Directions for questions 21 to 24:
In Spectras Academy, there are 200 students, 100 are learning Quant, 50 Verbal, 60 LR. 30
students are learning both Quant and Verbal, 35 students are learning both Verbal and LR, and
45 students are learning both Quant and LR.
21.
22.
23.
24.
What is the maximum number of students learning atleast one thing?
1. 100
2. 130
3. 60
4. 120
What is the maximum number of students learning everything?
1. 20
2. 15
3. 30
4. 40
What is the minimum number of students learning atleast one thing?
1. 100
2. 130
3. 120
4. 60
What is the minimum number of students learning everything?
1. 20
2. 15
3. 30
4. 40
Directions for questions 25 and 26:
In Therons Workshop, where participating in atleast one activity is compulsory, 78% of the
participants participate in Chattersport, 69% participate in Hittersport and 87% participate in
Shuddersport.
25.
What is the maximum percentage of participants participating in all three sports?
1. 67
2. 66
3. 34
4. 74
26.
What is the minimum percentage of participants participating in all three sports?
1. 66
2. 34
3. 44
4. 21
APPENDIX I: THE GREAT LIMITS APPROACH
I have noticed a lot of people asking exactly how to solve problems based on minima-maxima of venn
diagram based problems directly through venn diagram, rather than using equations that tend to be
messy and prone to errors.
I reveal an analysis of sets that you may or may not have read elsewhere, but will definitely not harm
your preparation.
In the fig alongside, the venn diagram of three sets
have been shown: set A, set B, set C.
The intersecting areas have been marked
accordingly acc to standard nomenclature
applicable.
X -> only (single set)
Y -> exactly (two sets)
Z -> all (three sets)
a -> set A
b -> set B
c -> set C
Notice that areas Xa, Xb, Xc and Z have been marked in green: AREA 1
that areas Yab, Ybc, Yac have been marked in blue: AREA 2
This is because these are the interconnected regions in the venn diagram. Any change in either part of
these two areas causes a direct change in other parts of the same area, and inverse change in parts of
the other area.
Example, if Z were to increase by 10 => 1. Xa, Xb and Xc increase by 10 respectively &
2. Yab, Ybc, and Yac decrease by 10 respectively
Notice that there are 4 parts to area 1 and 3 parts to area 2. That means the behavior (increase or
decrease) of the venn is determined completely by AREA 1.
Continuing our example from above, if Z were to increase by 10, there would be an overall change in
number of elements in atleast one set by
= (effect of area 1)*4 + (effect of area2)*3 = (effect of area1)*4 – (effect of area1)*3 = (effect of area1)
Thus, as per our example, if Z increases by 10, the number of elements in atleast 1 set increases by 10.
The only limits to be kept in mind are that the minimum Yab,Ybc,Yca can be 0 and Xa,Xb, Xc can be 0
respectively.
What this effectively means is that, you can simply adjust the values of Z to get the limits of
1. elements belonging to atleast one set
2. elements belonging to all sets.
This also means that when you calculate the maximum number of elements in all three sets, you get the
maximum value of elements in atleast one set.
Similarly, when you calculate the minimum number of elements in all three sets, you get the minimum
value of elements in atleast one set.
Given n(A), n(B), n(C), n(A and B), n(B and C), n(C and A) 1. To find the maximum no of elements in all three sets:
We have to take Z to the maximum. This means that as Z increases, each of the Ys will decrease
by an equivalent amount each. But Z can only increase to a maximum value where atleast one of
the Ys = 0
This means that
=> To find the maximum no of elements belonging to all 3 sets: Assume Z = Min(n(A and B), n(B
and C), n(C and A)) and calculate the remaining values in the venn diagram using the given
values.
Tip: If given the total number of elements in atleast one set, calculate this value with your
assumed Z in the venn, and make sure this does not exceed the given value (data). If it does,
reduce Z by the difference between your calculated union value and the given union value.
2. To find the maximum no of elements in atleast one set:
We have found the Maximum Value of Z in the above step. We also know that the maximum
number of elements in all three sets gives the maximum value of elements in atleast one set.
Thus, simply calculate the values of the rest of the parts in the venn diagram and add them all
up to get this value.
Tip: Make sure this does not exceed the given value (total). If it does, reduce Z by the difference
between your calculated union value and the given union value.
3. To find the minimum no of elements in all three sets:
We have already found the max value of Z. We also know that a decrease in the value of Z, will
cause a corresponding equal decrease in the Xs, and a similar increase in the Ys.
So we can decrease the value of Z till we get one of the Xs = 0
This means that
=> To find the minimum no of elements belonging to atleast one set: Decrease the calculated
Zmax until calculated values of min(Xa,Xb,Xc) = 0. This will be the minimum value.
4. To find the minimum no of elements in atleast one set:
We have found the Minimum Value of Z in the above step. We also know that the minimum
number of elements in all three sets gives the minimum value of elements in atleast one set.
Thus, simply calculate the values of the rest of the parts in the venn diagram and add them all
up to get this value.
Explanations for Q21. To 24.
The basic Venn diagram is shown below, and we have utilized the given values in the question as
follows:
According to Step 1: =======================
Assume Z = min(30,35,45) = 30
We will obtain the following scenario – (bottom)
As we see in the fig to the left, if we go any
further above Z=30, Yqv < 0 (unacceptable)
Zmax = 30
Max n(Atleast one set) = 100+10+5+15 = 130
Since 130 < 200, this value is acceptable. So in
one iteration, we have obtained the maximum values for elements belonging to (atleast one) and (all
three) sets.
Now, let us move onto the next step.
We notice that min(Xq,Xv,Xl) = 10
If we were to reduce Zmax by 10, Xv = 0. If Z<20, Xv<0 (unacceptable)
Thus we have obtained Zmin = 20
We will obtain the following scenario: -------
Thus, we have obtained by calculating values as
Min n(Atleast one set) = 100+5+15 = 120
Notice the correlation between the numbers as
indicated in the analysis of behavior of the
venn!
MAXIMUM
MINIMUM
ALL THREE
30
20
ATLEAST ONE
130
120
As indicated by behavior analysis, an increase
in (common ie all three) leads to a
corresponding increase in the behavior of the
entire venn! (atleast one)
Hope this helped you understand and clear a lot of doubts!
APPENDIX II: THE GREAT BAR GRAPH APPROACH
For problems where the values given are not complete, ie no n(A and B) etc. the conventional venn
approach may not work. Hence, we need to adopt a newer method to beat it! This is where the bar
graph approach can work wonders!
There will be no analysis, but a simple and effective answer to a question.
Explanation for Q 25:
Explanation for Q 26: