MathsCBSE Class 10 · Basic · Board exam 2025, set 430/3/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 36

Set in the CBSE Class 10 board exam of 2025, paper 430/3/1, 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

In a society, a yoga instructor was hired to train the people of the society to live a healthy lifestyle. Yoga sessions were held daily from 5 p.m. to 7 p.m. in the society park. On day one, 5 people joined the yoga session, on day two, 3 more people joined, on day three, another 3 people joined and in this manner every next day, 3 more people kept on joining.

Based on the given information, answer the following questions:

(i)
On which day did 59 people join the yoga session?
1 mark
(ii)
How many people joined the yoga session on the 31st day?
1 mark
(iii)
(a) The yoga instructor was paid ₹100 for each person attending the yoga session. On which day would he earn ₹5,000?
2 marks
OR
(iii)
(b) What was the total amount earned by the yoga instructor in 16 days?
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/3/1.

Hints

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

Hint 1

Write down the number of people on days 1, 2, 3, … and look at the pattern.

Hint 2

The numbers form an AP with first term 5 and common difference 3. Use $a_{n} = a + (n - 1) d$ .

Hint 3

For the earnings, convert rupees into the number of persons. For the total over 16 days, use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ and multiply by ₹100.

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 number of people attending on successive days is 5, 8, 11, … This is an AP with $a = 5$ and $d = 3$ , so $a_{n} = 5 + (n - 1) 3$ .
(i) $5 + (n - 1) \times 3 = 59 \Rightarrow 3 (n - 1) = 54 \Rightarrow n - 1 = 18 \Rightarrow n = 19$ .
So 59 people attended on the 19th day.
1 mark
(ii) $a_{31} = 5 + 30 \times 3 = 5 + 90 = 95$ .
So 95 people attended on the 31st day.
1 mark
(iii)(a) To earn ₹5,000 at ₹100 per person, the number of persons needed $= \frac{5000}{100} = 50$ .
1 mark
(iii)(a) $5 + (n - 1) \times 3 = 50 \Rightarrow 3 (n - 1) = 45 \Rightarrow n = 16$ .
He earns ₹5,000 on the 16th day.
1 mark
(iii)(b) $S_{16} = \frac{n}{2} [2 a + (n - 1) d] = \frac{16}{2} [2 \times 5 + 15 \times 3]$
1 mark
(iii)(b) $= 8 [10 + 45] = 8 \times 55 = 440$ persons in total over 16 days.
½ mark
(iii)(b) Total amount earned $= 440 \times ₹100 = ₹44, 000$ .
½ mark

Answer: (i) 19th day; (ii) 95 people; (iii)(a) 16th day; (iii)(b) ₹44,000

Where the marks are

(i) Uses $5 + (n - 1) 3 = 59$ to get $n = 19$ (1 mark)
(ii) $a_{31} = 95$ (1 mark)
(iii)(a) Number of persons $= 5000/100 = 50$ (1 mark)
(iii)(a) Solves $5 + (n - 1) 3 = 50$ to get $n = 16$ (1 mark)
(iii)(b) Correct substitution in $S_{16} = \frac{16}{2} [10 + 15 \times 3]$ (1 mark)
(iii)(b) $S_{16} = 440$ (½ mark)
(iii)(b) Amount $= ₹44, 000$ (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 430/3/1 (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 36

Hints

Step-by-step solution

Revise the chapter

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