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

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HarveyKlaus
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The 20 people at a party are divided into n mutually exclusi [#permalink] New post   Updated on: 14 Nov 2018, 13:18
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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.
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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
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 18:38
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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
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   24 Mar 2017, 15:32
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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
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AE
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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?
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 11:52
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Brent showed you two approaches.

I do not think there is one more.

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Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   14 Nov 2018, 13:21
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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] 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.
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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.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   04 Aug 2022, 18:13
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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.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   10 Aug 2022, 18:54
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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.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink] New post   13 Jan 2024, 20:39
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