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
How Mensuration Appears in RRB Group D
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.
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?
- A.0.83 mCorrect
- B.1.76 m
- C.1.57 m
- 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.
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:
- A.10
- B.11
- C.12Correct
- 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.
If the volume of a hemisphere is 19404 cm³, then its radius is:
- A.21 cmCorrect
- B.14 cm
- C.28 cm
- 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
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?
- A.180 m³Correct
- B.30 m³
- C.60 m³
- 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.
Find the volume of a right circular cylinder having base radius 14 cm and height 20 cm
- A.12320 cm³Correct
- B.12120 cm³
- C.12770 cm³
- D.12520 cm³
Solution
Volume of cylinder = πr²h = (22/7) × (14)² × 20 = (22/7) × 196 × 20 = 12320 cm³.
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:
- A.5 cm
- B.6 cmCorrect
- C.7 cm
- 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.
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.
- A.1694 m²Correct
- B.1540 m²
- C.1700 m²
- 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
A circular track has an outer radius of 28 m and an inner radius of 14 m. Find the area of the track
- A.1804 m²
- B.1936 m²
- C.1848 m²Correct
- D.1892 m²
Solution
Area of circular ring = π(R² - r²) = (22/7) × (28² - 14²) = (22/7) × (784 - 196) = (22/7) × 588 = 1848 m².
From a square sheet of side 28 cm, four circles of maximum possible size are cut. Find the area of the remaining sheet.
- A.168 sq cmCorrect
- B.616 sq cm
- C.84 sq cm
- 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.
A circle is inscribed in a square of side 14 cm. Find the area of the circle.
- A.154 sq cmCorrect
- B.616 sq cm
- C.308 sq cm
- 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.
Find the volume of a right circular cylinder having base radius 28 cm and height 40 cm
- A.98360 cm³
- B.98560 cm³Correct
- C.98760 cm³
- D.99010 cm³
Solution
Volume of cylinder = πr²h = (22/7) × (28)² × 40 = (22/7) × 784 × 40 = 98560 cm³.
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.
- A.12 cm
- B.14 cmCorrect
- C.16 cm
- 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.
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.