MathsCBSE Class 10 · Standard · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2023-24: question 34(b)

From the official CBSE sample question paper for 2023-24, section D, question 34(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(b) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes

A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, find the area of the canvas required for making the tent, keeping a provision of $26 m^{2}$ of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of ₹ 500 per $m^{2}$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

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

Hint 1

Only the curved surfaces need canvas. There is no base or top circle.

Hint 2

First find the height of the conical part, then its slant height using $l = h^{2} + r^{2}$ .

Hint 3

Add the two curved surface areas and the 26 m², then multiply by ₹500.

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.

Radius $r = 14$ m, height of the cylinder $= 3$ m.
Height of the cone $h = 13.5 - 3 = 10.5$ m
½ mark
Slant height $l = h^{2} + r^{2} = (10.5)^{2} + 1 4^{2} = 110.25 + 196 = 306.25 = 17.5$ m
1 mark
CSA of the cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 14 \times 3 = 264 m^{2}$
1 mark
CSA of the cone $= π r l = \frac{22}{7} \times 14 \times 17.5 = 770 m^{2}$
1 mark
Total curved surface (canvas for the tent) $= 264 + 770 = 1034 m^{2}$
½ mark
Canvas to be purchased $= 1034 + 26 = 1060 m^{2}$
½ mark
Cost $= 1060 \times ₹500 = ₹5, 30, 000$
½ mark

Answer: Canvas required = 1060 m²; cost = ₹5,30,000

Where the marks are

Height of the cone = 10.5 m (½ mark)
Slant height = 17.5 m (1 mark)
CSA of the cylinder = 264 m² (1 mark)
CSA of the cone = 770 m² (1 mark)
Total CSA = 1034 m² (½ mark)
Adds the wastage: 1060 m² (½ mark)
Cost = ₹5,30,000 (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 2023-24 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2023-24: question 34(b)

Hints

Step-by-step solution

Revise the chapter

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