A management education program offers courses in four main disciplines

Question

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 disciplines or in none of the four disciplines. 50% of students concentrated in Marketing, 25% concentrated in Finance, 30% concentrated in Operations, 60% concentrated in Strategy.

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

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

Select the appropriate values of X and Y from the following options:

AdarshSambare · Answer

Total counts = 50 + 25 + 30 + 60 = 165 (for 100 students)

Extra counts beyond 100 = 65 → represent overlaps.
X (max students in 3+ disciplines)
To maximize 3+ overlap, we pack overlaps efficiently into 3-discipline students.
Max feasible = 25%
Y (min students in 2+ disciplines)
To minimize 2+ students, we push overlap into fewer people but cannot avoid at least 30% being in 2+ disciplines.

Final Answer:
 X = 25% 
 Y = 30%

Adit_ · Answer

It is not that simple! Give it another try.

bharathhr · Answer

got to know 1st part , max value of X is 50, but didnt quite get Y value as 22

Adit_ · Answer

It is a great problem to rehearse indeed! What did you get as the Y value?

bharathhr · Answer

Initially got 65, then it turned out to be wrong .

suntprovident · Answer

If only 25% students concentrated in finance, how can we have 50% students finance in all 4 ?

Adit_ · Answer

The question states: " three or more " and not ALL 4 particularly.

SomyaMBA2025 · Answer

There is no clear explanation or solution of this question.

Cooley · Answer

X should be 52.5 percent. Y should be 21.66%.

for x you can have ---
Marketing + Finance + Strategy: 22.5% 
Marketing + Operations + Strategy: 27.5% 
Finance + Operations + Strategy: 2.5% 
Strategy only: 7.5% 
No Concentration: 40%

Sajal193 · Answer

I got x as 50, need an explanation for Y

I took the min value of highest 3, subtracted the value from each , then took remaining elements and added the min value common to remaining three and added to previous value. This way got X as 50.
 Adit_  please provide solution for Y. You may DM me if required. Thanks

Adit_ · Answer

What are the equations that you got?

Firstly you have x+2a+3b+4c = 165.
Where x is single items, a is two items, b is three items and c is 4 items.
Why do we get such a large number as 165? Because we double,triple and quadrupe count the items.

What is the minimum?
Minimize the remaining portion:
x+2a = 0 let's say.

Questions asks us to maximize 3b+4c = 165 after all.
What do you do now?
put c = 0 as well and we get b = 55.
Now notice that in a 4 way venn diagram we have 4C3 = 4 three-item overlap.
Meaning 55/4 = 14 lets say, and we just need to confirm if all the 4 courses have more than 14% here as we need to sum the 4 three-item overlaps giving us 55.
So 55 is valid.
But 55 is not in the list, thus  50  is our answer for the max value.

For the least value :
Use the same equation but we have a constaint where x+a+b+c = 100 which is the max value.

Thus when you subtract the two equations you have:
a+2b+3c = 65.
How do you minimize this equation?
Put a+2b as 0.
3c = 65 and c =  22  around and hence our answer.

========================================================

Bunuel  Would appreciate your inputs on this tricky problem as well.

egmat · Answer

Hi Sajal193,

Nice work getting  X = 50%  on your own. For  Y , the key is to flip your thinking: instead of  spreading  overlaps out, you want to  cram  them into as few students as possible.

Start with the same total you already used:

- Marketing  50  + Finance  25  + Operations  30  + Strategy  60  =  165 memberships  across  100  students.

Those  165  "slots" have to land on  100  people. A student counted in  2 or more  disciplines is exactly the kind of person who soaks up extra slots. So to  minimize  how many such people exist, give each of them the maximum load -  4  disciplines each - and let everyone else hold at most  1 .

Setting up the bound

Let  t  = number of students with  2+  concentrations. The most memberships you can possibly cover is:

-  4·t  from the heavy students, plus  1·(100 - t)  from the rest.

That must reach  165 :

4t + (100 - t) ≥ 165  →  3t + 100 ≥ 165  →  3t ≥ 65  →  t ≥ 21.7

Since  t  must be a whole number,  t ≥ 22 . You can't dip below  22  - there simply aren't enough "slots per person" to cover  165  otherwise.

Why 22 actually works

The bound is only useful if  22  is reachable. Put  22  students into all four disciplines (that uses  22  of each - fine, since Finance has  25  and Operations has  30  available). That leaves Marketing  28 , Finance  3 , Operations  8 , Strategy  38  =  77  leftover memberships, which you hand out  one each  to  77  of the remaining  78  students. Everyone outside the  22  now has exactly  1  concentration, so nobody new joins the "2+" group.

Everything balances, so  Y = 22% .

The one habit to carry forward: to  minimize  the overlap count, maximize the load per overlapping student; to  maximize  it (like your X), you spread thin instead.

Answer: Column 1 (X) = 50%; Column 2 (Y) = 22%

sagregas · Answer

Great response by chatgpt

egmat · Answer

sagregas  - Please do not use Chatgpt for learning.. at least use a proper model. At e-GMAT, we use proprietary agents to help us. Did you see a problem with the response?

-Rajat

pizzapalooza · Answer

Let n be number of people with concentration in no discipline
x be number of people with one discipline
a be number of people with two disciplines
b with three
c with four

I) To maximise 3 or more disciplines,

Here, it makes sense to put some people in the 'none'/'n' category, so that all the tickets can be distributed amongst people with 3 or more disciplines and in turn maximise it (Otherwise these tickets would've gone to x/a and would've consumed some tickets out of the 165)
 The limiting equation becomes x + 2a + 3b + 4c = 165 , where x and a should be 0 and even amongst variables contributing to 3 or more disciplines, we can maximise the number of people by giving them 3 tickets each and taking c=0.
Hence, b=55 and the next number available is 50.

II) To minimise 2 or more disciplines,

Here, there's no need to put anything in 'n' since we want to distribute more tickets in other variables to minimise 2 or more.
 The limiting equations become x + 2a + 3b + 4c = 165 & x + a + b + c = 100 
Which gives us a + 2b + 3c = 65 and we get c~22 on taking a and b as 0.

A management education program offers courses in four main disciplines

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