Test: Venn Diagrams- 3 - CAT MCQ

Let total number of people in the town = 100x
People who have both brown hair and eyes = b = 12x
⇒ People who have brown hair = a+ b = 40x
⇒ a = 28x
Similarly, c = 18x
∴ People who have neither brown hair nor brown eyes = 100x -(12x + 18x + 28x) = 42%

Students who play all the three games are 4, and with the rest information given in the question, we can at best draw the venn diagram as:

Now, total students = 40.

So, Chess+ 5 + 8 + 15 - x = 40

=> 18 + 28 - x = 40

=> 46 - x = 40

∴ x = 6

Given, a + b + c = 50 and a + b + c + x + y + z = 190 => x + y + z = 140.

Also, let E1 = k => E2 = 2k and E3 = 3k

E1 + E2 + E3 = 6k = 190 + 50 = 240 => k = 40.

Option A: If the number of students choosing only E2, the number of students
choosing both E2 and E3, are given then the number of students who choose E2 and E1, E1 and E3 can be found. From this only E1, only E3 can be calculated.

Option B: knowing the number of students choosing both E1 and E2, the number of students
choosing both E2 and E3, and a number of students choosing both E3 and E1 is insufficient. This information is not enough to calculate the number of students who choose only E1, only E2, and only E3.

Option C: If x and c are known, we can’t find y and z.

Option D is not sufficient.

Given, a + b + c = 50 and a + b + c + x + y + z = 190 => x + y + z = 140.

Also, let E1 = k => E2 = 2k and E3 = 3k

E1 + E2 + E3 = 6k = 190 + 50 = 240 => k = 40.

Option A,  the number of students choosing E1 is 40.

Option B, number of students choosing either E1 or E2 or both, but not E3 = total – E3 = 190-120 = 70.

Option C, number of students choosing both E1 and E2 => this can not be obtained.

Option D, number of students choosing E3 = 3x = 120.

Given Readership of newspaper X = 8.7

only X + 2.5 + 0.5 + 1 = 8.7

only X = 4.7

Given Readership of newspaper Y = 9.1

only Y + 2.5 + 1.5 + 0.5 = 9.1

only Y = 4.6

Given Readership of newspaper Z = 5.6

only Z + 0.5 + 1 + 1.5 = 5.6

only Z = 2.6

Number of people who read only one newspaper = 4.7 + 4.6 + 2.6 = 11.9 lakhs

Exactly two consecutive months include both July-August and August-September. We cannot include July-September, as these months are not consecutive.

N – none of the three months

Number of people who read in July and August only = 7

Number of people who read in August and September only = 2

Therefore, the number of surveyed people who have read exactly for two consecutive months = 7+2 = 9

It is given that 200 candidates scored above 90th percentile overall in CET. Let the following Venn diagram represent the number of persons who scored above 80 percentile in CET in each of the three sections:

From 1, h = 0.
From 2, d + e + f = 150
From 3, a = b = c
Since there are a total of 200 candidates, 3a + g = 200 – 150 = 50
From 4, (2a + f) : (2a + e) : (2a + d) = 4 : 2 : 1
Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1 = 7.
Since d + e + f = 150, 6a + 150 is divisible by 7, i.e., 6a + 3 is divisible by 7. Hence, a = 3, 10, 17, . . .
Further, since 3a + g = 5 0, a must be less than 17. Therefore, only two cases are possible for the value of a, i.e., 3 or 10.
We can calculate the values of the other variables for the two cases.
a = 3 or 10
d = 18 or 10
e = 42 or 40
f = 90 or 100
g = 41 or 20

The number of candidates sitting for separate test for BIE who were at or above 90th percentile in CET (a) is either 3 or 10.

Given e = 8

Number of students shortlisted for A and B is 14

e + b = 14

b = 6

Number of students shortlisted for A and C is 12

e + d = 12

d = 4

Number of students shortlisted for B and C is 15

e + f = 15

f = 7

Given

Number of students shortlisted for B is 26

b + f + e + c = 26

6 + 7 + 8 + c = 26

c = 5

Number of students shortlisted for C is 29

d + e + f + g = 29

g = 10

Number of students shortlisted for A is 21

b + e + d + a = 21

a = 3

Number of students shortlisted for only B = c = 5

Number of students shortlisted for only C = g = 10

Ratio = 5:10 = 1:2

The following Venn diagram may represent the given information, where T = tea and C = coffee.

• Number of students who like only tea = 60
• Number of students who like only coffee = 40
• Number of students who like at least one of tea or coffee = n (only Tea) + n (only coffee) + n (both Tea & coffee) . So, 60 + 40 + 80 = 180

Anyone who does Maths does Physics also.

Maths is a subset of Physics. Now, let us build on this.
No one does maths and chemistry, 16 do physics and chemistry.

Number outside is 0 as all the students do at least one of the three subjects
a + b + c +16 = 60, or a + b + c = 44

The number of people who do exactly one of the three is more than the number who do more than one of the three. => a + b > c + 16
So, we have a + b + c = 44 and a + b > c + 16
We need to find the maximum and minimum possible values of b.
Let us start with the minimum. Let b = 0, a + c = 44. a > c + 16. We could have
a = 40, c = 4. So, b can be 0.
Now, thinking about the maximum value. b = 44, a = c = 0 also works.

So, minimum value = 0, maximum value = 44.

Hence, the answer is "44 and 0".
Choice D is the correct answer.