There are 101 students in a school. The students are split into 3 team

We Have,

a + d + e + g = 70
b + d + f + g = 75
c + e + f + g = 80

Adding them we get,

a + b + c + 2(d + e + f) + 3g = 225
also,
a + b + c + d + e + f + g = 101

Subtracting them,

(d+e+f) + 2g = 124

To maximize g,
d + e + f = 0 --> 2g = 124

Hence, g = 62

Correct Ans: D

We need to maximise the overlap of all three teams such that all 101 students are in at least one team.

101 = 70 + 75 + 80 - Overlap in exactly 2 teams - 2*Overlap in all 3 teams

Overlap in exactly 2 teams + 2*Overlap in all 3 teams = 124

To maximise overlap in all 3 teams, make overlap in exactly 2 teams = 0

Overlap in all 3 teams = 62

Answer (D)
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wrong question. There should be 101 students in order to support the official Answer

Though I have changed the total to 101,
Why can't the total be 100?

My Logic:

Number the students from 1 to 100

A,B and C have students 1-62 in common.
B,C have student 63 in common.

Now there are 100-(62+1)=37 students to be split.

A can have 8 students, B can have 12 students and C can have 17 students. So in total, there are 62 students common to A,B,C.

Total Students: 101

Team A: 70 ; Team B: 75 ; Team C: 80

Maximum number of students present in all teams: 70.

Team B: 75 - 70 = 5 extra

Team C: 80 - 70 = 10 extra

101 - 70 - 5 - 10 = 16 extra

8 goes to Team B and 8 goes to team C, hence 70 - 8 = 62

62 maximum common students

Answer D

A useful formula for 3 overlapping groups:

T = A + B + C - (AB + AC + BC) - 2(ABC)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in A, everyone in B, and everyone in C:
Those in exactly 2 of the groups (AB+AC+BC) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (ABC) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.



Applying the formula above, we get:
101 = 70 + 75 + 80 - (AB+AC+BC) - 2(ABC)

To MAXIMIZE ABC, we must MINIMIZE (AB+AC+BC).
If AB+AC+BC=0, we get:
101 = 70 + 75 + 80 - 0 - 2(ABC)
101 = 225 - 2(ABC)
2(ABC) = 124
ABC = 62


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VeritasKarishma,

How is this problem different from one of the problems in your post about finding the maximum number of all three (TV, MP3 and Cellphone) [I've included the attachment here]

If we have 70 as the maximum number of all 3, we would have 5 only in one team, 10 in another team, 70 in all three teams and 16 is neither. There is nothing in the question stem that says that all students are in at least one team making neither 0. Some insight on this is greatly appreciated

Good question SiddharthR
In this question (teams), we ARE given that none must be 0. We are given that every student must be in at least 1 team.
"There are 101 students in a school. The students are split into 3 teams."
implies that the 101 students are split into 3 teams. So I need to ensure that none is 0.
If some students could belong to none, it would be a simple case of putting one circle inside the other.

But now, say I put down the 80 circle first. I still have to put 21 students in at least 1 group. If I take away all these 21 from 75, I would be left with 54 for the overlap. So instead, I should take some away from 70 and some away from 75 so that I have a higher number is available for overlap.
Let's take 5 away from 75 so that now I am left with 16 students whom I must put in some group. From both 70s now, I can take away 8 each to be left with 62 for the overlap of all three.
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Hello VeritasKarishma

Thank you so much for your reply.

In the above explanation you stated that In this question (teams), we ARE given that none cannot be 0 but while solving this problem algebraically using the formula, you put neither / none = 0

--> 101 = 70 + 75 + 80 - Overlap in exactly 2 teams - 2*Overlap in all 3 teams

This would imply that each of the student is in at least 1 team. Could you please clarify ?

VeritasKarishma

I think I get it now. I realized if I have 13 + 70 + 18, I may have 101 and none would be zero but in this case Team B would have a number greater than 75 (83) and Team C a number greater than 80 (88) which goes against the stem of the problem.

Hence we reduce the common 3, reducing the numbers in B and C to their given values while slowly adding the corresponding number for “A only” to get to a sum total of 101.

I hope my understanding is accurate.

Posted from GMAT ToolKit

No SiddharthR, what I meant was that we are given that None MUST BE 0. (I reversed it in the explanation)

Look at the next line given "We are given that every student must be in at least 1 team."

"There are 101 students in a school. The students are split into 3 teams."
implies that the 101 students are split into 3 teams. That every student must belong to at least one of the teams.

*Editing above to avoid confusion.
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Think about this: You have 101 people inside a rectangle. You also have 3 circles which can hold 70, 75 and 80 people each. How will you place the 3 circles inside the rectangle such that all 101 are a part of at least 1 circle and that the overlap of all 3 maximises.

I will start with putting the 80 circle covering 80 people. Now I still need to cover 21 people with a circle. So should I throw the 75 circle on these 21 such that I have only 54 left for the overlap? No. I want to maximise the overlap so I would like to cover these 21 with a bit of the 75 circle and a bit of the 70 circle so that larger parts are available for overlap.
So, say I cover 5 out of the 21 with the 75 circle. I am still left with 16 uncovered people. My 75 circle has a capacity for 70 more people and I still have my entire 70 circle.
So now, to leave maximum for overlap, I cover 8 people each with these two circles so that I have 62 available in each circle for the overlap. This gives me a maximum overlap of 62.

This is a holistic approach involving a bit of visualisation and a tab difficult in the beginning. Just that, once you get used to such approaches, you can never go back to formulas! If you do wish to internalise them, try the same process by starting with the 70 circle.
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VeritasKarishma

Yes visualization like you have done is a little tough for me right now. I feel like the method I came up with, starting with 70 as common and working backwards worked better for me.

I will keep rereading your replies to hopefully get a better grasp of it in the future. Thank you so much for your help !

Posted from GMAT ToolKit

I think the Question premise has left some space for ambiguity .
It is not mentioned if all the students are part of one or the other team .

In that case the answer will be 70.

SiddharthR

You have used visualisation too. You have started with the maximum possible overlap and then adjusted it to ensure that none is 0 in your own way. To be honest, I am not exactly sure of the logic you have used but I could do the same by thinking this way.

Say the overlap of all 3 is 70 (all circles inside each other).
Then the 75 circle has 5 extra and 80 circle has 10 extra. The 70 circle has 0 extra.
This adds up to 5 + 10 + 70 which is less than 101. So you take away from overlap 70 slowly and add 3 to the total sum
6 + 11 + 1 + 69
7 + 12 + 2 + 68
...
13 + 18 + 8 + 62 = 101
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Hello VeritasKarishma

Thank you so much for your reply.

My logic is similar to what you just posted here but in a more abridged version. I guess there are multiple ways to solve this problem visually. The method you have mentioned here is what I was able to think of to solve this problem !

You have to find maximum A&B&C, meaning you have to minimise "exactly 2" i.e only A&B + only B&C + only C&A

Total = A + B + C - A&B - B&C - C&A + A&B&C

exactly 2 = A&B + B&C + C&A - 3(A&B&C)
=> A&B + B&C + C&A = exactly 2 + 3(A&B&C)

Hence the first equation can be written as

Total = A + B + C - A&B - B&C - C&A + A&B&C
=> Total = A + B + C - exactly 2 - 3(A&B&C) + A&B&C
=> Total = A + B + C - exactly 2 - 2(A&B&C)

This is how you get the formula

Now replacing values
101 = 70 + 75 + 80 - exactly 2 - 2(A&B&C)
=> 124 = exactly 2 + 2(A&B&C)

To maximise A&B&C, replace exactly 2 as 0

124 = 0 + 2(A&B&C)
=> (A&B&C) = 62

Hi Karishma, why is 70 not an answer here?




Posted from my mobile device

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Because each student needs to be in at least one team. When you put all circles one inside the other, you are left with 21 students in no team. But all 101 students are split in the three teams so that is not possible. 
I have discussed this issue here in my webinar:
https://youtu.be/oLKbIyb1ZrI

Alternatively, you can check out my Sets (including maximizing and minimizing) module, video and practice questions for free tomorrow under the Super Sundays Program. Details here:
https://youtu.be/gN_vlDpUflo

 
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