A tent house owner provides furniture on rent. He stacks chairs in his shop to save space. In the diagram, the height of seat of chair from ground is represented by h_1, h_2, h_3, …. The height of first seat is 44 cm from ground level and gap between every two seats is 10 cm. (i) Write the values of h_1, h_2, h_3, h_4 and h_5 in this order only. (ii) Show that the above values form an A.P. Write its first term and common difference. (iii) (a) If chairs can be stacked up to the maximum height of 160 cm, then find the maximum number of chairs in a stack. (iii) (b) Is it possible to stack 15 chairs if maximum height of the stack can not be more than 180 cm? Justify your answer.

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 38

Set in the CBSE Class 10 board exam of 2025, paper 430/5/1, 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-basedArithmetic Progressions

A tent house owner provides furniture on rent. He stacks chairs in his shop to save space. In the diagram, the height of seat of chair from ground is represented by $h_{1}, h_{2}, h_{3}, \dots$ . The height of first seat is 44 cm from ground level and gap between every two seats is 10 cm.

(i)
Write the values of $h_{1}, h_{2}, h_{3}, h_{4}$ and $h_{5}$ in this order only.
1 mark
(ii)
Show that the above values form an A.P. Write its first term and common difference.
1 mark
(iii)
(a) If chairs can be stacked up to the maximum height of 160 cm, then find the maximum number of chairs in a stack.
2 marks
OR
(iii)
(b) Is it possible to stack 15 chairs if maximum height of the stack can not be more than 180 cm? Justify your answer.
2 marks

Figure in the paper: Left: a stack of chairs. Right: five stacked chair seats drawn as trapezoids, with the bottom seat's height from the ground marked as 44 cm and the vertical gap between successive seats marked as 10 cm. Heights from ground are labelled h1 (lowest), h2, h3, h4, h5 (highest) with arrows from the ground line. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/5/1.

Hints

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

Hint 1

Each new chair raises the seat by 10 cm above the previous one, starting from 44 cm.

Hint 2

Check the difference between consecutive heights to decide whether they form an A.P.

Hint 3

Use aₙ = a + (n − 1)d. Set it equal to the height limit and remember that n must be a whole number.

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) Each seat is 10 cm above the one below it. So $h_{1} = 44$ cm, $h_{2} = 54$ cm, $h_{3} = 64$ cm, $h_{4} = 74$ cm and $h_{5} = 84$ cm.
1 mark
(ii) $h_{2} - h_{1} = h_{3} - h_{2} = h_{4} - h_{3} = h_{5} - h_{4} = 10$ . The difference between consecutive terms is constant, so the heights form an A.P.
½ mark
(ii) First term $a = 44$ cm and common difference $d = 10$ cm.
½ mark
(iii)(a) The seat height of the $n$ th chair is $a_{n} = a + (n - 1) d$ . For the maximum height of 160 cm: $44 + (n - 1) \times 10 = 160$ .
1 mark
(iii)(a) This gives $(n - 1) \times 10 = 116$ , so $n - 1 = 11.6$ and $n = 12.6$ .
½ mark
(iii)(a) $n$ must be a whole number and the height must not exceed 160 cm, so at most 12 chairs can be stacked.
½ mark
(iii)(b) $h_{15} = a + 14 d = 44 + 14 \times 10 = 184$ cm.
1 mark
(iii)(b) $184$ cm $> 180$ cm.
½ mark
(iii)(b) Hence it is not possible to stack 15 chairs.
½ mark

Answer: (i) 44, 54, 64, 74, 84 cm; (ii) common difference 10, so an A.P. with a = 44, d = 10; (iii)(a) 12 chairs; (iii)(b) No, since h₁₅ = 184 cm > 180 cm

Where the marks are

(i) Writes 44, 54, 64, 74, 84 (1 mark)
(ii) Shows the common difference is constant, hence an A.P. (½ mark)
(ii) a = 44, d = 10 (½ mark)
(iii)(a) Forms 44 + (n − 1) × 10 = 160 (1 mark)
(iii)(a) Gets n = 12.6 (½ mark)
(iii)(a) Concludes maximum 12 chairs (½ mark)
(iii)(b) Finds h₁₅ = 44 + 14 × 10 = 184 cm (1 mark)
(iii)(b) Compares 184 > 180 (½ mark)
(iii)(b) Concludes 15 chairs cannot be stacked (½ mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)
Topic: Finding nth Term from the Sum of n Terms

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 430/5/1 (Basic): 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 Basic 2025: question 38

Hints

Step-by-step solution

Revise the chapter

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