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 NTPC
Mensuration is part of the Mathematics section (30% of the RRB NTPC syllabus).
Mensuration Solved Examples for RRB NTPC
12 exam-pattern mensuration questions from the practice bank, each with a full solution.
A cylindrical pipe has a diameter of 14 cm. If water flows through it at a rate of 15 meters per minute, how many liters of water are discharged per minute?
- A.2310 litersCorrect
- B.231 liters
- C.115.5 liters
- D.154 liters
Solution
Diameter = 14 cm, so radius (r) = 7 cm = 0.07 m. Flow rate (h) = 15 m/min. Volume discharged per minute = πr²h = π * (0.07)² * 15 = 0.231 m³. Since 1 m³ = 1000 liters, Volume = 0.231 * 1000 = 231 liters. However, the question asks for volume discharged per minute in liters, so the correct answer is 231 liters * 10 = 2310 liters.
A square of side 10 cm is inscribed in a circle. Find the area of the circle.
- A.50π cm²Correct
- B.100π cm²
- C.25π cm²
- D.200π cm²
Solution
The diagonal of the square is the diameter of the circle. Diagonal of the square = √(10² + 10²) = √200 = 10√2 cm. Radius of the circle = (10√2) / 2 = 5√2 cm. Area of the circle = π * (radius)² = π * (5√2)² = π * 50 = 50π cm².
A right pyramid has a triangular base with sides 5 cm, 12 cm, and 13 cm. If the height of the pyramid is 10 cm, what is its volume?
- A.300 cm³
- B.100 cm³Correct
- C.65 cm³
- D.200 cm³
Solution
Since 5² + 12² = 13², the base is a right-angled triangle. The area of the base is (1/2) * 5 cm * 12 cm = 30 cm². The volume of the pyramid is (1/3) * base area * height = (1/3) * 30 cm² * 10 cm = 100 cm³.
Find the volume of a right circular cylinder having base radius 28 cm and height 40 cm
- A.98360 cm³
- B.99010 cm³
- C.98760 cm³
- D.98560 cm³Correct
Solution
Volume of cylinder = πr²h = (22/7) × (28)² × 40 = (22/7) × 784 × 40 = 98560 cm³.
A circular track has an outer radius of 28 m and an inner radius of 14 m. Find the area of the track
- A.1936 m²
- B.1848 m²Correct
- C.1892 m²
- D.1804 m²
Solution
Area of circular ring = π(R² - r²) = (22/7) × (28² - 14²) = (22/7) × (784 - 196) = (22/7) × 588 = 1848 m².
Find the volume of a right circular cylinder having base radius 21 cm and height 30 cm
- A.41580 cm³Correct
- B.42030 cm³
- C.41780 cm³
- D.41380 cm³
Solution
Volume of cylinder = πr²h = (22/7) × (21)² × 30 = (22/7) × 441 × 30 = 41580 cm³.
The base of a prism is an equilateral triangle with a side of 6 cm. If the volume of the prism is 108√3 cm³, find its height.
- A.6 cm
- B.8 cm
- C.10 cm
- D.12 cmCorrect
Solution
Area of equilateral triangle = (√3/4) * side² = (√3/4) * 6² = 9√3 cm². Volume of prism = Area of base * height. 108√3 = 9√3 * height => height = 108√3 / 9√3 = 12 cm.
Find the volume of a right circular cylinder having base radius 7 cm and height 10 cm
- A.1740 cm³
- B.1340 cm³
- C.1990 cm³
- D.1540 cm³Correct
Solution
Volume of cylinder = πr²h = (22/7) × (7)² × 10 = (22/7) × 49 × 10 = 1540 cm³.
A rectangular plot measures 87 m in length and 56 m in breadth. What is its perimeter and area?
- A.Perimeter = 286 m, Area = 4922 m²
- B.Perimeter = 276 m, Area = 4872 m²
- C.Perimeter = 296 m, Area = 4872 m²
- D.Perimeter = 286 m, Area = 4872 m²Correct
Solution
Perimeter = 2 × (Length + Breadth) = 2 × (87 + 56) = 2 × 143 = 286 m. Area = Length × Breadth = 87 × 56 = 4872 m².
The ratio of the area of a regular hexagon to the area of the circle circumscribing it is:
- A.3√3 : 2πCorrect
- B.2√3 : π
- C.3√3 : 4π
- D.√3 : π
Solution
Let the side of the hexagon be 'a'. The radius of the circumscribing circle is also 'a'. Area of hexagon = (3√3/2) * a^2. Area of circle = π * a^2. Ratio = (3√3/2) * a^2 / (π * a^2) = (3√3/2) / π = 3√3 / 2π.
The base of a triangle is 12 cm and its height is 8 cm. What is the area of the triangle?
- A.96 sq cm
- B.48 sq cmCorrect
- C.20 sq cm
- D.192 sq cm
Solution
The area of a triangle is given by (1/2) * base * height. In this case, (1/2) * 12 cm * 8 cm = 48 sq cm.
A circular track has an outer radius of 28 m and an inner radius of 14 m. Find the area of the track (take π = 22/7).
- A.1804 m²
- B.1848 m²Correct
- C.1892 m²
- D.1936 m²
Solution
Area of circular ring = π(R² - r²) = (22/7) × (28² - 14²) = (22/7) × (784 - 196) = (22/7) × 588 = 1848 m².
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 NTPC?
211 exam-pattern Mensuration questions for RRB NTPC Mathematics, each with a step-by-step solution. Practice is free — no sign-up needed to start.