MathsCBSE Class 10 · Standard · Sample paper, session 2024-25 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard sample paper 2024-25: question 38

From the official CBSE sample question paper for 2024-25, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes

Metallic silos are used by farmers for storing grains. Farmer Girdhar has decided to build a new metallic silo to store his harvested grains. It is in the shape of a cylinder mounted by a cone.

Dimensions of the conical part of a silo is as follows: Radius of base = $1.5$ m, Height = $2$ m.

Dimensions of the cylindrical part of a silo is as follows: Radius = $1.5$ m, Height = $7$ m.

On the basis of the above information answer the following questions.

(i)
Calculate the slant height of the conical part of one silo.
1 mark
(ii)
Find the curved surface area of the conical part of one silo.
1 mark
(iii)(A)
Find the cost of metal sheet used to make the curved cylindrical part of 1 silo at the rate of ₹ $2000$ per $m^{2}$ .
2 marks
OR
(iii)(B)
Find the total capacity of one silo to store grains.
2 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

The slant height, radius and height of a cone form a right triangle.

Hint 2

The curved surface of a cone is $π r l$ , and that of a cylinder is $2 π r h$ .

Hint 3

Capacity means volume: add the volume of the cylinder and the volume of the cone.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Slant height $l = r^{2} + h^{2} = (1.5)^{2} + 2^{2} = 2.25 + 4$
½ mark
(i) $l = 6.25 = 2.5$ m
½ mark
(ii) CSA of cone $= π r l = \frac{22}{7} \times 1.5 \times 2.5$
½ mark
(ii) CSA of cone $= \frac{82.5}{7} \approx 11.79$ m $^{2}$
½ mark
(iii)(a) CSA of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66$ m $^{2}$
1 mark
(iii)(a) Cost of metal sheet $= 66 \times 2000 =$ ₹ $1, 32, 000$
1 mark
(iii)(b) Volume of cylinder $= π r^{2} h = \frac{22}{7} \times (1.5)^{2} \times 7 = 49.5$ m $^{3}$
½ mark
(iii)(b) Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (1.5)^{2} \times 2 \approx 4.71$ m $^{3}$
1 mark
(iii)(b) Total capacity $= 49.5 + 4.71 = 54.21$ m $^{3}$ (approx.)
½ mark

Answer: (i) 2.5 m; (ii) ≈ 11.79 m²; (iii)(a) ₹1,32,000; (iii)(b) ≈ 54.21 m³

Where the marks are

(i) Uses $l = r^{2} + h^{2}$ with the values substituted (½ mark)
(i) Gets $l = 2.5$ m (½ mark)
(ii) Writes $π r l$ with substitution (½ mark)
(ii) Gets $\approx 11.79$ m $^{2}$ (½ mark)
(iii)(a) Gets CSA of cylinder $= 66$ m $^{2}$ (1 mark)
(iii)(a) Gets cost ₹1,32,000 (1 mark)
(iii)(b) Gets volume of cylinder $= 49.5$ m $^{3}$ (½ mark)
(iii)(b) Gets volume of cone $\approx 4.71$ m $^{3}$ (1 mark)
(iii)(b) Adds to get total $\approx 54.21$ m $^{3}$ (½ mark)

Revise the chapter

Chapter: Surface Areas and Volumes (notes, formulas, where marks are lost and every past question)
Topic: Surface Area and Volume Formulae Revision

Practise this chapterBoard-style questions from Surface Areas and Volumes, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard sample paper 2024-25: question 38

Hints

Step-by-step solution

Revise the chapter

More questions from Surface Areas and Volumes