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Intern Joined: 22 Aug 2016
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The 20 people at a party are divided into n mutually exclusi [#permalink] 00:00

Question Stats: 47% (01:15) correct 52% (00:54) wrong based on 53 sessions
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.
[Reveal] Spoiler: OA

Last edited by Carcass on 14 Nov 2018, 13:18, edited 1 time in total.
Edited by Carcass GRE Instructor Joined: 10 Apr 2015
Posts: 2021
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
1
KUDOS
Expert's post
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
[Reveal] Spoiler:
D

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com Director Joined: 09 Nov 2018
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
Is there any easy way?
Founder  Joined: 18 Apr 2015
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
Expert's post
Brent showed you two approaches.

I do not think there is one more.

Regards
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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
Expert's post
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
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
Expert's post
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
_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Manager Joined: 22 Feb 2018
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
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.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
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. Re: The 20 people at a party are divided into n mutually exclusi   [#permalink] 20 Nov 2018, 08:03
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