CAT Venn Diagrams Questions

Let n, s, d and t be the number of students who likes none of the drinks, exactly one drink, exactly 2 drinks and all three drinks, respectively.

It is given, 

n + s + d + t = 100 ...... (1)

s + 2d + 3t = 73 + 80 + 52

s + 2d + 3t = 205 ...... (2)

(2)-(1), we get

d + 2t - n = 105

Maximum value t can take is 52, i.e. t = 52, d = 1 and n = 0

Minimum value t can take is 5, i.e. t = 5, d = 95 and n = 0 (This also satisfies equation (1))

Difference = 52 - 5 = 47

The answer is option A.

Now 23 students choose maths as one of their subject.

This means (MPC)+ (MC) + (PC)=23 where MPC denotes students who choose all the three subjects maths, physics and chemistry and so on.

So MC + PM =5 Similarly we have PC+ MP =7 

We have to find the smallest number of students choosing chemistry

For that in the first equation let PM=5 and MC=0. In the second equation this PC=2

Hence minimum number of students choosing chemistry will be (18+2)=20 Since 18 students chose all the three subjects.

Assume the number of members who can play exactly 1 game = I

The number of members who can play exactly 2 game = II

The number of members who can play exactly 3 game = III

I+2II+3III=144+123+132=399....(1)

I+II+III=256......(2)

subtracting (1) from (2), we get

=> II+2III=143.....(3)

Also, II+3III=58+25+63=146 ......(4)

subtracting (3) from (4), we get

=> III = 3 (From 3 and 4)

=> II =137

=> I = 116

The members who play only tennis = 123-58-25+3 = 43

 P = {1,2,3,4} and  Q = {2,3,5,6,}
PΔQ = {1, 4, 5, 6}
R = {1,3,7,8,9} and S = {2,4,9,10}
RΔS = {1, 2, 3, 4, 7, 8, 10}
(PΔQ)Δ(RΔS) = {2, 3, 5, 6, 7, 8, 10}
Thus, there are 7 elements in (PΔQ)Δ(RΔS) .
hence, 7 is the correct answer.

Let us draw a Venn diagram using the information present in the question. 

It is given that the number of students studying H equals that studying E. 

Let 'x' be the total number of students who studied H, and H and P but mot E.We can also say that the same will be the number of students who studied E, and E and P but not H.Therefore,

x + 20 + 10 + x = 74 

x = 22

Hence, the number of students studying H = 22 + 10+ 20 = 52

It has been given that among 200 students, 105 like pizza and 134 like burger.
The question asks us to find out the number of students who can be liking only burgers among the given values. 

The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too. 
In this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.

The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible. 
There are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95. 

The number of students who like only burger should lie between 29 and 95 (both the values are included). 
93 is the only value among the given options that satisfies this condition and hence, option D is the right answer. 

If we carefully observe set A, then we find that $$6^{2n} -35n - 1$$ is divisible by 35. So, set A contains multiples of 35. However, not all the multiples of 35 are there in set A, for different values of $$n$$.
For $$n = 1$$, the value is 0, for $$n = 2$$, the value is 1225 which is the 35th multiple of 3.
If we observe set B, it consists of all the multiples of 35 including 0.
So, we can say that every member of set A will be in B while every member of set B will not necessarily be in set A.
Hence, option A is the correct answer.

Let the areas be labelled as shown in the diagram above.

The number of people corresponding to "none of the three months" is 24. So, H is 24.

Only September is 18. So, G = 18

September but not August is 23. So, G + D = 23.

Hence, D = 23 - 18 = 5.

We know that September and July is 8. So, D + E = 8

This implies E = 3.

September = 28. So, D + E + F + G = 28.

So, F = 28 - 5 - 3 - 18 = 2.

July and August = 10.

So, B + E = 10.

E = 3. So, B = 7.

July = 48.

So, A + B + D + E = 48

A = 48 - 7 - 5 - 3 = 33.

There are 100 people in total. So, C = 100 - A - B - D - E - F - G - H = 100 - 33 - 7 - 5 - 3 - 2 - 18 - 24 = 8

So, number of people who read exactly two consecutive issues = (July & August) + (August & September) = B + F = 7 + 2 = 9

Total number of volunteers are 37

=> 2x+y+x+y+x+y+z+10-z-x+2x-1=37

=> 5x+2y+9=37

=> 5x+2y=28

Also we know that 3x + 2y = 20.

We get x=4, => y=4

We need to find the minimum value of 6-z, and it is given FR get the most number of volunteers, We get that z cannot be more than 2 because if it is 3 or above ER will have the maximum number of volunteers.

We can get the information mentioned in options B and C using the data given in the passage.

But, we need the information in option A to find the exact number of volunteers in various projects.

Hence, option A is the answer.

FR is greater than ER, thus z is 4,5,6.

If z=4,

FR=20, ER= 18

If z=5,

FR=21, ER= 17

If z=6,

FR=22, ER= 16

Now even if we transfer one of the volunteers opted out of the TR project thus he will now be handling FR and ER, and one opted out of the ER project will be handling TR and FR, while the remaining ones involved in all the three projects opted out of the FR project.

The worst-case scenario will be when z=4. Now 2 are out of FR, 1 from ER and 1 from TR.

FR=20-2=18 and ER=18-1=17

So we can see that even if volunteers are withdrawn, the number of volunteers in FR are more as in, total if we calculate comes out to be more. Hence option B. 

Consider p volunteers be added to TR project and q be added to each of FR and ER projects.

Then, 7 + p = 8 + q => p = q + 1

Also, Number of volunteers working on TR = 7 + q + 1 + 4 + 5 = 17 + q 

Number of volunteers working on FR = 17 + q 

Number of volunteers working on ER = 18 + q.

So if we take any values we get ER  greater than both FR and TR. 

Let the number of total days=N
They played tennis for=N-14  days
They did yoga for =N-24 days

And the question says that total days when they did yoga or played tennis are 22

which means

N-14 + N-24 = 22

2N - 38 = 22

2N = 60

N = 30

Hence total days they stayed together were 30

6+x+22+x+6y/5 = 50
=> 5x + 3y = 55
Since y/5 is even, y should be a multiple of 10
The only possible value is 10
So, y = 10 and x = 5
No. of oak leaves = 17

Let the distribution of votes for each of the proposal be as given below.
From the information given, we know that 

a+b+c+d+e+f+g = 78 --- (1)
a+b+e+f = 50  ---- (2)
b+c+f+g = 30 ----  (3)
e+f+g+d = 20 ---- (4) and 
f = 5 --- (5)

We need to find b+e+g+f = ?
In the above equations, (2)+(3)+(4) - (1) implies

(a+b+e+f)+(b+c+f+g)+(e+f+g+d) - (a+b+c+d+e+f+g) = 50+30+20-78 = 22
Or, b+e+g+2f=22.
As, f = 5, it implies that b+e+g+f=17

Let a be the number who watch only one channel, b be the number who watch only 2 channels and c be the number who watch all channels.

a+b+c = 100

a+2b+3c = 80+22+15  =117

Subtracting both equations,

b+2c = 117-100 = 17

Maximum c occurs when b = 0

2c = 17

c = 8.5

Applying AUBUC formula

Let x be the number who watch BBC and CNN and y be the number who watch all three channels.

100 = 80+22+15-(10+5+x)+y

x-y = 2

Hence only 2% people watch BBC and CNN only.

Applying AUBUC formula

Let x be the number who watch BBC and CNN and y be the number who watch all three channels.

100 = 80+22+15-(10+5+x)+y

x-y = 2

We cannot find the exact value of y.

Hence, the answer is "cannot be determined".

Let the distribution of people having cable TV and VCR be as given in the diagram above.
Hence, $$a+c =\frac{2}{3}$$
$$b+c=\frac{1}{5}$$
and $$c=\frac{1}{10}$$

We need to find $$a+b+c=?$$
This equals $$(a+b)+(b+c)-c = \frac{2}{3}+\frac{1}{5}-\frac{1}{10}$$
Which equals $$\frac{20+6-3}{30}=\frac{23}{30}$$

Number of people owning exactly 2 articles = 9

Number of people owning exactly 3 articles = 1

Applying AUBUC formula, we get

AUBUC = 22+15+14 - 9 -2*(1)=40

Number of people who do not own any article = 50-40 = 10

We know that x + y + z = T and x + 2y + 3z = R, where
x = number of members belonging to exactly 1 set = 70
y = number of members belonging to exactly 2 sets
z = number of members belonging to exactly 3 sets = 10
T = Total number of members
R = Repeated total of all the members = (40+50+60) = 150
Thus we have two equations and two unknowns. Solving this we get y = 25

So, 25 people belong to exactly 2 clubs.

The diagram for this question has been shown:

Total number of schools having either or LAB or LIB or both = a+b+x/2 - y + y + 3x = 7x/2 + a + b = 35

Here a = b = y = 0

7x/2 = 35

x = 10

Total number of schools having at least one of PG, LIB or LAB = 30+2x+x+x/2 = 30+3x+x/2 = 30+30+5 = 65

Number of schools having neither of the three = 100-65 = 35

The diagram for this question has been shown:

Total number of schools having either or LAB or LIB or both = a+b+x/2 - y + y + 3x = 7x/2 + a + b = 35

It has been given that the schools having playground don't have a Library or Laboratory. 

Hence  a = b = y = 0

7x/2 = 35

x = 10

Required ratio = 25:15 = 5:3