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All questions of Mensuration: Volume, Surface Area & Solid Figures for UPSC CSE Exam

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Cylindrical cans of cricket balls are to be packed in a box. Each can has a radius of 7 cm and height of 30 cm. Dimension of the box is l = 76 cm, b = 46 cm, h = 45 cm. What is the maximum number of cans that can fit in the box?
  • a)
    15
  • b)
    17
  • c)
    22
  • d)
    21
Correct answer is option 'D'. Can you explain this answer?

Naroj Boda answered
This question requires a good deal of visualization. Since, both the box and cans are hard solids, simply dividing the volume won’t work because the shape can’t be deformed. 

Each cylindrical can has a diameter of 14 cm and while they are kept erect in the box will occupy height of 30 cm 

Number of such cans that can be placed in a row

Number of such rows that can be placed

Thus 5 x 3 = 15 cans can be placed in an erect position.

However, height of box = 45cm and only 30 cm has been utilized so far 

Remaining height = 15 cm > 14 cm (Diameter of the can)

So, some cans can be placed horizontally on the base.

Number of cans in horizontal row
Number of such rows
∴ 2 x 3 = 6 cans can be placed horizontally

∴ Maximum number of cans = 15+6 = 21 

Choice (D) is therefore, the correct answer.
 
 
 
 

PQRST is a pentagon in which all the interior angles are unequal. A circle of radius ‘r’ is inscribed in each of the vertices. Find the area of portion of circles falling inside the pentagon. 
  • a)
    πr2
  • b)
    1.5πr2
  • c)
    2πr2
  • d)
    1.25πr2
Correct answer is option 'B'. Can you explain this answer?

Preeti Khanna answered
Since neither angles nor sides are given in the question, immediately the sum of angles of pentagon should come in mind. To use it,

We know the area of the sectors of a circle is given as,
Note => The above concept is applicable for a polygon of n sides.

Choice (B) is therefore, the correct answer.

Correct Answer: 1.5πr2
 
 

PQRS is a circle and circles are drawn with PO, QO, RO and SO as diameters. Areas A and B are marked. A/B is equal to:
  • a)
    π
  • b)
    1
  • c)
    π/4
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Preeti Khanna answered
Such questions are all about visualization and ability to write one area in terms of others.

Here,

Let the radius of PQRS be 2r 

∴ Radius of each of the smaller circles = 2r/2 = r

∴ Area A can be written as 

A = π (2r)2 – 4 x π(r)2 (Area of the four smaller circles) + B (since, B has been counted twice in the previous subtraction)

=) A = 4πr2 - 4πr2 + B

=) A = B

=) A/B = 1

Choice (B) is therefore, the correct answer.

Correct Answer: 1

Figure above shows a box which has to be completely wrapped with paper. However, a single Sheet of paper need to be used without any tearing. The dimension of the required paper could be 
  • a)
    17 cm by 4 cm
  • b)
    12 cm by 6 cm
  • c)
    15 cm by 4 cm
  • d)
    13 cm by 4 cm
Correct answer is option 'B'. Can you explain this answer?

Naroj Boda answered
Total surface area of the box = 2(4x6 + 1x6 + 1x4)

= 2(24 + 6 + 4)

= 68 cm2

As the problem says the paper can’t be torn/cut a portion of paper will need to be fold, so, the area of paper required would be greater than 68 cm2. Only option b) gives the area greater 68 cm2

Choice (B) is therefore, the correct answer.

Correct Answer: 12 cm by 6 cm

John Nash, an avid mathematician, had his room constructed such that the floor of the room was an equilateral triangle in shape instead of the usual rectangular shape. One day he brought home a bird and tied it to one end of a string and then tied the other end of the string to one of the corners of his room. The next day, he untied the other end of the string from the corner of the room and tied it to a point exactly at the center of the floor of the room. Assuming that the dimensions of the room are relatively large compared to the length of the string, find the number of times, by which the maximum possible space in which the bird can fly, increase.
  • a)
    4
  • b)
    5
  • c)
    6
  • d)
    7
Correct answer is option 'B'. Can you explain this answer?

Ashwin Chawla answered
Consider the length of the string less than or equal to the inradius of the floor. At the comer of the floor you will find a sixth part of a hemisphere and at the centre it will be a hemisphere.

Find the number of spheres of the maximum volume that can be accommodated in the above region.
  • a)
    324
  • b)
    323
  • c)
    162
  • d)
    161
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
To find the maximum number of spheres that can be accommodated in a given region, we need to consider the volume of the region and the volume of each sphere.

Given information:
- The region is not specified, but we know it can accommodate spheres.
- The volume of each sphere is also not specified.

To solve this problem, we can follow these steps:

1. Determine the volume of the region:
- The volume of the region is not given in the question.
- Without the volume of the region, it is not possible to find the maximum number of spheres that can be accommodated.
- We need more information about the region to proceed.

2. Determine the volume of each sphere:
- The volume of each sphere is not given in the question.
- Without the volume of each sphere, it is not possible to find the maximum number of spheres that can be accommodated.
- We need more information about the spheres to proceed.

Since we do not have sufficient information about the region or the spheres, we cannot determine the maximum number of spheres that can be accommodated. Therefore, none of the provided options (a, b, c, d) can be considered as the correct answer.

To solve this problem, we would need additional information such as the volume of the region and/or the volume of each sphere. Without these details, it is not possible to find the maximum number of spheres that can be accommodated.

A right circular cone has height H and radius R. A small cone is cut off at the top by a plane parallel to the base. At what height above the base the section has been made?

Statement (I): H = 20 cm
Statement (II): Volume of small cone: volume of large cone : 1:15
  • a)
    If the question can be answered with statement I alone but not statement II alone, or can be answered with statement II alone but not statement I alone.
  • b)
    If the question cannot be answered with statement I alone or with statement II alone, but can be answered if both statements are used together.
  • c)
    If the question can be answered with either statement alone.
  • d)
    If the question cannot be answered with the information provided.
Correct answer is option 'B'. Can you explain this answer?

Ishani Rane answered
From statement I, we know that the height of the initial cone is 20cm. However, nothing is said about the small cone. Hence, we cannot answer the question using statement A. So, we can eliminate choices (A) and (D).

We are down to choices (A), (B) or (D).

From Statement II, we know that the ratio of the volume of the small cone to that of the large cone is 1 : 15.

i.e. *π*r2*h : *π*R2*H is 1 : 15 (r is the base radius of the smaller cone and h is the height of the smaller cone)
or r2 * h : R2 * H is 1 : 15

From this information, we will not be able to find the answer to h. Hence, we can eliminate choice (A). 

Combining the information in the two statements: 

When a section is made the two cones are similar triangles. so = 
R = 

We know H = 20 
h = * r

i.e., h3 = H3. Substituting H = 20, we can get the value for h.
Choice (B) is therefore, the correct answer.

Correct Answer: If the question cannot be answered with statement I alone or with statement II alone, but can be answered if both statements are used together.

 An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?
  • a)
    Rs. 4082.40
  • b)
    Rs. 3868.80
  • c)
    Rs. 4216.20
  • d)
    Rs. 3642.40
Correct answer is option 'A'. Can you explain this answer?

Aarav Sharma answered
Given:
- Breadth of the first carpet = 6 m
- Length of the first carpet = 1.44 times the breadth

To find:
- Cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet

Formula:
- Area of a rectangle = Length × Breadth

Calculation:
1. Length of the first carpet:
- Length = 1.44 × Breadth
- Length = 1.44 × 6
- Length = 8.64 m

2. Area of the first carpet:
- Area = Length × Breadth
- Area = 8.64 × 6
- Area = 51.84 sq m

3. Increased length and breadth of the second carpet:
- Length = 1.4 × Length of the first carpet
- Length = 1.4 × 8.64
- Length = 12.096 m
- Breadth = 1.25 × Breadth of the first carpet
- Breadth = 1.25 × 6
- Breadth = 7.5 m

4. Area of the second carpet:
- Area = Length × Breadth
- Area = 12.096 × 7.5
- Area = 90.72 sq m

5. Cost of the carpet:
- Cost per sq m = Rs. 45
- Cost of the first carpet = Area of the first carpet × Cost per sq m
- Cost of the first carpet = 51.84 × 45
- Cost of the first carpet = Rs. 2332.80
- Cost of the second carpet = Area of the second carpet × Cost per sq m
- Cost of the second carpet = 90.72 × 45
- Cost of the second carpet = Rs. 4082.40

Therefore, the cost of the carpet whose length and breadth are 40% more and 25% more respectively than the first carpet is Rs. 4082.40, which is option A.

A square PQRS has an equilateral triangle PTO inscribed as shown:
What is the ratio of AΔPQT to AΔTRU?
  • a)
    1 : 3
  • b)
    1 : √3
  • c)
    1 : √2
  • d)
    1 : 2
Correct answer is option 'D'. Can you explain this answer?

Gowri Chakraborty answered
Let PQ, a side of equilateral triangle be b


By symmetry QT=ST=z (say)


=) a^2 + z^2 – 2az = 2az (Please note how the solution is being managed here. You must always be aware of what you are looking for. Here, as equation -℗ we are looking for (a-z)2 in terms of az) 

A cuboid of length 20 m, breadth 15 m and height 12 m is lying on a table. The cuboid is cut into two equal halves by a plane which is perpendicular to the base and passes through a pair of diagonally opposite points of that surface. Then, a second cut is made by a plane which is parallel to the surface of the table again dividing the cuboid into two equal halves. Now this cuboid is divided into four pieces. Out of these four pieces, one piece is now removed from its place. What is the total surface area of the remaining portion of the cuboid?
  • a)
    1290 m2
  • b)
    1380 m2
  • c)
    1440 m2
  • d)
    Cannot be determined
Correct answer is option 'D'. Can you explain this answer?

Sagar Sharma answered

Explanation:

Given Parameters:
- Length of cuboid = 20 m
- Breadth of cuboid = 15 m
- Height of cuboid = 12 m

Step 1: Cutting the Cuboid into Two Equal Halves
- When the cuboid is cut into two equal halves by a plane perpendicular to the base and passing through a pair of diagonally opposite points, we get two equal halves.
- Each half has dimensions: Length = 20 m, Breadth = 15 m, Height = 6 m

Step 2: Cutting the Halves into Two Equal Parts Again
- When the halves are cut by a plane parallel to the table, we get four equal pieces.
- Each piece has dimensions: Length = 20 m, Breadth = 7.5 m, Height = 6 m

Step 3: Removing One Piece
- One piece is removed from the four pieces obtained from the second cut.

Calculating the Total Surface Area of the Remaining Portion
- The total surface area of the remaining portion can't be determined because the exact shape and dimensions of the remaining portion after removing one piece are not specified.

Therefore, the total surface area of the remaining portion of the cuboid cannot be determined.

The maximum distance between two points of the unit cube is
  • a)
    √2 + 1
  • b)
    √2
  • c)
    √3
  • d)
    √2 + √3
Correct answer is option 'C'. Can you explain this answer?

Ashwin Chawla answered
The distance from any vertex at the base of the cube to the vertex that is perpendicular along height to the diametrically opposite vertex is required.
We have to calculate DF.

Anil grows tomatoes in his backyard which is in the shape of a square. Each tomato takes 1 cm2 in his backyard. This year, he has been able to grow 131 more tomatoes than last year. The shape of the backyard remained a square. How many tomatoes did Anil produce this year?
  • a)
    4225
  • b)
    4096
  • c)
    4356
  • d)
    Insufficient Data
Correct answer is option 'C'. Can you explain this answer?

Naveen Jain answered
Let the area of backyard be x2 this year and y2 last year

∴ X2- Y2 = 131

=) (X+Y) (X-Y) = 131

Now, 131 is a prime number (a unique one too. Check out its properties on Google). Also, always identify the prime number given in a question. Might be helpful in cracking the solution.

=) (X+Y) (X-Y) = 131 x 1

=) X+Y = 131

X-Y = 1

=) 2X = 132 =) X = 66 

and Y = 65

∴ Number of tomatoes produced this year = 662 = 4356

Choice (C) is therefore, the correct answer.

Correct Answer: 4356

PQRS is a square of sides 2 cm & ST = 2 cm. Also, PT=RT. What is the area of ?PST?
  • a)
    2 cm2
  • b)
    √3 cm2
  • c)
    √2 cm2
  • d)
Correct answer is option 'C'. Can you explain this answer?

Rajeev Kumar answered
Although the figure looks like a 3D figure but on reading through the question, it is clear that the diagram is on a single plane.

 
 
 

Chapter doubts & questions for Mensuration: Volume, Surface Area & Solid Figures - CSAT Preparation 2025 is part of UPSC CSE exam preparation. The chapters have been prepared according to the UPSC CSE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for UPSC CSE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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