MathsCBSE Class 10 · Standard · Board exam 2025, set 30/1/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2025: question 36

Set in the CBSE Class 10 board exam of 2025, paper 30/1/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions

Case Study – 1
A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.

Based on the information given above, answer the following questions :

(i)
Write the fourth, fifth and sixth term of the Arithmetic Progression so formed.
1 mark
(ii)
Determine the distance of the $8^{t h}$ round.
1 mark
(iii)(a)
Find the total distance run after completing all 10 rounds.
2 marks
OR
(iii)(b)
If a runner completes only the first 6 rounds, what is the total distance run by the runner ?
2 marks

Figure in the paper: Illustration of ten coloured runner silhouettes running around a circular track. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/1/1.

Hints

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

Hint 1

The first term is 300 and each round is 50 m longer than the one before.

Hint 2

Use $a_{n} = a + (n - 1) d$ for one round's distance.

Hint 3

Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ for the total distance.

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.

The round distances form an AP: 300, 350, 400, …, with $a = 300$ and $d = 50$ .
(i) $a_{4} = 400 + 50 = 450$ m, $a_{5} = 500$ m, $a_{6} = 550$ m.
1 mark
(ii) $a_{n} = a + (n - 1) d$ , so $a_{8} = 300 + 7 \times 50$
½ mark
(ii) $a_{8} = 300 + 350 = 650$ m
½ mark
(iii)(a) $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , so $S_{10} = \frac{10}{2} [2 (300) + 9 (50)] = 5 (600 + 450)$
1 mark
(iii)(a) $S_{10} = 5 \times 1050 = 5250$ m
1 mark
(iii)(b) $S_{6} = \frac{6}{2} [2 (300) + 5 (50)] = 3 (600 + 250)$
1 mark
(iii)(b) $S_{6} = 3 \times 850 = 2550$ m
1 mark

Answer: (i) 450 m, 500 m, 550 m; (ii) 650 m; (iii)(a) 5250 m, (iii)(b) 2550 m

Where the marks are

(i) Writes the 4th, 5th and 6th terms as 450, 500 and 550 (1 mark)
(ii) Uses $a_{8} = 300 + 7 \times 50$ (½ mark)
(ii) Gets 650 m (½ mark)
(iii)(a) Substitutes correctly into $S_{10}$ (1 mark)
(iii)(a) Gets 5250 m (1 mark)
(iii)(b) Substitutes correctly into $S_{6}$ (1 mark)
(iii)(b) Gets 2550 m (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 paper2025 board paper, set 30/1/1 (Standard): 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 Standard 2025: question 36

Hints

Step-by-step solution

Revise the chapter

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