MathsCBSE Class 10 · Basic · Board exam 2026, set 430/5/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Basic 2026: question 28(b)

Set in the CBSE Class 10 board exam of 2026, paper 430/5/1, section C, question 28(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28(b) · 3 marks · Short answerArithmetic Progressions

How many 4-digit numbers are divisible by 7?

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/5/1.

Hints

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

Hint 1

Find the first 4-digit number and the last 4-digit number that are divisible by 7.

Hint 2

These multiples form an AP with common difference 7; use the nth-term formula to count them.

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.

Smallest 4-digit multiple of 7 is $1001$ ( $7 \times 143$ ); largest is $9996$ ( $7 \times 1428$ ). The numbers $1001, 1008, 1015, \dots, 9996$ form an AP with $a = 1001$ , $d = 7$ , $a_{n} = 9996$ .
1½ marks
$a_{n} = a + (n - 1) d \Rightarrow 9996 = 1001 + (n - 1) \times 7$
1 mark
$(n - 1) = \frac{8995}{7} = 1285 \Rightarrow n = 1286$
½ mark

Answer: 1286 four-digit numbers

Where the marks are

Identifies AP 1001, 1008, …, 9996 with a, d, $a_{n}$ (1½ marks)
Uses $a_{n} = a + (n - 1) d$ correctly (1 mark)
n = 1286 (½ 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 paper2026 board paper, set 430/5/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2026: question 28(b)

Hints

Step-by-step solution

Revise the chapter

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