at most, in which case 3 people can form 2 groups, n=9

(2) For the combination of 3 and 4, find the minimum and maximum values ​​of n. The minimum value is obtained when the combination of 4 is the most, and the maximum value is obtained when the combination of 3 is the most. After enumeration, there are only one combination of 3 and 4: 4 groups with 3 people and 2 groups with 4 people, n=6

D is the answer
Prep Club for GRE Bot
gmatclubot
Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
15 Nov 2024, 08:27
Moderators:
adewale223
GRE Instructor
88 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1115 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact