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CAT Practice Test: Week 12 - CAT MCQ


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30 Questions MCQ Test - CAT Practice Test: Week 12

CAT Practice Test: Week 12 for CAT 2024 is part of CAT preparation. The CAT Practice Test: Week 12 questions and answers have been prepared according to the CAT exam syllabus.The CAT Practice Test: Week 12 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CAT Practice Test: Week 12 below.
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CAT Practice Test: Week 12 - Question 1
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The height of a cylinder is 14 cm and its curved surface area is 264 sq. cm. Find the radius of its base.  

Detailed Solution for CAT Practice Test: Week 12 - Question 1

Curved surface area of a cylinder = 2πrh
Given: h = 14 cm and curved surface area = 264 cm2
∴ 2πrh = 264
r = 3 cm

CAT Practice Test: Week 12 - Question 2
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Three metal cubes having volumes 125 cubic cm, 64 cubic cm and 27 cubic cm are melted to form a new cube. Find the value of edge of the new cube.  

Detailed Solution for CAT Practice Test: Week 12 - Question 2

Since three metal cubes of volumes 125 cm3, 64 cm3 and 27 cm3 are melted to form a new cube,
Therefore, volume of the new cube = 125 + 64 + 27 = 216 cm3
Total volume = 216 cm3
a3 = 216 cm3
a3 = 63
On comparing, we get
a = 6
Edge = 6 cm

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CAT Practice Test: Week 12 - Question 3
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If the diameter of a cylindrical drum is 1.5 m and the height is 1.5 m, then maximum how many smaller aluminium cylindrical cans can be completely filled by the liquid in the cylindrical drum, if the radius and height of each can are 0.2 metres and 0.4 metres, respectively?

Detailed Solution for CAT Practice Test: Week 12 - Question 3

According to the question:
Volume of the drum = πr2h

Volume of small can

Divide both the above volumes to get the maximum number of cans as 52.73, i.e. maximum 52 cans can be completely filled. 

CAT Practice Test: Week 12 - Question 4
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A steel wire is bent in the shape of a square having semi-perimeter as 18 cm and an area of 81 cm2, and then the same wire is again bent to make a circle. Find the ratio of the area of square to that of circle.
 
Detailed Solution for CAT Practice Test: Week 12 - Question 4

Semi-perimeter (2a) = 18 cm
Area (a × a) = 81 cm2
Solving both, we get a = 9 cm.
Now, circumference = 2 × 18 = 36
So, r = 
Area of circle = 

CAT Practice Test: Week 12 - Question 5
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The curved surface areas of a right-circular cone and a hemisphere, having the same radius are in the ratio of  Find the ratio of the height of the cone to the radius of the hemisphere.
Detailed Solution for CAT Practice Test: Week 12 - Question 5


Here, 'l' is slant height,

As both have the same radius, therefore: 


Squaring both sides:

h/r = 3/1

CAT Practice Test: Week 12 - Question 6
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If a square is rotated along one of its diagonals, then find the curved surface area of the shape formed by it, given that one of its side is 9 metres.
Detailed Solution for CAT Practice Test: Week 12 - Question 6

According to the question, rotating square along a diagonal makes two right-circular cones.
As the diagonal makes a right-angled triangle by the sides of square, so
D = 9√2 m (Using Pythagoras theorem)

So, curved surface area of the cone = 2 πrl

CAT Practice Test: Week 12 - Question 7
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The percentage increase in the surface area of a cube, when each side is doubled, is  
Detailed Solution for CAT Practice Test: Week 12 - Question 7

Let each side of the cube be x cm.
Therefore, surface area = 6x2
Surface area of the cube after making the side of cube double = 6(2x)2 = 24x2
So, clearly, it becomes 4 times of the previous one.
Percentage increase 

CAT Practice Test: Week 12 - Question 8
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A circus tent is in the form of a cone over a cylinder. The diameter of the base is 6 m, the height of the cylindrical part is 15 m and the total height of the tent is 19 m. The canvas required for the tent is
Detailed Solution for CAT Practice Test: Week 12 - Question 8

Height of the cylinder = 15 m
Height of the cone, 'h' = 19 - 15 = 4 m
Radius =  6/2 = 3 m
Slant height of the cone = 
Canvas required = C.S.A. of (cone + cylinder)
= πrl + 2πrH
= πr(5 + 2 × 15)

= 330 m2

CAT Practice Test: Week 12 - Question 9
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A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel, partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Detailed Solution for CAT Practice Test: Week 12 - Question 9

Radius of the sphere = 6/2 = 3 cm
Radius of the cylinder = 12/2 = 6cm
Volume of the sphere 

New volume of the cylindrical vessel - Initial volume of the cylindrical vessel = Volume of the sphere
Initial height = h cm
Final height = H cm
πr2H - πr2h = 36π

CAT Practice Test: Week 12 - Question 10
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In a swimming pool measuring 90 m by 40 m, 150 men take a dip. If the average displacement of water by a man is 8 cubic metres, what will be the approximate rise in water level?
Detailed Solution for CAT Practice Test: Week 12 - Question 10

Volume of water displaced by 150 men = 150 × 8 = 1200 m3
Let the rise in water level be x.
90 × 40 × x = 1200
⇒ x = 12/36 = 1/3 or 0.33 m = 33.33 cm (Approx.)

CAT Practice Test: Week 12 - Question 11
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Directions: Read the following information and answer the given question.

A doctor wanted to know the weight of 5 boys, each weighing less than 55 kg(Consider all the weights are natural numbers). The children would not let the doctor measure their weights on the weighing machine unless they were weighed in pairs. The weights (in kg) thus obtained were: 90, 92, 93, 94, 95, 96, 97, 98, 100 and 101. The names of the 5 children were Tinu, Minu, Chinu, Tipu and Tilu. Also, their weights were in the order Tilu > Tipu > Tinu > Minu > Chinu. The total weight of Tilu and Tipu is 101 kg, that of Minu and Chinu is 90 kg, and that of Tinu and Chinu is 92 kg.

Q. How many of them weighed less than 48 kg?
Detailed Solution for CAT Practice Test: Week 12 - Question 11

Ranking of weights: Tilu > Tipu > Tinu > Minu > Chinu
Sum total of weights: Tilu + Tipu = 101  ... (1)
Sum total of weights: Minu + Chinu = 90   ... (2)
Sum total of weights: Tinu + Chinu = 92   ... (3)
Sum total of weights: 4(Tilu + Tipu + Tinu + Minu + Chinu) = 90 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 100 + 101
4(101 + Tinu + 90) = 956
Tinu = (956/4) - 191 = 48 kg
Hence, two people weighed less than 48 kg.

CAT Practice Test: Week 12 - Question 12
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Directions: Read the following information and answer the given question.

A doctor wanted to know the weight of 5 boys, each weighing less than 55 kg(Consider all the weights are natural numbers). The children would not let the doctor measure their weights on the weighing machine unless they were weighed in pairs. The weights (in kg) thus obtained were: 90, 92, 93, 94, 95, 96, 97, 98, 100 and 101. The names of the 5 children were Tinu, Minu, Chinu, Tipu and Tilu. Also, their weights were in the order Tilu > Tipu > Tinu > Minu > Chinu. The total weight of Tilu and Tipu is 101 kg, that of Minu and Chinu is 90 kg, and that of Tinu and Chinu is 92 kg.

Q. What was the weight of Tinu?
Detailed Solution for CAT Practice Test: Week 12 - Question 12

According to the question:
Ranking of weights: Tilu > Tipu > Tinu > Minu > Chinu
Sum total of weights: Tilu + Tipu = 101 kg
Sum total of weights: Minu + Chinu = 90 kg
Also, according to the question:
4(Tilu + Tipu + Tinu + Minu + Chinu) = 90 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 100 + 101
4(101 + Tinu + 90) = 956
Tinu = (956/4) - (101 + 90) = 48
Weight of Tinu = 48 kg

CAT Practice Test: Week 12 - Question 13
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Directions: There are 300 students in the 2017-18 batch of MBA of IIM, Bangalore. Each student is either a boy or a girl, either doing a 'correspondence course' or a 'regular course' and is either an engineer or an accountant. The two Venn diagrams (below) give information about the number of students of the entire batch falling in different categories.

Q. Which of the following represents the lowest value?

1. Number of accountants who are doing a regular course
2. Number of boys who are accountants
3. Number of boys who are engineers and are doing a regular course
4. Number of accountants who are doing a correspondence course
Detailed Solution for CAT Practice Test: Week 12 - Question 13

Option 1: Number of accountants who are doing a regular course = 17 + 9 = 26
Option 2: Number of boys who are accountants = 17 + 17 = 34
Option 3: Number of boys who are engineers and are doing a regular course = 69
Option 4: Number of accountants who are doing a correspondence course = 17 + 7 = 24

CAT Practice Test: Week 12 - Question 14
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Directions: Read the information carefully to answer the question that follows.

Arun, Varun, Karun, Bhim and Shina had some red balls with them. When the teacher asked them about the number of red balls, one boy gave the following information.
1. The total number of red balls with all five boys is divisible by 2.
2. The number of balls with each boy had exactly 2 factors.
3. None of the boys had more than 28 balls.
4. No two boys had the same number of balls.
5. The total number of balls with any two of the given 5 boys is non-prime.
6. Varun had more red balls than twice the number of balls with Karun.
7. Karun had less red balls than balls with Bhim, while Varun had more red balls than balls with Arun.
8. The number of red balls with Bhim was less than one third of the number of red balls with Shina.

Q. What was the total sum of red balls with Bhim and Varun?
Detailed Solution for CAT Practice Test: Week 12 - Question 14

Since the number of balls with each boy had exactly 2 factors, none of the boys had more than 28 balls and no two boys had same number of balls, so we have 2, 3, 5, 7, 11, 13, 17, 19 and 23 as options.
Since the total number of red balls with all five boys is divisible by 2, so one of them has 2 red balls.
Also, the total number of balls with any two of the given 5 boys is non-prime.
Now,
2 + 3 = 5 (Prime)
2 + 5 = 7 (Prime)
2 + 11 = 13 (Prime)
2 + 17 = 19 (Prime)
So, no boy had 3, 5, 11 and 17 balls.
Now, we have 2, 7, 13, 19 and 23 as options.
Now, the following is given.
Varun had more red balls than twice the balls with Karun.
The number of red balls with Bhim was less than one third the number of red balls with Shina.
Karun had less red balls than balls with Bhim.
The only condition which satisfies all the above conditions is Karun and Bhim having 2 and 7 balls.
So, Shina had 23 balls and since Varun had more balls than balls with Arun, so both of them have 19 and 13 balls each.
Required sum = 7 + 19 = 26

CAT Practice Test: Week 12 - Question 15
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Directions: Read the information carefully to answer the question that follows.
Arun, Varun, Karun, Bhim and Shina had some red balls with them. When the teacher asked them about the number of red balls, one boy gave the following information.
1. The total number of red balls with all five boys is divisible by 2.
2. The number of balls with each boy had exactly 2 factors.
3. None of the boys had more than 28 balls.
4. No two boys had the same number of balls.
5. The total number of balls with any two of the given 5 boys is non-prime.
6. Varun had more red balls than twice the number of balls with Karun.
7. Karun had less red balls than balls with Bhim, while Varun had more red balls than balls with Arun.
8. The number of red balls with Bhim was less than one third of the number of red balls with Shina.

Q. For how many pairs was the sum of red balls a perfect square?
Detailed Solution for CAT Practice Test: Week 12 - Question 15

Since the number of balls with each boy had exactly 2 factors, none of the boys had more than 28 balls and no two boys had same number of balls, so we have 2, 3, 5, 7, 11, 13, 17, 19 and 23 as options.
Since the total number of red balls with all five boys is divisible by 2, so one of them has 2 red balls.
Also, the total number of balls with any two of the given 5 boys is non-prime.
Now,
2 + 3 = 5 (Prime)
2 + 5 = 7 (Prime)
2 + 11 = 13 (Prime)
2 + 17 = 19 (Prime)
So, no boy had 3, 5, 11 and 17 balls.
Now, we have 2, 7, 13, 19 and 23 as options.
Now, the following is given.
Varun had more red balls than twice the balls with Karun.
The number of red balls with Bhim was less than one third the number of red balls with Shina.
Karun had less red balls than balls with Bhim.
The only condition which satisfies all the above conditions is Karun and Bhim having 2 and 7 balls.
So, Shina had 23 balls and since Varun had more balls than balls with Arun, so both of them have 19 and 13 balls each.
2 + 7 = 9
2 + 23 = 25
13 + 23 = 36
For 3 pairs, sum was a perfect square.

CAT Practice Test: Week 12 - Question 16
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Directions: Read the information carefully to answer the question that follows.
Out of Raja, Deepak and Uday, one person was to be selected for the managing director's position. It was decided that all the employees of the company, excluding the candidates, will vote for them. Each employee could vote for a minimum of one and maximum of all three candidates. After the voting, the following was observed.
1. The difference between the number of employees who voted only for Raja and number of employees who voted for Deepak, but not for Uday, was zero.
2. 28 employees voted for only 2 contestants, while 19 employees voted for all 3 contestants.
3. If we subtract the number of employees who voted for Raja, but not for Deepak from the number of employees who voted for at least 2 contestants, we get 20 as the answer.
4. The number of employees who voted for both Deepak and Uday was 33, while the number of employees who voted for Uday, but not for Raja was 12 more than the number of employees who voted for all three contestants.
5. The number of employees who voted for Deepak was the same as the number of employees who voted of at least two contestants.

Q. How many employees voted for both Raja and Uday, but not for Deepak?
Detailed Solution for CAT Practice Test: Week 12 - Question 16

According to given condition,
g - (c + f) = 0
g = c + f ...(1)
b + d + f = 28 ...(2)
e = 19 ...(3)
g + d = e + b + d + f - 20 ...(4)
b + e + f - g = 20
33 - c = 20
c = 13
b + e = 33 ...(5)
a + b = 31 ...(6)
b = 14
a = 17
b + c + e + f = b + d + e + f ...(7)
c = d
Solving, we get

Required number = 13

CAT Practice Test: Week 12 - Question 17
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Directions: Read the information carefully to answer the question that follows.
Out of Raja, Deepak and Uday, one person was to be selected for the managing director's position. It was decided that all the employees of the company, excluding the candidates, will vote for them. Each employee could vote for a minimum of one and maximum of all three candidates. After the voting, the following was observed.
1. The difference between the number of employees who voted only for Raja and number of employees who voted for Deepak, but not for Uday, was zero.
2. 28 employees voted for only 2 contestants, while 19 employees voted for all 3 contestants.
3. If we subtract the number of employees who voted for Raja, but not for Deepak from the number of employees who voted for at least 2 contestants, we get 20 as the answer.
4. The number of employees who voted for both Deepak and Uday was 33, while the number of employees who voted for Uday, but not for Raja was 12 more than the number of employees who voted for all three contestants.
5. The number of employees who voted for Deepak was the same as the number of employees who voted of at least two contestants.
Detailed Solution for CAT Practice Test: Week 12 - Question 17

:

According to given condition,
g - (c + f) = 0
g = c + f … (1)
b + d + f = 28 … (2)
e = 19 ……(3)
g + d = e + b + d + f - 20 = 27 …..(4)
b + e = 33 ….(5)
a + b = 31 …..(6)
b + c + e + f = 47 …(7)
Solving, we get

Required number = 13 + 19 = 32

CAT Practice Test: Week 12 - Question 18
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Directions: The question is based on the following table and information.

1. In 1984-85, the value of exports of manufactured articles exceeds the value of exports of raw materials by 100%.
2. In 1985-86, the ratio of % of exports of raw material to that of exports of manufactured articles is 3 : 4.
3. Exports of food in 1985-86 exceeds the 1984-85 figure by Rs. 1006 crore.

Percentage of Total Value of Exports in India

Q. The change in value of exports of manufactured articles from 1984-85 to 1985-86 is
Detailed Solution for CAT Practice Test: Week 12 - Question 18

Value of food exports in 1985-86 = 0.23 × Rs. 25,800 crore = Rs. 5934 crore
Let the value of exports of raw material be 3y and that of manufactured articles be 4y.
So, 7y + Rs. 5934 crore = Rs. 25,800 crore
Or, y = Rs. 2838 crore
Thus, value of exports of manufactured articles in 1985-86 = 4 × Rs. 2838 crore = Rs. 11,352 crore
Value of food exports in 1984-85 = Rs. 5934 crore - Rs. 1006 crore = Rs. 4928 crore
Now, let the value of raw material exports in 1984-85 be z.
Thus, value of exports of manufactured articles in 1984-1985 be 2z.
Now, Rs. 4928 crore + 3z = Rs. 22,400 crore
Or, z = Rs. 5824 crore
Value of exports of manufactured articles in 1984-85 = 2z = 2 × Rs. 5824 crore = Rs. 11,648 crore
Thus, required value = Rs. (11648 - 11352) crore = Rs. 296 crore
Hence, option 1 is correct.

CAT Practice Test: Week 12 - Question 19
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Directions: The question is based on the following table and information.

1. In 1984-85, the value of exports of manufactured articles exceeds the value of exports of raw materials by 100%.
2. In 1985-86, the ratio of % of exports of raw material to that of exports of manufactured articles is 3 : 4.
3. Exports of food in 1985-86 exceeds the 1984-85 figure by Rs. 1006 crore.

Percentage of Total Value of Exports in India

Q. During 1984-85, how much more raw material was exported as compared to food in terms of value?
Detailed Solution for CAT Practice Test: Week 12 - Question 19

Food related exports in (1984-85) = food related exports in (1985-86) - Rs. 1006 crore = Rs. 0.23 × 25,800 crore - Rs. 1006 crore = Rs. 4928 crore
In 1984-85, exports of raw materials + exports of manufactured goods = Rs. (22400 - 4928) crore = Rs. 17,472 crore
And exports in manufactured goods = 2 × (exports of raw materials)
Exports of raw materials in 1984-85 = Rs. (17472/3) crore = Rs. 5824 crore
Difference between exports of raw materials and food exports = Rs. (5824 - 4928) crore = Rs. 896 crore
Hence, option (2) is correct.

CAT Practice Test: Week 12 - Question 20
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Directions: Read the information carefully to answer the question that follows.

The premises of a school are to be renovated. The renovation is in terms of flooring. Certain areas are to be floored either with marble or wood. All classrooms/halls and pantry are rectangular. The area to be renovated comprises of a big hall measuring 25 m × 25 m, principal's room measuring 13 m × 17 m, a pantry measuring 14 m × 13 m, a recordkeeping cum server room measuring 21 m × 13 m and locker area measuring 29 m × 21 m. The total area of the school is 2000 square metres. The cost of wooden flooring is Rs. 100 per square metre and the cost of marble flooring is Rs. 150 per square metre. The locker area, recordkeeping cum server room and pantry are to be floored with marble. The principal's room and the hall are to be floored with wood. No other area is to be renovated in terms of flooring.

Q. As per given information, how much money will be spent on marble?
Detailed Solution for CAT Practice Test: Week 12 - Question 20

Area to be floored with marble = Area of Pantry + Area of Recordkeeping cum server room + Area of Locker = (14 × 13) + (21 × 13) + (29 × 21) = 1064 m2
Total cost of renovation (Marble) = 1064 × 150 = Rs. 1,59,600

CAT Practice Test: Week 12 - Question 21
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

Who is in charge of a museum?
Detailed Solution for CAT Practice Test: Week 12 - Question 21

A curator is a professional responsible for the care and management of a museum's collection. They play a crucial role in selecting, organizing, and interpreting the artworks or artifacts within the museum.

CAT Practice Test: Week 12 - Question 22
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What is a government by the nobles called?
Detailed Solution for CAT Practice Test: Week 12 - Question 22

Aristocracy is a form of government where power is held by the nobility or a privileged upper class. It is characterized by hereditary rule and often associated with social and economic elitism.

CAT Practice Test: Week 12 - Question 23
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Direction: In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What term is used for someone who is honourably discharged from service?
Detailed Solution for CAT Practice Test: Week 12 - Question 23

The term "emeritus" is commonly used to denote an individual who has been honorably discharged from service, typically after a distinguished career. It is often associated with academic or professional titles.

CAT Practice Test: Week 12 - Question 24
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What is the term for presenting opposing arguments or evidence?
Detailed Solution for CAT Practice Test: Week 12 - Question 24

To rebut is to provide evidence or arguments that contradict or oppose a statement or claim made by someone else. It involves offering a counterargument to refute or disprove the opposing position.

CAT Practice Test: Week 12 - Question 25
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What is the policy of extending a country's empire and influence?
Detailed Solution for CAT Practice Test: Week 12 - Question 25

Imperialism is a policy or ideology of extending a country's power and influence through territorial acquisition, colonization, or other means. It involves the expansion of a nation's control over other territories.

CAT Practice Test: Week 12 - Question 26
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What term describes something of outstanding significance?
Detailed Solution for CAT Practice Test: Week 12 - Question 26

"Monumental" is used to describe something of great significance, importance, or size. It often implies a sense of grandeur or impressiveness.

CAT Practice Test: Week 12 - Question 27
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What is the code of diplomatic etiquette and precedence called?
Detailed Solution for CAT Practice Test: Week 12 - Question 27

Protocol refers to the established code of diplomatic etiquette and precedence followed in official ceremonies, meetings, and interactions between diplomats and governments.

CAT Practice Test: Week 12 - Question 28
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What is a fixed orbit in space in relation to Earth?
Detailed Solution for CAT Practice Test: Week 12 - Question 28

A geo-stationary orbit is a specific type of orbit where a satellite remains fixed in relation to a point on the Earth's surface. This orbit is directly above the equator.

CAT Practice Test: Week 12 - Question 29
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What term describes something that cannot be believed?
Detailed Solution for CAT Practice Test: Week 12 - Question 29

"Incredible" is used to describe something so extraordinary or unbelievable that it may be difficult to accept or comprehend.

CAT Practice Test: Week 12 - Question 30
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Direction:In questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

What is the term for issuing a thunderous verbal attack?
Detailed Solution for CAT Practice Test: Week 12 - Question 30

To fulminate is to issue a vehement and thunderous verbal attack or condemnation. It implies expressing strong disapproval or criticism with great intensity.

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