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Test Level 2: Set Theory - 2 - CAT MCQ


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20 Questions MCQ Test - Test Level 2: Set Theory - 2

Test Level 2: Set Theory - 2 for CAT 2024 is part of CAT preparation. The Test Level 2: Set Theory - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 2: Set Theory - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 2: Set Theory - 2 below.
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Test Level 2: Set Theory - 2 - Question 1
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In a school, 100 students were offered Hindi as an optional subject and 70 students were offered French and Sanskrit each. Out of these, 63 were offered Hindi only, 47 were offered French only, 12 were offered Hindi and French only and 6 were offered French and Sanskrit only. The total number of students in the school is  

Detailed Solution for Test Level 2: Set Theory - 2 - Question 1


Total number of students = n(H ∪ S ∪ F)
= n(H) + n(S) + n(F) - n(H ∪ S) - n(S ∪ F) - n(H ∪ F) + n(H ∩ S ∩ F)
= 100 + 70 + 70 - 17 - 25 - 11 + 5 = 192
Or adding the various numbers in different regions in the Venn diagram, total number of students
= 63 + 12 + 47 + 25 + 6 + 39 = 192

Test Level 2: Set Theory - 2 - Question 2
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In a group of 1000 people, 750 people can speak Hindi and 400 people can speak English. In the group, all the people speak at least one out of the two languages. How many people can speak only Hindi?  

Detailed Solution for Test Level 2: Set Theory - 2 - Question 2


The Venn diagram for the situation is as shown above.
In the Venn diagram, H denotes the set of people who speak Hindi and E denotes the set of people who speak English.
So, a + x = 750 ... (i)
x + b = 400 ... (ii)
Adding equations (i) and (ii), we get a + x + b + x = 1150 ... (iii)
But a + x + b = 1000 ... (iv)
Putting this value of a + x + b from equation (iv) into equation (iii), we get
1000 + x = 1150
Or x = 150 ... (v)
Putting this value of x from equation (v) into equation (i) and solving for a, we get

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Test Level 2: Set Theory - 2 - Question 3
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The shaded area in the given diagram represents

Detailed Solution for Test Level 2: Set Theory - 2 - Question 3

Test Level 2: Set Theory - 2 - Question 4
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Out of 100 students in Holy Christ School of Science, 40 study Physics and 70 study Mathematics. If there are at least 20 students who study neither Mathematics nor Physics, what is the possible range of students who study both Mathematics and Physics?
Detailed Solution for Test Level 2: Set Theory - 2 - Question 4

Total students = Physics + Mathematics – Both 'Mathematics' and 'Physics' + Neither 'Mathematics' nor 'Physics'
Here's what we know:
Total = 100
Physics = 40
Mathematics = 70
Both (B) = Unknown
Neither (N) = Unknown
or we have
100 = 40 + 70 - Both + Neither
100 = 110 - B + N (Let's isolate 'B' since we are solving for possible values of 'B')
B = N + 10
Now, the minimum value of N is 20 and the maximum value can be 30.
Thus, the minimum value of B is 30 and the maximum value of B is 40.

Test Level 2: Set Theory - 2 - Question 5
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In a survey of population of 450 people, it is found that 205 people can speak English, 210 people can speak Hindi and 120 people can speak Tamil. If 100 people can speak both English and Hindi, 80 people can speak both English and Tamil, 35 people can speak Hindi and Tamil, and 20 people can speak all the three languages, then find the number of people who cannot speak English, Hindi, or Tamil.  
Detailed Solution for Test Level 2: Set Theory - 2 - Question 5

n(E ∪ H ∪ T) = n(E) + n(H) + n(T) - n(E ∩ H) - n(E ∩ T) - n(H ∩ T) + n(E ∩ H ∩ T)
= 205 + 210 + 120 - 100 - 80 - 35 + 20
= 340
Number of people who cannot speak English, Hindi, or Tamil = 450 - 340 = 110

Test Level 2: Set Theory - 2 - Question 6
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Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts; 210 bags contain almonds. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. How many bags contain only one kind of item?  
Detailed Solution for Test Level 2: Set Theory - 2 - Question 6


There are a total of 435 bags and 210 bags contain almonds.
From the diagram,
20y = 5x
y = x/4
Now, 435 = {Almonds} + 10x + y + x
435 = 210 + 10x + (x/4) + x
x = 20
Hence, number of bags that contain only one kind of item = 10x + x + 5x = 16x
= 320

Test Level 2: Set Theory - 2 - Question 7
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A merchant surveyed 400 people to determine the way they learned about an upcoming sale. The survey showed that 180 learned about the sale from radio, 190 from television, 190 from newspaper, 80 from radio and television, 90 from radio and newspaper, 50 from television and newspaper and 30 from all the three. How many people did not come to know about the sale from any of these sources?  
Detailed Solution for Test Level 2: Set Theory - 2 - Question 7

We can calculate by using the formula,
n(a) + n(b) + n(c) - n(a ∩ b) - n(b ∩ c) - n(c ∩ a) + n(a ∩ b ∩ c)
n(a ∪ b ∪ c) = 180 + 190 + 190 - 80 - 90 - 50 + 30 = 370
Number of People did not come to know about the sale from any of these sources = (400 - 370) = 30
Thus, 30 people did not come to know about the sale from any of these sources.

Test Level 2: Set Theory - 2 - Question 8
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In a survey at Massachusetts Institute of Technology, 80 students were asked whether they used laptop or desktop or both. Of these, 50 students said they used desktop and 70 said they used laptop. What could be the respective minimum and maximum numbers of students who used both desktop and laptop?
Detailed Solution for Test Level 2: Set Theory - 2 - Question 8

n (A ∩ B) = n(A) + n(B) - n (A ∪ B)

For min n (A ∩ B), n (A ∪ B) should be maximum, which is 80.
Hence, n(A ∩ B) min = 50 + 70 - 80 = 40
Also, A ∩ B is maximum when all desktop users are also laptop users. Hence, 50.

Test Level 2: Set Theory - 2 - Question 9
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An investigator interviewed 100 students to determine their preferences for the three drinks: milk (M), coffee (C) and tea (T). He reported the following: 10 students liked M, C and T, 20 liked M and C only, 30 liked C and T, 25 liked M and T, 12 liked M only, 5 liked C only and 8 liked T only. How many did not like any of the three drinks?
Detailed Solution for Test Level 2: Set Theory - 2 - Question 9


Number of people who liked at least one drink = 12 + 15 + 20 + 10 + 20 + 5 + 8 = 90
Number of people who liked none of the drinks = 100 - 90 = 10

Test Level 2: Set Theory - 2 - Question 10
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The result of an exam is given below.
Out of 1000 students who appeared- 
(i) 658 failed in Physics
(ii) 166 failed in Physics and Chemistry
(iii) 372 failed in Chemistry and 434 failed in Physics and Maths
(iv) 590 failed in Maths and 126 failed in Maths and Chemistry

Find the number of students who failed in Physics but not in Chemistry or Maths.
Detailed Solution for Test Level 2: Set Theory - 2 - Question 10


Let number of students who failed in all the subjects be x.
1000 = 658 + 372 + 590 - 166 - 126 - 434 + x
x = 106
Number of students who failed in Physics but not in Chemistry or Maths = 658 - 60 - 106 - 328 = 164

Test Level 2: Set Theory - 2 - Question 11
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A survey was conducted to know the number of households who owned either car or bike or both car and bike. It was determined that out of 200 households that were surveyed, 80 used neither bike nor car, 60 used only bike, and for every household that used both car and bike, 3 used only car. How many of the 200 households surveyed used both car and bike?"
Detailed Solution for Test Level 2: Set Theory - 2 - Question 11

Total = 200
Neither having car nor bike = 80
Only bike = 60
Only car = 3 (using both car and bike)
Only bike + Only car + Neither of both + Both car and bike = Total
60 + 3 x {Both car and bike} + 80 + {Both car and bike} = 200
4 x {Both car and bike} = 60
Therefore, having both car and bike = 15

Test Level 2: Set Theory - 2 - Question 12
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Visitors to a modern art museum who watched a certain Picasso painting were asked to fill in a short questionnaire, indicating whether they had enjoyed looking at the picture, and whether they felt they had understood it. According to the results of the survey, all 120 visitors who did not enjoy the painting also did not feel they had understood the painting, and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting. If 3/4th of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting, how many visitors answered the questionnaire? 
Detailed Solution for Test Level 2: Set Theory - 2 - Question 12

Let the total number of people who filled the questionnaire be x. 
If 120 didn't enjoy; then x - 120 enjoyed. 
Also (3x/4) = those who enjoyed and understood 
Total = x - 120 + x - 120 - (3x/4) + 120 = x 
⇒ (5x/4) - 120 = x 
⇒ (x/4) = 120 
⇒ x = 480

Test Level 2: Set Theory - 2 - Question 13
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In a survey of 100 people about their movie-going habits, it was found that 55 people had seen movie X, 45 people had seen movie Y, 50 people had seen movie Z, 10 people had not seen any of the movies and 20 people had seen all the three movies (X, Y and Z). How many people had seen only two of the movies?
Detailed Solution for Test Level 2: Set Theory - 2 - Question 13

{Total} = {Movie X} + {Movie Y} + {Movie Z} - {Exactly 2 movies} - 2 × {All 3 movies} + {No movies}
Let 'n' denote the number of people who had seen only two of the movies.
Therefore,
100 = 55 + 45 + 50 - n - (2 × 20) + 10
Or, n = 20
Thus, number of people who had seen only two of the movies is 20.

Test Level 2: Set Theory - 2 - Question 14
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In a survey, 500 people were asked two questions. They were simply to answer 'yes' or 'no'. Of these, 410 people said yes to the first question and 220 people said yes to the second. What is the least number that said yes to both?
Detailed Solution for Test Level 2: Set Theory - 2 - Question 14


n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
= 410 + 220 - n(A ∪ B)
To minimise intersection, we need to maximise union, which could be most 500. Putting the value,
We get n(A ∩ B) = 410 + 220 - 500 = 130

Test Level 2: Set Theory - 2 - Question 15
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Every student of a certain school must take one and only one elective course. Last year, 1/2 of the students took biology as an elective subject, 1/3 of the students took chemistry, and the rest took physics. This year, 1/3 of the students who took biology and 1/4 of the students who took chemistry left school. No other student left and no fresh student was admitted. What fraction of all students took biology and chemistry?
Detailed Solution for Test Level 2: Set Theory - 2 - Question 15

If total = 1, then:
Last year
B = 1/2
C = 1/3
P = 1/6
This year
B = (1/2) x (2/3) = 1/3
C = (1/3) x (3/4) = 1/4
Total = (1/3) + (1/4) = 7/12

Students left this year:
B = (1/2) x (1/3) = (1/6)
C = (1/3) x (1/4) = (1/12)
Total = (1/6 + 1/12) = (1/4)

Students at the school this year
= Number of students last year - Number of students left this year
= 1 - (1/4)
= (3/4)
Fraction of all students who took biology and chemistry = (7/12) ÷ (3/4) = 7/9

Test Level 2: Set Theory - 2 - Question 16
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In a survey on political preference, 78% of those asked were in favour of at least one of the proposals I, II and III. 50% of those asked favoured proposal I, 30% of them favoured proposal II, and 20% of them favoured proposal III. If 5% of them favoured all three proposals, then what percentage of those asked favoured more than one of the three proposals?
Detailed Solution for Test Level 2: Set Theory - 2 - Question 16


Let the total number of people surveyed be 100.
Number of people in favour of at least one proposal = 78
From Venn diagram:
Number of people in favour of exactly two proposals = x + y + z
Number of people in favour of all three proposals = 5
Number of people in favour of only proposal I = 50 - (x + y + 5)
Number of people in favour of only proposal II = 30 - (y + z + 5)
Number of people in favour of only proposal III = 20 - (x + z + 5)
According to the question:
78 = 50 - (x + y + 5) + 30 - (y + z + 5) + 20 - (x + z + 5) + (x + y + z) + (5)
⇒ 78 = 100 - (x + y + z) - 10
⇒ (x + y + z) = 12
Number of people who favoured more than one proposal = Number of people in favour of exactly two proposals +
Number of people in favour of all three proposals
= x + y + z + 5
= 12 + 5 = 17
Required percentage = 17%

Test Level 2: Set Theory - 2 - Question 17
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If X = {8n - 7n - 1 : n ∈ N} and Y = {49(n - 1) : n ∈ N}, then
Detailed Solution for Test Level 2: Set Theory - 2 - Question 17

We have: 8n - 7n - 1 = (7 + 1)n - 7n - 1 = (nC2 72 + nC3 73 + … nCn 7n)
= 49(nC2 + nC3 7 + … + nCn 7n - 2) for n ≥ 2. For n = 1, 8n - 7n - 1 = 0.
Thus, 8n - 7n - 1 is a multiple of 49 for n ≥ 2 and 0 for n = 1.
Hence, X consists of all positive integral multiples of 49 of the form 49 Kn, where Kn = nC2 + nC3 7 + … + nCn 7n - 2 together with zero.
Also, Y consists of all positive integral multiple of 49, including zero. Therefore, X ⊂ Y.

Test Level 2: Set Theory - 2 - Question 18
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Directions: Study the given information and answer the following question.
In a survey of 500 customers, the numbers of customers who buy different grades of rice are given below:

Calculate the approximate percentage of customers who buy all three grades of rice.
Detailed Solution for Test Level 2: Set Theory - 2 - Question 18


The Venn diagram for the same is as shown.
Percentage of consumers who buy all the 3 grades 

Test Level 2: Set Theory - 2 - Question 19
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Directions: Study the given information and answer the following question.
In a survey of 500 customers, the numbers of customers who buy different grades of rice are given below:

The number of customers who buy A or B, but not C, is
Detailed Solution for Test Level 2: Set Theory - 2 - Question 19


Number of customers who buy A or B, but not C = 90 + 50 + 0 = 140

Test Level 2: Set Theory - 2 - Question 20
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Directions: Read the given information and answer the following question. 

Ghosh Babu is staying at Ghosh Housing Society, Aghosh Colony, Dighoshpur, Kolkata. In Ghosh Housing Society, 6 persons daily read Ganashakti and 4 persons read Anand Bazar Patrika. In this society, there is no person who reads both the newspapers. The total numbers of persons who read these two newspapers in Aghosh Colony and in Dighoshpur are 52 and 200, respectively. The numbers of persons who read Ganashakti in Aghosh Colony and in Dighoshpur are 33 and 121, respectively, while the numbers of persons who read Anand Bazar Patrika in Aghosh Colony and in Dighoshpur are 32 and 117, respectively.

Q. The number of persons in Dighoshpur who read only Ganashakti is
Detailed Solution for Test Level 2: Set Theory - 2 - Question 20



Ganashakti ∩ Anand Bazar Patrika = Ganashakti + Anand Bazar Patrika - [Ganashakti ∩ Anand Bazar Patrika]
Ganashakti ∩ Anand Bazar Patrika = 121 + 117 - 200 = 38

Ganashakti ∩ Anand Bazar Patrika = Ganashakti + Anand Bazar Patrika - [Ganashakti ∩ Anand Bazar Patrika]
Ganashakti ∩ Anand Bazar Patrika = 33 + 32 - 52 = 13
So, the number of persons in Dighoshpur who read only Ganashakti = 83

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