(A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

A number ending in 0 must have both 2 and 5 as prime factors.
$4^{n} = 2^{2 n}$ , and by uniqueness of prime factorisation it has no other prime factor, so 5 never divides it.
So A is true, R is true and R explains A.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q91 markMCQReal Numbers CBSE 2026 · Basic 430/4/2

HCF of two consecutive natural numbers is :

(A)2
(B)1
(C)0
(D)smaller number

Show answer & solution

Answer: (B) 1

Any common factor of n and n + 1 also divides their difference, 1.
So the HCF is 1.

Q11 markMCQReal Numbers CBSE 2026 · Basic 430/5/1

$7 \times 11 \times 13 + 5$ is

(A)a prime number.
(B)an odd number.
(C)a composite number.
(D)a multiple of 5.

Show answer & solution

Answer: (C) a composite number.

$7 \times 11 \times 13 = 1001$ , so the number is $1001 + 5 = 1006$ .
$1006 = 2 \times 503$ , so it has a factor other than 1 and itself.
It is even (not odd) and does not end in 0 or 5 (not a multiple of 5).
Hence it is a composite number.

Also asked in: 2026 Basic 430/5/2

Q71 markMCQReal Numbers CBSE 2026 · Basic 430/5/3

$17 \times 11 \times 13 + 11$ is

(A)a prime number.
(B)multiple of 17.
(C)a composite number.
(D)an odd number.

Show answer & solution

Answer: (C) a composite number.

$17 \times 11 \times 13 + 11 = 11 (17 \times 13 + 1) = 11 \times 222 = 2442$
It has factors 11 and 222 besides 1 and itself, so it is composite.
2442 is even (not odd) and 2442 = 17 × 143 + 11, so it is not a multiple of 17.

Q11 markMCQReal Numbers CBSE 2026 · Standard 30/1/1

The HCF of 960 and 432 is :

(A)48
(B)54
(C)72
(D)36

Show answer & solution

Answer: (A) 48

$960 = 2^{6} \times 3 \times 5$ and $432 = 2^{4} \times 3^{3}$ .
HCF = product of the smallest powers of common primes = $2^{4} \times 3 = 48$ .

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q21 markMCQReal Numbers CBSE 2026 · Standard 30/1/1

The natural number 2 is :

(A)a prime number
(B)a composite number
(C)prime as well as composite
(D)neither prime nor composite

Show answer & solution

Answer: (A) a prime number

2 has exactly two factors, 1 and 2.
So 2 is a prime number (the only even prime).

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q31 markMCQReal Numbers CBSE 2026 · Standard 30/1/1

For any natural number $n$ , $6^{n}$ ends with the digit :

(A)0
(B)6
(C)3
(D)2

Show answer & solution

Answer: (B) 6

$6^{1} = 6$ , $6^{2} = 36$ , $6^{3} = 216$ , ...
Any power of a number ending in 6 also ends in 6, since $6 \times 6 = 36$ ends in 6.
So $6^{n}$ always ends with the digit 6.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q11 markMCQReal Numbers CBSE 2026 · Standard 30/2/1

The LCM of 960 and 240 is :

(A)960
(B)240
(C)60
(D)15

Show answer & solution

Answer: (A) 960

Since $960 = 4 \times 240$ , 240 is a factor of 960.
So the smallest common multiple of 960 and 240 is 960 itself.

Also asked in: 2026 Standard 30/2/3

Q21 markMCQReal Numbers CBSE 2026 · Standard 30/2/1

The natural number 1 is :

(A)a prime number.
(B)a composite number.
(C)prime as well as composite.
(D)neither prime nor composite.

Show answer & solution

Answer: (D) neither prime nor composite.

A prime number has exactly two factors and a composite number has more than two factors.
1 has only one factor (itself), so it is neither prime nor composite.

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

Q31 markMCQReal Numbers CBSE 2026 · Standard 30/2/1

For any natural number n, $5^{n}$ ends with the digit :

(A)0
(B)5
(C)3
(D)2

Show answer & solution

Answer: (B) 5

$5^{1} = 5$ , $5^{2} = 25$ , $5^{3} = 125$ , ...
Every power of 5 ends with the digit 5.

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

Q21 markMCQReal Numbers CBSE 2026 · Standard 30/2/2

There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

(A)144
(B)2
(C)420
(D)272

Show answer & solution

Answer: (C) 420

The required number is the LCM of 28 and 30.
$28 = 2^{2} \times 7$ , $30 = 2 \times 3 \times 5$
LCM $= 2^{2} \times 3 \times 5 \times 7 = 420$

Q41 markMCQReal Numbers CBSE 2026 · Standard 30/3/1

$(3 \times 11 \times 13 + 3)$ is :

(A)a prime number
(B)divisible by 13
(C)a composite number
(D)an odd number

Show answer & solution

Answer: (C) a composite number

$3 \times 11 \times 13 + 3 = 3 (11 \times 13 + 1) = 3 \times 144 = 432$
It has factors other than 1 and itself, so it is composite
(432 is even and $432 = 13 \times 33 + 3$ , so it is not odd and not divisible by 13)

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

Q201 markAssertion–ReasonReal Numbers CBSE 2026 · Standard 30/4/1

Assertion (A) : $(3 + 5)$ is an irrational number.
Reason (R) : Sum of the any two irrational numbers is always irrational.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (C) A is true, but R is false.

If $3 + 5 = r$ (rational), then $5 = r^{2} - 2 r 3 + 3$ , giving $3 = \frac{r ^{2} - 2}{2 r}$ , rational: a contradiction. So A is true.
R is false: e.g. $2 + (- 2) = 0$ is rational.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q201 markAssertion–ReasonReal Numbers CBSE 2026 · Standard 30/5/1

Assertion (A): H.C.F. $(36 m^{2}, 18 m) = 18 m$ , where m is a prime number.
Reason (R): H.C.F. of two numbers is always less than or equal to the smaller number.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

$36 m^{2} = 2^{2} \times 3^{2} \times m^{2}$ and $18 m = 2 \times 3^{2} \times m$ .
HCF $= 2 \times 3^{2} \times m = 18 m$ (18m divides $36 m^{2}$ ). A is true.
The HCF divides both numbers, so it cannot exceed the smaller one. R is true.
But R does not explain why the HCF here equals 18m; that follows from 18m being a factor of $36 m^{2}$ . So the answer is (B).

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

Q11 markMCQReal Numbers CBSE 2025 · Basic 430/1/1

If the HCF of two positive integers $a$ and $b$ is 1, then their LCM is :

(A) $a + b$
(B) $a$
(C) $b$
(D) $ab$

Show answer & solution

Answer: (D)

ab

HCF × LCM = product of the numbers.
$1 \times LCM = ab$ , so LCM $= ab$ .

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

Q21 markMCQReal Numbers CBSE 2025 · Basic 430/1/1

The number $3 + 2$ is :

(A)a rational number
(B)an irrational number
(C)an integer
(D)a natural number

Show answer & solution

Answer: (B) an irrational number

$2$ is irrational.
The sum of a rational number (3) and an irrational number is irrational, so $3 + 2$ is irrational.

Q191 markAssertion–ReasonReal Numbers CBSE 2025 · Basic 430/1/1

Assertion (A) : For any two natural numbers a and b, the HCF of a and b is a factor of the LCM of a and b.
Reason (R) : HCF of any two natural numbers divides both the numbers.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

R is true: the HCF is a common factor, so it divides both numbers.
The HCF divides $a$ , and $a$ divides the LCM, so the HCF divides the LCM. A is true.
This follows directly from R, so R correctly explains A.

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

Q141 markMCQReal Numbers CBSE 2025 · Basic 430/1/2

$\frac{3 - 3}{3}$ is :

(A)a rational number
(B)an irrational number
(C)an integer
(D)a natural number

Show answer & solution

Answer: (B) an irrational number

$\frac{3 - 3}{3} = 1 - \frac{3}{3} = 1 - 3$ .
1 is rational and $3$ is irrational, so $1 - 3$ is irrational.

Q71 markMCQReal Numbers CBSE 2025 · Basic 430/1/3

$(2 + 2)^{2}$ is :

(A)a rational number
(B)an irrational number
(C)an integer
(D)a natural number

Show answer & solution

Answer: (B) an irrational number

$(2 + 2)^{2} = 4 + 42 + 2 = 6 + 42$ .
$42$ is irrational, and a rational number plus an irrational number is irrational, so the value is irrational.

Q11 markMCQReal Numbers CBSE 2025 · Basic 430/2/1

The value of (HCF – LCM) for the two numbers 3 and 5 is :

(A) $2$
(B) $4$
(C) $14$
(D) $- 14$

Show answer & solution

Answer: (D)

- 14

3 and 5 are both prime, so HCF = 1 and LCM = $3 \times 5 = 15$ .
HCF – LCM $= 1 - 15 = - 14$ .

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

Q21 markMCQReal Numbers CBSE 2025 · Basic 430/2/1

The number $2^{n}$ , where $n$ is a natural number, cannot end with the digit :

(A) $4$
(B) $6$
(C) $2$
(D) $0$

Show answer & solution

Answer: (D) 0

A number ending in 0 is divisible by 10, so it must have both 2 and 5 as prime factors.
The only prime factor of $2^{n}$ is 2, so $2^{n}$ can never end with the digit 0.
(Its last digits cycle 2, 4, 8, 6.)

Q191 markAssertion–ReasonReal Numbers CBSE 2025 · Basic 430/2/1

Assertion (A) : The prime numbers which divide 36 also divide 6.
Reason (R) : Any number which divides $p^{2}$ also divides p.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (C) Assertion (A) is true, but Reason (R) is false.

$36 = 2^{2} \times 3^{2}$ , so the primes dividing 36 are 2 and 3, and both divide 6. A is true.
R is false as stated: for example, 4 divides $2^{2} = 4$ but 4 does not divide 2 (also $p^{2}$ divides $p^{2}$ but not p).
The true result is: if a prime divides $a^{2}$ , then it divides a.
So A is true and R is false.

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

Q141 markMCQReal Numbers CBSE 2025 · Basic 430/2/2

The number $3^{n}$ , where $n$ is a natural number, cannot end with the digit :

(A) $1$
(B) $3$
(C) $5$
(D) $7$

Show answer & solution

Answer: (C) 5

A number ending in 5 is divisible by 5, so 5 would have to be a prime factor.
The only prime factor of $3^{n}$ is 3, so $3^{n}$ cannot end with 5.
(Its last digits cycle 3, 9, 7, 1.)

Q71 markMCQReal Numbers CBSE 2025 · Basic 430/2/3

If the number $a^{n}$ , where $n$ is a natural number, always ends with digit a, then the possible value of ‘a’ is :

(A) $2$
(B) $4$
(C) $6$
(D) $8$

Show answer & solution

Answer: (C) 6

Last digits of powers: 2 → 2, 4, 8, 6; 4 → 4, 6; 6 → 6, 6, 6, ...; 8 → 8, 4, 2, 6.
Only $6^{n}$ always ends with 6.
So a = 6.

Q11 markMCQReal Numbers CBSE 2025 · Basic 430/3/1

If $p = 2^{3} \times 3^{2} \times 5$ and $q = 2^{2} \times 3^{3}$ , then the LCM of $p$ and $q$ is :

(A) $2^{3} \times 3^{3}$
(B) $2^{2} \times 3^{2}$
(C) $2^{2} \times 3^{2} \times 5$
(D) $2^{3} \times 3^{3} \times 5$

Show answer & solution

Answer: (D)

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

LCM = product of the highest powers of all prime factors.
Highest powers: $2^{3}$ , $3^{3}$ and $5$ .
LCM $= 2^{3} \times 3^{3} \times 5$ .

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

Q21 markMCQReal Numbers CBSE 2025 · Basic 430/3/1

$3^{n}$ , where $n$ is a natural number, cannot end with the digit :

(A)3
(B)5
(C)7
(D)9

Show answer & solution

Answer: (B) 5

The only prime factor of $3^{n}$ is 3.
A number ending in 5 must be divisible by 5, so $3^{n}$ can never end with 5.
(The last digits of $3^{n}$ cycle through 3, 9, 7, 1.)

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

Q31 markMCQReal Numbers CBSE 2025 · Basic 430/3/1

A prime number has :

(A)exactly two prime factors
(B)exactly one prime factor
(C)at least one prime factor
(D)at least two prime factors

Show answer & solution

Answer: (B) exactly one prime factor

A prime number $p$ has only the factors 1 and $p$ .
1 is not prime, so its only prime factor is $p$ itself.
Hence a prime number has exactly one prime factor.

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

Q11 markMCQReal Numbers CBSE 2025 · Basic 430/4/1

If HCF $(x, 20) = 2$ and LCM $(x, 20) = 60$ , then value of $x$ is :

(A)3
(B)6
(C)20
(D)10

Show answer & solution

Answer: (B) 6

HCF × LCM = product of the two numbers.
$2 \times 60 = 20 x$
$x = \frac{120}{20} = 6$ .

Q61 markMCQReal Numbers CBSE 2025 · Basic 430/4/3

The LCM of two numbers is 3600. Which of the following can not be their HCF ?

(A)600
(B)400
(C)500
(D)150

Show answer & solution

Answer: (C) 500

The HCF of two numbers always divides their LCM.
$3600 \div 600 = 6$ , $3600 \div 400 = 9$ , $3600 \div 150 = 24$ , but $3600 \div 500 = 7.2$ .
So 500 can not be the HCF.

Q11 markMCQReal Numbers CBSE 2025 · Basic 430/5/1

$(2 - 53)^{2}$ is

(A)a negative integer
(B)an irrational number
(C)a rational number
(D)a positive integer

Show answer & solution

Answer: (B) an irrational number

$(2 - 53)^{2} = 4 - 203 + 75 = 79 - 203$ .
$3$ is irrational, so $203$ is irrational and $79 - 203$ is irrational.

Also asked in: 2025 Basic 430/5/2

Q181 markMCQReal Numbers CBSE 2025 · Basic 430/5/3

$2 (2 - 1)$ is

(A)an integer
(B)a rational number
(C)an irrational number
(D)equal to 1

Show answer & solution

Answer: (C) an irrational number

$2 (2 - 1) = 2 - 2$ .
$2$ is irrational, so $2 - 2$ is irrational.

Q191 markAssertion–ReasonReal Numbers CBSE 2025 · Basic 430/6/1

Assertion (A) : $(a + b) \cdot (a - b)$ is a rational number, where a and b are positive integers.
Reason (R) : Product of two irrationals is always rational.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (C) Assertion (A) is true, but Reason (R) is false.

$(a + b) (a - b) = a^{2} - b$ , an integer, so it is rational. A is true.
R is false: e.g. $2 \times 3 = 6$ is irrational.

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q61 markMCQReal Numbers CBSE 2025 · Standard 30/1/1

If $HCF (98, 28) = m$ and $LCM (98, 28) = n$ , then the value of $n - 7 m$ is :

(A) $0$
(B) $28$
(C) $98$
(D) $198$

Show answer & solution

Answer: (C)

98

$98 = 2 \times 7^{2}$ and $28 = 2^{2} \times 7$ .
$m = HCF = 2 \times 7 = 14$ .
$n = LCM = 2^{2} \times 7^{2} = 196$ .
$n - 7 m = 196 - 98 = 98$ .

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q91 markMCQReal Numbers CBSE 2025 · Standard 30/1/1

If $(- 1)^{n} + (- 1)^{8} = 0$ , then $n$ is :

(A)any positive integer
(B)any negative integer
(C)any odd number
(D)any even number

Show answer & solution

Answer: (C) any odd number

$(- 1)^{8} = 1$ , so $(- 1)^{n} = - 1$ .
$(- 1)^{n} = - 1$ exactly when $n$ is odd.

Also asked in: 2025 Standard 30/1/3

Q141 markMCQReal Numbers CBSE 2025 · Standard 30/1/1

Which of the following is a rational number between $3$ and $5$ ?

(A) $1.4142387954012 \dots$
(B) $2.32 6$
(C) $π$
(D) $1.857142$

Show answer & solution

Answer: (D)

1.857142

$3 \approx 1.732$ and $5 \approx 2.236$ .
(A) is non-terminating and non-repeating, so it is irrational (and less than $3$ ).
(B) $2.32 6$ is rational but greater than $5$ .
(C) $π$ is irrational.
(D) $1.857142$ is a terminating decimal, hence rational, and $1.732 < 1.857142 < 2.236$ .

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q141 markMCQReal Numbers CBSE 2025 · Standard 30/1/2

The greatest number which divides $70$ and $125$ , leaving remainders $5$ and $8$ respectively, is :

(A) $13$
(B) $65$
(C) $875$
(D) $1750$

Show answer & solution

Answer: (A)

13

The required number divides $70 - 5 = 65$ and $125 - 8 = 117$ exactly.
$65 = 5 \times 13$ and $117 = 3^{2} \times 13$ .
$HCF (65, 117) = 13$ .

Q41 markMCQReal Numbers CBSE 2025 · Standard 30/2/1

The least number which is a perfect square and is divisible by each of 16, 20 and 50, is :

(A)1200
(B)100
(C)3600
(D)2400

Show answer & solution

Answer: (C) 3600

$16 = 2^{4}$ , $20 = 2^{2} \times 5$ , $50 = 2 \times 5^{2}$
LCM $= 2^{4} \times 5^{2} = 400$ , which is itself a perfect square ( $2 0^{2}$ )
So the true least such number is 400, which is not among the options.
Checking the options: 1200 and 2400 are not perfect squares; 100 is not divisible by 16; $3600 = 6 0^{2}$ and is divisible by 16, 20 and 50.
So the intended option is 3600

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

Q101 markMCQReal Numbers CBSE 2025 · Standard 30/2/1

The sum of the exponents of prime factors in the prime factorisation of 4004 is :

(A)5
(B)4
(C)3
(D)2

Show answer & solution

Answer: (A) 5

$4004 = 4 \times 1001 = 2^{2} \times 7 \times 11 \times 13$
Sum of exponents $= 2 + 1 + 1 + 1 = 5$

Also asked in: 2025 Standard 30/2/3

Q181 markMCQReal Numbers CBSE 2025 · Standard 30/2/1

The HCF of 40, 110 and 360 is :

(A)40
(B)110
(C)360
(D)10

Show answer & solution

Answer: (D) 10

$40 = 2^{3} \times 5$ , $110 = 2 \times 5 \times 11$ , $360 = 2^{3} \times 3^{2} \times 5$
HCF $= 2 \times 5 = 10$

Also asked in: 2025 Standard 30/2/2

Q121 markMCQReal Numbers CBSE 2025 · Standard 30/2/2

If $1080 = 2^{p} \times 3^{q} \times 5$ , then $(p - q)$ is equal to :

(A)6
(B) $- 1$
(C)1
(D)0

Show answer & solution

Answer: (D) 0

$1080 = 8 \times 135 = 2^{3} \times 27 \times 5 = 2^{3} \times 3^{3} \times 5$
$p = 3$ , $q = 3$ , so $p - q = 0$

Q41 markMCQReal Numbers CBSE 2025 · Standard 30/2/3

If $a^{b} = 32$ , where ‘ $a$ ’ and ‘ $b$ ’ are positive integers, then the value of $b^{ab}$ is :

(A)72
(B) $5^{10}$
(C) $2^{10}$
(D) $5^{12}$

Show answer & solution

Answer: (B)

5^{10}

$32 = 2^{5}$ , so $a = 2$ , $b = 5$
$ab = 10$ , so $b^{ab} = 5^{10}$
(The trivial pair $a = 32$ , $b = 1$ gives $1^{32} = 1$ , which is not an option.)

Q21 markMCQReal Numbers CBSE 2025 · Standard 30/3/1

If $x$ is the LCM of 4, 6, 8 and $y$ is the LCM of 3, 5, 7 and $p$ is the LCM of $x$ and $y$ , then which of the following is true ?

(A) $p = 35 x$
(B) $p = 4 y$
(C) $p = 8 x$
(D) $p = 16 y$

Show answer & solution

Answer: (A)

p = 35 x

$x = LCM (4, 6, 8) = 2^{3} \times 3 = 24$ .
$y = LCM (3, 5, 7) = 105$ .
$p = LCM (24, 105) = 2^{3} \times 3 \times 5 \times 7 = 840$ .
$840 = 35 \times 24$ , so $p = 35 x$ .

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

Q11 markMCQReal Numbers CBSE 2025 · Standard 30/4/1

If $x = a b^{3}$ and $y = a^{3} b$ , where $a$ and $b$ are prime numbers, then $[HCF (x, y) - LCM (x, y)]$ is equal to :

(A) $1 - a^{3} b^{3}$
(B) $ab (1 - ab)$
(C) $ab - a^{4} b^{4}$
(D) $ab (1 - ab) (1 + ab)$

Show answer & solution

Answer: (D)

ab (1 - ab) (1 + ab)

$x = a \times b^{3}$ , $y = a^{3} \times b$
HCF $= ab$ , LCM $= a^{3} b^{3}$
HCF $-$ LCM $= ab - a^{3} b^{3} = ab (1 - a^{2} b^{2}) = ab (1 - ab) (1 + ab)$

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

Q21 markMCQReal Numbers CBSE 2025 · Standard 30/4/1

$(1 + 3)^{2} - (1 - 3)^{2}$ is :

(A)a positive rational number.
(B)a negative integer.
(C)a positive irrational number.
(D)a negative irrational number.

Show answer & solution

Answer: (C) a positive irrational number

$(1 + 3)^{2} = 4 + 23$ and $(1 - 3)^{2} = 4 - 23$
Difference $= 43$ , which is positive and irrational.

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

Q191 markAssertion–ReasonReal Numbers CBSE 2025 · Standard 30/4/1

Assertion (A) : $4^{n}$ ends with digit 0 for some natural number $n$ .
Reason (R) : For a number ' $x$ ' having 2 and 5 as its prime factors, $x^{n}$ always ends with digit 0 for every natural number $n$ .

(A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both, Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.

Show answer & solution

Answer: (D) Assertion (A) is false but Reason (R) is true.

$4^{n} = 2^{2 n}$ has only the prime factor 2; to end in 0 it must have 5 as a factor. So A is false.
If 2 and 5 are prime factors of $x$ , then $10 ∣ x$ , so $10 ∣ x^{n}$ and $x^{n}$ ends in 0. So R is true.

Also asked in: 2025 Standard 30/4/2

Q201 markAssertion–ReasonReal Numbers CBSE 2025 · Standard 30/4/3

Assertion (A) : Unit digit of $3^{n}$ cannot be an even number for any natural number $n$ .
Reason (R) : 2 is not a prime factor of $3^{n}$ for any natural number $n$ .

(A)Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both, Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.

Show answer & solution

Answer: (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

$3^{n}$ has only the prime factor 3 (uniqueness of prime factorisation), so 2 is not a factor: R is true.
A number with an even unit digit is divisible by 2. Since 2 does not divide $3^{n}$ , its unit digit is odd: A is true and follows from R.

Q11 markMCQReal Numbers CBSE 2025 · Standard 30/5/1

$(3 + 2)^{2} + (3 - 2)^{2}$ is a/an

(A)positive rational number
(B)negative rational number
(C)positive irrational number
(D)negative irrational number

Show answer & solution

Answer: (A) positive rational number

$(3 + 2)^{2} = 3 + 43 + 4 = 7 + 43$
$(3 - 2)^{2} = 3 - 43 + 4 = 7 - 43$
Sum $= 14$ , which is a positive rational number.

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

Q21 markMCQReal Numbers CBSE 2025 · Standard 30/5/1

Let $x = a^{2} b^{3} c^{n}$ and $y = a^{3} b^{m} c^{2}$ , where $a, b, c$ are prime numbers. If LCM of $x$ and $y$ is $a^{3} b^{4} c^{3}$ , then the value of $m + n$ is

(A)10
(B)7
(C)6
(D)5

Show answer & solution

Answer: (B) 7

LCM takes the highest power of each prime.
Power of $b$ : $max (3, m) = 4 \Rightarrow m = 4$
Power of $c$ : $max (n, 2) = 3 \Rightarrow n = 3$
$m + n = 7$

Q31 markMCQReal Numbers CBSE 2025 · Standard 30/5/1

For any prime number $p$ , if $p$ divides $a^{2}$ , where $a$ is any real number then $p$ also divides

(A) $a$
(B) $a^{\frac{1}{2}}$
(C) $a^{\frac{3}{2}}$
(D) $a^{\frac{1}{8}}$

Show answer & solution

Answer: (A)

a

Theorem: if a prime $p$ divides $a^{2}$ (for a positive integer $a$ ), then $p$ divides $a$ .
Hence $p$ divides $a$ .

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

Q41 markMCQReal Numbers CBSE 2025 · Standard 30/5/2

Let $a = p^{2} q^{3} r^{n}$ and $b = p^{3} q^{m} r^{2}$ , where $p, q, r$ are prime numbers. If LCM of $a$ and $b$ is $p^{3} q^{4} r^{3}$ , then the value of $3 n - 2 m$ is

(A) $- 1$
(B)1
(C)3
(D) $- 3$

Show answer & solution

Answer: (B) 1

LCM takes the highest power of each prime.
Power of $q$ : $max (3, m) = 4 \Rightarrow m = 4$
Power of $r$ : $max (n, 2) = 3 \Rightarrow n = 3$
$3 n - 2 m = 9 - 8 = 1$

Q111 markMCQReal Numbers CBSE 2025 · Standard 30/5/3

Let $p = x^{2} y^{3} z^{n}$ and $q = x^{3} y^{m} z^{2}$ , where $x, y, z$ are prime numbers. If LCM $(p, q) = x^{3} y^{4} z^{3}$ , then the value of $(2 m + 3 n)$ is

(A)18
(B)17
(C)15
(D)14

Show answer & solution

Answer: (B) 17

LCM takes the highest power of each prime.
Power of $y$ : $max (3, m) = 4 \Rightarrow m = 4$
Power of $z$ : $max (n, 2) = 3 \Rightarrow n = 3$
$2 m + 3 n = 8 + 9 = 17$

Q11 markMCQReal Numbers CBSE 2025 · Standard 30/6/1

$0.4$ is a/an

(A)natural number
(B)integer
(C)rational number
(D)irrational number

Show answer & solution

Answer: (D) irrational number

$0.4 = \frac{4}{10} = \frac{2}{5}$ , so $0.4 = \frac{2}{5} = \frac{10}{5}$
$10$ is irrational, so $\frac{10}{5}$ is irrational.

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

Q21 markMCQReal Numbers CBSE 2025 · Standard 30/6/1

Which of the following cannot be the unit digit of $8^{n}$ , where $n$ is a natural number ?

(A)4
(B)2
(C)0
(D)6

Show answer & solution

Answer: (C) 0

Unit digits of $8^{1}, 8^{2}, 8^{3}, 8^{4}$ are 8, 4, 2, 6 and then they repeat.
For the unit digit to be 0, $8^{n} = 2^{3 n}$ would need 5 as a prime factor, which is impossible.
So 0 can never be the unit digit.

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

Q191 markAssertion–ReasonReal Numbers CBSE 2025 · Standard 30/6/1

Assertion (A) : For any two prime numbers $p$ and $q$ , their HCF is 1 and LCM is $p + q$ .
Reason (R) : For any two natural numbers, HCF $\times$ LCM = product of numbers.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (D) Assertion (A) is false, but Reason (R) is true.

For distinct primes $p$ and $q$ , HCF $= 1$ and LCM $= pq$ , not $p + q$ (e.g. 2 and 3: LCM $= 6 \neq = 5$ ). So A is false.
HCF $\times$ LCM $=$ product of the two numbers holds for any two natural numbers. So R is true.

Q191 markAssertion–ReasonReal Numbers CBSE 2025 · Standard 30/6/2

Assertion (A) : For two prime numbers $x$ and $y$ ( $x < y$ ), HCF $(x, y) = x$ and LCM $(x, y) = y$ .
Reason (R) : HCF $(x, y) \leq$ LCM $(x, y)$ , where $x$ , $y$ are any two natural numbers.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (D) Assertion (A) is false, but Reason (R) is true.

For two distinct primes $x$ and $y$ : HCF $= 1$ and LCM $= x y$ (e.g. 2 and 3: HCF $= 1 \neq = 2$ , LCM $= 6 \neq = 3$ ). So A is false.
HCF divides both numbers and LCM is a multiple of both, so HCF $\leq$ LCM always. R is true.

Q201 markAssertion–ReasonReal Numbers CBSE 2025 · Standard 30/6/3

Assertion (A) : For two odd prime numbers $x$ and $y$ , ( $x \neq = y$ ), LCM $(2 x, 4 y) = 4 x y$
Reason (R) : LCM $(x, y)$ is a multiple of HCF $(x, y)$ .

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

$2 x = 2 \times x$ and $4 y = 2^{2} \times y$ , where $x$ , $y$ are distinct odd primes.
LCM $= 2^{2} \times x \times y = 4 x y$ . So A is true.
HCF divides both numbers, and LCM is a multiple of both, so LCM is a multiple of HCF. R is true.
A follows from prime factorisation, not from R, so R is not the correct explanation of A.

Q11 markMCQReal Numbers CBSE 2024 · Basic 430/1/2

LCM (850, 500) is :

(A) $850 \times 50$
(B) $17 \times 500$
(C) $17 \times 5^{2} \times 2^{2}$
(D) $17 \times 5^{3} \times 2$

Show answer & solution

Answer: (B)

17 \times 500

$850 = 2 \times 5^{2} \times 17$ and $500 = 2^{2} \times 5^{3}$ .
LCM $= 2^{2} \times 5^{3} \times 17 = 8500$ .
$17 \times 500 = 8500$ , so option (B).

Q51 markMCQReal Numbers CBSE 2024 · Basic 430/1/3

HCF (132, 77) is :

(A)11
(B)77
(C)22
(D)44

Show answer & solution

Answer: (A) 11

$132 = 2^{2} \times 3 \times 11$ and $77 = 7 \times 11$ .
HCF $= 11$ .

Q11 markMCQReal Numbers CBSE 2024 · Basic 430/2/1

The HCF of smallest 2 – digit number and the smallest composite number is :

(A)2
(B)20
(C)40
(D)4

Show answer & solution

Answer: (A) 2

Smallest 2-digit number = 10 and smallest composite number = 4.
$10 = 2 \times 5$ , $4 = 2^{2}$ .
HCF = 2.

Also asked in: 2024 Basic 430/2/2

Q91 markMCQReal Numbers CBSE 2024 · Basic 430/2/3

If n is any natural number, then which of the following numbers ends with digit 0 ?

(A) $(3 \times 2)^{n}$
(B) $(5 \times 2)^{n}$
(C) $(6 \times 2)^{n}$
(D) $(4 \times 2)^{n}$

Show answer & solution

Answer: (B)

(5 \times 2)^{n}

A number ends with 0 only if its prime factorisation contains both 2 and 5.
$(5 \times 2)^{n} = 1 0^{n}$ contains both, so it ends with 0.
The others have no factor 5.

Q201 markAssertion–ReasonReal Numbers CBSE 2024 · Basic 430/3/1

Assertion (A) : $2 (5 - 2)$ is an irrational number.
Reason (R) : Product of two irrational numbers is always irrational.

(A)Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true. Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (C) Assertion (A) is true, but Reason (R) is false.

$2 (5 - 2) = 52 - 2$ , which is irrational because $52$ is irrational. So A is true.
$2 \times 2 = 2$ is rational, so R is false.

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q131 markMCQReal Numbers CBSE 2024 · Basic 430/4/1

If $1080 = 2^{x} \times 3^{y} \times 5$ , then $(x - y)$ is equal to :

(A) $6$
(B) $- 1$
(C) $1$
(D) $0$

Show answer & solution

Answer: (D)

0

$1080 = 8 \times 135 = 2^{3} \times 27 \times 5 = 2^{3} \times 3^{3} \times 5$ .
So $x = 3$ , $y = 3$ and $x - y = 0$ .

Also asked in: 2024 Basic 430/4/3

Q21 markMCQReal Numbers CBSE 2024 · Basic 430/4/2

If $2800 = 2^{x} \times 5^{y} \times 7$ , then the value of $(x + y)$ is :

(A)5
(B)4
(C)8
(D)6

Show answer & solution

Answer: (D) 6

$2800 = 28 \times 100 = 2^{2} \times 7 \times 2^{2} \times 5^{2} = 2^{4} \times 5^{2} \times 7$ .
So $x = 4$ , $y = 2$ and $x + y = 6$ .

Q11 markMCQReal Numbers CBSE 2024 · Basic 430/5/1

HCF $\times$ LCM for the numbers 40 and 30 is :

(A)12
(B)120
(C)1200
(D)40

Show answer & solution

Answer: (C) 1200

For two numbers, HCF $\times$ LCM = product of the numbers.
HCF $\times$ LCM $= 40 \times 30 = 1200$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q71 markMCQReal Numbers CBSE 2024 · Basic 430/5/1

Prime factorisation of 424 is :

(A) $2 \times 53 \times 4$
(B) $2 \times 53 \times 2$
(C) $2^{3} \times 53$
(D) $2^{4} \times 53$

Show answer & solution

Answer: (C)

2^{3} \times 53

$424 = 2 \times 212 = 2 \times 2 \times 106 = 2 \times 2 \times 2 \times 53$ .
53 is prime, so $424 = 2^{3} \times 53$ .

Also asked in: 2024 Basic 430/5/3

Q81 markMCQReal Numbers CBSE 2024 · Basic 430/5/1

If x is a whole number, then $8^{x}$ ends with an even digit, except for which value of x ?

(A)6
(B)4
(C)2
(D)0

Show answer & solution

Answer: (D) 0

For $x \geq 1$ , $8^{x}$ is even, so it ends with an even digit.
For $x = 0$ , $8^{0} = 1$ , which ends with an odd digit.
So the exception is $x = 0$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q181 markMCQReal Numbers CBSE 2024 · Basic 430/5/2

Prime factorisation of 882 is :

(A) $2^{2} \times 3^{2} \times 7$
(B) $2^{3} \times 3 \times 7^{2}$
(C) $2 \times 3^{2} \times 7^{2}$
(D) $2^{2} \times 3^{3} \times 7$

Show answer & solution

Answer: (C)

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

$882 = 2 \times 441 = 2 \times 9 \times 49$ .
$882 = 2 \times 3^{2} \times 7^{2}$ .

Q51 markMCQReal Numbers CBSE 2024 · Standard 30/1/1

If two positive integers $p$ and $q$ can be expressed as $p = 18 a^{2} b^{4}$ and $q = 20 a^{3} b^{2}$ , where $a$ and $b$ are prime numbers, then LCM $(p, q)$ is :

(A) $2 a^{2} b^{2}$
(B) $180 a^{2} b^{2}$
(C) $12 a^{2} b^{2}$
(D) $180 a^{3} b^{4}$

Show answer & solution

Answer: (D)

180 a^{3} b^{4}

$p = 2 \times 3^{2} \times a^{2} b^{4}$ and $q = 2^{2} \times 5 \times a^{3} b^{2}$ .
LCM takes the highest power of each prime factor: $2^{2} \times 3^{2} \times 5 \times a^{3} b^{4} = 180 a^{3} b^{4}$ .

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

Q51 markMCQReal Numbers CBSE 2024 · Standard 30/2/1

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

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

Show answer & solution

Answer: (A) 1650

HCF $\times$ LCM = product of the numbers.
$40 \times 252 k = 2520 \times 6600$
$k = \frac{2520 \times 6600}{40 \times 252} = 10 \times 165 = 1650$

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

Q51 markMCQReal Numbers CBSE 2024 · Standard 30/3/1

If $a = 2^{2} \times 3^{x}$ , $b = 2^{2} \times 3 \times 5$ , $c = 2^{2} \times 3 \times 7$ and LCM (a, b, c) = 3780, then $x$ is equal to

(A)1
(B)2
(C)3
(D)0

Show answer & solution

Answer: (C) 3

$3780 = 2^{2} \times 3^{3} \times 5 \times 7$ .
LCM (a, b, c) $= 2^{2} \times 3^{x} \times 5 \times 7$ (taking the highest power of each prime, with $x \geq 1$ ).
Comparing the powers of 3: $3^{x} = 3^{3}$ , so $x = 3$ .

Also asked in: 2024 Standard 30/3/2

Q131 markMCQReal Numbers CBSE 2024 · Standard 30/3/1

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

Show answer & solution

Answer: (A) 1650

HCF $\times$ LCM = product of the two numbers.
$40 \times 252 k = 2520 \times 6600$ .
$k = \frac{2520 \times 6600}{40 \times 252} = 10 \times 165 = 1650$ .

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

Q141 markMCQReal Numbers CBSE 2024 · Standard 30/3/1

A pair of irrational numbers whose product is a rational number is :

(A) $(16, 4)$
(B) $(5, 2)$
(C) $(3, 27)$
(D) $(36, 2)$

Show answer & solution

Answer: (C)

(3, 27)

$16 = 4$ , $4 = 2$ and $36 = 6$ are rational, so (A) and (D) are not pairs of irrationals.
$5 \times 2 = 10$ is irrational.
$3$ and $27 = 33$ are irrational and $3 \times 27 = 81 = 9$ is rational.

Also asked in: 2024 Standard 30/3/3

Q151 markMCQReal Numbers CBSE 2024 · Standard 30/3/3

The smallest irrational number by which $20$ should be multiplied so as to get a rational number, is :

(A) $20$
(B) $2$
(C)5
(D) $5$

Show answer & solution

Answer: (D)

5

$20 = 25$ .
$25 \times 5 = 10$ , which is rational.
5 is not irrational, $20 \times 2 = 210$ is irrational, and $5 < 20$ .
So the smallest such irrational number is $5$ .

Q21 markMCQReal Numbers CBSE 2024 · Standard 30/4/1

The HCF of two numbers 65 and 104 is 13. If LCM of 65 and 104 is $40 x$ , then the value of $x$ is :

(A)5
(B)13
(C)40
(D)8

Show answer & solution

Answer: (B) 13

HCF $\times$ LCM = product of the numbers.
LCM = $\frac{65 \times 104}{13} = 520$ .
$40 x = 520$ , so $x = 13$ .

Also asked in: 2024 Standard 30/4/3

Q161 markMCQReal Numbers CBSE 2024 · Standard 30/4/2

If $3825 = 3^{x} \times 5^{y} \times 1 7^{z}$ , then the value of $x + y - 2 z$ is :

(A)0
(B)1
(C)2
(D)3

Show answer & solution

Answer: (C) 2

$3825 = 3 \times 1275 = 3^{2} \times 425 = 3^{2} \times 5^{2} \times 17$ .
So $x = 2$ , $y = 2$ , $z = 1$ .
$x + y - 2 z = 2 + 2 - 2 = 2$ .

Q101 markMCQReal Numbers CBSE 2024 · Standard 30/5/1

The greatest number which divides 281 and 1249, leaving remainder 5 and 7 respectively, is :

(A)23
(B)276
(C)138
(D)69

Show answer & solution

Answer: (C) 138

The number divides $281 - 5 = 276$ and $1249 - 7 = 1242$ exactly.
$276 = 2^{2} \times 3 \times 23$ and $1242 = 2 \times 3^{3} \times 23$ .
HCF $= 2 \times 3 \times 23 = 138$ (and 138 > 7, so the remainders are valid).

Also asked in: 2024 Standard 30/5/3

Q131 markMCQReal Numbers CBSE 2024 · Standard 30/5/1

The LCM of three numbers 28, 44, 132 is :

(A)258
(B)231
(C)462
(D)924

Show answer & solution

Answer: (D) 924

$28 = 2^{2} \times 7$ , $44 = 2^{2} \times 11$ , $132 = 2^{2} \times 3 \times 11$ .
LCM $= 2^{2} \times 3 \times 7 \times 11 = 924$ .

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

Q141 markMCQReal Numbers CBSE 2024 · Standard 30/5/1

If the product of two co-prime numbers is 553, then their HCF is :

(A)1
(B)553
(C)7
(D)79

Show answer & solution

Answer: (A) 1

Co-prime numbers have no common factor other than 1, so their HCF is 1.

Also asked in: 2024 Standard 30/5/2

Q121 markMCQReal Numbers CBSE 2024 · Standard 30/5/2

If the prime factorisation of 2520 is $2^{3} \times 3^{a} \times b \times 7$ , then the value of $a + 2 b$ is :

(A)12
(B)10
(C)9
(D)7

Show answer & solution

Answer: (A) 12

$2520 = 2^{3} \times 315 = 2^{3} \times 3^{2} \times 5 \times 7$ .
So $a = 2$ , $b = 5$ and $a + 2 b = 2 + 10 = 12$ .

Q11 markMCQReal Numbers CBSE 2024 · Standard 30/5/3

The LCM of the smallest prime number and the smallest odd composite number is :

(A)10
(B)6
(C)9
(D)18

Show answer & solution

Answer: (D) 18

Smallest prime number = 2; smallest odd composite number = 9.
$2$ and $9 = 3^{2}$ have no common factor, so LCM $= 2 \times 9 = 18$ .

Q11 markMCQReal Numbers CBSE 2023 · Basic 430/1/1

The prime factorisation of natural number 288 is

(A) $2^{4} \times 3^{3}$
(B) $2^{4} \times 3^{2}$
(C) $2^{5} \times 3^{2}$
(D) $2^{5} \times 3^{1}$

Show answer & solution

Answer: (C)

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

$288 = 2 \times 144 = 2 \times 2 \times 72 = 2 \times 2 \times 2 \times 36$
$36 = 2 \times 2 \times 3 \times 3$
So $288 = 2^{5} \times 3^{2}$ .

Q41 markMCQReal Numbers CBSE 2023 · Basic 430/1/2

The prime factorisation of 432 is :

(A) $2^{3} \times 3^{4}$
(B) $2^{4} \times 3^{3}$
(C) $2^{3} \times 3^{3}$
(D) $2^{4} \times 3^{4}$

Show answer & solution

Answer: (B)

2^{4} \times 3^{3}

$432 = 2 \times 216 = 2 \times 2 \times 108 = 2 \times 2 \times 2 \times 54 = 2 \times 2 \times 2 \times 2 \times 27$
$27 = 3^{3}$
So $432 = 2^{4} \times 3^{3}$ .

Q21 markMCQReal Numbers CBSE 2023 · Basic 430/1/3

The prime factorisation of 1728 is

(A) $2^{5} \times 3^{3}$
(B) $2^{5} \times 3^{4}$
(C) $2^{6} \times 3^{3}$
(D) $2^{6} \times 3^{2}$

Show answer & solution

Answer: (C)

2^{6} \times 3^{3}

$1728 = 2 \times 864 = 2^{2} \times 432 = 2^{3} \times 216 = 2^{4} \times 108 = 2^{5} \times 54 = 2^{6} \times 27$
$27 = 3^{3}$
So $1728 = 2^{6} \times 3^{3}$ .

Q71 markMCQReal Numbers CBSE 2023 · Basic 430/2/1

If the HCF of 360 and 64 is 8, then their LCM is :

(A)2480
(B)2780
(C)512
(D)2880

Show answer & solution

Answer: (d) 2880

HCF $\times$ LCM = product of the numbers
LCM $= \frac{360 \times 64}{8} = 360 \times 8 = 2880$

Also asked in: 2023 Basic 430/2/3

Q131 markMCQReal Numbers CBSE 2023 · Basic 430/2/2

If the HCF of 72 and 234 is 18, then the LCM (72, 234) is :

(A)936
(B)836
(C)324
(D)234

Show answer & solution

Answer: (a) 936

LCM $= \frac{72 \times 234}{18} = 4 \times 234 = 936$

Q21 markMCQReal Numbers CBSE 2023 · Basic 430/4/1

(HCF $\times$ LCM) for the numbers 70 and 40 is :

(A)10
(B)280
(C)2800
(D)70

Show answer & solution

Answer: (C) 2800

For two numbers, HCF $\times$ LCM = product of the numbers.
$70 \times 40 = 2800$ .

Also asked in: 2023 Basic 430/4/3

Q41 markMCQReal Numbers CBSE 2023 · Basic 430/4/1

The number $(5 - 35 + 5)$ is :

(A)an integer
(B)a rational number
(C)an irrational number
(D)a whole number

Show answer & solution

Answer: (C) an irrational number

$5 - 35 + 5 = 5 - 25$ .
Since $5$ is irrational, $25$ is irrational and so is $5 - 25$ .

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q81 markMCQReal Numbers CBSE 2023 · Basic 430/4/2

(HCF $\times$ LCM) for the numbers 30 and 70 is :

(A)2100
(B)21
(C)210
(D)70

Show answer & solution

Answer: (A) 2100

For two numbers, HCF $\times$ LCM = product of the numbers.
$30 \times 70 = 2100$ .

Q61 markMCQReal Numbers CBSE 2023 · Basic 430/5/1

LCM of $(2^{3} \times 3 \times 5)$ and $(2^{4} \times 5 \times 7)$ is :

(A)40
(B)560
(C)1680
(D)1120

Show answer & solution

Answer: (C) 1680

LCM = product of the highest powers of all prime factors.
LCM $= 2^{4} \times 3 \times 5 \times 7 = 16 \times 105 = 1680$ .

Also asked in: 2023 Basic 430/5/2

Q171 markMCQReal Numbers CBSE 2023 · Basic 430/5/3

HCF of $(3^{4} \times 2^{2} \times 7^{3})$ and $(3^{2} \times 5 \times 7)$ is :

(A)630
(B)63
(C)729
(D)567

Show answer & solution

Answer: (B) 63

HCF = product of the smallest powers of the common prime factors.
Common primes are 3 and 7: HCF $= 3^{2} \times 7 = 63$ .

Q51 markMCQReal Numbers CBSE 2023 · Basic 430/6/1

If HCF (72, 120) = 24, then LCM (72, 120) is

(A)72
(B)120
(C)360
(D)9640

Show answer & solution

Answer: (C) 360

HCF $\times$ LCM = product of the numbers.
LCM $= \frac{72 \times 120}{24} = 360$ .

Also asked in: 2023 Basic 430/6/3

Q91 markMCQReal Numbers CBSE 2023 · Basic 430/6/1

The prime factorisation of the number 2304 is

(A) $2^{8} \times 3^{2}$
(B) $2^{7} \times 3^{3}$
(C) $2^{8} \times 3^{1}$
(D) $2^{7} \times 3^{2}$

Show answer & solution

Answer: (A)

2^{8} \times 3^{2}

$2304 = 2 \times 1152 = 2^{2} \times 576 = \dots = 2^{8} \times 9$ .
So $2304 = 2^{8} \times 3^{2}$ (check: $256 \times 9 = 2304$ ).

Also asked in: 2023 Basic 430/6/2

Q101 markMCQReal Numbers CBSE 2023 · Basic 430/6/1

If n is a natural number, then $8^{n}$ cannot end with digit

(A)0
(B)2
(C)4
(D)6

Show answer & solution

Answer: (A) 0

$8^{n} = 2^{3 n}$ has only the prime factor 2.
A number ending in 0 must have both 2 and 5 as prime factors, so $8^{n}$ cannot end with 0.
(Its last digits cycle 8, 4, 2, 6.)

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q11 markMCQReal Numbers CBSE 2023 · Basic 430/6/2

The HCF of the smallest 2-digit number and the smallest composite number is

(A)4
(B)20
(C)2
(D)10

Show answer & solution

Answer: (C) 2

Smallest 2-digit number = 10; smallest composite number = 4.
$10 = 2 \times 5$ , $4 = 2^{2}$ , so HCF $= 2$ .

Q11 markMCQReal Numbers CBSE 2023 · Basic 430/6/3

The prime factorisation of the number 5488 is

(A) $2^{3} \times 7^{3}$
(B) $2^{4} \times 7^{3}$
(C) $2^{4} \times 7^{4}$
(D) $2^{3} \times 7^{4}$

Show answer & solution

Answer: (B)

2^{4} \times 7^{3}

$5488 = 2 \times 2744 = 2^{2} \times 1372 = 2^{3} \times 686 = 2^{4} \times 343$ .
$343 = 7^{3}$ , so $5488 = 2^{4} \times 7^{3}$ .

Q201 markAssertion–ReasonReal Numbers CBSE 2023 · Standard 30/1/1

Assertion (A) : The perimeter of $△ A B C$ is a rational number.
Reason (R) : The sum of the squares of two rational numbers is always rational.

Diagram for CBSE 2023 Class 10 Maths question 20

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.

Show answer & solution

Answer: (D) A is false but R is true.

$∠ B = 9 0^{\circ}$ , so $A C = 2^{2} + 3^{2} = 13$ cm.
Perimeter $= 2 + 3 + 13 = 5 + 13$ cm, which is irrational. A is false.
If a, b are rational, $a^{2} + b^{2}$ is rational. R is true.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q41 markMCQReal Numbers CBSE 2023 · Standard 30/2/1

If ‘p’ and ‘q’ are natural numbers and ‘p’ is the multiple of ‘q’, then what is the HCF of ‘p’ and ‘q’ ?

(A) $pq$
(B) $p$
(C) $q$
(D) $p + q$

Show answer & solution

Answer: (C)

q

Since $p$ is a multiple of $q$ , $q$ divides both $p$ and $q$ .
No number larger than $q$ divides $q$ , so HCF $= q$ .

Also asked in: 2023 Standard 30/2/2

Q11 markMCQReal Numbers CBSE 2023 · Standard 30/2/3

If ‘n’ is a natural number, then which of the following numbers end with zero ?

(A) $(3 \times 2)^{n}$
(B) $(2 \times 5)^{n}$
(C) $(6 \times 2)^{n}$
(D) $(5 \times 3)^{n}$

Show answer & solution

Answer: (B)

(2 \times 5)^{n}

A number ends with zero only if its prime factorisation contains both 2 and 5.
Only $(2 \times 5)^{n} = 2^{n} \times 5^{n} = 1 0^{n}$ has both, so it always ends with zero.

Q11 markMCQReal Numbers CBSE 2023 · Standard 30/4/1

The ratio of HCF to LCM of the least composite number and the least prime number is :

(A)1:2
(B)2:1
(C)1:1
(D)1:3

Show answer & solution

Answer: (A) 1:2

Least composite number = 4, least prime number = 2.
HCF(4, 2) = 2 and LCM(4, 2) = 4.
HCF : LCM = 2 : 4 = 1 : 2

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q201 markAssertion–ReasonReal Numbers CBSE 2023 · Standard 30/5/1

Assertion (A) : The number $5^{n}$ cannot end with the digit 0, where n is a natural number.
Reason (R): Prime factorisation of 5 has only two factors, 1 and 5.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (C) Assertion (A) is true, but Reason (R) is false.

To end in 0 a number must have both 2 and 5 as prime factors.
$5^{n}$ has only the prime 5 in its factorisation, so it never ends in 0. A is true.
The prime factorisation of 5 is just 5; 1 is not a prime factor. So R, as stated, is false (and it does not explain A anyway).

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

Q11 markMCQReal Numbers CBSE 2023 · Standard 30/6/1

If $p^{2} = \frac{32}{50}$ , then p is a/an

(A)whole number
(B)integer
(C)rational number
(D)irrational number

Show answer & solution

Answer: (C) rational number

$p^{2} = \frac{32}{50} = \frac{16}{25}$ .
$p = \pm \frac{4}{5}$ , which is a rational number (not an integer).

Also asked in: 2023 Standard 30/6/3

Q61 markMCQReal Numbers CBSE 2023 · Standard 30/6/2

The LCM of smallest 2-digit number and smallest composite number is

(A)12
(B)4
(C)20
(D)40

Show answer & solution

Answer: (C) 20

Smallest 2-digit number = 10 $= 2 \times 5$ ; smallest composite number = 4 $= 2^{2}$ .
LCM $= 2^{2} \times 5 = 20$

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