Given HCF(2520, 6600) = 40, LCM(2520, 6600) = 252 × k, then the value of k is: (A) 1650 (B) 1600 (C) 165 (D) 1625

MathsCBSE Class 10 · Standard · Board exam 2024, set 30/3/1 · 1 mark · MCQ

CBSE Class 10 Mathematics Standard 2024: question 13

Set in the CBSE Class 10 board exam of 2024, paper 30/3/1, section A, question 13, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q13 · 1 mark · MCQReal Numbers

Given $HCF (2520, 6600) = 40$ , $LCM (2520, 6600) = 252 \times k$ , then the value of $k$ is:

(A)1650
(B)1600
(C)165
(D)1625

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/3/1.

Hints

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

Hint 1

Recall the relation between HCF, LCM and the product of two numbers.

Hint 2

Note that 2520 = 252 × 10.

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.

Use HCF × LCM = product of the numbers: LCM $= \frac{2520 \times 6600}{40} = 2520 \times 165$ .
$252 k = 2520 \times 165$ , so $k = 10 \times 165 = 1650$ . This is option (A).
1 mark

Answer: $k = 1650$ , option (A)

Where the marks are

Uses HCF × LCM = product to get $k = 1650$ , option (A) (1 mark)

Revise the chapter

Chapter: Real Numbers (notes, formulas, where marks are lost and every past question)
Topic: HCF and LCM by Prime Factorisation

Practise this chapterBoard-style questions from Real Numbers, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 30/3/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2024: question 13

Hints

Step-by-step solution

Revise the chapter

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