Geometry & Mensuration PYQs for UPSC (2011-2024) | Solved Questions & PDF Download

Master Geometry & Mensuration for UPSC with solved Previous Year Questions (PYQs) from 2011 to 2024. This page provides detailed solutions, practice questions, and a free downloadable PDF to help you ace your UPSC preparation.

Why Geometry & Mensuration Matter in UPSC?

Geometry and Mensuration are crucial topics in the UPSC CSAT (Civil Services Aptitude Test). They test your ability to:

Understand shapes, sizes, and properties of geometric figures.
Calculate areas, volumes, and perimeters.
Solve complex spatial reasoning problems.

This page breaks down the concepts and provides year-wise solved questions to help you master this topic.

Key Concepts in Geometry & Mensuration

Before diving into the questions, let’s quickly revise the key concepts:

1. Geometry

➡️ Triangles: Types, properties, and theorems (e.g., Pythagoras theorem).
➡️ Circles: Radius, diameter, circumference, and area.
➡️ Quadrilaterals: Properties of squares, rectangles, parallelograms, and trapeziums.

2. Mensuration

➡️ 2D Shapes: Area and perimeter of triangles, rectangles, circles, etc.
➡️ 3D Shapes: Volume and surface area of cubes, cylinders, spheres, etc.

Year-Wise Geometry & Mensuration PYQs (2011-2024)

FORMULAE OF GEOMETRY AND MENSURATION

Q1. A village having population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. The water of this tank will last for [CSAT 2011]

(a) 2 days

(b) 3 days

(d) 5 days

Solution:

Given that,

Population of village requires 150 litres per day = 4000

Length of tank = 20 m

Breadth = 15 m

Height = 6 m

Now,

Volume of tank = 20 x 15 x 6 = 1800 m³ = 18 x 10⁵ litres

Village population requires 150 litres per day per head = 4000 x 150 = 6 x 10⁵ litres/ per day

Water last for w days

So,

w x 6 x 10⁵ = 18 x 10⁵

w = 3 days

Hence option (b) is correct

Q2. Consider the following figure and answer the item that follows:

A square is divided into four rectangles as shown above.

The lengths of the side of rectangle are natural numbers. The areas of two rectangles are indicated in the figure. What is the length of each side of the square? [CSAT 2011]

(a) 10

(b) 11

(d) Cannot be determined as the given data are insufficient

Solution:

Given that,

Area of smaller rectangle = 15 units

Factors are 1 x 15 and 3 x 5

1 x 15 is not possible to form a bigger square. So, 3 and 5 are the factors

Area of bigger rectangle = 48

Factors = 1 x 48, 24 x 2, 12 x 4 and 8 x 6

However, 1 x 48 and 24 x 2 is not possible as they cannot form square

Taking 12 x 4

Now

adding the length and breadth of smaller and bigger rectangles

3 + 12 = 15 and 4 + 5 = 9

Thus cannot form a square

Taking 8 + 3 = 11 and 6 + 5 = 11

Thus formed a square

So sides of a square are 11 x 11

Hence option (b) is correct

Q3. In a plane, line X is perpendicular to line Y and parallel to line Z; line U is perpendicular to both lines V and W; line X is perpendicular to line V. [CSAT 2015]

Which of following statement is correct?

(a) Z, U and W are parallel

(b) X, V and Y are parallel

(d) Y, V and W are parallel

Solution:

Given that,

line X is perpendicular to line Y and parallel to line Z

line U is perpendicular to both lines V and W

line X is perpendicular to line V

Thus from the figure Y, V and W are parallel

Hence option (d) is correct

Q4. An agricultural field is in the form of a rectangle having length x₁ meters and breadth x₂meters (x₁ and x₂ are variable). If x₁ + x₂ = 40 meters, then the area of the agricultural field will not exceed which one of the following values? [CSAT 2016]

(a) 400 sq. m

(b) 300 sq. m

(d) 80 sq. m

Solution:

Given that,

Length of field = X₁

Breadth = X₂

Area = X₁ x X₂

For maximum area the rectangle must be a square. So, X₁= X₂

Thus X₁ = 20 ( As sum is equal to 40)

Area = 20 x 20 = 400 m²

Hence option (a) is correct

Q5. AB is vertical trunk of huge tree with A being the point where base of trunk touches the ground. Due to a cyclone, the trunk has been broken at C which is at height of 12 meters, broken part is partially attached to vertical portion of trunk at C. If the end of broken part B touches the ground at D which is at distance of 5 meters from A, then the original height of trunk is: [CSAT 2016]

(a) 20M

(b) 25M

(d) 35M

Solution:

Given that,

B touches the ground at D which is at distance of 5 meters from A

So, AD = 5 m

AC = 12 m

By Pythagoras theorem

CD² = 12² + 5² = 13²

CD = 13 m

Thus the original height = CD + AC = 12 + 13 = 25 m

Hence option (b) is correct

Q6. A piece of tin is in the form of rectangle having length 12 cm and width 08 cm. This is used to construct a closed cube. The side of the cube is: [CSAT 2016]

(a) 2cm

(b) 3cm

(d) 5cm

Solution:

Given that,

Length of tin = 12cm

Breadth = 8cm

Area of rectangle = 12 x 8 = 96 cm²

To construct a cube the area of rectangle must be equal to total surface area of cube (Total surface area of the cube = 6a²)

96 = 6 x a²

a² = 16

a = 4 cm

Hence option (c) is correct

Q7. A cylindrical overhead tank of radius 2m and height 7m is to be filled from an underground tank of size 5.5m×4m×6m. How much portion of the underground tank is still filled with water after filling the overhead tank completely? [CSAT 2016]

(a) 1/3

(b) 1/2

(d) 1/6

Solution:

Given that,

Radius of overhead tank = 2m

Height of overhead tank = 7m

Underground Tank dimensions = 5.5m×4m×6m

Now,

Volume of cylindrical overhead tank = (22/7) x 2² x 7 = 88 m³ (volume of the cylinder = )

Volume of underground tank = 5.5 x 4 x 6 = 132 m³

132 - 88 = 44 m³ still left

Portion of underground still left = 44/132 = 1/3

Hence option (a) is correct

Q8. There are three pillars X,Y and Z of different heights. Three spiders A, B and C start to climb on these pillars simultaneously. In one chance, A climbs on X by 6cm but slips down 1cm. B climbs on Y by 7 cm but slips down 3 cm. C climbs on Z by 6.5 cm but slips down 2 cm. If each of them requires 40 chances to reach the top of the pillars, what is height of shortest pillar? [CSAT 2017]

(a) 161CM

(b) 163CM

(d) 210CM

Solution:

Given that,

In one chance, A climbs on X by 6cm but slips down 1cm.

B climbs on Y by 7 cm but slips down 3 cm.

C climbs on Z by 6.5 cm but slips down 2 cm.

Total chance 40

Now,

For A:

Climbs 6cm and slips 1 cm so in one chance it climbs = 6 - 1 = 5cm

So, height of pillar for A = 39 x 5 + 6 (last attempt) = 201 cm = X

For B:

7 - 3 = 4cm

Height of Y = 39 x 4 + 7 = 163 cm

For C:

6.5 - 2 = 4.5 cm

Height of Z = 39 x 4.5 + 6.5 = 182 cm

Therefore the shortest pillar height = 163 cm

Hence option (b) is correct

Q9. Two walls and ceiling of a room meet at right angles at a point P. A fly is in the air 1m from one wall, 8 m from the other wall and 9 m from point P. How many meters is fly from ceiling? [CSAT 2017]

(a) 4

(b) 6

(d) 15

Solution:

Given that,

Distance of fly from P = 9m

Distance of fly from one wall = 1 m

Distance of fly from other wall = 8 m

Also, two walls and ceiling of a room meet at right angles at a point P

Taking three dimensional axis where ceiling is on Z axis

The distance from P

Distance formula =

(X,Y,Z) = P = (0,0,0) and (X1,Y1,Z1) = (1,8,Z’)

9² = (-1)² + (-8)² + (0 - Z')²

81 = 1 + 64 + (Z')²

Z^' = 4 m

Distance from ceiling = 4m

Hence option (a) is correct

Q10. There are 24 equally spaced points lying on the circumference of a circle. What is the maximum number of equilateral triangles that can be drawn by taking sets of three points as the vertices? [CSAT 2018]

(a) 4

(b) 6

(d) 12

Solution:

Given that,

There are 24 equally spaced points lying on the circumference of a circle.

Angle of complete circle = 360 degree

Now,

24 equally divided among circumference = 360/24 = 15 degrees at the centre of circle

Two vertices of equilateral triangle = 120 degrees

So total number of triangles that can be formed = 120/15 = 8

Hence option (c) is correct

Q11. P, Q and R are three towns. The distance between P and Q is 60 km, whereas the distance between P and R is 80 km. Q is in the West of P and R is in the South of P. What is the distance between Q and R? [CSAT 2019]

(a) 140 km

(b) 130 km

(d) 100 km

Solution:

Given that,

The distance between P and Q is 60 km,

The distance between P and R is 80 km.

Q is in the West of P and R is in the South of P.

Now,

Using Pythagoras theorem

(QR)² = 60² + 80²

(QR)² = 3600 + 6400 = 10000

QR = 100 KM

Hence option (d) is correct

Q12. Consider the following statements:

1. The minimum number of points of intersection of a square and a circle is 2.

2. The maximum number of points of intersection of a square and a circle is 8.

Which of the above statements is/are correct? [CSAT 2020]

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Statement 1: The minimum number of points of intersection of a square and a circle is 2.

The minimum intersection between square and circle can be zero as the circle can be entirely into the square (that means circle is smaller than square). Thus, with this possibility the number of intersection is zero.

Hence statement 1 is incorrect

Statement 2: The maximum number of points of intersection of a square and a circle is 8.

The maximum intersection between square and circle is 8, two on each side of a square.

Hence statement 2 is correct

Q13. Let x, y be the volumes; m, n be the masses of two metallic cubes P and Q respectively. Each side of Q is two times that of P and the mass of Q is two times that of P. Let u=m/x and v=n/y. Which one of the following is correct? [CSAT 2020]

(a) u = 4v

(b) u = 2v

(d) v = 4u

Solution:

Given that,

Let x, y be the volumes; m, n be the masses of two metallic cubes P and Q respectively.

Each side of Q is two times that of P and the mass of Q is two times that of P.

Let u=m/x and v=n/y.

Now,

Let the side of a cube P be S

Side of cube Q = 2S

Volume of cube P = S³

Volume of cube Q = (2S)³ = 8S³

Let Mass of cube P be M

Q = 2M

So,

u = m/x = M/ S³

v = n/y = 2M/ 8S³

u/v = 4/1

u = 4v

Hence option (a) is correct

Q14. There are three points P, Q and Ron a straight line such that PQ: QR = 3: 5. If n is the number of possible values of PQ: PR, then what is n equal to? [CSAT 2021]

(a) 1

(b) 2

(d) 4

Solution:

Given that,

PQ : QR = 3 : 5

Possibility 1:

If PQ = 3x and QR = 5x and PR = PQ + QR

PQ : PR = 3x : (5x + 3x) = 3x : 8x = 3:8

Possibility 2x:

If PQ = 3x and QR = 5x and PR = QR - PQ

PQ : PR = 3x : (5x – 3x) = 3x : 2x = 3:2

Therefore n = 2

Hence option (b) is correct

Q15. A cubical vessel of side 1 m is filled completely with water. How many millilitres of water are contained in it (neglect thickness of the vessel)? [CSAT 2021]

(a) 1000

(b) 10000

(d) 1000000

Solution:

Given that,

A cubical vessel of side 1 m is filled completely with water.

With a side length of 1 meter:

Volume = 1³ = 1 cubic meter

Since 1 cubic meter is equal to 1,000 liters, and 1 liter is equal to 1,000 milliliters:

So, 1000 x 1000 = 10⁶

Hence option (d) is correct

Q16. There are eight equidistant points on a circle. How many right-angled triangles can be drawn using these points as vertices and taking the diameter as one side of the triangle? [CSAT 2022]

(a) 24

(b) 16

(d) 8

Solution:

Given that,

There are eight equidistant points on a circle.

Now,

Considering all the points on the circle P,Q,R,S,T,U,V & W

Considering PT as a diameter and one side of the triangle. So, on one side of the semicircle there are 3 right angled triangles

Thus, in a circle there are 6 right angled triangles (taking PT as a diameter)

Similarly, there are 3 more diameters that can be formed

The diameters are : QU, RV & SW.

Therefore, total right angled triangles = 6 x 4 = 24

Hence option (a) is correct

Q17. Consider the following statements in respect of a rectangular sheet of length 20 cm and breadth 8 cm: [CSAT 2022]

1. It is possible to cut the sheet exactly into 4 square sheets.

2. It is possible to cut the sheet into 10 triangular sheets of equal area.

Which of the above statements is/are correct?

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

Rectangular sheet of length = 20 cm

Breadth = 8 cm

Statement 1: It is possible to cut the sheet exactly into 4 square sheets.

Two sheets of 4 x 4 and two sheets of 8 x 8

Hence statement 1 is correct

Statement 2: It is possible to cut the sheet into 10 triangular sheets of equal area.

Area = 20 x 8 = 160

Cutting into 10 triangular sheet is possible 160/10 = 16 cm² area each

Hence statement 2 is correct

Hence option (c) is correct

Q18. A rectangular floor measures 4 m in length and 2.2 m in breadth. Tiles of size 140 cm by 60 cm have to be laid such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. What is the maximum number of tiles that can be accommodated on the floor? [CSAT 2023]

(a) 6

(b) 7

(d) 9

Solution:

Given that,

A rectangular floor measures 4 m in length and 2.2 m in breadth.

Tiles of size 140 cm by 60 cm have to be laid such that the tiles do not overlap.

A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor.

Now,

According the given question the figure formed is

Therefore, from the given figure it is clear that the maximum number of tiles that can be accommodated on the floor is 9

Hence option (d) is correct

ANSWER KEY

1. B

2. B

3. D

4. A

5. B

6. C

7. A

8. B

9. A

10. C

11. D

12. B

13. A

14. B

15. D

16. A

17. C

18. D

Download Geometry & Mensuration PYQs PDF

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FAQs on Geometry & Mensuration for UPSC

1. Is Geometry & Mensuration important for UPSC?

➡️ Yes, Geometry & Mensuration are crucial topics for the UPSC CSAT paper. They test your spatial reasoning and problem-solving skills.

2. How to prepare for Geometry & Mensuration in UPSC?

➡️ Understand basic concepts like area, perimeter, volume, and theorems.
➡️ Practice PYQs regularly.
➡️ Use reliable resources like NCERT books and online platforms like iassetu.com.

3. What are the most frequently asked Geometry & Mensuration questions in UPSC?

➡️ Area and perimeter of 2D shapes
➡️ Volume and surface area of 3D shapes
➡️ Theorems and properties of triangles and circles

Conclusion

Practicing Geometry & Mensuration PYQs is essential for cracking the UPSC CSAT paper. This page provides year-wise solved questions, practice problems, and a free PDF download to help you prepare effectively. Bookmark this page and revisit it regularly for updates.

For more UPSC preparation resources, explore iassetu.com. Happy learning!