# Miscellaneous - Previous Year CAT/MBA Questions

The best way to prepare for Miscellaneous is by going through the previous year **Miscellaneous omet questions**.
Here we bring you all previous year Miscellaneous omet questions along with detailed solutions.

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It would be best if you clear your concepts before you practice previous year Miscellaneous omet questions.

**XAT 2021 VARC | Miscellaneous omet Question**

Which of the following sentences uses a WRONG tag-question?

- (a)
He has few reasons for saying no to the match, has he?

- (b)
You like to play, don’t you?

- (c)
There’s little point in doing anything about the match, is there?

- (d)
Moreover, he plays well, isn’t it?

- (e)
Nobody has called for you, have they?

Answer: Option D

**Explanation** :

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**XAT 2020 QADI | Miscellaneous omet Question**

If A ʘ B = (A + B) × B, then what is (5 ʘ 2) ʘ 5 ?

- (a)
95

- (b)
275

- (c)
125

- (d)
74

- (e)
200

Answer: Option A

**Explanation** :

Given, A ʘ B = (A + B) × B

∴ (5 ʘ 2) = (5 + 2) × 2 = 14

Now, (5 ʘ 2) ʘ 5 = 14 ʘ 5

⇒ 14 ʘ 5 = (14 + 5) × 5 = 19 × 5 = 95

Hence, option (a).

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**IIFT 2019 QA | Miscellaneous omet Question**

At what time between 2.00 pm and 3.00 pm, the two arms of a watch are completely opposite to each other?

- (a)
2:40 pm

- (b)
2:44 pm

- (c)
2:45 pm

- (d)
2:47 pm

Answer: Option B

**Explanation** :

Since, the two arms are completely opposite to each other, if the hour hand is between 2 and 3, the minute hand has to be between 8 and 9.

Thus, the time (in minutes) has to be between 40 and 45.

Hence, from among the options, the time should be 2:44 pm

Hence, option (b).

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**XAT 2018 QADI | Miscellaneous omet Question**

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

Answer: Option E

**Explanation** :

There are two cases to be considered.

Since there is only one mandatory course, all students will register for this course.

Since each student registers for exactly 3 courses, it means that sum of registrations for all four courses will be = 3 × (number of students).

**Case 1**: The mandatory course is one of the courses with registrations 45, 55 or 70.

The only possibility is that the mandatory course has 70 registrations, i.e., total number of students is 70.

∴ The total number of students = 70 and the number of student-course combinations = 3 × 70 = 210

∴ 70 + 55 + 45 + X = 210

∴ X = 40

Therefore, the minimum registrations for a course = 40.

**Case 2**: The mandatory course is not one with registrations 45, 55 or 70.

If the total number of students is X, the mandatory course will have X registrations.

Now, X will definitely be greater than or equal to 70.

∴ Minimum number of registrations in this case = 45

Hence, option (e).

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**XAT 2017 QADI | Miscellaneous omet Question**

In a True/False quiz, 4 marks are awarded for each correct answer and 1 mark is deducted for each wrong answer. Amit, Benn and Chitra answered the same 10 questions, and their answers are given below in the same sequential order.

AMIT T T F F T T F T T F

BENN T T T F F T F T T F

CHITRA T T T T F F T F T T

If Amit and Benn both score 35 marks each then Chitra’s score will be:

- (a)
10

- (b)
15

- (c)
20

- (d)
25

- (e)
None of the above

Answer: Option A

**Explanation** :

The correct answer fetches 4 marks whereas the wrong answer fetches -1 marks.

Amit scored 35 marks each in these 10 questions.

Let the number of correct answers be ‘c’ and wrong answers be ‘w’

4c – w = 35

c + w = 10

⇒ c = 9, w = 1, which means 9 correct answers and 1 wrong.

∴ Both Amit and Ben gave 1 wrong answer each.

Amit and Ben marked different answers for 3^{rd} and 5^{th} questions.

⇒ Amit and Ben would’ve answered 3^{rd} or 5^{th} question wrong in any order.

∴ They marked all other answers correctly.

⇒ Chitra marked correct answers for 3 questions (i.e., 1^{st}, 2^{nd} and 9^{th})

Out of 3^{rd} and 5^{th} questions marked by Chitra one would be correct and the other wrong.

Hence, Chitra marks 4 answers correctly and remaining 6 were wrong.

∴ Chitra’s score = 4 × 4 – 6 × 1 = 10.

Hence, option (a).

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**Answer the next 2 questions based on the following information:**

In an innings of a T20 cricket match (a team can bowl for 20 overs) 6 bowlers bowled from the fielding side, with a bowler allowed maximum of 4 overs. Only the three specialist bowlers bowled their full quota of 4 overs each, and the remaining 8 overs were shared among three non-specialist bowlers. The economy rates of four bowlers were 6, 6, 7 and 9 respectively. (Economy rate is the total number of runs conceded by a bowler divided by the number of overs bowled by that bowler). This however, does not include the data of the best bowler (lowest economy rate) and the worst bowler (highest economy rate). The number of overs bowled and the economy rate of any bowler are in integers.

**XAT 2017 QADI | Miscellaneous omet Question**

Read the two statements below:

S1: The worst bowler did not bowl the minimum number of overs.

S2: The best bowler is a specialist bowler.

Which of the above statements or their combinations can help arrive at the minimum number of overs bowled by a non-specialist bowler?

- (a)
S1 only

- (b)
S2 only

- (c)
Either S1 or S2

- (d)
S1 and S2 in combination

- (e)
The minimum number of overs can be determined without using S1 or S2

Answer: Option E

**Explanation** :

There are 3 specialist bowlers and 3 non- specialists bowlers.

The three specialist bowlers bowled 4 overs each and 3 non-specialist bowlers would have bowled 8 overs.

8 overs can be bowled by 3 non-specialist bowlers when they bowl 3, 3 and overs.

This is the only possibility since maximum overs a non-specialist can bowl is 3 overs.

So, the least number of overs bowled by a non-specialist bowler would be 2 only.

So, none of the S1 and S2 required to answer the question.

Hence, option (e).

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**XAT 2017 QADI | Miscellaneous omet Question**

Read the two statements below:

S1: The economy rates of the specialist bowlers are lower than that of the non-specialist bowlers.

S2: The cumulative runs conceded by the three non-specialist bowlers were 1 more than those conceded by the three specialist bowlers.

Which of the above statements or their combinations can help arrive at the economy rate of the worst bowler?

- (a)
S1 only

- (b)
S2 only

- (c)
Either S1 or S2

- (d)
S1 and S2 in combination

- (e)
The economy rate can be calculated without using S1 or S2.

Answer: Option D

**Explanation** :

S1: The given economy rates of 4 bowlers are 6, 6, 7 and 9. So, the non-specialist bowlers would have 7, 9, x as their economy rates and specialist bowlers would have y, 6, 6 as their economy rates.

S2: The overs bowled by specialist bowlers would be 4, 4 and 4 each. The number of overs bowled by non-specialist bowlers would in any combination of 3, 3 and 2 each.

The runs given by specialist bowlers would be 6 × 4 + 6 × 4 + 4y = 48 + 4y …(1)

**Case 1**: For non-specialist bowlers, overs bowled are 3, 3, 2 and economy rate is 7, 9 and x respectively.

∴ The runs given by non-specialist bowlers = (7 × 3 + 9 × 3 + x × 2) = 48 + 2x ...(2)

According to the question: (2) – (1) = 1

∴ 2x – 4y = 1

Now, 2x – 4y cannot give an odd value.

**Case 2**: For non-specialist bowlers, overs bowled are 3, 3, 2 and economy rate is x, 7 and 9 respectively.

The runs given by non-specialist bowlers would be (9 × 2 + 7 × 3 + 3x) = 39 + 3x

According to the question: (2) – (1) = 1

∴ 39 + 3x = 48 + 4y + 1

This is satisfied for x =10 and y = 5

Thus, we need both the statements to get to worst economy rate of the bowler.

Hence, option (d).

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**XAT 2016 QADI | Miscellaneous omet Question**

Each day on Planet M is 10 hours, each hour 60 minutes and each minute 40 seconds. The inhabitants of Planet M use 10 hour analog clock with an hour hand, a minute hand and a second hand. If one such clock shows 3 hours 42 minutes and 20 seconds in a mirror what will be the time in Planet M exactly after 5 minutes?

- (a)
6 hours 18 minutes 20 seconds

- (b)
6 hours 22 minutes 20 seconds

- (c)
6 hours 23 minutes 20 seconds

- (d)
7 hours 17 minutes 20 seconds

- (e)
7 hours 23 minutes 20 seconds

Answer: Option B

**Explanation** :

For hours there are 10 division and 10 and 5 are opposite to each other.

When the mirror shows 3 hours 42 minutes and 20 seconds, the hour hand is between 3 and 4.

∴ The hour hand must be between 6 and 7

For minutes there are 60 division hence 60 and 30 minute mark will be opposite each other.

In the mirror the minute hand is between 42 minute mark and 43 minute mark.

∴ The minute hand must be between 17 minute mark and 18 minute mark

For seconds there are 40 divisions hence 40 and 20 second mark will be opposite each other.

In the mirror the second hand is at 20.

∴ The second hand will be at 20.

So, the actual time must be 6 hours 17 minutes and 20 seconds

So, after five minutes, the time will be 6 hours 22 minutes and 20 seconds.

**Alternately,**

The sum of the time visible in the mirror and the actual time should be equal to 10:00:00 hours.

10:00:00 – 03:42:20 = 06:17:20 hours

Thus, actual time is 6 hours 17 minutes and 20 seconds.

So, after 5 minutes the time will be 6 hours 22 minutes and 20 seconds

Hence, option (b).

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