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The 20 people at a party are divided into n mutually exclusi
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Updated on: 14 Nov 2018, 13:18
Updated on: 14 Nov 2018, 13:18
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51% (01:45) correct
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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.
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.
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.
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.
Last edited by Carcass on 14 Nov 2018, 13:18, edited 1 time in total.
Edited by Carcass
Re: The 20 people at a party are divided into n mutually exclusi
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14 Nov 2018, 18:38
14 Nov 2018, 18:38
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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.
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.
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
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24 Mar 2017, 15:32
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
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
