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MathsCBSE Class 10 · Basic · Board exam 2025, set 430/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2025: question 32(b)

Set in the CBSE Class 10 board exam of 2025, paper 430/4/1, section D, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerQuadratic Equations

Three consecutive positive integers are such that the sum of the square of smallest and product of other two is 67. Find the numbers, using quadratic equation.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/4/1.

Hints

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

Hint 1

Call the smallest integer ; what are the next two?

Hint 2

Square the smallest and add the product of the other two, then set the result equal to 67.

Hint 3

Factorise the quadratic and discard any root that is not a positive integer.

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.

  1. Let the three consecutive positive integers be , and .

    1 mark
  2. According to the question, .

    1 mark
  3. 1 mark
  4. , so or . Reject since must be a positive integer, so .

    1 mark
  5. The numbers are 5, 6 and 7. Check: .

    1 mark

Answer: 5, 6 and 7

Where the marks are

  • Takes the integers as (1 mark)
  • Forms the equation (1 mark)
  • Reduces to (1 mark)
  • Solves to get , rejecting (1 mark)
  • States the numbers 5, 6, 7 (1 mark)

Revise the chapter

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.