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]

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.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 .......

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

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)...

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

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!

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%.

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!

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!