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Sets - Maximum & Minimum

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Sets - Maximum & Minimum

by gmatrant » Sun Jun 20, 2010 6:55 am
In a college where every student follows atleast one of the three activities - drama, sports, or arts - 65% follow drama, 86% follow sports, and 57% follow arts. What can be the maximum and minimum percentage of students who follow

a) all three activities (max & min)
b) exactly two activities (max & min)
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by mj78ind » Sun Jun 20, 2010 7:40 am
Refer the attached document, we have the following equations:

x+a+b+d = 65 (1)
y+a+c+d = 86 (2)
z+b+c+d = 56 (3)
a+b+c+x+y+z+d = 100 (4)

Adding the above equation 1,2 and 3 we have 2(a+b+c) + (x+y+z) + 3d = 207
Substituting from eqtn 4, we have a+b+c + 2d = 107

Hence max d can be 53, when a+b+c = 1 (which also gives the min a or b or c which is 0,0 and 1 or some combinaiton of that)and min d = 0, a+ b +c = 107, which gives max a or b or c = 107 when the other two are zero[/img]
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by hardik.jadeja » Fri Jun 25, 2010 10:27 pm
mj78ind wrote:Refer the attached document, we have the following equations:

x+a+b+d = 65 (1)
y+a+c+d = 86 (2)
z+b+c+d = 56 (3)
a+b+c+x+y+z+d = 100 (4)

Adding the above equation 1,2 and 3 we have 2(a+b+c) + (x+y+z) + 3d = 207
Substituting from eqtn 4, we have a+b+c + 2d = 107

Hence max d can be 53, when a+b+c = 1 (which also gives the min a or b or c which is 0,0 and 1 or some combinaiton of that)and min d = 0, a+ b +c = 107, which gives max a or b or c = 107 when the other two are zero[/img]
first of all, z+b+c+d should be 57 not 56.

That gives us,
a+b+c+2d = 108

If a+b+c = 0, then that gives us d=54, x=11, y=32, z=3, a=0, b=0, c=0.
If we plugin the above values we get x+y+z+a+b+c+d = 100. So its a valid solution.

So maximum number of student following all three activities is 54.

But to find the minimum number of student following all three activities, if we take d=0 just as it is done in the above solution, then we get a+b+c = 108. This cant be true because if x+y+z+a+b+c+d = 100, then a+b+c cannot be 108. All x,y,z,a,b,c and d are integers greater or equal to 0.

Anyone has a solution for this question?
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by mj78ind » Sat Jun 26, 2010 1:15 am
hardik.jadeja wrote:
mj78ind wrote:Refer the attached document, we have the following equations:

x+a+b+d = 65 (1)
y+a+c+d = 86 (2)
z+b+c+d = 56 (3)
a+b+c+x+y+z+d = 100 (4)

Adding the above equation 1,2 and 3 we have 2(a+b+c) + (x+y+z) + 3d = 207
Substituting from eqtn 4, we have a+b+c + 2d = 107

Hence max d can be 53, when a+b+c = 1 (which also gives the min a or b or c which is 0,0 and 1 or some combinaiton of that)and min d = 0, a+ b +c = 107, which gives max a or b or c = 107 when the other two are zero[/img]
first of all, z+b+c+d should be 57 not 56.

That gives us,
a+b+c+2d = 108

If a+b+c = 0, then that gives us d=54, x=11, y=32, z=3, a=0, b=0, c=0.
If we plugin the above values we get x+y+z+a+b+c+d = 100. So its a valid solution.

So maximum number of student following all three activities is 54.

But to find the minimum number of student following all three activities, if we take d=0 just as it is done in the above solution, then we get a+b+c = 108. This cant be true because if x+y+z+a+b+c+d = 100, then a+b+c cannot be 108. All x,y,z,a,b,c and d are integers greater or equal to 0.

Anyone has a solution for this question?
@hardik ........... good observation. I think either I am doing dome math wrong for max of a,b,c or the question has issues. I mean a simple way is to just say x+y+z+a+b+c+d = 100, hence make x,y,z and d = 0 hence a+b+c = 100, but this does not fit with the algebra shown above. And I think the algebra does not have an issue.......... Hence I would go with max of a,b and c = 100.

May be the experts can guide here .......
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by vivekmalhotra » Sat Jun 26, 2010 8:45 pm
a+b+c+2d = 108
to maximize d; a+b+c = 0; d = 54
to minimize d; a+b+c should be maximum but such that a+b+c+d = 100
call a+b+c = M
TO minimize d; we have two equations
M+ 2d= 108
M + d = 100
==> d = 8; M = 92

So following should be the answer:

Max for Both Programs = 92
Min for both programs= 0
Max for all three programs = 54
Min for all three programs = 8
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by gmat1011 » Sun Jun 27, 2010 12:29 am
Thanks.

But x+y+z + a+b+c+d = 100

For getting the min value of d as 8 how did you take a+b+c+d=100? What about the people who only do one of the activities (x,y or z)...
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by vivekmalhotra » Sun Jun 27, 2010 12:19 pm
To maximize the number of persons who were in both the programs, the number of persons who were in only one program should be zero
So, x+y+z should be equal to zero
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by albatross86 » Mon Jun 28, 2010 9:09 am
In the attached diagram: D - Only Drama; A - Only Arts; S - Only Sports; X - All 3.

Let total number of students be 100
=> D + S + A + DS + AS + AD + X = 100 .... Eqn. 1

Question stem tells us:

D + DS + AD + X = 65
S + DS + AS + X = 86
A + AD + AS + X = 57

We need max and min values for :
1. X
2. AD + DS + AS

Let's add up all the above equations.

D + S + A + 2*(DS + AS + AD) + 3*X = 208 ... Eqn 2

Eqn 2 - Eqn 1:

(DS + AS + AD) + 2*X = 108

Term A + 2*Term B = 108

This is basically our equation which we will use to maximize and minimize our two values of concern.

1. Maximize Term A and Minimize Term B.

Your first impulse will be to do this:

Set Term B = 0 and Term A = 108

This is incorrect!

Remember, you cannot have a 108 students out of a 100 in 3 separate regions of that venn diagram. You can only distribute a maximum of 100 people.

i.e. The limitation is: Term A + Term B <= 100

Taking the max case of Term A + Term B = 100 and substituting in our "concern terms" equation:

We get Term B = 8, so Term A = 92.


2. Minimize Term A and Maximize Term B

Set Term A = 0 and Term B = 54 (This does not violate our limitation)

Great sum!
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by outreach » Mon Jun 28, 2010 12:02 pm
65-d
86-s
57-a

Our job is to minimize the number of people who love all 3; to do so, we want to maximize the number of people who love exactly 2 of the 3 and minimize the number of people who love exactly 1 of the 3.

now =d+s+a=208
so we have 208 students who r enrolled in different activities out of total 100. that is 108 more participants in activites among 100 people

The first equation is derived from the triple-overlapping set equation:
True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

100=65+86+57-(ds+da+sa)-2(dsa)
ds+da+sa+2dsa=108 (1)


The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)
We set the "total in exactly 1 group" to 0, so we get:
100=0+ds+da+sa+dsa (2)

solve (1) and (2)
dsa=8(minimum who follow 3 activities)
92(maximum who follow 2 activities)



above solution i worked out based on stuarts solution in below forum
https://www.beatthegmat.com/least-value- ... 46464.html



below solution found in another forum
adding 1 element to intersection of three sets give a surplus of 2 sets, adding 54% to intersection of three sets will give a surplus of 108%. Therefore, the maximum value of students who follow all three activities is 54%. In this case the percentage of students who follow exactly two activities will be minimum = 0%.
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by raunakrajan » Tue Jun 29, 2010 10:21 am
albatross86 wrote:In the attached diagram: D - Only Drama; A - Only Arts; S - Only Sports; X - All 3.

Let total number of students be 100
=> D + S + A + DS + AS + AD + X = 100 .... Eqn. 1

Question stem tells us:

D + DS + AD + X = 65
S + DS + AS + X = 86
A + AD + AS + X = 57

We need max and min values for :
1. X
2. AD + DS + AS

Let's add up all the above equations.

D + S + A + 2*(DS + AS + AD) + 3*X = 208 ... Eqn 2

Eqn 2 - Eqn 1:

(DS + AS + AD) + 2*X = 108

Term A + 2*Term B = 108

This is basically our equation which we will use to maximize and minimize our two values of concern.

1. Maximize Term A and Minimize Term B.

Your first impulse will be to do this:

Set Term B = 0 and Term A = 108

This is incorrect!

Remember, you cannot have a 108 students out of a 100 in 3 separate regions of that venn diagram. You can only distribute a maximum of 100 people.

i.e. The limitation is: Term A + Term B <= 100

Taking the max case of Term A + Term B = 100 and substituting in our "concern terms" equation:

We get Term B = 8, so Term A = 92.


2. Minimize Term A and Maximize Term B

Set Term A = 0 and Term B = 54 (This does not violate our limitation)

Great sum!

thanks a ton Abhay extremely well put
boy 780 doesnt seem too far for you mate!
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by albatross86 » Tue Jun 29, 2010 10:25 am
Thanks Raunak! I am glad my explanation was useful.

Let's see how the real thing goes, I'll be happy if I'm not rudely surprised!
~Abhay

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