MathsCBSE Class 10 · Basic · Board exam 2026, set 430/1/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2026: question 37

Set in the CBSE Class 10 board exam of 2026, paper 430/1/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions

Case Study – 2

In a garden, saplings of rose flowers were planted at equal intervals to form a spiral pattern. The spiral is made up of successive semicircles, with centres alternatively at A and B, starting with centre at A, of radii 50 cm, 100 cm, 150 cm, ....... as shown in the figure given below. Spiral 1 has 10 flowers, Spiral 2 has 20 flowers, Spiral 3 has 30 flowers and so on.

Based on the above information, answer the following questions :

(i)
What is the radius of the $1 3^{t h}$ spiral ?
1 mark
(ii)
If the radius of the $n^{t h}$ spiral is 500 cm, find the value of n.
1 mark
(iii)
(a) Find the total number of saplings till the $1 1^{t h}$ spiral.
2 marks
OR
(iii)
(b) Till which spiral, will there be a total of 450 saplings ?
2 marks

Figure in the paper: Spiral made of successive semicircles on a horizontal line with points A and B on it, centres alternating. Arcs labelled l1 (small, above line, around A-B), l2 (below line), l3 (above, larger), l4 (below, larger), and further arcs continuing outward as a dotted curve on the right. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/1/1.

Hints

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

Hint 1

Both the radii and the numbers of flowers form arithmetic progressions.

Hint 2

Use $a_{n} = a + (n - 1) d$ for the radius and $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ for the total number of saplings.

Hint 3

For (iii)(b), set $S_{n} = 450$ and solve the quadratic, rejecting any negative value.

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) The radii 50, 100, 150, … form an AP with $a = 50$ and $d = 50$ , so $r_{n} = 50 n$ . Hence $r_{13} = 50 + 12 \times 50 = 650$ cm.
1 mark
(ii) $50 + (n - 1) 50 = 500 \Rightarrow 50 n = 500 \Rightarrow n = 10$ .
1 mark
(iii)(a) The numbers of flowers 10, 20, 30, … form an AP with $a = 10$ and $d = 10$ . Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ : $S_{11} = \frac{11}{2} [2 (10) + 10 (10)]$ .
1 mark
(iii)(a) $S_{11} = \frac{11}{2} \times 120 = 660$ saplings.
1 mark
(iii)(b) $\frac{n}{2} [20 + (n - 1) 10] = 450 \Rightarrow 5 n (n + 1) = 450 \Rightarrow n^{2} + n - 90 = 0$ .
1 mark
(iii)(b) $(n + 10) (n - 9) = 0 \Rightarrow n = 9$ , since $n = - 10$ is rejected because $n$ must be positive. So there are 450 saplings till the 9th spiral.
1 mark

Answer: (i) 650 cm; (ii) n = 10; (iii)(a) 660 saplings; (iii)(b) 9th spiral

Where the marks are

(i) Uses the AP of radii to get $r_{13} = 650$ cm (1 mark)
(ii) Solves $50 n = 500$ to get $n = 10$ (1 mark)
(iii)(a) Correct substitution in $S_{11}$ (1 mark)
(iii)(a) Total = 660 saplings (1 mark)
(iii)(b) Forms $n^{2} + n - 90 = 0$ (1 mark)
(iii)(b) Solves to get $n = 9$ , rejecting the negative root (1 mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 430/1/1 (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 2026: question 37

Hints

Step-by-step solution

Revise the chapter

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