MathsCBSE Class 10 · Basic · Board exam 2025, set 430/2/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 37

Set in the CBSE Class 10 board exam of 2025, 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

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/2/1.

Hints

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

Hint 1

The instalments increase by the same amount each month, so they form an AP. Identify $a$ and $d$ .

Hint 2

Use $a_{n} = a + (n - 1) d$ for a single instalment and $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ for the total returned.

Hint 3

Divide the equation by 5000 to get a simple quadratic in $n$ , then factorise and reject 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.

The instalments 10000, 15000, 20000, ... form an AP with $a = 10000$ and $d = 5000$ .
(i) $a_{10} = a + 9 d = 10000 + 9 \times 5000 = 55000$
So the tenth instalment is ₹55,000.
1 mark
(ii) $a_{n} = a + (n - 1) d \Rightarrow 40000 = 10000 + (n - 1) 5000$
$(n - 1) = 6 \Rightarrow n = 7$
She paid ₹40,000 in the 7th instalment.
1 mark
(iii)(a) Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ :
$1150000 = \frac{n}{2} [20000 + (n - 1) 5000]$
1 mark
(iii)(a) This gives $2300000 = n (15000 + 5000 n)$ , that is, $5 n^{2} + 15 n - 2300 = 0$ , or $n^{2} + 3 n - 460 = 0$ .
½ mark
(iii)(a) $(n - 20) (n + 23) = 0 \Rightarrow n = 20$ or $n = - 23$ (rejected, since $n$ is a natural number).
So she paid 20 instalments.
½ mark
(iii)(b) $325000 = \frac{n}{2} [20000 + (n - 1) 5000]$
1 mark
(iii)(b) This gives $650000 = n (15000 + 5000 n)$ , that is, $5 n^{2} + 15 n - 650 = 0$ , or $n^{2} + 3 n - 130 = 0$ .
½ mark
(iii)(b) $(n - 10) (n + 13) = 0 \Rightarrow n = 10$ or $n = - 13$ (rejected).
She had returned ₹3,25,000 by the 10th instalment.
½ mark

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

Where the marks are

(i) Uses $a_{10} = a + 9 d$ to get ₹55,000 (1 mark)
(ii) Solves $40000 = 10000 + (n - 1) 5000$ to get $n = 7$ (1 mark)
(iii)(a) Sets up $S_{n} = 1150000$ correctly (1 mark)
(iii)(a) Reduces to $n^{2} + 3 n - 460 = 0$ (½ mark)
(iii)(a) Gets $n = 20$ and rejects $- 23$ (½ mark)
(iii)(b) Sets up $S_{n} = 325000$ correctly (1 mark)
(iii)(b) Reduces to $n^{2} + 3 n - 130 = 0$ (½ mark)
(iii)(b) Gets $n = 10$ and rejects $- 13$ (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 430/2/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2025: question 37

Hints

Step-by-step solution

Revise the chapter

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