A management education program offers courses in 5

­Hi GMATCoachBen
Can you help me in reaching the solution for this one please-

Question Source: Expert's Global

A management education program offers courses in four main disciplines - Marketing, Finance, Operations, and Strategy. A student who opts for three
or more courses in one discipline is awarded a concentration in that discipline. A student may, therefore, concentrate in multiple as well as none of the
four disciplines. 50% students concentrated in Marketing, 25% students concentrated in Finance, 30% students concentrated in Operations, and 60%
students concentrated in Strategy.

X represents the maximum possible percentage of students who may have concentrated in three or more of the four disciplines.

Y represents the minimum possible percentage of students who may have concentrated in two or more of the four disciplines.

Choices - 22, 25, 30, 50, 65

­The Venn diagram attached isn't a perfect representation of the 5 sets as it doesn't include a few cases (such as cases where students have elected for only strategy and finance - there are no areas where only S and F intersect), but it is good enough to help us solve this problem. I have made the 5th object a triangle so that I can get the cases needed to solve Y (it is not possible when using 5 circles).

X: maximum possible percentage of students who may have opted for at least one elective in each five discipline.
Students who have opted for at least one elective in each five discipline = the common area where all 5 objects intersect. We need to maximise this blue region. Now, maximum possible value of S = 70%. So the blue area which falls within S, cannot exceed 70%. So max possible % that the blue area can contain = 70%. Logically thinking, if we assume that number of students who have opted for at least one elective in each five discipline is 80%, that means 80% of the students have opted for S, F, M, B, O. While, this can hold true for M, F, B, O, but it is not possible for S as S = 70% given.

Y: minimum possible percentage of students who may have opted for at least one elective in each five discipline.
We need to minimise the blue area in this case. The above statement can also be read as complement of maximum % of students who have not opted for at least one elective. Let this be A.
A will include: (1) - students who have opted for 4 electives and have not opted for 1 elective + (2) - students who have opted for 3 electives and have not opted for 2 electives ... + (5) students who have not opted for any electives.
We need to maximise A.
If we maximise part (1) of A, such that there are no overlaps among students who have not chosen 2 / 3 / 4 / 5 electives then (2).. to (5) all are 0. By this we mean that, if a student has not chosen finance, then there are no other electives that he/she has not chosen (means he / she must have chosen all other 4 electives). 
Now, A = students who have chosen all other 4, but not chosen finance + Students who have chosen all other 4, but not chosen M + chosen all other 4, but not chosen B + chosen all other 4, but not chosen O + chosen all other 4, but not chosen S
Max possible value of chosen all 4, but not chosen finance = 1 - (Number of students who have chosen finance) = 1 - 95% = 5%.
Similarly, max % of chosen all 4, but not chosen M = 10%.
Thus, A  = 5% + 10% + 15% + 20% + 30% = 80%.
Y = 1 - A = 20%

If this method is a bit difficult to grasp, another method which is more visual is given below:

Yellow colour represents students who chose all 4 electives but didn't choose Strategy. Maximum % of yellow can be 30% (as 70% chose Strategy).
Similarly, Green colour represents students who chose all 4 electives but didn't choose Marketing. Max % of green can be 10%.
Orange = Chose all 4 but didn't choose Operations, max % = 20%.
Pink = Chose all 4 but didn't chose org Behaviour, max % = 15%.
Now, areas coloured Grey have to be 0%. Because M contains 90%, while outside of M i.e. those who have not chosen M are all present in Green. So, all these areas other than M and Green should be 0%.
Also, area coloured purple needs to be 0%. As S contains 70%, while outside of S, i.e. those who have not chosen S are all present in Yellow. So, all areas other than S and Yellow, should be 0%.

Now, if you look at F as a circle, it contains 95% of students. F contains: Grey + Purple + Blue + Yellow + Green + Orange + Pink. In this, all colours except Blue are maximised. So, if we solve for Blue, we will get the minimum possible value for Blue.
0 + 0 + Blue (min.) + 30% + 10% + 20% + 15% = 95%
Thus, Blue (min.) = 20% = Y­

 
­An accurate representation of the above problem in Venn diagram wouldn't be using 4 circles (see attachment 1 called 4 sets) as it doesn't include a few cases (such as cases where students concentrate in only finance and marketing disciplines OR cases where students concentrate in only strategy and operation disciplines). A much accurate representation would be attachment 2 (called 4 Sets Accurate) - orange region refers to students who have concentrated in finance and marketing, yellow shows students in strategy and operation. But for answering X and Y, the 1st attachment is enough so we will use that.

X = Max % for students concentration in 3 or more disciplines. = Max % for 3 disciplines + Max % for 4 disciplines.
In attachment 1, 4 sets, the top half of the image shows X = Maximise red (3 disciplines) + Maximise blue (4 disciplines)

Now, if we maximise blue first, it will be common for all 4 disciplines and thus will reduce the amount we will have available for red regions. Example, let's give blue maximum possible amount = 25% as F can't have more than 25%. Then the red areas of 1, 2, and 3 will have to be 0 (as they all fall in F and can't have more students). Red area 4, can have 5% students as O can have max 30% (25% of O already comes from blue area). So total of blue and red = 25 + 5 = 30%. M and S will get their remaining students from the green area (20% for M remaining, 30% for S remaining; distribution doesn't matter to us)

But, if we maximise the red regions first, we will have more students left to distribute later. We will have to go from lower circle totals to higher ones.
If we give red area 1 max = 25%, then red area 2 = red area 3 = 0 (as F = 25% max). Next, maximise red area 4 = 5% (as O can have max 30 and red area 1 already gives O 25%). Blue has to be 0 as both F and O are at max. So total of blue and red = 30%
Same thing happens when we give red area 2 max = 25%, we get total blue + red = 30%.
For red area 3, look at attachment 1, 4 sets, the bottom half. The green area (red area 3) is maximised first. It can hold 25% as F = 25%. Red area 1 and 2 are 0 as both are in F. Next, maximise orange area (red area 4) = 30% (as O can have max 30%). But this brings the total of M to be 55%, which isn't possible as M = 50%. So, orange can be max 25%, the remaining 5% of O will be distributed in the pink region (we can't give anything to blue area as it is a part of F and O and both are saturated). The remaining 10% of S, goes to yellow as intersection of M and S can't take any more students due to M being saturated. So, total of blue and red = 50%.
For red area 4, we cannot start with this as it is not a part of F. It will be impractical to assign values to other areas later as F is the lowest total.

Thus, X = 50%

Y = Min % of students concentrated in two or more of the four disciplines = blue and yellow parts in attachment 3, 4 sets part 2 = Complement of Max % of students concentrated in one or less disciplines = B (say)
B = Max % concentrated in 1 discipline + Max % concentrated in 0 disciplines. Let's assume students who concentrated in 0 disciplines is 0% (grey area is 0). So, we only need to maximise % concentrated in 1 discipline. In attachment 3, 4 sets part 2, this is represented by the red areas.

Ideally, I want to say that all the students concentrated only in 1 discipline, so blue and yellow parts = 0%. But then the total of the red areas = total of all 4 circles = 25 + 30 + 50 + 60 = 165% which is not possible as total number of students is 100% (assumed that no students have concentrated in 0 disciplines, grey area is 0). Thence, we need to add students to the blue and yellow parts so that the total of red and blue and yellow areas comes to be 100% instead of 165% (red + blue and yellow = 165 + 0). If we assign students to the yellow areas then they will only be counted in 2 or 3 of the disciplines, so we will need to remove more students from the red areas. However, if we assign students to the blue areas, then they will be counted in all the 4 disciplines, so we need to remove lesser number of students from the red areas [remember our goal is to maximise red areas]. We should thus only assign students to the blue area and check for possibilities.

Let us assign x students to the blue area. The total of red and blue and yellow area should be 100 (as we have assumed that number of students who have concentrated in 0 disciplines is 0, grey area is 0).
Red area 1 + red area 2 + red area 3 + red area 4 + blue area + yellow area = 100
(25 - x) + (30 - x) + (60 - x) + (50 - x) + x + 0 = 100
165 - 3x = 100
3x = 65, thus x = 21.7 = 22 approx.
So, when blue = 22, yellow = 0, then the red areas are maximised. So minimum value of blue and yellow area is 22% = Y

Quick verification, B = red area 1 + 2 + 3 + 4 = (25-21.7) + (30-21.7) + (60-21.7) + (50-21.7) = 3.3 + 8.3 + 28.3 + 38.3 = 78.2% = 78% approx.
Y = 100 - B = 100 - 78 = 22%

Clarification: if we assume that number of students who concentrated in 0 electives is more than 0, then let grey area = 1%. Then, red area 1 + 2 + 3 + 4 + blue + yellow = 99% i.e. 165 - 3x = 99 i.e. 3x = 66, thus x=22 (same answer). [Verify Y = 100 - B = 100 - (max of only 1 concentration + max of 0 concentration) = 100 - red areas - grey area = 100 - (3+8+28+38) - 1 = 100 - 77 - 1 = 22%] 
If grey area = 2%, then 165 - 3x = 98 i.e. 3x = 67, thus x = 22.33% (approx 22, so same answer).
If grey area = 3%, then 165 - 3x = 97 i.e. 3x = 68, thus x = 22.67% (approx 23, so now Y = 23, which is not the minimum value for Y). We can notice a pattern: when we increase grey area, value of Y increases and doesn't remain minimum. Logically, when we increase grey, we are taking away students from red areas, so for that we need to increase students in blue (or yellow areas) to ensure that the circular totals are maintained.­

­For calculating the minimum number of students who have opted for atleast 1 elective in all the 5 subjects:
Consider 100 students.
If 90 have taken marketing that means 10 have not taken.
Similarly 5 have not taken finance, 20 have not taken operations, similarly 30 and 15 for the rest of the subjects.

Now we want to find minimum possible students that must have taken all the subjects. If we can find what is the maximum number of students who haven't taken at least one subject, the remaining out of 100 would have surely taken all 5 subjects.

• If we consider 10 non marketing students that tells that surely 10 students haven't taken all the 5 subjects.
• Now 5 students of finance may be among these 10 marketing students or maybe not. But we want the maximum students who have not taken all the 5 subjects, so we consider them 5 seperate students. Now surely 10+5 = 15 students have not taken at least one subject.
• Similarly 20 students have not taken operations. Some of these might/might not be a part of the previous 15 students who haven't taken atleast 1 subject. Since we want to maximize the number of students not taking all the 5 subjects we consider these 20 seperately. Hence now 10 + 5 + 20 students surely haven't taken atleast one subject.
• Similarly considering the remaining subjects, 10 + 5 + 20 + 30 + 15 = 80 haven't taken atleast 1 subject. Thus the remaining would have taken all the subjects.

Remaining = 100 - 80 = 20.­­

90%: M
95%: F
80%: O
70%: S
85%: OB

X max % selecting all 5 => 70% (the smallest % selecting at least 1 of 5 disciplines - Strategy)


Y min % selecting all 5
­
(___________________90%____________________)(__10%__)
(________________85%_________________)(_5%_)(__10%__)
(__________65%____________)(___20%___)(_5%_)(__10%__)
(_______30%______)(__35%__)(___20%___)(_5%_)(__10%__)
(___20%___)(_15%_)(__35%__)(___20%___)(_5%_)(__10%__)

=> 20%­

can you please elaborate this.

For those who want to use arithmetic method to solve this problems:

In the minimum case, suppose total students = 100. There are 5 types of students: Choose 1 elective, 2 electives, 3 electives, 4 electives, 5 electives call them A,B,C,D,E.

=> A + B + C + D + E = 100 (1)

Total slots in the class that has been registered by these students: 90 + 95 + 80 + 70 + 85 = 420 slots

=> A + 2B + 3C + 4D + 5E = 420 (2)

To minimum E, we have to maximize (A + B + C + D) or (A + 2B + 3C + 4D) => Should max D > C > B > A => let A,B = 0
Max D = 80 (the same method in case maximum)

(1) becomes: C + E = 20
(2) becomes: 3C + 5D = 100

Solve 2 equations => E = 20

Hi,

For the max% case, this scenario allows for the maximum overlap among the five categories. Here, individuals who choose S (70% - least % of all 5) also choose M, F, O, and OB. To visualize:

(___________________90%____________________)(____10%_)
(___________________ 95%_______________________)(_ 5%_)
(___________________80%______________)(__________20%_)
(___________________70%________)(________________30%_)
(___________________85%__________________)(______15%_)

For the min% case, the goal is the opposite—we aim to minimize the overlap among the five categories.

Hope it helps.

Hi, could you please help me understand this shouldn't the yellow region (S) add up to 70%?