RRB NTPC Mensuration Questions and Solutions PDF

Candidates preparing for the Railway Recruitment Board NTPC know the importance of the Mensuration section. This is a common section in both the stages of the CBT conducted in the NTPC exam. Candidates prepared for this section can easily get 3-4 questions already done for them. To get full marks in this section, candidates must study all topics in the syllabus. To help you get exam-ready, we have prepared some practice questions using which you can assess your preparation and the level you stand now. These practice questions will give candidates an idea of the types of questions they can expect in the RRB NTPC Examination.

RRB NTPC Mensuration Question

Mensuration is a lengthy and important topic for aspirants preparing for the RRB NTPC Exam, covering both 2D and 3D shapes. The RRB NTPC Syllabus includes finding the volume, height, and surface area of different shapes like cylinders, cubes, rhombuses, spheres, and more. We have provided RRB NTPC Mensuration Questions for practice below.

Q1. The length of diagonals of a rhombus is 16 cm and 12 cm respectively. Find the length of each of its sides.
(a) 10
(b) 11
(c) 12
(d) 13

Q2. The two parallel sides of a trapezium are 13 cm and 7 cm respectively. If the height of the trapezium is 4 cm, then what is its area?
(a) 100 cm²
(b) 40 cm²
(c) 60 cm²
(d) 80 cm²

Q3. A rectangular lawn 30m × 20 m has two roads each 5 m wide running between the park, one is parallel to width and the other is parallel to length. The cost of gravelling is 30 paise /m². Find the total cost of gravelling.
(a) Rs. 78.5
(b) Rs. 65.5
(c) Rs. 67.5
(d) Rs. 85.5

Q4. The perimeter of a square is the same as the perimeter of a rectangle. The side of the square is 120 m. If its breadth is three-fifths of its length, then the area in m2 of the rectangle is:

(a) 12350 m2
(b) 13400 m2
(c) 14600 m2
(d) 13500 m2

Q5. The area of the ring between two concentric circles, whose circumference are 264cm and 396cm is.

(a) 7800 cm²
(b) 6930 cm²
(c) 6650 cm²
(d) 7215 cm²

Q6. The radii of three concentric circles are in the ratio 2: 4: 5. What is the ratio of the area between the two inner circles to that of between the two outer circles?
(a) 6 : 5
(b) 5 : 6
(c) 3 : 4
(d) 4 : 3

Q7. A circular wire of radius 63 cm is reshaped into a semi-circular wire. Find the diameter of the semi-circular wire (in meters)?
(a) 2.62 m
(b) 1.54 m
(c) 3.08 m
(d) 1.96 m

Q8. The sides of a triangular park are 72m, 128m and 100m. The cost of levelling the park at the rate of Rs. 5.50/m² is –
(a) Rs. 15825
(b) Rs. 19731
(c) Rs. 81815
(d) Rs. 65195

Q9. The length and breadth of a rectangle are increased by 8% and 5%, respectively. By how much percentage will the area of the rectangle increase?
(a) 13.4%
(b) 15.4%
(c) 12.4%
(d) 16.4%

Q10. Three circles of radius 6 cm are kept touching each other. The string is tightly tied around these three circles. What is the length of the string?
(a) 36 + 12π cm
(b) 36 +18π cm
(c) 24 +36π cm
(d) 36 + 20π cm

Q11. The area of a sector of a circle is 308cm², with the central angle measuring 45°. The radius of the circle is:
(a) 14 cm
(b) 21 cm
(c) 7 cm
(d) 28 cm

Q12. The lateral surface area of a cone is 924 cm², its slant height is 28 cm. the radius of the base of the cone is:
(a) 11 cm
(b) 9 cm
(c) 10.5 cm
(d) 8 cm

Q13. What will be the area of base of Prism if volume is 30 cm³ and height is 15 cm?
(a) 2 cm²
(b) 3 cm²
(c) 4 cm²
(d) 5 cm²

Q14. The volumes of two cubes are in the ratio of 27 : 64. If the volume of the Larger cube is 296 cm³ more than that of the smaller cube. Find the surface area of the smaller cube.
(a) 384 cm²
(b) 216 cm²
(c) 512 cm²
(d) 144 cm²

RRB NTPC Mensuration Question Solutions

For all the questions above, we have provided the best possible solutions in the attached RRB NTPC Mensuration Question Solutions PDF. However, candidates should first try to solve these questions on their own. If they face any difficulty, they can refer to the solutions in the PDF below.

S1. Ans.(a)
Sol. Diagonals of a rhombus bisect each other at a right angle and divide it in the ratio of 1: 1
Therefore,
In ∆ADB, OA = 8 cm and OB = 6 cm
by Pythagoras theorem, (OA)² + (OB)² = (AB)²
⟹ (8)² + (6)² = (AB)²
⟹ 64 + 36 = AB
⟹ 100 = AB
⟹ 10 cm

S2. Ans.(b)

S3. Ans.(c)
Sol.

30 × 5 + 20 × 5 – 5 × 5 = 250 – 25 = 225 m
Cost of Gravelling = 225 × 0.30
= Rs. 67.5

S4. Ans.(d)
Sol.
Let the breadth of the rectangle = 3x and length = 5x
Perimeter of Square = Perimeter of rectangle
4 × Side = 2 (length + breadth)
4 × 120 = 2(3x + 5x)
x = 30
Length = 3x = 3 × 30 = 90m
Breadth = 5x = 5 × 30 = 150m
Area of Rectangle = 90 × 150 = 13500m²

S5. Ans.(b)

S6. Ans.(d)
Sol. Radius of two inner circles are 2x and 4x
Radius of two outer circles are 4x and 5x
A.T.Q
(4x)² – (2x)² : (5x)² – (4x)²
16x² – 4x² : 25x² – 16x²
12x² : 9x²
4 : 3

S7. Ans(b)

S8. Ans.(b)

S9. Ans.(a)

S10. Ans.(a)

S11. Ans(d)

S12. Ans.(c)

S13. Ans.(a)

S14. Ans.(a)
Sol. Ratio of volume = 27 : 64
According to Question, 64x – 27x = 296
Volume of smaller cube = 37 × 8 = 296cm³
Side = 8 cm, Surface area = 6 × (side)²
= 6 × (8)² = 384cm²

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