# Miscellaneous - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Miscellaneous. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2017 QA Slot 2 | Miscellaneous**

The numbers 1,2, …, 9 are arranged in a 3 × 3 square grid in a such way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.

If the top left and the top right entries of the grid are 6 and 2, respectively, then the bottom middle entry is

Answer: 3

**Explanation** :

If entries along each of the rows, columns or diagonals adds upto the same number, it implies that the averages of each row, columns or diagonal will be 45 ÷ 3 or 15.

(∵ there are 3 rows and 3 columns we divide the total of the numbers 1 to 9 i.e. 45 by 3)

In the 1st row the top left cell contains the number 6 and the top right cell contains the number 2.

Therefore the middle element in the first row will be 15 – 6 – 2 = 7. Further elements of each row, columns or diagonal adds upto 15. Now this is only possible if the middle element in the entire grid i.e. the element corresponding to the 2nd row and 2nd column is the average of all 9 numbers i. e., 5. Now if the middle element is 5, then the bottom element in the 2nd column will be 15 – 7 – 5 or 3.

Hence, 3.

Workspace:

**Answer the following question based on the information given below.**

Five horses, Red, White, Grey, Black and Spotted participated in a race. As per the rules of the race, the persons betting on the winning horse get four times the bet amount and those betting on the horse that came in second get thrice the bet amount. Moreover, the bet amount is returned to those betting on the horse that came in third, and the rest lose the bet amount. Raju bets Rs. 3000, Rs. 2000 Rs. 1000 on Red, White and Black horses respectively and ends up with no profit and no loss.

**CAT 2008 QA | Miscellaneous**

Suppose, in addition, it is known that Grey came in fourth. Then which of the following cannot be true?

- A.
Spotted came in first

- B.
Red finished last

- C.
White came in second

- D.
Black came in second

- E.
There was one horse between Black and White

Answer: Option C

**Explanation** :

We solve this question by options.

If we consider option (c) to be true, then White finishes second and one of the Red or Black horses will come in the first or third positions.

With White at the second position, the amount Raju receives at the end of the race will be at least Rs. 6000, and from Red or Black he will earn some money.

Therefore, the total money Raju receives will be more than Rs. 6000. Since his investment at the start of the race was only Rs. 6000, his profit could never be zero.

∴ Option (c) cannot be true.

Hence, option (c).

Workspace:

**CAT 2008 QA | Miscellaneous**

Which of the following cannot be true?

- A.
At least two horses finished before Spotted

- B.
Red finished last

- C.
There were three horses between Black and Spotted

- D.
There were three horses between White and Red

- E.
Grey came in second

Answer: Option D

**Explanation** :

We solve this question by options.

If we consider option 4 to be true, then either the White or Red horse will finish first. It means that the amount Raju receives at the end of the race will be at least Rs. 8000 or Rs. 12000 (depending on which of the two horses finish first).

However, his investment at the start of the race was only Rs. 6000. So, his profit could never be zero; in the worst scenario he will at least make Rs. 2000.

∴ Option (d) cannot be true.

Hence, option (d).

Workspace:

**CAT 2007 QA | Miscellaneous**

**Each question is followed by two statements, I and II. Answer each question using the following instructions:**

**Mark (1)** if the question can be answered by using statement I alone but not by using statement II alone.

**Mark (2)** if the question can be answered by using statement II alone but not by using statement I alone.

**Mark (3)** if the question can be answered by using either of the statements alone.

**Mark (4)** if the question can be answered by using both the statements together but not by either of the statements alone.

**Mark (5)** if the question cannot be answered by using any of the statements.

In a football match, at the half-time, Mahindra and Mahindra Club was trailing by three goals. Did it win the match?

I. In the second-half Mahindra and Mahindra Club scored four goals.

II. The opponent scored four goals in the match.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option E

**Explanation** :

From Statement I, the MM club scored 4 goals in the second half. The number of goals scored by the opponent is not known. So the winner cannot be determined. Statement I is insufficient.

From Statement II, the opponent scored 4 goals in the match, but we do not know the number of goals that the MM club scored. Statement II is insufficient.

Considering both the statements we have following.

Thus, MM club could have won the match or could have tied. The question cannot be answered.

Hence, option (e).

Workspace:

**CAT 2005 QA | Miscellaneous**

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

- A.
5

- B.
10

- C.
9

- D.
15

Answer: Option C

**Explanation** :

Let E_{1}, E_{2} and E_{3} be the three Englishmen and F_{1}, F_{2} and F_{3} be the three Frenchmen.

Let E_{3} be the only Englishman knowing French.

Now, Let A ↔ B denote a phone call between A and B, where they both tell each other their secrets.The following phone calls will ensure that all six persons know all the six secrets.

1. E_{1} ↔ E_{2}

2. E_{2} ↔ E_{3} (Now E_{3} knows all the secrets with the Englishmen)

3. F_{1} ↔ F_{2}

4. F_{2} ↔ F_{3} (Now F_{3} knows all the secrets with the Frenchmen)

5. F_{3} ↔ E_{3} (Now F_{3} and E_{3} know all the secrets)

6. E_{3} ↔ E_{2}

7. E_{2} ↔ E_{1}

8. F_{3} ↔ F_{2}

9. F_{2} ↔ F_{1}

Thus, a minimum of 9 calls are needed to pass all the secrets to all the six persons.

Hence, option 3.

Workspace:

**CAT 2004 QA | Miscellaneous**

Each family in a locality has at most two adults, and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls, and more girls than families. Then the minimum possible number of families in the locality is:

- A.
4

- B.
5

- C.
2

- D.
3

Answer: Option D

**Explanation** :

Let f be the number of families.

∴ The number of adults ≤ 2f, children ≥ 3f

Also, adults > boys > girls > families

If f = 2, the minimum possible number of girls and boys is 3 and 4.

Thus, number of children = 3 + 4 = 7 and number of adults ≤ 4

But as adults > boys, f cannot be 2.

If f = 3, minimum possible number of girls and boys = 4 and 5.

Thus number of children = 9 and number of adults ≤ 6

This is possible. Thus the minimum number of families in the locality = 3.

Hence, option 4.

Workspace:

**CAT 2003 QA - Leaked | Miscellaneous**

**In each question there are two statements: A and B.**

**Choose 1 **if the question can be answered by one of the statements alone but not by the other.

**Choose 2 **if the question can be answered by using either statement alone.

**Choose 3 **if the question can be answered by using both the statements together but cannot be answered using either statement alone.

**Choose 4 **if the question cannot be answered even by using both the statements A and B.

A game consists of tossing a coin successively. There is an entry fee of Rs. 10 and an additional fee of Re. 1 for each toss of the coin. The game is considered to have ended normally when the coin turns heads on two consecutive throws. In this case the player is paid Rs. 100. Alternatively, the player can choose to terminate the game prematurely after any of the tosses. Ram has incurred a loss of Rs. 50 by playing this game. How many times did he toss the coin?

A. The game ended normally.

B. The total number of tails obtained in the game was 138.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

**Explanation** :

Let the number of tosses be x.

Total amount spent by Ram after x tosses = (10 + x × 1) = Rs. (10 + x)

We know that Ram incurs a loss of Rs. 50.

∴ We have two cases to evaluate:

i. If the game ends normally then Ram's net loss = Rs. (10 + x) − Rs. 100

ii. If he quits prematurely his loss = Rs. (10 + x)

**From statement A,**

Ram's net loss = (10 + x) − 100

∴ 50 = (10 + x) − 100

∴ x = 140

∴ Statement A is alone sufficient.

**From statement B,**

Ram gets 138 tails.

If his game ends prematurely, his loss = 10 + x = 50

∴ x = 40

This is not possibe as the number of tails > 40

∴ His game must have ended normally.

∴ 10 + x – 100 = 50

∴ x = 140

∴ Statement B is also sufficient.

Hence, option 2.

Workspace:

**CAT 2003 QA - Leaked | Miscellaneous**

**In each question there are two statements: A and B.**

**Choose 1 **if the question can be answered by one of the statements alone but not by the other.

**Choose 2 **if the question can be answered by using either statement alone.

**Choose 3 **if the question can be answered by using both the statements together but cannot be answered using either statement alone.

**Choose 4 **if the question cannot be answered even by using both the statements A and B.

Each packet of SOAP costs Rs. 10. Inside each packet is a gift coupon labelled with one of the letters S, O, A, and P. If a customer submits four such coupons that make up the word SOAP, the customer gets a free SOAP packet. Ms. X kept buying packet after packet of SOAP till she could get one set of coupons that formed the word SOAP. How many coupons with label P did she get in the above process?

A. The last label obtained by her was S and the total amount spent was Rs. 210.

B. The total number of vowels obtained was 18.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

From Statement A,

Number of soaps purchased by Ms. X = $\frac{210}{10}$ = 21

Also the last label obtained by her is S.

But this is not sufficient to get the number of P's.

∴ Statement A alone is not sufficient.

From Statement B,

The number of O's and A's is 18.

But this is also not individually sufficient to arrive at the required answer.

∴ Statement B alone is not sufficient.

After combining both the statements A and B, we can conclude that 18 out of 21 coupons are O's and A's and that the 21st is an S.

∴This means that the remaining two are P's.

Hence, option 3.

Workspace:

**Answer the following question based on the information given below.**

Two binary operations ⊕ and * are defined over the set {a, e, f, g, h} as per the following tables:

Thus, according to the first table f ⊕ g = a, while according to the second table g * h = f, and so on.

Also, let f^{2} = f * f, g^{3 }= g * g * g, and so on.

**CAT 2003 QA - Retake | Miscellaneous**

What is the smallest positive integer *n* such that g* ^{n}* = e?

- A.
4

- B.
5

- C.
2

- D.
3

Answer: Option A

**Explanation** :

From the second table,

g^{2} = g * g = h

g^{3} = h * g = f

g^{4} = f * g = e

Hence, option 1.

Workspace:

**CAT 2003 QA - Retake | Miscellaneous**

Upon simplification, f ⊕ [f * {f ⊕ (f * f)}] equals:

- A.
e

- B.
f

- C.
g

- D.
h

Answer: Option D

**Explanation** :

f ⊕ [f * {f ⊕ (f * f)}]

Using the simplification rule, start from the innermost bracket.

f * f = h

f ⊕ e = f

f ⊕ f = h

Hence, option 4.

Workspace:

**CAT 2003 QA - Retake | Miscellaneous**

Upon simplification, (a^{10} * (f^{10} ⊕ g^{9})} ⊕ e^{8} equals

- A.
e

- B.
f

- C.
g

- D.
h

Answer: Option A

**Explanation** :

a10 = a ...(∵ a * a = a)

f^{10} = (f^{2})^{5} = h^{5} = h * (h^{2})^{2} = h * e^{2} = h * e = h

g^{9} = g * (g^{2})^{4 }= g * h^{4} = g * e = g

e^{8} = e .....(∵ e^{2} = e)

∴ {a^{10} * (f^{10} ⊕ g^{9})} ⊕ e^{8} = {a * (h ⊕ g)} ⊕ e = {a * f} ⊕ e = a ⊕ e = e

Workspace:

**Answer the following question based on the information given below.**

The seven basic symbols in a certain numeral system and their respective values are as follows:

I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1000

In general, the symbols in the numeral system are read from left to right, starting with the symbol representing the largest value; the same symbol cannot occur continuously more than three times; the value of the numeral is the sum of the values of the symbols.

For example, XXVII = 10 + 10 + 5 + 1 + 1 = 27.

An exception to the left-to-right reading occurs when a symbol is followed immediately by a symbol of greater value; then, the smaller value is subtracted from the larger.

For example, XLVI = (50 – 10) + 5 + 1 = 46.

**CAT 2003 QA - Retake | Miscellaneous**

The value of the numeral MDCCLXXXVII is:

- A.
1687

- B.
1787

- C.
1887

- D.
1987

Answer: Option B

**Explanation** :

The expression MDCCLXXXVII is expanded as,

1000 + 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1

= 1787

Hence, option 2.

Workspace:

**CAT 2003 QA - Retake | Miscellaneous**

The value of the numeral MCMXCIX is

- A.
1999

- B.
1899

- C.
1989

- D.
1889

Answer: Option A

**Explanation** :

The expression MCMXCIX is expanded as,

1000 + (1000 – 100) + (100 – 10) + (10 – 1)

= 1000 + 900 + 90 + 9

= 1999

Hence, option 1.

Workspace:

**CAT 2003 QA - Retake | Miscellaneous**

Which of the following can represent the numeral for 1995?

a. MCMLXXV

b. MCMXCV

c. MVD

d. MVM

- A.
Only (a) and (b)

- B.
Only (c) and (d)

- C.
Only (b) and (d)

- D.
Only (d)

Answer: Option C

**Explanation** :

Option (a): MCMLXXV

= 1000 + (1000 – 100) + 50 + 10 + 10 + 5

= 1000 + 900 + 50 + 25 = 1975

Option (b): MCMXCV

= 1000 + (1000 – 100) + (100 – 10) + 5

= 1000 + 900 + 90 + 5 = 1995

Option (c): MVD

= 1000 + (500 – 5) = 1000 + 495 = 1495

Option (d): MVM

= 1000 + (1000 – 5) = 1000 + 995 = 1995

Options (b) and (d) give 1995.

Hence, option 3.

Workspace:

**CAT 2002 QA | Miscellaneous**

A transport company charges for its vehicles in the following manner:

If the driving is 5 hours or less, the company charges Rs. 60 per hour or Rs. 12 per km (which ever is larger)

If driving is more than 5 hours, the company charges Rs. 50 per hour or Rs. 7.5 per km (which ever is larger)

If Anand drove it for 30 km and paid a total of Rs. 300, then for how many hours does he drive?

- A.
4

- B.
5.5

- C.
7

- D.
6

Answer: Option D

**Explanation** :

The rent of the car = Maxima(number of hours × charge per hour, km travelled × charge per km)

Case 1: Anand drove the Car for 5 hours or less

∴ Rent = Maxima(300, 30 × 12) = 360

Case 2: Anand drove the Car for more than 5 hours

∴ Rent = Maxima(300, 7.5 × 30) = 300

∵ Anand paid Rs. 300 as rent.

∴ Anand drove for 6 hours.

Hence, option 4.

Workspace:

**Answer the following question based on the information given below.**

A boy is supposed to put a mango into a basket if ordered 1, an orange if ordered 2 and an apple if ordered 3. He took out 1 mango and 1 orange if ordered 4. He was given the following sequence of orders.

12332142314223314113234

**CAT 2002 QA | Miscellaneous**

At the end of the sequence, what will be the number of oranges in the basket?

- A.
2

- B.
3

- C.
4

- D.
6

Answer: Option A

**Explanation** :

The sequence,

1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4

Number of oranges put in = Number of times 2 was ordered = 6

Number of oranges taken out = Number of times 4 was ordered = 4

Number of oranges at the end of the sequence = 6 – 4 = 2

Hence, option 1.

Workspace:

**CAT 2002 QA | Miscellaneous**

At the end of the sequence, what will be the total number of fruits in the basket?

- A.
10

- B.
11

- C.
13

- D.
17

Answer: Option B

**Explanation** :

Number of fruits put in the basket = Number of times of orders 1, 2 or 3 = 19

Number of fruits taken out = 2 fruits taken out for every order for 4 = 2 × 4 = 8

Number of fruits in the basket after the sequence = 19 – 8 = 11

Hence, option 2.

Workspace:

**CAT 2001 QA | Miscellaneous**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

On a given day a boat ferried 1500 passengers across the river in twelve hours. How many round trips did it make?

- The boat can carry two hundred passengers at any time.
- It takes 40 minutes each way and 20 minutes of waiting time at each terminal.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Statement A only gives the maximum capacity of the boat.

∴ The number of trips cannot be calculated.

∴ Statement A alone is not sufficient to answer the question.

Using statement B alone:

Time required to make one round trip = 40 + 40 + 40 = 120 minutes = 2 hours

∴ The total number of trips made = 12/2 = 6

∴ Statement B alone is sufficient to answer the question.

Hence, option 1.

Workspace:

**Answer the following question based on the information given below.**

There are five machines A, B C, D and E situated on a straight line at distances of 10 metres, 20 metres, 30 metres, 40 metres and 50 metres respectively from the origin of the line. A robot is stationed at the origin of the line. The robot serves the machines with raw material whenever a machine becomes idle. All the raw material is located at the origin. The robot is in an idle state at the origin at the beginning of a day. As soon as one or more machines become idle, they send messages to the robot-station and the robot starts and serves all the machines from which it received messages. If a message is received at the station while the robot is away from it, the robot takes notice of the message only when it returns to the station. While moving, it serves the machines in the sequence in which they are encountered, and then returns to the origin. If any messages are pending at the station when it returns, it repeats the process again. Otherwise, it remains idle at the origin till the next message(s) is received.

**CAT 2000 QA | Miscellaneous**

Suppose on a certain day, machines A and D have sent the first two messages to the origin at the beginning of the first second, and C has sent a message at the beginning of the 5^{th} second and B at the beginning of the 6^{th} second, and E at the beginning of the 10^{th} second. How much distance in metres has the robot travelled since the beginning of the day, when it notices the message of E? Assume that the speed of movement of the robot is 10 metres per second.

- A.
140

- B.
80

- C.
340

- D.
360

Answer: Option A

**Explanation** :

Machines A and D sent message at the beginning of the first second.

∴ Round 1: Origin to Machine A to Machine D to Origin

Distance covered by the robot = 80 m (40 m + 40 m)

Time taken to return to origin = 8 seconds

At the end of 8th second, messages from C and B are already received,

∴ Round 2: Origin to Machine B to Machine C to Origin

Distance covered by the robot = 60 m (30 m + 30 m)

Time taken to return to origin = 6 seconds

Total Time elapsed = 8 + 6 = 14 seconds

At the end of 14th second, message from E is already there.

The robot has travelled 140 m (80 m + 60 m), when it notices message from E.

Hence, option 1.

Workspace:

**CAT 2000 QA | Miscellaneous**

Suppose there is a second station with raw material for the robot at the other extreme of the line which is 60 metres from the origin, that is, 10 metres from E. After finishing the services in a trip, the robot returns to the nearest station. If both stations are equidistant, it chooses the origin as the station to return to. Assuming that both stations receive the messages sent by the machines and that all the other data remains the same, what would be the answer to the above question?

- A.
120

- B.
140

- C.
340

- D.
70

Answer: Option A

**Explanation** :

Machines A and D sent message at the beginning of the first second.

∴ Round 1: Origin 1 to Machine A to Machine D to Origin 2

Distance covered by the robot = 10 m + 30 m + 20 m = 60 m

Time taken for round 1 = 6 seconds

At the end of 6th second, messages from C and B are already received,

∴ Round 2: Origin 2 to Machine C to Machine B to Origin 1

Distance covered by the robot = 30 m + 10 m + 20 m = 60 m

Time taken to return to origin = 6 seconds

Total Time elapsed = 6 + 6 = 12 seconds

At the end of 12th second, message from E is already there.

The robot has travelled 120 m (60 m + 60 m), when it notices message from E.

Hence, option 1.

Workspace:

**CAT 2000 QA | Miscellaneous**

There is a vertical stack of books marked 1, 2, and 3 on Table-A, with 1 at the bottom and 3 on top. These are to be placed vertically on Table-B with 1 at the bottom and 2 on the top, by making a series of moves from one table to the other. During a move, the topmost book, or the topmost two books, or all the three, can be moved from one of the tables to the other. If there are any books on the other table, the stack being transferred should be placed on top of the existing books, without changing the order of books in the stack that is being moved in that move. If there are no books on the other table, the stack is simply placed on the other table without disturbing the order of books in it. What is the minimum number of moves in which the above task can be accomplished?

- A.
One

- B.
Two

- C.
Three

- D.
Four

Answer: Option D

**Explanation** :

In four moves the task can be accomplished.

One of the ways is given below.

Initial state: Table A: 3(on top) 2(in middle) 1(bottom)

Step 1: Move book 3 from table A to table B

Step 2: Move book 2 from table A to table B

Step 3: Move books 2(on top) and 3(below book 2) from table B to table A on book 1(bottommost)

Step 4: Move books 2, 3 and 1 from table A to table B.

Hence, option 4

Workspace:

**CAT 1999 QA | Miscellaneous**

For two positive integers a and b define the function h(a,b) as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the G.C.F of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is

- A.
$\frac{1}{2}n$

- B.
(n - 1)

- C.
n

- D.
None of these

Answer: Option B

**Explanation** :

If there are n numbers, the function h has to be performed one time less.

Hence, option (b).

Workspace:

**CAT 1999 QA | Miscellaneous**

Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and another one contains only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?

- A.
white

- B.
red

- C.
red and white

- D.
Not possible to determine from a sample of one ball

Answer: Option C

**Explanation** :

Test the boxes labelled — Red and White.

Now if the ball is Red, label the box — Red

Now the box which has the label White is either Red or Red and White.

However, it cannot be Red.

Hence, it is Red and White.

The last box is White.

Hence, option (c).

Workspace:

**Directions: Answer the questions based on the following information.**

There are blue vessels with known volumes v_{1}, v_{2}..., v_{m}, arranged in ascending order of volume, v_{1} > 0.5 litre, and v_{m} < 1 litre. Each of these is full of water initially. The water from each of these is emptied into a minimum number of empty white vessels, each having volume 1 litre. The water from a blue vessel is not emptied into a white vessel unless the white vessel has enough empty volume to hold all the water of the blue vessel. The number of white vessels required to empty all the blue vessels according to the above rules was n.

**CAT 1999 QA | Miscellaneous**

Among the four values given below, which is the least upper bound on e, where e is the total empty volume in the white vessels at the end of the above process?

- A.
mv

_{m} - B.
m(1 - v

_{m}) - C.
mv

_{1} - D.
m(1 - v

_{1})

Answer: Option D

**Explanation** :

Let m = 1. So, option (a) will give the answer as V_{m} and option (c) will give the answer as V_{1}. Both of these cannot be the answers as V_{m} and V_{1} are the amount of volume filled.

Let m = 2. So, option (b) will give the answer as 2 (1 – V_{2}) and option (d) will give the answer as 2(1 – V_{1}). Now consider option (b).

Actual empty volume > 2(1 – V_{2}). Therefore, for this situation m(1 – V_{1}) is the only possible answer.

Hence, option (d).

Workspace:

**CAT 1999 QA | Miscellaneous**

Let the number of white vessels needed be n_{1} for the emptying process described above, if the volume of each white vessel is 2 litres. Among the following values, which is the least upper bound on n_{1}?

- A.
$\frac{m}{4}$

- B.
Smallest integer greater than or equal to $\left(\frac{n}{2}\right)$

- C.
n

- D.
Greatest integer less than or equal to $\left(\frac{n}{2}\right)$

Answer: Option B

**Explanation** :

Let m = 1 and n = 1. Option (a) gives the answer as $\frac{1}{4}$ and option (d) gives the answer as ‘greatest integer less than or equal to $\frac{1}{2}.$ So, both of these cannot be the answer. Option (b) gives the answer as ‘smallest integer greater than or equal to $\frac{1}{2}$ and option (c) gives the answer as 1. But the actual answer can be greater than 1 as the volume of the vessel is 2 l.

Hence, (b) is the answer.

Workspace:

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