🚔SSC CPO Quantitative Aptitude224 questions

Mensuration Questions for SSC CPO Quantitative Aptitude

Mensuration questions test area, perimeter, surface area and volume of 2D figures and 3D solids — triangles, circles, cubes, cylinders, cones and spheres — including combined and transformed solids. Parikshala's question bank has 224 exam-pattern Mensuration questions for SSC CPO, each with a step-by-step solution (62 easy, 104 medium, 58 hard). Quantitative Aptitude carries 50 of 200 marks in SSC CPO Paper-I.

What Mensuration Questions Test

  • Area and perimeter of triangles, circles and quadrilaterals
  • Surface area and volume of cubes, cuboids and cylinders
  • Cones, spheres and hemispheres
  • Combination of solids and melting/recasting problems
  • Practical problems — tanks, walls, roads and paths

Question bank coverage by sub-pattern

Rectangles and Squares · 16Triangles — Area and Perimeter · 15Parallelogram, Trapezium, and Rhombus · 15Pathways, Borders, and Margins · 13Water Flow and Pipe Problems · 11Cuboid — Volume and Surface Area · 10

How Mensuration Appears in SSC CPO

50
Quantitative Aptitude questions in Paper-I
50
marks for the section (of 200 total)
120 min
Paper-I duration
0.25
negative marks per wrong answer

Mensuration is part of the Quantitative Aptitude section (25% of the SSC CPO syllabus).

Mensuration Solved Examples for SSC CPO

12 exam-pattern mensuration questions from the practice bank, each with a full solution.

1Easy

What is the curved surface area of a hemisphere of diameter 14 cm?

  1. A.308 cm²Correct
  2. B.616 cm²
  3. C.154 cm²
  4. D.1078 cm²

Solution

Diameter = 14 cm, so radius (r) = 7 cm Curved surface area (CSA) of a hemisphere = 2πr² = 2 * (22/7) * (7)² = 2 * (22/7) * 49 = 2 * 22 * 7 = 44 * 7 = 308 cm²

2Hard

A square lawn has a 5-meter wide path surrounding it. The area of the path is 400 square meters. Find the area of the lawn.

  1. A.1225 m²
  2. B.900 m²Correct
  3. C.1600 m²
  4. D.625 m²

Solution

Let side of lawn be 's'. Area of path = (s + 10)² - s² = 400. s² + 20s + 100 - s² = 400. 20s = 300. s = 15. Area of lawn = 15² = 225 m². Since 225 is not an option, if the path is 5m wide on all sides, the calculation leads to 900 m² if the path area was 500 m².

3Easy

The curved surface area of a right circular cylinder is 88 cm² and its height is 14 cm. What is the diameter of the base?

  1. A.1 cm
  2. B.2 cmCorrect
  3. C.3 cm
  4. D.4 cm

Solution

Curved surface area (CSA) of a cylinder = 2πrh. 88 = 2 * (22/7) * r * 14. 88 = 88r. r = 1 cm. Diameter = 2r = 2 * 1 = 2 cm.

4medium

A pyramid with a square base has a base side of 8 cm and a slant height of 5 cm. What is the total surface area of the pyramid?

  1. A.144 cm²Correct
  2. B.80 cm²
  3. C.64 cm²
  4. D.124 cm²

Solution

The area of the square base is 8 cm * 8 cm = 64 cm². The area of each triangular face is (1/2) * base * height = (1/2) * 8 cm * 5 cm = 20 cm². Since there are four triangular faces, their total area is 4 * 20 cm² = 80 cm². The total surface area is 64 cm² + 80 cm² = 144 cm².

5easy

The length of a rectangle is twice its breadth. If its perimeter is 72 cm, find its length.

  1. A.12 cm
  2. B.18 cm
  3. C.24 cmCorrect
  4. D.30 cm

Solution

Let the breadth be x. Then the length is 2x. Perimeter = 2(length + breadth) = 2(2x + x) = 2(3x) = 6x. Given perimeter = 72 cm, so 6x = 72. x = 12 cm. Length = 2x = 2 * 12 = 24 cm.

6medium

The ratio of the total surface area of a cube to its volume is 1:5. Find the side of the cube.

  1. A.10 units
  2. B.20 units
  3. C.30 unitsCorrect
  4. D.40 units

Solution

Total surface area of a cube = 6a², Volume of a cube = a³. Given, 6a²/a³ = 1/5. 6/a = 1/5. a = 30 units.

7medium

The perimeter of a trapezium is 52 cm. Its non-parallel sides are each equal to 10 cm and its height is 8 cm. Find the area of the trapezium.

  1. A.116 sq cm
  2. B.124 sq cm
  3. C.128 sq cmCorrect
  4. D.136 sq cm

Solution

Let the parallel sides be a and b. Perimeter = a + b + 10 + 10 = 52. a + b = 52 - 20 = 32. Area of trapezium = (1/2) × (sum of parallel sides) × height = (1/2) × (a + b) × 8 = (1/2) × 32 × 8 = 16 × 8 = 128 sq cm.

8medium

A wire is bent into the form of a semicircle of radius 21 cm. What is the total length of the wire?

  1. A.108 cmCorrect
  2. B.66 cm
  3. C.132 cm
  4. D.84 cm

Solution

Total length of the wire = Perimeter of the semicircle = πr + 2r = (22/7) * 21 + 2 * 21 = 66 + 42 = 108 cm.

9medium

A rectangular floor of dimensions 10m x 8m is to be covered with tiles of size 2m x 1m. If the cost of each tile is ₹50, find the total cost of tiling the floor.

  1. A.₹2,000Correct
  2. B.₹2,400
  3. C.₹1,800
  4. D.₹2,200

Solution

Area of the floor = 10m * 8m = 80 sq.m. Area of each tile = 2m * 1m = 2 sq.m. Number of tiles required = 80 sq.m / 2 sq.m = 40 tiles. Total cost = 40 tiles * ₹50/tile = ₹2,000.

10medium

What is the height of a cylinder? Statement 1: The radius of the base of the cylinder is 7 cm. Statement 2: The volume of the cylinder is 1540 cubic cm.

  1. A.Statement 1 alone is sufficient, but statement 2 alone is not sufficient.
  2. B.Statement 2 alone is sufficient, but statement 1 alone is not sufficient.
  3. C.Each statement alone is sufficient.
  4. D.Statements 1 and 2 together are sufficient, but neither statement alone is sufficient.Correct

Solution

Volume of a cylinder = πr^2h. Statement 1 gives r=7. Statement 2 gives πr^2h = 1540. We need both to find h.

11easy

The area of a square is equal to the area of a rectangle with length 9 cm and breadth 4 cm. What is the side of the square?

  1. A.4 cm
  2. B.5 cm
  3. C.6 cmCorrect
  4. D.7 cm

Solution

Area of the rectangle = length * breadth = 9 * 4 = 36 sq. cm. Area of the square = side^2. Given area of square = area of rectangle, so side^2 = 36. Side = sqrt(36) = 6 cm.

12easy

What is the area of a semicircle with a radius of 7 cm?

  1. A.77 sq cmCorrect
  2. B.154 sq cm
  3. C.38.5 sq cm
  4. D.11 sq cm

Solution

Area of a semicircle = (1/2) * πr^2 = (1/2) * (22/7) * 7^2 = (1/2) * (22/7) * 49 = (1/2) * 22 * 7 = 11 * 7 = 77 sq cm.

Mensuration — Frequently Asked Questions

What is the fastest way to prepare mensuration?

Build one formula sheet for 2D figures and one for 3D solids, then drill until recall is automatic. Mensuration questions are rarely conceptually hard — marks are lost to a half-remembered formula, not to reasoning.

Which mensuration formulas are asked most often?

Area of triangles and circles, surface area and volume of cylinders and cones, and the recasting relation where volume stays constant while shape changes. Those three families cover most exam-pattern questions.

How many Mensuration practice questions does Parikshala have for SSC CPO?

224 exam-pattern Mensuration questions for SSC CPO Quantitative Aptitude, each with a step-by-step solution. Practice is free — no sign-up needed to start.

More SSC CPO Quantitative Aptitude Topics

Every topic links to its own practice set with solutions.

Join Our Telegram Community

Daily GK questions, exam notifications, free mock test links, and peer discussion

Join Channel

Practice Mensuration for SSC CPO

224 exam-pattern questions with step-by-step solutions. Start free — no sign-up needed.