Question 1

Time: 00:00:00
If ∪ = {1, 3, 5, 7, 9, 11, 13}, then which of the following are 
subsets of U.

B = {2, 4}

A = {0}

C = {1, 9, 5, 13}

D = {5, 11, 1}

E = {13, 7, 9, 11, 5, 3, 1}

F = {2, 3, 4, 5}

CDE

CDE

DEF

DEF

BAD

BAD

ADF

ADF

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Question 2

Time: 00:00:00
In a group, there were 115 people whose proofs of identity were 
being verified. Some had passport, some had voter id and some
had both. If 65 had passport and 30 had both, how many had
voter id only and not passport?

30

30

50

50

80

80

None of the above

None of the above

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Question 3

Time: 00:00:00
Among a group of people, 40% liked red, 30% liked blue and 30% 
liked green. 7% liked both red and green, 5% liked both red and
blue, 10% liked both green and blue. If 86% of them liked at
least one color, what percentage of people liked all three?

0

0

6

6

8

8

None

None

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Question 4

Time: 00:00:00
In a class of 120 students numbered 1 to 120, all even numbered 
students opt for Physics, those whose numbers are divisible by
5 opt for Chemistry and those whose numbers are divisible by
7 opt for Math. How many opt for none of the three subjects?

19

19

41

41

21

21

57

57

26

26

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PrepInsta User

In this question, we have to find out the number of students who took at least one of the three subjects and subtract that number from the overall 120 to get the number of students who did not opt for any of the three subjects. Number of students who took at least one of the three subjects can be found by finding out A U B U C, Where A is the set of those who took Physics, B the set of those who took Chemistry and C the set of those who opted for Math. Now, AUBUC = A + B + C - (A n B + B n C + C n A) + (A n B n C) A is the set of those who opted for Physics = 120/2 = 60 students B is the set of those who opted for Chemistry = 120/5 = 24 C is the set of those who opted for Math = 120/7 = 17 The 10th, 20th, 30th..... numbered students would have opted for both Physics and Chemistry. Therefore, A n B = 120/10 = 12 The 14th, 28th, 42nd..... Numbered students would have opted for Physics and Math. Therefore, C n A = 120/14 = 8 The 35th, 70th.... numbered students would have opted for Chemistry and Math. Therefore, B n C = 120/35 = 3 And the 70th numbered student would have opted for all three subjects. Therefore, AUBUC = 60 + 24 + 17 - (12 + 8 + 3) + 1 = 79 Now, the number of students who opted for none of the three subjects = 120 - 79 = 41

Question 5

Time: 00:00:00
Of 60 students in a class, anyone who has chosen to study maths 
elects to do physics as well. But no one does maths and chemistry,
16 do physics and chemistry. All the students do at least one of
the three subjects and the number of people who do exactly one of
the three is more than the number who do more than one of the three.
What are the maximum and a minimum number of people who could
have done Chemistry only?

40, 0

40, 0

28, 0

28, 0

38, 2

38, 2

44, 0

44, 0

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PrepInsta User

put 0 and that satisfies min and put 44 i.e max

Question 6

Time: 00:00:00
John was born on Feb 29th of 2012 which happened to be a Wednesday.
If he lives to be 101 years old, how many birthdays would he
celebrate on a Wednesday?

3

3

4

4

5

5

1

1

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PrepInsta User

The answer is 4. As we know that the 29th February came after every four years.  So that 26 total birthday will be celebrated in 101 years. Also after every years day are moved by 1 so in every birthday  date will be moved by 5 days . This means wednesday will came  on every 7th birthday of the john . That day will be on 1st , 8th , 15th and 22th birthday.

PrepInsta User

He celebrates his birthday every 4 years. So totally 26 birthdays (including the original). Also, every birthday will be moved 5 days from the previous one (1 day per year and an extra for the leap year). So every 7th birthday will come back to Wednesday. So he celebrates birthday number 1, 8, 15, 22 i.e. 4 days on Wednesday (including the first)

Question 7

Time: 00:00:00
How many of the following statements have to be true?

i. No year can have 5 Sundays in the month of May and
5 Thursdays in the month of June.
ii. If Feb 14th of a certain year is a Friday, May 14th of
the same year cannot be a Thursday
iii. If a year has 53 Sundays, it can have 5 Mondays in the
month of May.

0

0

1

1

2

2

3

3

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Question 8

Time: 00:00:00
Set P comprises all multiples of 4 less than 500. Set Q comprises 
all odd multiples of 7 less than 500, Set R comprises all multiples
of 6 less than 500. How many elements are present in P∪Q∪R?

202

202

243

243

228

228

186

186

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Question 9

Time: 00:00:00
Set A = {2, 3, 5, 6, 7}, Set B = {a, b, c}. How many onto functions
can be defined from Set B to Set A?

2

2

3

3

4

4

None of the above

None of the above

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Question 10

Time: 00:00:00
In a class of 345 students, the students who took English, Math,
and Science are equal in number. There are 30 students who took
both English and math, 26 who took both math and science,
28 who took science and English and 14 who took all the 3 subjects.
Answer the following question according to the data given above.
How many students have taken only one subject?

286

286

124

124

246

246

108

108

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["0","40","60","80","100"]
["Need more practice! \r\n \r\n","Keep trying! \r\n \r\n","Not bad! \r\n \r\n","Good work! \r\n \r\n","Perfect! \r\n \r\n"]

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