MathsCBSE Class 10 · Standard · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2025-26: question 34(A)

From the official CBSE sample question paper for 2025-26, section D, question 34(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(A) · 5 marks · Long answerSurface Areas and Volumes

From a solid right circular cone, whose height is 6cm and radius of base is 12cm, a right circular cylindrical cavity of height 3cm and radius 4cm is hollowed out such that bases of cone and cylinder form concentric circles. Find the surface area of the remaining solid in terms of $π$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

List every surface you could touch on the remaining solid: outside, bottom and inside the hole.

Hint 2

The bottom loses a circle of radius 4 cm, but the top of the cavity adds an equal circle back.

Hint 3

Total = $π R L + π R^{2} + 2 π r h$ .

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.

Given: cone with radius $R = 12$ cm and height $H = 6$ cm. Cylindrical cavity with radius $r = 4$ cm and height $h = 3$ cm, hollowed out from the base. The two bases are concentric circles.
Slant height of the cone: $L = R^{2} + H^{2} = 1 2^{2} + 6^{2} = 180 = 65$ cm
½ mark
Curved surface area of the cone $= π R L = π \times 12 \times 65 = 725 π$ cm²
1 mark
Area of the base. The exposed base is a ring: the outer circle minus the inner circle. The top face of the cavity, a circle of radius 4 cm, is also exposed. The two circles of radius 4 cm cancel, so:
$π R^{2} - π r^{2} + π r^{2} = π R^{2} = π (12)^{2} = 144 π$ cm²
1 mark
Inner curved surface of the cylindrical cavity $= 2 π r h = 2 π \times 4 \times 3 = 24 π$ cm²
1 mark
Total surface area of the remaining solid
$=$ CSA of cone $+$ base area $+$ CSA of cylinder
$= 725 π + 144 π + 24 π$
1 mark
$= (168 + 725) π$ cm²
This is the same as $(168 + 3620) π$ cm².
½ mark

Answer: $(168 + 725) π$ cm² (same as $(168 + 3620) π$ cm²)

Where the marks are

Finds slant height $65$ cm (½ mark)
Finds CSA of cone $725 π$ cm² (1 mark)
Finds base area $144 π$ cm² after allowing for the cavity's top face (1 mark)
Finds CSA of cylinder $24 π$ cm² (1 mark)
Adds the correct three parts for the total surface area (1 mark)
Final answer $(168 + 725) π$ cm² (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 2025-26 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2025-26: question 34(A)

Hints

Step-by-step solution

Revise the chapter

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