CBSE Maths › Class 10 PYQs › Real Numbers

Real Numbers: 3 marks Questions (CBSE Class 10)

49 different 3 marks questions on Real Numbers from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (104)2 marks (48)3 marks (49)4 marks (2)

Q313 marksShort AnswerReal Numbers CBSE 2026 · Standard 30/4/1

Find the greatest number less than 10,000 which is exactly divisible by 48, 60 and 65.

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Answer: 9360

$48 = 2^{4} \times 3$ , $60 = 2^{2} \times 3 \times 5$ , $65 = 5 \times 13$
LCM $= 2^{4} \times 3 \times 5 \times 13 = 3120$
Required number is the greatest multiple of 3120 below 10,000.
$3 \times 3120 = 9360$ and $4 \times 3120 = 12480 > 10000$
Answer: 9360

Q263 marksShort AnswerReal Numbers CBSE 2026 · Standard 30/4/2

A trader has three different types of oils of volume 870 $l$ , 812 $l$ and 638 $l$ . Find the least number of containers of equal size required to store all the oil without getting mixed.

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Answer: 40 containers (each of 58

l

)

$870 = 2 \times 3 \times 5 \times 29$ , $812 = 2^{2} \times 7 \times 29$ , $638 = 2 \times 11 \times 29$
HCF $= 2 \times 29 = 58$ , so each container holds 58 $l$ .
Number of containers $= \frac{870}{58} + \frac{812}{58} + \frac{638}{58} = 15 + 14 + 11 = 40$

Q283 marksShort AnswerReal Numbers CBSE 2026 · Standard 30/4/3

Find the greatest number which divides 764 and 1198, leaving remainders 8 and 10 respectively.

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Answer: 108

The number divides 764 $-$ 8 = 756 and 1198 $-$ 10 = 1188 exactly.
$756 = 2^{2} \times 3^{3} \times 7$ , $1188 = 2^{2} \times 3^{3} \times 11$
HCF $= 2^{2} \times 3^{3} = 108$
Required number = 108 (it exceeds both remainders).

Q313 marksShort AnswerReal Numbers CBSE 2026 · Standard 30/5/1

The dimensions of a window are 156 cm × 216 cm. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

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Answer: Side = 12 cm; number of squares = 234

$156 = 2^{2} \times 3 \times 13$ , $216 = 2^{3} \times 3^{3}$
HCF $= 2^{2} \times 3 = 12$ , so the side of each square is 12 cm.
Number of squares $= \frac{156}{12} \times \frac{216}{12} = 13 \times 18 = 234$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/1/1

The factor tree of a number $x$ is shown below :
Find the values of $x$ , $y$ , $a$ and $b$ . Hence, write the product of the prime factors of the number $x$ so obtained.

Diagram for CBSE 2025 Class 10 Maths question 26 (OR)

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Answer:

b = 7

a = 3

y = 420

x = 840

;

840 = 2^{3} \times 3 \times 5 \times 7

$35 = 5 \times b$ , so $b = 7$ .
$210 = a \times 70$ , so $a = 3$ .
$y = 2 \times 210 = 420$ .
$x = 2 \times y = 840$ .
$x = 2 \times 2 \times 3 \times 2 \times 5 \times 7 = 2^{3} \times 3 \times 5 \times 7$ .

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/2/1

State the “Fundamental Theorem of Arithmetic” and use it to find LCM of 36 and 54.

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Answer: LCM = 108

Fundamental Theorem of Arithmetic: every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
$36 = 2^{2} \times 3^{2}$ and $54 = 2 \times 3^{3}$ .
LCM = product of the greatest power of each prime factor $= 2^{2} \times 3^{3} = 4 \times 27 = 108$ .

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/3/1

Find which among the following numbers a, b and c is/are composite numbers.
$a = 7 \times 11 \times 13 + 13$
$b = 6 \times 5 \times 4 + 4$
$c = 7 \times 13 + 6$

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Answer: a and b are composite numbers (c = 97 is prime).

$a = 13 (7 \times 11 + 1) = 13 \times 78 = 1014$ , which has factors other than 1 and itself, so $a$ is composite.
$b = 4 (6 \times 5 + 1) = 4 \times 31 = 124$ , so $b$ is composite.
$c = 91 + 6 = 97$ , which is not divisible by 2, 3, 5 or 7 ( $7 \times 14 = 98 > 97$ ), so $c$ is prime.
Hence $a$ and $b$ are composite.

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q293 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/5/1

The traffic lights at three different road crossings change after every 45 seconds, 75 seconds and 60 seconds respectively. If they change together at 5.00 a.m., then at what time they will change together next ?

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Answer: 5:15 a.m. (after 900 seconds = 15 minutes)

$45 = 3^{2} \times 5$ , $75 = 3 \times 5^{2}$ , $60 = 2^{2} \times 3 \times 5$ .
LCM $= 2^{2} \times 3^{2} \times 5^{2} = 900$ seconds $= 15$ minutes.
They will change together next at 5:15 a.m.

Also asked in: 2025 Basic 430/5/3

Q263 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/5/2

Find the smallest number which when increased by 20, is exactly divisible by 72, 90 and 150.

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Answer: 1780

$72 = 2^{3} \times 3^{2}$ , $90 = 2 \times 3^{2} \times 5$ , $150 = 2 \times 3 \times 5^{2}$ .
LCM $= 2^{3} \times 3^{2} \times 5^{2} = 1800$ .
The smallest number divisible by all three is 1800, so the required number is $1800 - 20 = 1780$ .

Q283 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/6/1

Three measuring rods are of lengths 120 cm, 100 cm and 150 cm. Find the least length of a fence that can be measured an exact number of times, using any of the rods. How many times each rod will be used to measure the length of the fence ?

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Answer: Least length = 600 cm (6 m); the rods are used 5, 6 and 4 times respectively.

$120 = 2^{3} \times 3 \times 5$ , $100 = 2^{2} \times 5^{2}$ , $150 = 2 \times 3 \times 5^{2}$ .
LCM $= 2^{3} \times 3 \times 5^{2} = 600$ cm.
Rod of 120 cm: $600 \div 120 = 5$ times; rod of 100 cm: $600 \div 100 = 6$ times; rod of 150 cm: $600 \div 150 = 4$ times.

Q263 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/6/2

Three friends plan to go for a morning walk. They step off together and their steps measures 48 cm, 52 cm and 56 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ten times ?

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Answer: LCM = 4368 cm; covering it ten times, each should walk 43680 cm = 436.8 m.

$48 = 2^{4} \times 3$ , $52 = 2^{2} \times 13$ , $56 = 2^{3} \times 7$ .
LCM $= 2^{4} \times 3 \times 7 \times 13 = 4368$ cm, the least distance each can cover in complete steps.
To do this ten times, distance $= 10 \times 4368 = 43680$ cm $= 436.8$ m.

Q263 marksShort AnswerReal Numbers CBSE 2025 · Basic 430/6/3

Given that $5$ is an irrational number, prove that $2 + 35$ is an irrational number.

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Answer: Proved.

Suppose $2 + 35$ is rational, say $2 + 35 = r$ where $r$ is rational.
Then $5 = \frac{r - 2}{3}$ .
The right side is rational (rationals are closed under subtraction and division by a non-zero rational), so $5$ would be rational.
This contradicts the fact that $5$ is irrational.
Hence $2 + 35$ is irrational.

Q293 marksShort AnswerReal Numbers CBSE 2025 · Standard 30/1/1

Prove that $\frac{1}{5}$ is an irrational number.

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Answer: Proved.

First, $5$ is irrational: suppose $5 = \frac{a}{b}$ with $a, b$ coprime integers, $b \neq = 0$ . Then $a^{2} = 5 b^{2}$ , so 5 divides $a^{2}$ and hence 5 divides $a$ . Write $a = 5 c$ : $25 c^{2} = 5 b^{2}$ , so $b^{2} = 5 c^{2}$ and 5 divides $b$ . Then 5 is a common factor of $a$ and $b$ , a contradiction. So $5$ is irrational.
Now suppose $\frac{1}{5}$ is rational, say $\frac{1}{5} = \frac{p}{q}$ with $p, q$ integers, $p \neq = 0, q \neq = 0$ .
Then $5 = \frac{q}{p}$ , which is rational. This contradicts the irrationality of $5$ .
Hence $\frac{1}{5}$ is irrational.

Q273 marksShort AnswerReal Numbers CBSE 2025 · Standard 30/1/2

Three sets of Physics, Chemistry and Mathematics books have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The number of Physics books is $144$ , the number of Chemistry books is $180$ and the number of Mathematics books is $192$ . Assuming that the books are of same thickness, determine the number of stacks of Physics, Chemistry and Mathematics books.

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Answer: Physics:

12

stacks, Chemistry:

15

stacks, Mathematics:

16

stacks (12 books in each stack)

For the least number of stacks of equal height, the number of books in each stack is the HCF of 144, 180 and 192.
$144 = 2^{4} \times 3^{2}$ , $180 = 2^{2} \times 3^{2} \times 5$ , $192 = 2^{6} \times 3$ .
HCF $= 2^{2} \times 3 = 12$ books per stack.
Physics: $\frac{144}{12} = 12$ stacks; Chemistry: $\frac{180}{12} = 15$ stacks; Mathematics: $\frac{192}{12} = 16$ stacks.

Q273 marksShort AnswerReal Numbers CBSE 2025 · Standard 30/3/1

Prove that $(53 + \frac{2}{3})$ is an irrational number given that $3$ is an irrational number.

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Answer: Proved.

Assume, to the contrary, that $53 + \frac{2}{3} = r$ , where $r$ is rational.
Then $3 = \frac{1}{5} (r - \frac{2}{3}) = \frac{3 r - 2}{15}$ .
Since $r$ is rational, $\frac{3 r - 2}{15}$ is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $53 + \frac{2}{3}$ is irrational.

Also asked in: 2025 Standard 30/3/3

Q263 marksShort AnswerReal Numbers CBSE 2025 · Standard 30/3/2

Prove that $(42 + \frac{5}{3})$ is an irrational number given that $2$ is an irrational number.

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Answer: Proved.

Assume, to the contrary, that $42 + \frac{5}{3} = r$ , where $r$ is rational.
Then $2 = \frac{1}{4} (r - \frac{5}{3}) = \frac{3 r - 5}{12}$ .
Since $r$ is rational, $\frac{3 r - 5}{12}$ is rational, so $2$ would be rational.
This contradicts the fact that $2$ is irrational.
Hence $42 + \frac{5}{3}$ is irrational.

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2025 · Standard 30/4/1

Let $x$ and $y$ be two distinct prime numbers and $p = x^{2} y^{3}$ , $q = x y^{4}$ , $r = x^{5} y^{2}$ . Find the HCF and LCM of $p$ , $q$ and $r$ . Further check if $HCF (p, q, r) \times LCM (p, q, r) = p \times q \times r$ or not.

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Answer: HCF

= x y^{2}

, LCM

= x^{5} y^{4}

; HCF

\times

LCM

\neq = p \times q \times r

HCF = product of smallest powers: $x^{1} y^{2} = x y^{2}$
LCM = product of greatest powers: $x^{5} y^{4}$
HCF $\times$ LCM $= x^{6} y^{6}$
$p \times q \times r = x^{2 + 1 + 5} y^{3 + 4 + 2} = x^{8} y^{9}$
$x^{6} y^{6} \neq = x^{8} y^{9}$ , so HCF $\times$ LCM $\neq = p \times q \times r$ .

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2025 · Standard 30/5/1

State true or false for each of the following statements and justify in each case :
(i) $2 \times 3 \times 5 \times 7 + 7$ is a composite number.
(ii) $2 \times 3 \times 5 \times 7 + 1$ is a composite number.

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Answer: (i) True (ii) False

(i) $2 \times 3 \times 5 \times 7 + 7 = 7 (2 \times 3 \times 5 + 1) = 7 \times 31 = 217$ .
It has factors other than 1 and itself, so it is composite. True.
(ii) $2 \times 3 \times 5 \times 7 + 1 = 211$ .
211 is not divisible by any prime $\leq 211$ (2, 3, 5, 7, 11, 13), so 211 is prime. False.

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2025 · Standard 30/6/1

Let $p$ , $q$ and $r$ be three distinct prime numbers.
Check whether $p \cdot q \cdot r + q$ is a composite number or not.
Further, give an example for 3 distinct primes $p$ , $q$ , $r$ such that
(i) $p \cdot q \cdot r + 1$ is a composite number.
(ii) $p \cdot q \cdot r + 1$ is a prime number.

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Answer: Composite, since

pq r + q = q (p r + 1)

. (i) e.g.

3 \cdot 5 \cdot 7 + 1 = 106 = 2 \times 53

(ii) e.g.

2 \cdot 3 \cdot 5 + 1 = 31

$p \cdot q \cdot r + q = q (p r + 1)$
Both factors $q \geq 2$ and $p r + 1 \geq 7$ exceed 1, so the number has a factor other than 1 and itself: it is composite.
(i) $p = 3, q = 5, r = 7$ : $105 + 1 = 106 = 2 \times 53$ , composite.
(ii) $p = 2, q = 3, r = 5$ : $30 + 1 = 31$ , prime.

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q293 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/1/2

Two alarm clocks ring their alarms at regular intervals of 20 minutes and 25 minutes respectively. If they first beep together at 12 noon, at what time will they beep again together next time ?

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Answer: 1:40 p.m.

$20 = 2^{2} \times 5$ and $25 = 5^{2}$ .
LCM $= 2^{2} \times 5^{2} = 100$ minutes $= 1$ hour 40 minutes.
They beep together next at 12 noon + 1 h 40 min = 1:40 p.m.

Also asked in: 2024 Basic 430/1/3

Q283 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/2/1

Prove that $7 - 35$ is an irrational number, given that $5$ is an irrational number.

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Answer: Proved.

Suppose $7 - 35$ is rational, say $7 - 35 = r$ where r is rational.
Then $5 = \frac{7 - r}{3}$ .
Since 7, r and 3 are rational, $\frac{7 - r}{3}$ is rational, so $5$ would be rational.
This contradicts the fact that $5$ is irrational.
Hence $7 - 35$ is irrational.

Q313 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/2/3

Prove that $52 - 3$ is an irrational number, given that $2$ is an irrational number.

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Answer: Proved.

Suppose $52 - 3$ is rational, say $52 - 3 = r$ where r is rational.
Then $2 = \frac{r + 3}{5}$ .
Since r, 3 and 5 are rational, $\frac{r + 3}{5}$ is rational, so $2$ would be rational.
This contradicts the fact that $2$ is irrational.
Hence $52 - 3$ is irrational.

Q263 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/3/1

Find LCM and HCF of two numbers 336 and 54, using prime-factorisation method.

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Answer: HCF = 6, LCM = 3024

$336 = 2^{4} \times 3 \times 7$
$54 = 2 \times 3^{3}$
HCF = product of smallest powers of common primes $= 2 \times 3 = 6$ .
LCM = product of greatest powers of all primes $= 2^{4} \times 3^{3} \times 7 = 3024$ .
Check: $6 \times 3024 = 18144 = 336 \times 54$ .

Q293 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/3/2

Find the HCF and LCM of 96 and 404, using prime-factorisation method.

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Answer: HCF = 4, LCM = 9696

$96 = 2^{5} \times 3$
$404 = 2^{2} \times 101$
HCF $= 2^{2} = 4$ .
LCM $= 2^{5} \times 3 \times 101 = 9696$ .
Check: $4 \times 9696 = 38784 = 96 \times 404$ .

Q303 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/3/3

Find the HCF and LCM of 260 and 910 by prime-factorisation method.

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Answer: HCF = 130, LCM = 1820

$260 = 2^{2} \times 5 \times 13$
$910 = 2 \times 5 \times 7 \times 13$
HCF $= 2 \times 5 \times 13 = 130$ .
LCM $= 2^{2} \times 5 \times 7 \times 13 = 1820$ .
Check: $130 \times 1820 = 236600 = 260 \times 910$ .

Q303 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/4/1

Prove that $2 + 53$ is an irrational number, if it is given that $3$ is an irrational number.

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Answer: Proved.

Suppose $2 + 53$ is rational, say $2 + 53 = r$ , $r$ rational.
Then $3 = \frac{r - 2}{5}$ , which is rational.
This contradicts the fact that $3$ is irrational.
Hence $2 + 53$ is irrational.

Also asked in: 2024 Basic 430/4/3

Q273 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/4/2

Check whether the number $4^{n}$ can end with digit 0 for any natural number $n$ . Give reasons for your answer.

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Answer: No,

4^{n}

can never end with digit 0.

A number ending with 0 is divisible by 10, so its prime factorisation must contain both 2 and 5.
$4^{n} = (2^{2})^{n} = 2^{2 n}$ ; its only prime factor is 2.
By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), 5 is not a factor of $4^{n}$ .
Hence $4^{n}$ cannot end with digit 0 for any natural number $n$ .

Q263 marksShort AnswerReal Numbers CBSE 2024 · Basic 430/5/1

Prove that $6 + 32$ is an irrational number, given that $2$ is an irrational number.

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Answer: Proved.

Suppose $6 + 32$ is rational, say $6 + 32 = r$ , where r is rational.
Then $2 = \frac{r - 6}{3}$ .
Since r is rational, $\frac{r - 6}{3}$ is rational, so $2$ would be rational.
This contradicts the fact that $2$ is irrational.
Hence $6 + 32$ is irrational.

Also asked in: 2024 Basic 430/5/3

Q273 marksShort AnswerReal Numbers CBSE 2024 · Standard 30/1/1

In a teachers' workshop, the number of teachers teaching French, Hindi and English are 48, 80 and 144 respectively. Find the minimum number of rooms required if in each room the same number of teachers are seated and all of them are of the same subject.

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Answer: 17 rooms

The number of teachers per room must divide 48, 80 and 144; for the fewest rooms it is their HCF.
$48 = 2^{4} \times 3$ , $80 = 2^{4} \times 5$ , $144 = 2^{4} \times 3^{2}$ , so HCF $= 2^{4} = 16$ .
Number of rooms $= \frac{48 + 80 + 144}{16} = \frac{272}{16} = 17$ .

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q303 marksShort AnswerReal Numbers CBSE 2024 · Standard 30/4/1

Prove that $\frac{2 - 3}{5}$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Suppose $\frac{2 - 3}{5}$ is rational, say $\frac{2 - 3}{5} = r$ with $r$ rational.
Then $2 - 3 = 5 r$ , so $3 = 2 - 5 r$ .
$2 - 5 r$ is rational (rationals are closed under subtraction and multiplication), so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $\frac{2 - 3}{5}$ is irrational.

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2024 · Standard 30/5/1

Prove that $(2 + 3)^{2}$ is an irrational number, given that $6$ is an irrational number.

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Answer: Proved.

$(2 + 3)^{2} = 2 + 3 + 26 = 5 + 26$ .
Suppose $5 + 26 = r$ , a rational number.
Then $6 = \frac{r - 5}{2}$ , which is rational.
This contradicts the fact that $6$ is irrational.
Hence $(2 + 3)^{2}$ is irrational.

Also asked in: 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q303 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/1/1

Prove that $3 + 72$ is an irrational number, given that $2$ is an irrational number.

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Answer: Proved.

Assume, to the contrary, that $3 + 72$ is rational, say $3 + 72 = r$ , where r is rational.
Then $2 = \frac{r - 3}{7}$ .
Since r is rational, $\frac{r - 3}{7}$ is rational, so $2$ would be rational.
This contradicts the fact that $2$ is irrational.
Hence $3 + 72$ is irrational.

Also asked in: 2023 Basic 430/1/3

Q263 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/1/2

Prove that $5 - 3$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Assume, to the contrary, that $5 - 3$ is rational, say $5 - 3 = r$ , where r is rational.
Then $3 = 5 - r$ .
Since r is rational, $5 - r$ is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $5 - 3$ is irrational.

Q273 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/2/1

Prove that $4 + 23$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Suppose $4 + 23$ is rational, say $4 + 23 = \frac{p}{q}$ , where p, q are integers and $q \neq = 0$ .
Then $23 = \frac{p}{q} - 4 = \frac{p - 4 q}{q}$ , so $3 = \frac{p - 4 q}{2 q}$
The right side is rational since p, q are integers, so $3$ would be rational.
This contradicts the given fact that $3$ is irrational.
Hence $4 + 23$ is irrational.

Q303 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/2/2

Prove that $2 - 35$ is an irrational number, given that $5$ is an irrational number.

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Answer: Proved.

Suppose $2 - 35$ is rational, say $2 - 35 = \frac{p}{q}$ , where p, q are integers and $q \neq = 0$ .
Then $35 = 2 - \frac{p}{q} = \frac{2 q - p}{q}$ , so $5 = \frac{2 q - p}{3 q}$
The right side is rational since p, q are integers, so $5$ would be rational.
This contradicts the given fact that $5$ is irrational.
Hence $2 - 35$ is irrational.

Q313 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/2/3

Prove that $10 + 23$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Suppose $10 + 23$ is rational, say $10 + 23 = \frac{p}{q}$ , where p, q are integers and $q \neq = 0$ .
Then $23 = \frac{p}{q} - 10 = \frac{p - 10 q}{q}$ , so $3 = \frac{p - 10 q}{2 q}$
The right side is rational since p, q are integers, so $3$ would be rational.
This contradicts the given fact that $3$ is irrational.
Hence $10 + 23$ is irrational.

Q273 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/5/1

Prove that $(7 - 23)$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Suppose $7 - 23$ is rational, say $7 - 23 = r$ where $r$ is rational.
Then $3 = \frac{7 - r}{2}$ .
Since $r$ is rational, $\frac{7 - r}{2}$ is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $7 - 23$ is irrational.

Q293 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/5/2

Prove that $8 + 55$ is an irrational number, given that $5$ is an irrational number.

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Answer: Proved.

Suppose $8 + 55$ is rational, say $8 + 55 = r$ where $r$ is rational.
Then $5 = \frac{r - 8}{5}$ .
Since $r$ is rational, $\frac{r - 8}{5}$ is rational, so $5$ would be rational.
This contradicts the fact that $5$ is irrational.
Hence $8 + 55$ is irrational.

Q313 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/5/3

Prove that $11 + 32$ is an irrational number, given that $2$ is an irrational number.

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Answer: Proved.

Suppose $11 + 32$ is rational, say $11 + 32 = r$ where $r$ is rational.
Then $2 = \frac{r - 11}{3}$ .
Since $r$ is rational, $\frac{r - 11}{3}$ is rational, so $2$ would be rational.
This contradicts the fact that $2$ is irrational.
Hence $11 + 32$ is irrational.

Q303 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/6/1

Prove that $3 - 25$ is an irrational number, given that $5$ is an irrational number.

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Answer: Proved.

Suppose $3 - 25$ is rational, say $3 - 25 = r$ where $r$ is rational.
Then $5 = \frac{3 - r}{2}$ .
The right side is rational (rationals are closed under subtraction and division by a non-zero rational), so $5$ would be rational.
This contradicts the fact that $5$ is irrational. Hence $3 - 25$ is irrational.

Also asked in: 2024 Basic 430/5/2

Q263 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/6/2

Prove that $5 + 23$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Suppose $5 + 23$ is rational, say $5 + 23 = r$ where $r$ is rational.
Then $3 = \frac{r - 5}{2}$ .
The right side is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational. Hence $5 + 23$ is irrational.

Q263 marksShort AnswerReal Numbers CBSE 2023 · Basic 430/6/3

Prove that $7 + 45$ is an irrational number, given that $5$ is an irrational number.

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Answer: Proved.

Suppose $7 + 45$ is rational, say $7 + 45 = r$ where $r$ is rational.
Then $5 = \frac{r - 7}{4}$ .
The right side is rational, so $5$ would be rational.
This contradicts the fact that $5$ is irrational. Hence $7 + 45$ is irrational.

Q273 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/1/1

Prove that $5$ is an irrational number.

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Answer: Proved.

Suppose $5$ is rational. Then $5 = \frac{a}{b}$ with integers a, b, $b \neq = 0$ , having no common factor other than 1.
Squaring, $a^{2} = 5 b^{2}$ , so 5 divides $a^{2}$ , hence 5 divides a (5 is prime).
Let $a = 5 c$ . Then $25 c^{2} = 5 b^{2}$ , so $b^{2} = 5 c^{2}$ ; 5 divides $b^{2}$ , hence 5 divides b.
So 5 is a common factor of a and b, a contradiction.
Hence $5$ is irrational.

Q263 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/2/1

Find by prime factorisation the LCM of the numbers 18180 and 7575. Also, find the HCF of the two numbers.

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Answer: LCM = 90900, HCF = 1515

$18180 = 2^{2} \times 3^{2} \times 5 \times 101$
$7575 = 3 \times 5^{2} \times 101$
LCM $= 2^{2} \times 3^{2} \times 5^{2} \times 101 = 90900$
HCF $= 3 \times 5 \times 101 = 1515$
Check: $1515 \times 90900 = 18180 \times 7575$

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/2/1

Three bells ring at intervals of 6, 12 and 18 minutes. If all the three bells rang at 6 a.m., when will they ring together again ?

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Answer: 6:36 a.m.

$6 = 2 \times 3$ , $12 = 2^{2} \times 3$ , $18 = 2 \times 3^{2}$
LCM $= 2^{2} \times 3^{2} = 36$
They ring together again after 36 minutes, i.e. at 6:36 a.m.

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q303 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/4/2

Prove that $3$ is an irrational number.

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Answer: Proved.

Assume $3$ is rational. Then $3 = \frac{a}{b}$ with $a$ , $b$ co-prime integers, $b \neq = 0$ .
Squaring: $a^{2} = 3 b^{2}$ , so 3 divides $a^{2}$ , hence 3 divides $a$ (3 is prime).
Let $a = 3 c$ . Then $9 c^{2} = 3 b^{2} \Rightarrow b^{2} = 3 c^{2}$ , so 3 divides $b^{2}$ and hence $b$ .
So 3 is a common factor of $a$ and $b$ , contradicting that they are co-prime.
Hence $3$ is irrational.

Also asked in: 2026 Standard 30/1/1, 2026 Standard 30/1/3, 2026 Standard 30/3/2, 2025 Standard 30/2/1, 2025 Standard 30/2/2, 2025 Standard 30/2/3, 2025 Standard 30/5/1, 2025 Standard 30/5/2, 2025 Standard 30/5/3, 2025 Basic 430/1/1, 2025 Basic 430/1/2, 2025 Basic 430/1/3, 2025 Basic 430/4/1, 2024 Standard 30/5/1, 2024 Standard 30/5/2, 2024 Standard 30/5/3, 2023 Standard 30/5/1, 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q26 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/5/1

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change together next ?

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Answer: At 7:07:12 a.m.

$48 = 2^{4} \times 3$ , $72 = 2^{3} \times 3^{2}$ , $108 = 2^{2} \times 3^{3}$ .
LCM $= 2^{4} \times 3^{3} = 432$ seconds = 7 minutes 12 seconds.
They change together next at 7 hours 7 minutes 12 seconds a.m.

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q273 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/6/1

Find the HCF and LCM of 26, 65 and 117, using prime factorisation.

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Answer: HCF = 13, LCM = 1170

$26 = 2 \times 13$
$65 = 5 \times 13$
$117 = 3^{2} \times 13$
HCF = 13
LCM = $2 \times 3^{2} \times 5 \times 13 = 1170$

Also asked in: 2023 Standard 30/6/2, 2023 Standard 30/6/3

Q27 (OR) (OR)3 marksShort AnswerReal Numbers CBSE 2023 · Standard 30/6/1

Prove that $2$ is an irrational number.

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Answer: Proved.

Assume $2$ is rational, so $2 = \frac{a}{b}$ where a, b are co-prime integers and $b \neq = 0$ .
Squaring: $2 b^{2} = a^{2}$ , so 2 divides $a^{2}$ , hence 2 divides a.
Let a = 2c. Then $2 b^{2} = 4 c^{2}$ , so $b^{2} = 2 c^{2}$ ; 2 divides $b^{2}$ , hence 2 divides b.
So 2 is a common factor of a and b, contradicting that they are co-prime.
Hence $2$ is irrational.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/3/3, 2026 Basic 430/5/1, 2026 Basic 430/5/2, 2026 Basic 430/5/3, 2025 Standard 30/4/1, 2025 Standard 30/4/2, 2025 Standard 30/4/3, 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2025 Basic 430/3/3, 2025 Basic 430/4/3, 2023 Standard 30/6/2, 2023 Standard 30/6/3

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