WBBSE Class 10 Maths Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid Multiple Choice Questions
Example 1. The inner volume of a cuboidal box is 440 c.c. and the area of the inner base is 88 sq. The inner height is
- 4 cm
- 5 cm
- 3 cm
- 6 cm
Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions
Solution: base x height = volume
∴ height = \(\frac{440}{80}\) = 5
∴ The correct answer is 2. 5 cm

Example 2. The length, breadth and height of a cuboidal hole are 40 m, 12 m, and 16 m respectively. The number of planes having a height of 5 m, a breadth of 4 m and a thickness of 2 m can be kept in the hole is
- 190
- 192
- 184
- 180
Solution: Number of Planks = \(\frac{\text { volume of the hole }}{\text { volume of a plank }}=\frac{40 \times 12 \times 16}{5 \times 4 \times 2}=192\)
∴ The correct answer is 2. 192
Surface Areas And Volumes Mcqs Class 10
Example 3. The total surface area of a cube is 384 sq. m, the volume of the cube is
- 64 cu.m
- 216 cu.m
- 256 cu.m
- 512 cu.m
Solution: 6a2 = 384
or, a2 = \(\frac{384}{6}\) = 64
∴ a = 8
volume = 83 cu.m = 512 cu. m
∴ The correct answer is 4. 512 cu.m
Example 4. The ratio of the volume of the two cubes is 1: 27, and the ratio of the total surface area is
- 1: 3
- 1: 8
- 1: 7
- 1: 9
Solution: Let the lengths be a1, and a2 units respectively
∴ \(\frac{a_1^3}{a_2^3}=\frac{1}{27}\)
⇒ or, \(\frac{a_1}{a_2}=\frac{1}{3}\)
⇒ or, \(\frac{4 a_1^2}{4 a_2^2}=\frac{4}{4}: \frac{1}{9}=1: 9\)
The correct answer is 4. 1: 9
Surface Areas And Volumes Mcqs Class 10
Example 5. If the total surface area of a cube is S sq. unit and the length of the diagonal is d unit, then the relation between S and d is
- S = 6d2
- 3S = 7d
- S3 = d2
- d2 = \(\frac{S}{2}\)
Solution: 6a2 = S, √3a = d
∴ \(6\left(\frac{d}{\sqrt{3}}\right)^2=\mathrm{S}\)
⇒ or, \(\frac{6 d^2}{3}=\mathrm{S}\)
⇒ or, 2d2 = S
⇒ or, \(d^2=\frac{S}{2}\)
∴ The correct answer is 4. \(d^2=\frac{S}{2}\)
Example 6. If length of the diagonal of each surface of a cube is 8√2 cm, then the length of the diagonal of the cube is
- 5√3 cm
- 6√3 cm
- 7√3 cm
- 8√3 cm
Solution: side x √2 = 8√2 cm or, side = 8cm
⇒ Length of the diagonal = 8√3 cm
∴ The correct answer is 4. 8√3 cm
Example 7. Sum of the edges of a cube is 60 cm. Volume will be
- 85 cu. cm
- 110 cu. cm
- 125 cu. cm
- 100 cu. cm
Solution: 12a = 60, a = 5
∴ Volume = 53 cu. cm = 125 cu. cm
∴ The correct answer is 3. 125 cu. cm
Class 10 Mensuration Chapter 1 Mcqs With Answers
Example 8. Length of a diagonal of a surface of a cuboid is \(\sqrt{a^2+b^2}\) unit and the height is c unit. Length of the diagonal of the cuboid is
- \(\sqrt{a^2+b^2+c^2}\) unit
- \(\sqrt{abc}\)
- (a2 + b2 + c2) unit
- abc unit
Solution: Length of the diagonal = \(\sqrt{\text { length }^2+(\text { breadth })^2+(\text { height })^2}\)
= \(\sqrt{a^2+b^2+c^2}\) unit
∴ The correct answer is 1. \(\sqrt{a^2+b^2+c^2}\) unit
Example 9. No of vertices, edges and surfaces of a cuboid are x, y, and z respectively. Then x – y + z =
- 1
- 2
- 3
- 4
Solution: x = 8, y = 12, z = 6
∴ x-y + z = 8-12 + 6 = 2
∴ The correct answer is 2. 2
Surface Areas And Volumes Mcqs With Solutions
Example 10. The length of the edges of a cubical tin pot is 30 cm. What maximum quantity of water in litres will it contain?
- 27
- 27000
- 1000
- None of these
Solution: Volume = (30 cm)3
= 27000 cu.cm = 27 litres
∴ The correct answer is 1. 27