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CAT Exam  >  CAT Tests  >  Logical Reasoning (LR) and Data Interpretation (DI)  >  Test: Venn Diagrams- 3 - CAT MCQ

Test: Venn Diagrams- 3 - CAT MCQ


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10 Questions MCQ Test Logical Reasoning (LR) and Data Interpretation (DI) - Test: Venn Diagrams- 3

Test: Venn Diagrams- 3 for CAT 2024 is part of Logical Reasoning (LR) and Data Interpretation (DI) preparation. The Test: Venn Diagrams- 3 questions and answers have been prepared according to the CAT exam syllabus.The Test: Venn Diagrams- 3 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Venn Diagrams- 3 below.
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Test: Venn Diagrams- 3 - Question 1
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A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is

Detailed Solution for Test: Venn Diagrams- 3 - Question 1

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

The number of members who can play exactly 1 game = II

The number of members who can play exactly 1 game = III

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

I + II + III = 256……(2)

⇒ II + 2III = 143…..(3)

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

⇒ III = 3 (From 3 and 4)

⇒ II =137

⇒ I = 116

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

Test: Venn Diagrams- 3 - Question 2
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Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18, choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is

Detailed Solution for Test: Venn Diagrams- 3 - Question 2

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.

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Test: Venn Diagrams- 3 - Question 3
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In a certain town, 40% of people have brown hair, 30% of people have brown eyes and 12% have both brown hair and brown eyes. How many people in town have neither brown hair nor brown eyes?

Detailed Solution for Test: Venn Diagrams- 3 - Question 3


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%

Test: Venn Diagrams- 3 - Question 4
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Each student in a class of 40 plays at least one indoor game-chess, carom and scrabble. 18 play chess, 20 play scrabble and 27 play carom. 7 play both chess and scrabble, 12 play both scrabble and carom and 4 play all 3 games. The number of players who play chess and carom but not scrabble is
Detailed Solution for Test: Venn Diagrams- 3 - Question 4

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

Test: Venn Diagrams- 3 - Question 5
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Read the information given below and answer the 2 associated questions.

190 students have to choose at least one elective and at most two electives from a list of three electives: E1, E2 and E3. It is found that the number of students choosing E1 is half the number of students choosing E2, and one third the number of students choosing E3.
Moreover, the number of students choosing two electives is 50.

In addition to the given information, which of the following information is NECESSARY and SUFFICIENT to compute the number of students choosing only E1, only E2 and only E3?
Detailed Solution for Test: Venn Diagrams- 3 - Question 5

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.

Test: Venn Diagrams- 3 - Question 6
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Read the information given below and answer the 2 associated questions.

190 students have to choose at least one elective and at most two electives from a list of three electives: E1, E2 and E3. It is found that the number of students choosing E1 is half the number of students choosing E2, and one third the number of students choosing E3.
Moreover, the number of students choosing two electives is 50.

Which of the following CANNOT be obtained from the given information?
Detailed Solution for Test: Venn Diagrams- 3 - Question 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,  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.

Test: Venn Diagrams- 3 - Question 7
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The number of people (in lakhs) who read only one newspaper is
Detailed Solution for Test: Venn Diagrams- 3 - Question 7

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

Test: Venn Diagrams- 3 - Question 8
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A survey was conducted of 100 people whether they have read recent issues of ‘Golmal’, a monthly magazine. Summarized information is presented below :
Only September: 18
September but not August: 23
September and July: 8
September: 28
July: 48
July and August: 10
None of the three months: 24
What is the number of surveyed people who have read exactly for two consecutive months?
Detailed Solution for Test: Venn Diagrams- 3 - Question 8

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

Test: Venn Diagrams- 3 - Question 9
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During the placement season of a class, 21 students got shortlisted for company A, 26 got shortlisted for Company B and 29 got shortlisted for company C and 14 students got shortlisted for both A and B, 12 students got shortlisted for A and C and 15 for both B and C. All the companies shortlisted 8 students from the class. Then what is the ratio of number of students who got shortlisted for only B and number of students who got shortlisted for only C?
Detailed Solution for Test: Venn Diagrams- 3 - Question 9


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

Test: Venn Diagrams- 3 - Question 10
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In a college, 200 students are randomly selected. 140 like tea, 120 like coffee, and 80 like both tea and coffee.

  • How many students like at least one of the beverages?
Detailed Solution for Test: Venn Diagrams- 3 - Question 10

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
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