CBSE Class 10 Mathematics Basic 2025: question 37
Set in the CBSE Class 10 board exam of 2025, 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)1 mark
What is the radius of the 13th spiral ?
- (ii)1 mark
If the radius of the nth spiral is 500 cm, find the value of n.
- (iii)(a)2 marks
Find the total number of saplings till the 11th spiral.
OR
(iii)(b)2 marksTill which spiral, will there be a total of 450 saplings ?
Figure in the paper: A horizontal line with points A and B close together near the centre. Semicircular arcs labelled l1 (above, small, centred at A), l2 (below, centred at B), l3 (above, larger), l4 (below, larger) forming an expanding spiral; a dotted continuation arc is shown at the upper right. Photograph of rose saplings in a spiral pattern is also shown. See it in the official paper.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
Both the radii and the numbers of flowers increase by a fixed amount each time, so think of them as APs.
Hint 2
For the radius use . For the total number of saplings use .
Hint 3
In (iii)(b), set and solve the resulting quadratic in , rejecting the negative root.
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.
- 1 mark
(i) The radii 50, 100, 150, … form an AP with and .
So the radius of the 13th spiral is 650 cm.
- 1 mark
(ii) Set :
- 1½ marks
(iii)(a) The numbers of saplings 10, 20, 30, … form an AP with and . Then
- ½ mark
(iii)(a)
So there are 660 saplings till the 11th spiral.
- 1 mark
(iii)(b) With and , set :
- ½ mark
(iii)(b) Simplify: , so .
- ½ mark
(iii)(b) Factorise: , so ( is rejected because must be a natural number).
So there are 450 saplings in total till the 9th spiral.
Answer: (i) 650 cm; (ii) n = 10; (iii)(a) 660 saplings; (iii)(b) 9th spiral
Where the marks are
- (i) Finds cm using , (1 mark)
- (ii) Solves to get (1 mark)
- (iii)(a) Writes (1½ marks)
- (iii)(a) Gets 660 saplings (½ mark)
- (iii)(b) Forms (1 mark)
- (iii)(b) Reduces it to (½ mark)
- (iii)(b) Gets (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
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