MathsCBSE Class 10 · Basic · Board exam 2026, set 430/2/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2026: question 37

Set in the CBSE Class 10 board exam of 2026, paper 430/2/1, 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

A woman borrowed ₹ 10,00,000 from her friend and promised to return the borrowed money in monthly instalments beginning from the next month. After one month, she returned ₹ 10,000, the next month she returned ₹ 15,000, the third month she returned ₹ 20,000 and so on, thereby increasing the monthly instalment uniformly.

Based on the above information, answer the following questions :

(i)
Find the amount of instalment paid in the tenth month.
1 mark
(ii)
In which instalment did she pay ₹ 40,000 ?
1 mark
(iii)
(a) If she returned ₹ 11,50,000 in all, how many instalments did she pay ?
2 marks
OR
(iii)
(b) By which instalment has she returned a total amount of ₹ 3,25,000 ?
2 marks

Figure in the paper: Illustration of two women talking (one with a thought bubble showing money being handed over), a hand giving a banknote, and a hand holding a calendar with a clock. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/2/1.

Hints

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

Hint 1

Identify the first term and the common difference of the instalments.

Hint 2

Use $a_{n} = a + (n - 1) d$ for (i) and (ii).

Hint 3

For (iii), set $S_{n}$ equal to the given total and solve the quadratic, 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.

(i) The instalments form an AP with $a = 10000$ and $d = 5000$ . So $a_{10} = a + 9 d = 10000 + 9 \times 5000 = ₹55, 000$ .
1 mark
(ii) $a_{n} = 40000 \Rightarrow 10000 + (n - 1) 5000 = 40000 \Rightarrow n - 1 = 6 \Rightarrow n = 7$ . So she paid ₹40,000 in the 7th instalment.
1 mark
(iii)(a) $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [20000 + 5000 (n - 1)] = 2500 n (n + 3)$ . Setting $2500 n (n + 3) = 1150000$ gives $n^{2} + 3 n - 460 = 0$ .
1 mark
(iii)(a) $(n + 23) (n - 20) = 0 \Rightarrow n = 20$ (rejecting $n = - 23$ ). She paid 20 instalments.
1 mark
(iii)(b) $2500 n (n + 3) = 325000 \Rightarrow n^{2} + 3 n - 130 = 0 \Rightarrow (n + 13) (n - 10) = 0 \Rightarrow n = 10$ (rejecting $n = - 13$ ). She had returned ₹3,25,000 by the 10th instalment.
2 marks

Answer: (i) ₹55,000; (ii) 7th instalment; (iii)(a) 20 instalments; (iii)(b) 10th instalment

Where the marks are

(i) Uses $a_{10} = a + 9 d$ to get ₹55,000 (1 mark)
(ii) Solves $a_{n} = 40000$ to get $n = 7$ (1 mark)
(iii)(a) Forms $n^{2} + 3 n - 460 = 0$ from $S_{n}$ (1 mark)
(iii)(a) Solves to get $n = 20$ (1 mark)
(iii)(b) Forms $n^{2} + 3 n - 130 = 0$ and gets $n = 10$ (2 marks)

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 430/2/1 (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

More questions from Arithmetic Progressions