Top 10+ XAT Venn Diagram Questions with Solutions PDF

Question 1

In a business school, only three specializations are offered: marketing, finance and operations. A student can opt for either one, or two, or no specializations. In the current batch of 120 students, 60 are specializing in marketing, 50 are specializing in finance and 30 are specializing in operations. During the placement process, all students specializing either in marketing or in finance are shortlisted for consulting job interviews.
If 45 students are not shortlisted for consulting job interviews, what is the MINIMUM possible number of students specializing in both marketing and finance?

A

35

B

50

C

60

D

45

E

40

Video Solution

Solution

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In the above figure, let the number of students specializing in :

Only marketing = a, Only finance = b, Only operations = c

Both marketing and finance but not operations = d, Both marketing and operations but not finance = e, Both finance and operations but not marketing = f

All three subjects = g, None of the subjects = h

Now it is given that only students having an specialization in marketing or finance gets a consulting shortlist. So, we can say that students doing specialization only in operations and students not doing specialization in any of the 3 subjects will not get a shortlist.

Hence, c + h = 45 (given) $$\longrightarrow\ i$$

Number of students doing specialization in marketing = a + d + e + g = 60 (given) $$\longrightarrow\ ii$$

Number of students doing specialization in finance = b + d + g + f = 50 (given) $$\longrightarrow\ iii$$

Number of students doing specialization in operations = c + e + g + f = 30 (given) $$\longrightarrow\ iv$$

Number of students who received the consulting shortlist = a + b + d + e + f + g = 75 $$\longrightarrow\ v$$

By using equation iv and v, we get,

a + b + d + 30 - c = 75

a + b + d = 45 + c $$\longrightarrow\ vi$$

Add equation ii and iii,

a + b + d + (d + g) + e + f + g = 110 $$\longrightarrow\ vii$$

Now, using equation iv, vi and vii, we get,

45 + c + (d + g) + 30 - c = 110

d + g = 35

Hence, the number of students specializing in marketing and finance = 35

$$\therefore\ $$ The required answer is A.

Question 1

In a small college, students are allowed to take only one specialization. Traditionally, only two specializations are offered: Science and Arts. Students enrolled to specialize in Science must take Physics and Mathematics subjects, while students enrolled to specialize in Arts must take Economics and Political Science subjects. Students enrolled in Science are not allowed to take either Economics or Political Science, while students enrolled in Arts are not allowed to take either Physics or Mathematics.

Recently, the college has started a third specialization called MatEco that requires students to take Economics and Mathematics. However, MatEco students would not be allowed to take either Physics or Political Science. When the college opens this new specialization for enrolment, it allows students, originally enrolled in Science or Arts, to switch to MatEco.

From among the students originally enrolled in Arts, 20 students switch to MatEco. This makes the number of Science students twice the number of Arts students. After this, from among the students who originally enrolled in Science, 45 students switch to MatEco. This makes the number of Arts students twice the number of Science students.

In total, how many students, from among those originally enrolled in Science or Arts, are now taking Economics?

A

45

B

65

C

80

D

95

E

None of the remaining options is correct.

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Solution

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Let's assume that the number of students who initially took Arts is x and the number of students who initially took Science is y.

After 20 students shifted from Arts to MatEco, the number of Science students was twice that of Arts students.

That is y=2*(x-20) => 2x-y-40=0 -->1

Now, 45 of the students who originally enrolled in Science have switched to MatEco. This makes the number of Arts students twice the number of Science students.

So, x-20=2*(y-45) => x-2y+70=0 -->2

Now that we have two linear equations. Let's solve them.

Multiply Equation 1 with -2 and add it to Equation 2.

-4x+2y+80=0

x-2y+70=0

-3x+150=0

3x=150=>x=50

So, the original number of students enrolled in Arts is 50.

Substitute the value of x in equation 1 to get the y value.

y=60

There are 65 (20 from arts, 45 from Science who shifted to the MatEco) in MatEco. All these study Economics; apart from them, the people currently in Arts also study Economics. There are 50-20=30 people currently in Arts. So, the total number of people who study Economics is 65+30=95.

Instructions

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.

Question 1

Which of the following CANNOT be obtained from the given information?

A

Number of students choosing E1

B

Number of students choosing either E1 or E2 or both, but not E3

C

Number of students choosing both E1 and E2

D

Number of students choosing E3

E

Number of students choosing exactly one elective

Solution

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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

Also, E1 = (x+a+0+b) , E2 = (y+b+0+c) , E3 = (z+a+0+c) ----------------2

From 1, E1+E2+E3 = 6k and from 2, E1+E2+E3 = (x+y+z)+2(a+b+c) = 140 + 2*50 = 240

=> 6k = 240

=> k = 40

Thus our venn diagram will look like this -

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. Or, it is equal to (40-a-b) + b + (80-b-c) = 120 - a - b - c = 120 - (a+b+c), and (a+b+c) = 50. Thus, the number of students is = 120 - 50 = 70.

Option C, number of students choosing both E1 and E2 => this can not be obtained, because it is equal to b and the value of b is not known.

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

Option E, number of students choosing exactly one elective = Out of 190, 50 are choosing two electives, hence 190-50 = 140 are choosing exactly one elective. Or, it is equal to (40-a-b) + (80-b-c) + (120-a-c) = 240 - 2(a+b+c) = 240 - 2(50) = 140.

Instructions

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.

Question 2

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?

A

Number of students choosing only E2, and number of students choosing both E2 and E3

B

Number of students choosing both E1 and E2, number of students choosing both E2 and E3, and number of students choosing both E3 and E1

C

Number of students choosing only E1, and number of students choosing both E2 and E3

D

No extra information is necessary

E

Number of students choosing both E1 and E2

Solution

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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 ---------------1

Also, E1 = (x+a+0+b) , E2 = (y+b+0+c) , E3 = (z+a+0+c) ----------------2

From 1, E1+E2+E3 = 6k and from 2, E1+E2+E3 = (x+y+z)+2(a+b+c) = 140 + 2*50 = 240

=> 6k = 240

=> k = 40

Thus our venn diagram will look like this -

Option A: If the number of students choosing only E2 and the number of students choosing both E2 and E3 are given, then we know the value of (80-b-c) [ which will provide us with the value of (b+c) ], and the value of c. Since we know the value of c, we can get the value of b. And as the values of b and c are known, the value of a can also be known. Thus, we can find all the values.

Option B: Knowing the number of students choosing both E1 and E2, the number of students choosing both E2 and E3, and the number of students choosing both E3 and E1 is SUFFICIENT as we will get the values of a, b, and c. However, this information is not NECESSARY when calculating the number of students who choose only E1, E2, and E3 because we can calculate that with fewer conditions (like in option A).

Option C: Number of students choosing only E1, and number of students choosing both E2 and E3. Thus, we know the value of (40-a-b), which will provide us with the value of (a+b), and we know the value of c. However, the individual values of a and b cannot be known. Thus, we won't be able to find the number of students who choose only E1, only E2, and only E3.

Option D: Some extra information about a, b, and c is necessary. Thus, this option is incorrect.

Option E: Number of students choosing both E1 and E2. This will give us the value of b, and since we know the value of (a+b+c), thus we can find the value of a+c. But the individual values of a and c cannot be known. Thus, this is also incorrect.

Question 1

In the final semester, an engineering college offers three elective courses and one mandatory course. A student has to register for exactly three courses: two electives and the mandatory course. The registration in three of the four courses is: 45, 55 and 70. What will be the number of students in the elective with the lowest registration?

A

35

B

40

C

42

D

45

E

Either B or D

Video Solution

Question 1

In a class of 60, along with English as a common subject, students can opt to major in Mathematics, Physics, Biology or a combination of any two. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. In an English test, the average mark scored by students majoring in Mathematics is 45 and that of students majoring in Biology is 60. However, the combined average mark in English, of students of these two majors, is 50. What is the maximum possible number of students who major ONLY in Physics?

A

30

B

25

C

20

D

15

E

None of the above

Video Solution

Question 1

In an amusement park, a visitor gets to ride on three different rides (A, B and C) for free. On a particular day 77 opted for ride A, 55 opted for B and 50 opted for C; 25 visitors opted for both A and C, 22 opted for both A and B, while no visitor opted for both B and C. 40 visitors did not opt for ride A and B, or both. How many visited the amusement park on that day?

A

102

B

115

C

130

D

135

E

150

Video Solution

Question 1

70% of the students who joined XCRI last year play football, 75% play cricket, 80% play basketball and 85% play carrom. The minimum percentage of students who play all four games is:

A

5%

B

10%

C

15%

D

20%

E

None of the above

Video Solution

Question 1

There are 240 second year students in a B - School. The Finance area offers 3 electives in the second year. These are Financial Derivatives, Behavioural Finance, and Security Analysis. Four students have taken all the three electives, and 48 students have taken Financial Derivatives. There are twice as many students who study Financial Derivatives and Security Analysis but not Behavioural Finance, as those who study both Financial Derivatives and Behavioural Finance but not Security Analysis, and 4 times as many who study all the three. 124 students study Security Analysis. There are 59 students who could not muster courage to take up any of these subjects. The group of students who study both Financial Derivatives and Security Analysis but not Behavioural Finance, is exactly the same as the group made up of students who study both Behavioural Finance and Security Analysis. How many students study Behavioural Finance only?

A

29

B

30

C

32

D

35

E

None of the above

Video Solution

XAT Venn Diagrams Questions

XAT Venn Diagrams Questions

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