CBSE Class 10 · Chapter 1 · Number Systems · 2026-27

Real Numbers Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceFundamental Theorem of Arithmetic

The prime factorisation of $2340$ is

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Answer: (a)

2^{2} \times 3^{2} \times 5 \times 13

Why: Divide repeatedly by the smallest prime: 2340 = 2·2·3·3·5·13.

$2340 = 2 \times 1170 = 2^{2} \times 585 = 2^{2} \times 3^{2} \times 65 = 2^{2} \times 3^{2} \times 5 \times 13$ .

Q2

·1 mark·Multiple choiceHCF and LCM by prime factorisation

The HCF of $252$ and $588$ is

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Answer: (b)

84

Why: HCF takes the lowest power of each common prime.

$252 = 2^{2} \times 3^{2} \times 7$ , $588 = 2^{2} \times 3 \times 7^{2}$ . HCF $= 2^{2} \times 3 \times 7 = 84$ .

Q3

·1 mark·Multiple choiceHCF and LCM by prime factorisation

The LCM of $2^{3} \times 5^{2} \times 7$ and $2^{2} \times 3 \times 5^{3}$ is

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Answer: (c)

2^{3} \times 3 \times 5^{3} \times 7

Why: LCM takes the highest power of every prime that appears in either number.

Highest powers: $2^{3}$ , $3^{1}$ , $5^{3}$ , $7^{1}$ . LCM $= 2^{3} \times 3 \times 5^{3} \times 7$ .

Q4

·1 mark·Multiple choiceHCF and LCM by prime factorisation

The HCF and LCM of two numbers are $12$ and $504$ . If one of the numbers is $72$ , the other is

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Answer: (d)

84

Why: For two numbers, HCF × LCM = product of the numbers.

Other number $= \frac{12 \times 504}{72} = \frac{6048}{72} = 84$ . Check: $HCF (72, 84) = 12$ .

Q5

·1 mark·Multiple choiceIrrational numbers

To prove that $\sqrt{7}$ is irrational, Asha assumes $\sqrt{7} = \frac{p}{q}$ , where $p$ and $q$ are co-prime integers and $q \neq 0$ . Squaring gives $p^{2} = 7 q^{2}$ . Which conclusion gives the contradiction?

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Answer: (a)

7

divides both

p

and

q

Why: 7 | p² gives 7 | p (7 is prime); then 7 | q² gives 7 | q, so p and q share the factor 7.

$p^{2} = 7 q^{2}$ means $7 ∣ p^{2}$ , and as $7$ is prime, $7 ∣ p$ . Write $p = 7 r$ : $49 r^{2} = 7 q^{2} \Rightarrow q^{2} = 7 r^{2}$ , so $7 ∣ q$ . Then $7$ is a common factor of $p$ and $q$ , contradicting that they are co-prime. Hence $\sqrt{7}$ is irrational.

Q6

·1 mark·Multiple choiceFundamental Theorem of Arithmetic

For which natural number $k$ does $12^{k}$ end with the digit $0$ ?

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Answer: (b) For no natural number

k

Why: A number ending in 0 needs both 2 and 5 as prime factors; 12^k = 2^2k·3^k has no 5.

$12^{k} = 2^{2 k} \times 3^{k}$ . By the Fundamental Theorem of Arithmetic this factorisation is unique, so $5$ is never a factor and $12^{k}$ never ends in $0$ .

Q7

·1 mark·Multiple choiceHCF and LCM by prime factorisation

The smallest natural number divisible by each of $6$ , $8$ and $15$ is

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Answer: (d)

120

Why: The smallest common multiple is the LCM.

$6 = 2 \times 3$ , $8 = 2^{3}$ , $15 = 3 \times 5$ . LCM $= 2^{3} \times 3 \times 5 = 120$ .

Q8

·1 mark·Multiple choiceHCF and LCM by prime factorisation

The largest number that divides $223$ and $367$ , leaving a remainder of $7$ in each case, is

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Answer: (c)

72

Why: Subtract the remainder, then take the HCF of 216 and 360.

$223 - 7 = 216 = 2^{3} \times 3^{3}$ , $367 - 7 = 360 = 2^{3} \times 3^{2} \times 5$ . HCF $= 2^{3} \times 3^{2} = 72$ (and $72 > 7$ ).

Q9

·1 mark·Multiple choiceHCF and LCM by prime factorisation

If $p$ and $q$ are co-prime natural numbers, then $LCM (p, q)$ is

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Answer: (a)

p q

Why: Co-prime means HCF = 1, so LCM = product ÷ HCF = pq.

$HCF (p, q) \times LCM (p, q) = p q$ and $HCF (p, q) = 1$ , so $LCM (p, q) = p q$ .

Q10

·1 mark·Multiple choiceHCF and LCM by prime factorisation

If $p$ and $q$ are distinct primes, then $HCF (p^{3} q^{2}, p^{2} q^{4})$ is

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Answer: (a)

p^{2} q^{2}

Why: HCF uses the smaller power of each prime: p² and q².

Smaller powers: $p^{min (3, 2)} = p^{2}$ , $q^{min (2, 4)} = q^{2}$ . HCF $= p^{2} q^{2}$ .

Q11

·1 mark·Multiple choiceFundamental Theorem of Arithmetic

If $3087 = 3^{m} \times 7^{n}$ , where $m$ and $n$ are natural numbers, then $m + 2 n$ equals

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Answer: (c)

8

Why: 3087 = 9 × 343 = 3² × 7³, so m = 2, n = 3.

$3087 = 3 \times 1029 = 3^{2} \times 343 = 3^{2} \times 7^{3}$ . So $m = 2$ , $n = 3$ and $m + 2 n = 8$ .

Q12

·1 mark·Multiple choiceIrrational numbers

Exactly one of these four numbers is rational. Which one?

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Answer: (a)

(\sqrt{5} + 2) (\sqrt{5} - 2)

Why: (a + b)(a − b) = a² − b² removes the surd: 5 − 4 = 1.

$(\sqrt{5} + 2) (\sqrt{5} - 2) = 5 - 4 = 1$ . The others are $9 + 4 \sqrt{5}$ , $\sqrt{5} + 2$ and $1 + \frac{2 \sqrt{5}}{5}$ , all irrational.

Q13

·1 mark·Multiple choiceIrrational numbers

The sum of a rational number and an irrational number is

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Answer: (b) always irrational

Why: If r + x = s with r, s rational, then x = s − r would be rational: a contradiction.

Suppose $r + x = s$ with $r, s$ rational and $x$ irrational. Then $x = s - r$ is rational, a contradiction. So the sum is always irrational.

Q14

·1 mark·Multiple choiceHCF and LCM by prime factorisation

For the numbers $26$ and $91$ , $HCF \times LCM$ equals

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Answer: (b)

2366

Why: HCF × LCM of two numbers equals their product, 26 × 91.

$26 = 2 \times 13$ , $91 = 7 \times 13$ : HCF $= 13$ , LCM $= 2 \times 7 \times 13 = 182$ . $13 \times 182 = 2366 = 26 \times 91$ .

Q15

·1 mark·Multiple choiceFundamental Theorem of Arithmetic

The number $2^{4} \times 5^{3} \times 7$ ends with how many zeros?

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Answer: (d)

3

Why: Each zero needs one pair 2 × 5; there are only three 5s.

$2^{4} \times 5^{3} \times 7 = (2 \times 5)^{3} \times 2 \times 7 = 14 \times 10^{3} = 14000$ : three zeros.

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Bus depot timetable (4 marks)

Three city bus routes start from the same depot. A Route A bus leaves every $16$ minutes, a Route B bus every $20$ minutes and a Route C bus every $24$ minutes. At 7:00 a.m. one bus of each route leaves the depot together.

(i) Write

20

and

24

as products of their prime factors. [1 mark]

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Answer:

20 = 2^{2} \times 5

24 = 2^{3} \times 3

20 = 2^{2} \times 5

and

24 = 2^{3} \times 3

. A1

(ii) After how many minutes will a Route A bus and a Route B bus next leave together? [1 mark]

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Answer:

80

minutes

16 = 2^{4}

20 = 2^{2} \times 5

; LCM

= 2^{4} \times 5 = 80

minutes. A1

(iii) At what time will buses of all three routes next leave the depot together? [2 marks]

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Answer: 11:00 a.m.

LCM

(16, 20, 24) = 2^{4} \times 3 \times 5 = 240

minutes M1

= 4

hours, so at 11:00 a.m. A1

OR How many times from 7:00 a.m. to 7:00 p.m. (both included) do buses of all three routes leave together? [2 marks]

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Answer:

4

times

They leave together every LCM

(16, 20, 24) = 240

min

= 4

h M1: at 7 a.m., 11 a.m., 3 p.m. and 7 p.m., i.e.

4

times. A1

Case study 2: Stationery kits (4 marks)

A school club has $312$ pencils, $234$ erasers and $195$ sharpeners. It packs them into identical kits so that every kit has the same number of pencils, the same number of erasers and the same number of sharpeners, with nothing left over, and makes as many kits as possible.

(i) Find the HCF of

312

and

234

. [1 mark]

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Answer:

78

312 = 2^{3} \times 3 \times 13

234 = 2 \times 3^{2} \times 13

; HCF

= 2 \times 3 \times 13 = 78

. A1

(ii) What is the greatest number of kits that can be made? [1 mark]

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Answer:

39

195 = 3 \times 5 \times 13

; HCF

(312, 234, 195) = 3 \times 13 = 39

kits. A1

(iii) How many pencils, erasers and sharpeners are there in each kit? [2 marks]

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Answer:

8

pencils,

6

erasers,

5

sharpeners

Divide each total by

39

M1:

312 \div 39 = 8

234 \div 39 = 6

195 \div 39 = 5

. A1

OR Find the LCM of

234

and

195

and check that HCF

\times

LCM equals

234 \times 195

. [2 marks]

Show answer

Answer: LCM

= 1170

;

39 \times 1170 = 45630 = 234 \times 195

LCM

= 2 \times 3^{2} \times 5 \times 13 = 1170

A1; HCF

= 39

and

39 \times 1170 = 45630 = 234 \times 195

. A1

Next steps for Real Numbers

This free set is separate from the chapter's question bank. On the Real Numbers chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 40 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Real Numbers: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.