🛤️RRB Group D Mathematics224 questions

Mensuration Questions for RRB Group D Mathematics

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 RRB Group D, each with a step-by-step solution (166 easy, 58 medium). Mathematics carries 25 of 100 marks in RRB Group D CBT.

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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 RRB Group D

25
Mathematics questions in CBT
25
marks for the section (of 100 total)
90 min
CBT duration
—
negative marks per wrong answer

Mensuration is part of the Mathematics section (25% of the RRB Group D syllabus).

Mensuration Solved Examples for RRB Group D

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

1Medium

A rectangular tank, 20 m long and 15 m wide, contains water. How much will the water level rise if two cubes, each of edge 5 m, are completely submerged in the tank?

  1. A.0.83 mCorrect
  2. B.1.76 m
  3. C.1.57 m
  4. D.1.66 m

Solution

Volume of two cubes = 2 * 5³ = 250 m³. Area of tank = 20 * 15 = 300 m². Rise in water level = 250 / 300 = 5/6 ≈ 0.833 m.

2Easy

A wire of length 132 cm is bent to form a regular polygon. If the side of the polygon is 11 cm, then the number of sides of the polygon is:

  1. A.10
  2. B.11
  3. C.12Correct
  4. D.13

Solution

Perimeter of the polygon = Length of the wire = 132 cm. Side length = 11 cm. Number of sides = Perimeter / Side length = 132 / 11 = 12.

3Easy

If the volume of a hemisphere is 19404 cm³, then its radius is:

  1. A.21 cmCorrect
  2. B.14 cm
  3. C.28 cm
  4. D.7 cm

Solution

Volume of hemisphere = (2/3)πr³ 19404 = (2/3) * (22/7) * r³ r³ = (19404 * 3 * 7) / (2 * 22) = (19404 * 21) / 44 = 407484/44 = 9261 r = cube root of 9261 = 21 cm

4easy

Water flows through a rectangular channel 2 meters wide and 1 meter deep at a rate of 1.5 meters per second. What is the volume of water flowing through the channel in one minute?

  1. A.180 m³Correct
  2. B.30 m³
  3. C.60 m³
  4. D.90 m³

Solution

Width (w) = 2 meters, Depth (d) = 1 meter, Rate (h) = 1.5 meters/second, Time (t) = 1 minute = 60 seconds. Volume = w * d * h * t = 2 * 1 * 1.5 * 60 = 180 m³.

5easy

The adjacent sides of a parallelogram are 12 cm and 8 cm. If the distance between the longer sides is 4 cm, then the distance between the shorter sides is:

  1. A.5 cm
  2. B.6 cmCorrect
  3. C.7 cm
  4. D.8 cm

Solution

Area of parallelogram = base × height. Using the longer side as the base: Area = 12 × 4 = 48 sq cm. Using the shorter side as the base: Area = 8 × height = 48. Therefore, height = 48/8 = 6 cm.

6easy

A circular garden has a diameter of 70 m. A path of width 7 m is laid around it on the outside. Find the area of the path.

  1. A.1694 m²Correct
  2. B.1540 m²
  3. C.1700 m²
  4. D.1600 m²

Solution

Inner radius (r) = 70/2 = 35 m. Outer radius (R) = 35 + 7 = 42 m. Area of path = π(R² - r²) = π(42² - 35²) = (22/7) * (42+35) * (42-35) = (22/7) * 77 * 7 = 22 * 77 = 1694 m²

7easy

From a square sheet of side 28 cm, four circles of maximum possible size are cut. Find the area of the remaining sheet.

  1. A.168 sq cmCorrect
  2. B.616 sq cm
  3. C.84 sq cm
  4. D.252 sq cm

Solution

Side of the square sheet = 28 cm. Radius of each circle = 28/4 = 7 cm. Area of the square sheet = 28² = 784 sq cm. Area of four circles = 4 * π * r² = 4 * (22/7) * 7² = 4 * (22/7) * 49 = 4 * 22 * 7 = 616 sq cm. Area of the remaining sheet = 784 - 616 = 168 sq cm.

8easy

A circle is inscribed in a square of side 14 cm. Find the area of the circle.

  1. A.154 sq cmCorrect
  2. B.616 sq cm
  3. C.308 sq cm
  4. D.462 sq cm

Solution

The diameter of the inscribed circle is equal to the side of the square. So, diameter = 14 cm, and radius = 7 cm. Area of the circle = πr^2 = (22/7) * 7^2 = (22/7) * 49 = 22 * 7 = 154 sq cm.

9easy

The area of a parallelogram is 392 sq cm. A perpendicular from one vertex to the opposite side is 28 cm. Find the length of the corresponding side.

  1. A.12 cm
  2. B.14 cmCorrect
  3. C.16 cm
  4. D.18 cm

Solution

Area of parallelogram = base × height. Here, height = 28 cm and area = 392 sq cm. Therefore, base = Area / height = 392 / 28 = 14 cm.

10medium

A well is 28 m deep and has a diameter of 14 m. If the earth taken out of it is spread evenly to form a platform 10.5 m long and 8 m wide, find the height of the platform.

  1. A.8 m
  2. B.9 m
  3. C.10 m
  4. D.11 mCorrect

Solution

Volume of earth dug out = πr²h = (22/7) * (14/2)² * 28 = (22/7) * 49 * 28 = 22 * 7 * 28 = 4312 m³. Volume of platform = l * b * h = 10.5 * 8 * h = 84h. Equating the volumes, 84h = 4312 => h = 4312/84 = 51.33. Incorrect calculation. The correct calculation is Volume = (22/7)*49*28 = 4312. Area of platform = 10.5*8 = 84. Height = 4312/84 = 51.33. There seems to be an error in the question or answer. The correct height should be 51.33 m. Recalculating. The given options are wrong. Volume of earth = (22/7) * (7)^2 * 28 = 4312. Area of platform = 10.5 * 8 = 84. Height = 4312/84 = 51.33

11medium

The base of a triangular field is three times its altitude. If the cost of cultivating the field at ₹24.68 per hectare is ₹333.18, find its base and height.

  1. A.Base = 900m, Height = 300mCorrect
  2. B.Base = 300m, Height = 100m
  3. C.Base = 600m, Height = 200m
  4. D.Base = 1200m, Height = 400m

Solution

Total area = 333.18/24.68 = 13.5 hectares = 135000 sq m. Let altitude be x, then base = 3x. Area = 1/2 * 3x * x = 135000. 3x^2 = 270000. x^2 = 90000. x = 300m (height). Base = 3 * 300 = 900m.

12easy

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.

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 RRB Group D?

224 exam-pattern Mensuration questions for RRB Group D Mathematics, each with a step-by-step solution. Practice is free — no sign-up needed to start.

More RRB Group D Mathematics Topics

Every topic links to its own practice set with solutions.

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Practice Mensuration for RRB Group D

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