Here are 25 multiple-choice questions on Mensuration, suitable for JSSKB and similar competitive exams, with answers and explanations:
Section B: Mathematics – Mensuration
Q1. The area of a rectangle is 252 sq cm. Its length is 18 cm. What is its breadth?
(a) 12 cm
(b) 14 cm
(c) 16 cm
(d) 18 cm
Answer: (b)
Explanation: Area of rectangle = length × breadth. So, breadth = Area / length = 252 / 18 = 14 cm.
Q2. A square park has a perimeter of 320 m. What is the length of its side?
(a) 60 m
(b) 70 m
(c) 80 m
(d) 90 m
Answer: (c)
Explanation: Perimeter of a square = 4 × side. So, side = Perimeter / 4 = 320 / 4 = 80 m.
Q3. The radius of a circle is 7 cm. What is its circumference? (Use $\pi = \frac{22}{7}$)
(a) 22 cm
(b) 33 cm
(c) 44 cm
(d) 55 cm
Answer: (c)
Explanation: Circumference of a circle = $2\pi r$. So, Circumference = $2 \times \frac{22}{7} \times 7 = 44$ cm.
Q4. The volume of a cube is 1728 cubic cm. What is the length of its edge?
(a) 10 cm
(b) 11 cm
(c) 12 cm
(d) 13 cm
Answer: (c)
Explanation: Volume of a cube = (edge)$^3$. So, edge = $\sqrt[3]{1728} = 12$ cm.
Q5. What is the area of a triangle with a base of 10 cm and a height of 8 cm?
(a) 40 sq cm
(b) 60 sq cm
(c) 80 sq cm
(d) 100 sq cm
Answer: (a)
Explanation: Area of a triangle = $\frac{1}{2} \times$ base $\times$ height = $\frac{1}{2} \times 10 \times 8 = 40$ sq cm.
Q6. A cylinder has a radius of 3 cm and a height of 10 cm. Find its curved surface area. (Use $\pi = \frac{22}{7}$)
(a) $188.57$ sq cm (approx)
(b) $200.72$ sq cm (approx)
(c) $220.14$ sq cm (approx)
(d) $230.18$ sq cm (approx)
Answer: (a)
Explanation: Curved surface area of a cylinder = $2\pi rh = 2 \times \frac{22}{7} \times 3 \times 10 = \frac{1320}{7} \approx 188.57$ sq cm.
Q7. The side of a square field is 45 meters. What is the cost of fencing it at Rs. 4 per meter?
(a) Rs. 720
(b) Rs. 800
(c) Rs. 900
(d) Rs. 1000
Answer: (a)
Explanation: Perimeter of square = $4 \times 45 = 180$ meters. Cost of fencing = $180 \times 4 = Rs. 720$.
Q8. If the diameter of a circle is 14 cm, what is its area? (Use $\pi = \frac{22}{7}$)
(a) 154 sq cm
(b) 168 sq cm
(c) 182 sq cm
(d) 196 sq cm
Answer: (a)
Explanation: Radius = diameter / 2 = 14 / 2 = 7 cm. Area of circle = $\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154$ sq cm.
Q9. A rectangular box has length 5 cm, breadth 4 cm, and height 3 cm. What is its volume?
(a) 40 cubic cm
(b) 50 cubic cm
(c) 60 cubic cm
(d) 70 cubic cm
Answer: (c)
Explanation: Volume of a rectangular box (cuboid) = length × breadth × height = $5 \times 4 \times 3 = 60$ cubic cm.
Q10. The perimeter of a rectangle is 60 cm. If its length is 20 cm, what is its breadth?
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 25 cm
Answer: (a)
Explanation: Perimeter of rectangle = $2 \times$ (length + breadth). So, $60 = 2 \times (20 + breadth)$. $30 = 20 + breadth$. Breadth = $30 – 20 = 10$ cm.
Q11. What is the total surface area of a cube whose edge is 5 cm?
(a) 100 sq cm
(b) 125 sq cm
(c) 150 sq cm
(d) 175 sq cm
Answer: (c)
Explanation: Total surface area of a cube = $6 \times (\text{edge})^2 = 6 \times 5^2 = 6 \times 25 = 150$ sq cm.
Q12. A cone has a radius of 7 cm and a height of 24 cm. What is its slant height?
(a) 20 cm
(b) 22 cm
(c) 25 cm
(d) 28 cm
Answer: (c)
Explanation: Slant height ($l$) of a cone = $\sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm.
Q13. The area of a parallelogram is 84 sq cm. If its base is 12 cm, what is its height?
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Answer: (c)
Explanation: Area of a parallelogram = base × height. So, height = Area / base = $84 / 12 = 7$ cm.
Q14. A cylindrical pillar has a diameter of 70 cm and a height of 10 m. What is its volume in cubic meters? (Use $\pi = \frac{22}{7}$)
(a) $3.85$ cubic m
(b) $38.5$ cubic m
(c) $385$ cubic m
(d) $3850$ cubic m
Answer: (a)
Explanation: Radius = diameter / 2 = 70 cm / 2 = 35 cm = 0.35 m. Volume of cylinder = $\pi r^2 h = \frac{22}{7} \times (0.35)^2 \times 10 = \frac{22}{7} \times 0.1225 \times 10 = 22 \times 0.0175 \times 10 = 3.85$ cubic m.
Q15. The length of the diagonal of a square is $10\sqrt{2}$ cm. What is its side length?
(a) 5 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Answer: (c)
Explanation: Diagonal of a square = $a\sqrt{2}$, where ‘a’ is the side. So, $a\sqrt{2} = 10\sqrt{2}$. Therefore, $a = 10$ cm.
Q16. How many bricks, each measuring 25 cm x 10 cm x 5 cm, will be required to build a wall 5 m long, 2 m high, and 0.5 m thick?
(a) 4000
(b) 8000
(c) 16000
(d) 20000
Answer: (b)
Explanation: Volume of wall = $500 \text{ cm} \times 200 \text{ cm} \times 50 \text{ cm} = 5,000,000$ cubic cm.
Volume of one brick = $25 \times 10 \times 5 = 1250$ cubic cm.
Number of bricks = Volume of wall / Volume of one brick = $5,000,000 / 1250 = 4000$. (Wait, my calculation was wrong, let me recheck)
Number of bricks = $ (500 \times 200 \times 50) / (25 \times 10 \times 5) = 5000000 / 1250 = 4000$. Hmm, I made a mistake, 5m=500cm, 2m=200cm, 0.5m=50cm.
Okay, let’s re-calculate.
Volume of wall = $500 \text{ cm} \times 200 \text{ cm} \times 50 \text{ cm} = 5,000,000$ cubic cm.
Volume of one brick = $25 \text{ cm} \times 10 \text{ cm} \times 5 \text{ cm} = 1250$ cubic cm.
Number of bricks = $\frac{5,000,000}{1250} = \frac{50000}{12.5} = 4000$.
Wait, let’s look at the given options again. If the answer is 8000, something is off.
Let’s recheck units.
Wall: 5 m = 500 cm, 2 m = 200 cm, 0.5 m = 50 cm.
Brick: 25 cm, 10 cm, 5 cm.
Number of bricks = (500/25) (200/10) (50/5) = 20 20 10 = 400 * 10 = 4000.
The answer provided as 8000 suggests that perhaps the wall thickness was 1m or 100cm, or there’s a misunderstanding on my part.
Let me assume an error in my calculation OR the options given, and proceed with the calculated 4000.
If the option was 8000, it would mean the volume of the wall is twice as big, or bricks are half the size.
Let me assume the options are correct and my first calculation is correct based on the values. Perhaps the question intended a thickness of 1m.
Let’s consider the calculation again.
$ (500 \times 200 \times 50) / (25 \times 10 \times 5) = (500/25) \times (200/10) \times (50/5) = 20 \times 20 \times 10 = 4000$.
So the correct answer should be 4000 based on these numbers. Let me re-read the intent.
Ah, competitive exams often have tricky numbers. Let me re-evaluate based on the assumption that ‘8000’ is the correct answer and where might the discrepancy arise.
Could it be 1m thick? If thickness was 1m (100cm), then $20 \times 20 \times 20 = 8000$. This is a common exam trick based on similar numbers. Let’s adjust the question, or provide the answer given my calculation (4000). For the sake of matching my options, I will consider the thickness to be 1m or adjust the options.
Let’s assume the question meant 1m thick.
Wall: 5 m long, 2 m high, 1 m thick (100 cm).
Volume of wall = $500 \times 200 \times 100 = 10,000,000$ cubic cm.
Volume of one brick = $25 \times 10 \times 5 = 1250$ cubic cm.
Number of bricks = $10,000,000 / 1250 = 8000$.
So, let’s assume the thickness is 1m (100cm) for the answer to be 8000. If it’s truly 0.5m, then 4000 is the answer. For competitive exam questions, if 8000 is an option, it’s often because a round number like 1m was intended. I will proceed with 8000 and justify it by assuming 1m thickness.
Self-correction: If the question clearly states 0.5m, I must stick to 4000. Let me adjust the options so 4000 is available, or write a careful explanation.
Let’s re-state the question and options to reflect the correct calculation for 0.5m.
Q16. How many bricks, each measuring 25 cm x 10 cm x 5 cm, will be required to build a wall 5 m long, 2 m high, and 0.5 m thick?
(a) 2000
(b) 4000
(c) 6000
(d) 8000
Answer: (b)
Explanation: Volume of wall = $5m \times 2m \times 0.5m = 500 \text{ cm} \times 200 \text{ cm} \times 50 \text{ cm} = 5,000,000$ cubic cm.
Volume of one brick = $25 \text{ cm} \times 10 \text{ cm} \times 5 \text{ cm} = 1250$ cubic cm.
Number of bricks = Volume of wall / Volume of one brick = $5,000,000 / 1250 = 4000$.
Q17. A sphere has a radius of 21 cm. What is its volume? (Use $\pi = \frac{22}{7}$)
(a) 38808 cubic cm
(b) 39000 cubic cm
(c) 40000 cubic cm
(d) 41000 cubic cm
Answer: (a)
Explanation: Volume of a sphere = $\frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 21^3 = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 4 \times 22 \times 21 \times 21 = 88 \times 441 = 38808$ cubic cm.
Q18. The ratio of the radii of two circles is 2:3. What is the ratio of their areas?
(a) 2:3
(b) 3:2
(c) 4:9
(d) 9:4
Answer: (c)
Explanation: Area of a circle is proportional to the square of its radius ($\pi r^2$). So, the ratio of areas = $r_1^2 : r_2^2 = 2^2 : 3^2 = 4:9$.
Q19. A hollow cylinder has an external radius of 8 cm and an internal radius of 7 cm. If its height is 10 cm, what is the volume of the material used? (Use $\pi = \frac{22}{7}$)
(a) 471.43 cubic cm (approx)
(b) 500 cubic cm (approx)
(c) 520 cubic cm (approx)
(d) 550 cubic cm (approx)
Answer: (a)
Explanation: Volume of material = Volume of outer cylinder – Volume of inner cylinder = $\pi R^2 h – \pi r^2 h = \pi h (R^2 – r^2) = \frac{22}{7} \times 10 \times (8^2 – 7^2) = \frac{220}{7} \times (64 – 49) = \frac{220}{7} \times 15 = \frac{3300}{7} \approx 471.43$ cubic cm.
Q20. What is the area of a rhombus if its diagonals are 16 cm and 12 cm?
(a) 48 sq cm
(b) 96 sq cm
(c) 120 sq cm
(d) 192 sq cm
Answer: (b)
Explanation: Area of a rhombus = $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 16 \times 12 = 8 \times 12 = 96$ sq cm.
Q21. If the radius of a cylinder is doubled and its height is halved, by what factor does its volume change?
(a) No change
(b) 2 times
(c) 4 times
(d) 8 times
Answer: (b)
Explanation: Original volume $V_1 = \pi r^2 h$. New radius $r’ = 2r$, new height $h’ = h/2$.
New volume $V_2 = \pi (2r)^2 (h/2) = \pi (4r^2) (h/2) = 2\pi r^2 h = 2V_1$. The volume becomes 2 times the original.
Q22. A rectangular garden is 90 m long and 70 m wide. A path 5 m wide is built outside around it. What is the area of the path?
(a) 1200 sq m
(b) 1400 sq m
(c) 1600 sq m
(d) 1800 sq m
Answer: (d)
Explanation: Area of garden = $90 \times 70 = 6300$ sq m.
New length with path = $90 + 2 \times 5 = 100$ m. (5m on both sides)
New width with path = $70 + 2 \times 5 = 80$ m. (5m on both sides)
Area including path = $100 \times 80 = 8000$ sq m.
Area of path = Area including path – Area of garden = $8000 – 6300 = 1700$ sq m.
Self-correction: Options again. My calculation is 1700. Let me check for common scenarios.
If path is on one side, it’s not “around it”.
Let’s consider if the width refers to total width added, not from one side. Unlikely for “around it”.
Let me re-check numbers. $90+10 = 100$. $70+10 = 80$. $100 \times 80 = 8000$. $90 \times 70 = 6300$. $8000-6300=1700$.
The closest option is 1800. This could be a rounding issue or a slightly different interpretation of dimensions in some problem sets.
But let’s stick to the calculation explicitly. 1700 is the direct result. Let me re-phrase the options if necessary.
For competitive exams, sometimes they expect a slight approximation or there might be an error in provided options.
If the path area was calculated differently, e.g., $2 \times (\text{Length} + \text{Breadth} + 2 \times \text{path width}) \times \text{path width}$ (for outer perimeter approach)
Path area = $2 \times 5 \times (90+5+70+5) = 10 \times 170 = 1700$. No, this is for finding the perimeter.
Area of path = Outer Area – Inner Area
L_outer = L_inner + 2w = 90 + 2*5 = 100
W_outer = W_inner + 2w = 70 + 2*5 = 80
Area_outer = 100 * 80 = 8000.
Area_inner = 90 * 70 = 6300.
Area_path = 8000 – 6300 = 1700.
Okay, I will change option (d) to 1700.
Q22. A rectangular garden is 90 m long and 70 m wide. A path 5 m wide is built outside around it. What is the area of the path?
(a) 1200 sq m
(b) 1400 sq m
(c) 1600 sq m
(d) 1700 sq m
Answer: (d)
Explanation: Length of garden = 90 m, Width of garden = 70 m. Area of garden = $90 \times 70 = 6300$ sq m.
With a 5 m wide path around it, the new length and width will be:
New Length = $90 + (2 \times 5) = 100$ m
New Width = $70 + (2 \times 5) = 80$ m
Area including path = $100 \times 80 = 8000$ sq m.
Area of path = Area including path – Area of garden = $8000 – 6300 = 1700$ sq m.
Q23. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls and ceiling of the room at the rate of Rs. 20 per sq m.
(a) Rs. 1240
(b) Rs. 1480
(c) Rs. 1640
(d) Rs. 1800
Answer: (b)
Explanation: Area of walls = $2h(l+b) = 2 \times 3 \times (5+4) = 6 \times 9 = 54$ sq m.
Area of ceiling = $L \times B = 5 \times 4 = 20$ sq m.
Total area to be whitewashed = $54 + 20 = 74$ sq m.
Cost of whitewashing = $74 \times 20 = Rs. 1480$.
Q24. A well is dug 20 m deep and has a diameter of 7 m. What is the volume of earth taken out? (Use $\pi = \frac{22}{7}$)
(a) 770 cubic m
(b) 790 cubic m
(c) 800 cubic m
(d) 820 cubic m
Answer: (a)
Explanation: The well is a cylinder. Radius (r) = diameter / 2 = 7/2 m = 3.5 m. Height (h) = 20 m.
Volume of earth taken out = Volume of cylinder = $\pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 20 = \frac{22}{7} \times 12.25 \times 20 = 22 \times 1.75 \times 20 = 770$ cubic m.
Q25. If the diagonal of a square is ‘d’, its area is:
(a) $d^2$
(b) $\frac{d^2}{2}$
(c) $2d^2$
(d) $\frac{d^2}{\sqrt{2}}$
Answer: (b)
Explanation: Let ‘s’ be the side of the square. By Pythagoras theorem, $s^2 + s^2 = d^2$, so $2s^2 = d^2$.
Area of square = $s^2$. From $2s^2 = d^2$, we get $s^2 = \frac{d^2}{2}$. So, Area = $\frac{d^2}{2}$.
