A certain company has the following policy regarding members

Question

A certain company has the following policy regarding membership in committees A, B, C: each of the committees have exactly 40 members, and each employee must be a member of at least one of the committees. The company currently has 110 employees.

Let X be the current number of employees who are members of more than one of the committees. Based on the current total number of employees, select the least value of X that is compatible with the policy and select the greatest value of X that is compatible with the policy. Make only 2 selections one in each column.

Gemmie · Accepted Answer

#commitee/people           # of people       # committees
1                                     a                      a
2                                     b                      2b
3                                     c                      3c
Total                                110                  120 (=40*3)

X = b + c

a + b + c = 110
a + 2b + 3c = 120
=> b + 2c = 10
=> X + c = 10

X is highest when c = 0
=> b = 10
=> X = b + c = 10

X is lowest when c is highest => b = 0
=> c =  \frac{10-0}{2} = 5 
=> X = b + c = 5

egmat · Answer

This is an amazing Venn diagram based question in which one needs to very carefully process the information and use visualization skills to understand how to minimize and maximize the given qualities.

https://www.youtube.com/watch?v=ABNprkhYe6M

chetan2u · Answer

Responding to a PM

We could use Venn Diagrams or pure logic to answer.

Each team has 40 members, so total would have been 3*40 or 120. But the number of employees are 110. Surely, 120-110 are to be accounted by the overlap of members between the committees.

Each member in exactly two groups is counted twice, once in each group, while each member in exactly three groups is counted thrice, once in each group.

Largest:  The extra 10 could be members of exactly two groups, say A and B...... Answer:10 
 Lowest:  The max possible of extra 10 could be members of the three teams. One person in 'three group' accounts for increase of two as he is counted thrice. Thus, five person will account for 2*5 or 10 person...... Answer: 5

an1995

AnuK2222 · Answer

hi  chetan2u 
could you share the explanation using venn diagram please? still not very clear how we get 5 as the answer for first blank

sayan640 · Answer

To get the smallest count of people who were part of more than one group, divide the extra number of people i.e 10 amongst three groups.
A           B     C
-------------------
-------------------

------------------ ( Total count is 105 now..)

1         1       1
2         2       2
3         3       3
4         4       4
5         5       5  ( These 5 people are part of three groups)
---       ---     ----
40        40     40

Notice that each group contains 40 people now but total number of people used is 110 only. Five people were just re-used across three groups.
For the previous set of 35 rows , it's 105 and after that these 5 people were just used multiple times. Hence , 35*3 + 5 = 110 i.e total no of people used.
Hope it clarifies. Thank you  chetan2u

study1question · Answer

Hey, Im also not understanding the solutions for this one... Could someone help me out?

Thanks!

sayan640 · Answer

Which part ?

Posted from my mobile device

study1question · Answer

I understood and arrived at the rational of the overlapping of the 10 members, but then not on the least and greatest value for x, I get confused on that part.

AnuK2222 · Answer

yes, I get it now. I think I was confused tryig to use venn diagram. thank you very much

JoeKan1234 · Answer

This problem doesn't necessarily require us to use Venn diagram, which seems to complicate the solution.

The least value of X is when X takes up all three memberships.

Let total memberships from X = 3X and let N = the number of people who have one membership

Total memberships = 3*40 =120

3X + N = 120 .....(1)
X + N = 110.....(2)

Solve (1) and (2), you will get X = 5

Similarly, X reaches its max. value when X takes up 2 memberships.

Let total memberships from X = 2X and N = the number of people who have one membership

2X + N = 120 .....(1)
X + N = 110.....(2)

Upon solving, you will get X = 10

ubaid009 · Answer

Hope it helps
(d)min=0

SergejK · Answer

Using Venn diagram, we have 2 equations.
1: a+b+c=110 -> total number of  PEOPLE  in the group
2: a+2b+3c=120 -> total number of  MEMBERS :  1 person  can be simultaneously in  3 groups 
We are looking for X and X is defined as members in more than 1 committee, meaning b and c. X=b+c. Subtracting equation 1 from 2, we are left with b+2c=10, which is  NOT  X! We can insert different values for b and for c to arrive at 10. As both, b and c must be integers, we can have 6 pairings. If b=10 and c=0, this is our max of 10 (b+c=10). And if b=0 then c=5, leaving us with our min of 5 (b+c=5).

Nairobisha · Answer

I'm unable to understand the least value for x after going through the replies above. Can someone please explain?

__thisisjay · Answer

A B C

For Least

Take a ven Diagram

(A) = a + d(all common)
(B) = b + d(all common)
(C) = c +d(all common)

for least all have to be same , each should have same number minimize the load on common 
so 
from ven Diagram 
A + B + C = 110
(a +d) + (b + d )+ (c +d) - 2d = 110
a + b + c + d =110

all are same assuming common value as X

3X + d = 110 -------(i)

and vale of each total is exactly 40 so;

X + d = 40-------(ii)

comparing and solving both statement we get the answer

d = 5

SudsMeister · Answer

In questions where we aren't able to solve intuitively, instead of spending more time trying to find the Eureka moment, we should flesh it out in the simplest terms:

Quickly make a Venn diagram for 3 circles.

Attachment:

A.jpg [ 66.34 KiB | Viewed 5968 times ]

Each circle is a committee = 40 employees.
We need to find out the least and largest possible values of: (B + D + F + C) region.
Let's try to get these 4 regions in some mathematical equation.

We all know that all of the elements in the rectangle = total number of employees.
That implies:
Committee 1 + Committee 2 + Committee 3 - B - D - F - 2C + N (employees in no committee) = 110
40 + 40 + 40 - B - D - F - 2C + 0 = 110 (as all employees are part of at least 1 committee)
B + D + F + 2C = 10

To minimise (B + D + F + C) region, we must maximise the value of C in the above equation, let's make C = 5. Then, B + D + F = 0.
So, min(B + D + F + C) = 0 + 5 = 5.

To maximise (B + D + F + C) region, we must minimise the value of C in the above equation, let's make C = 0. Then, B + D + F = 10.
So, max(B + D + F + C) = 10 + 0 = 10.

Ans: Least is 5, Largest is 10.

chuting · Answer

Maybe we can make it clear with inadequaties.

There are totally 10 times that OVER-counted. since30*4=120,120-110=10

Given that A persons join 2 memberships. So there are A times OVER-counted . 
Given that B persons join 3 memberships. So there are 2B times OVER-counted .

We can know that 
A+2B=10, as totally 10 times.
And A+B=X.

The most important part is that A and B are both ≥0. 
A+2B=10, so A=10-2B
when A ≥0, so 10-2B≥0,then 5≥B≥0.

Then you can get X. And the same with B=1/2(10-A).

KarishmaB · Answer

Each committee must have 40 members and there are 3 committees so this means we will have total 120 instances. There are 110 employees so there are 10 instances extra. How do we distribute them? For least x - 2 instances each should be given to 5 people to distribute the 10 instances. Hence least value of x = 5 For greatest x - 1 instance each should be given to 10 people to distribute the 10 instances. Hence largest value of x = 10 Max min concept of sets is discussed here: https://anaprep.com/sets-statistics-min-max-in-sets/

Aboyhasnoname · Answer

Its a beautiful question to clarify further understanding of sets.....

We know here none is 0...

Now.....
110 = Exactly 1 + exactly 2 + Exactly 3 (No overlap) 
120 = Exactly 1 + 2(Exactly 2) + 3(Exactly 3) (with Overlaps)

Means there is a overlap of 10...and that could be with 2 cases... 
Exactly 2 overlap and exactly 3 overlap .....

Now for exactly 2..We double count it....and exactly 3 ..we triple count it.... 
No if we want to minimise the overlap..we put it in the values that gives maximum overlap...that means exactly 3..... 
Now...Exactly 2 = 0 ..Means exactly 3 = 10 ...
How many times has exactly 3 been overcounted...2 times....overlap is 10...means instead of 5..we took it 15.... So minimum will be 5....

Now same for exactly to we want to maximise the overlap...means minimise the extra counting should be as less as possible...and thats possible with exactly 2....
So...Exactly 3 be = 2 ....
means exactly 1 has been counted extra once....so....the maximum is 10......

For Simples Algebraic... 
110 = Exactly 1 + exactly 2 + Exactly 3 (No overlap) - (i)
120 = Exactly 1 + 2(Exactly 2) + 3(Exactly 3) (with Overlaps) - (ii)

Deducting (i) from (ii)
120 - 110 = Exactly 2 + 2(Exactly 3)

10 = Exactly 2 + 2(Exactly 3)

So....Maximum number that are doing more than 1...means minimum overcounting...So...Exactly 3 be = 0 
Exactly 2 = 10

Minimum number that are doing more than 1 ...means maximum overcounting...So exactly 2=0 
2*Exactly 3 = 10...Exactly 3 = 5....

A certain company has the following policy regarding members

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A certain company has the following policy regarding members