Real Numbers — Class 10 Maths PYQs

Previous-year board questions (2023–2026) with marking-scheme solutions, grouped by topic and marks.

Try each question first, then press (or tap Show Solution) to reveal the answer. Press again for the next question.

Year:Marks:Level:

Prime Factorization

1 Mark Questions
11 Mark · March 2024 · Standardopen ↗
If $3825 = 3^x \times 5^y \times 17^z$, then the value of $x + y - 2z$ is:
  • (a)$0$
  • (b)$1$
  • (c)$2$
  • (d)$3$
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(C) $2$
21 Mark · March 2024 · Standardopen ↗
If the prime factorisation of $2520$ is $2^3 \times 3^a \times b \times 7$, then the value of $a + 2b$ is:
  • (a)$12$
  • (b)$10$
  • (c)$9$
  • (d)$7$
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(A) $12$
31 Mark · July 2025 · Standardopen ↗
The total number of factors of the square of a prime number is :
  • (a)$1$
  • (b)$2$
  • (c)$3$
  • (d)$4$
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(C) $3$
41 Mark · March 2025 · Standardopen ↗
Which of the following is a rational number between $\sqrt{3}$ and $\sqrt{5}$?
  • (a)$1.4142387954012 \dots$
  • (b)$2.32\bar{6}$
  • (c)$\pi$
  • (d)$1.857142$
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(D) $1.857142$
51 Mark · March 2025 · Standardopen ↗
The sum of the exponents of prime factors in the prime factorisation of $4004$ is:
  • (a)$5$
  • (b)$4$
  • (c)$3$
  • (d)$2$
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(A) $5$
61 Mark · March 2025 · Standardopen ↗
If $1080 = 2^P \times 3^q \times 5$, then $(p - q)$ is equal to :
  • (a)$6$
  • (b)$-1$
  • (c)$1$
  • (d)$0$
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(D) $0$
71 Mark · March 2025 · Standardopen ↗
If $a^b = 32$, where 'a' and 'b' are positive integers, then the value of $b^{ab}$ is :
  • (a)$7^2$
  • (b)$5^{10}$
  • (c)$2^{10}$
  • (d)$5^{12}$
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(B) $5^{10}$
81 Mark · March 2025 · Basicopen ↗
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.
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(C) Assertion (A) is true, but Reason (R) is false.

Find lcm & hcf

1 Mark Questions
91 Mark · March 2023 · Standardopen ↗
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
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(c) q
101 Mark · March 2023 · Standardopen ↗
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)$1:3$
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(a) $1:2$
111 Mark · March 2024 · Standardopen ↗
Two positive integers $m$ and $n$ are expressed as $m = p^5q^2$ and $n = p^3q^4$, where $p$ and $q$ are prime numbers. The LCM of $m$ and $n$ is :
  • (a)$p^8q^6$
  • (b)$p^3q^2$
  • (c)$p^5q^4$
  • (d)$p^5q^2+ p^3q^4$
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(C) $p^5q^4$
121 Mark · July 2024 · Standardopen ↗
If $a = 2^4 \times 3^3$, $b = 2^3 \times 3^2 \times 5$, $c = 3^n \times 5^2$ and LCM $(a, b, c) = (5^2 \times 3^4 \times 2^4)$, then $n$ is :
  • (a)$1$
  • (b)$2$
  • (c)$3$
  • (d)$4$
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(D) $4$
131 Mark · March 2024 · Standardopen ↗
If two positive integers $p$ and $q$ can be expressed as $p = 18 a^2b^4$ and $q = 20 a^3b^2$, where $a$ and $b$ are prime numbers, then LCM $(p, q)$ is :
  • (a)$2a^2b^2$
  • (b)$180 a^2b^2$
  • (c)$12 a^2b^2$
  • (d)$180 a^3b^4$
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(d) $180 a^3b^4$
141 Mark · March 2024 · Standardopen ↗
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)$1625$
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(A) $1650$
151 Mark · March 2024 · Standardopen ↗
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)$0$
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(C) $3$
161 Mark · March 2024 · Standardopen ↗
The HCF of two numbers $65$ and $104$ is $13$. If LCM of $65$ and $104$ is $40x$, then the value of $x$ is:
  • (a)$5$
  • (b)$13$
  • (c)$40$
  • (d)$8$
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(B) $13$
171 Mark · March 2024 · Standardopen ↗
The LCM of three numbers $28, 44, 132$ is:
  • (a)$258$
  • (b)$231$
  • (c)$462$
  • (d)$924$
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(D) $924$
181 Mark · March 2024 · Standardopen ↗
If the product of two co-prime numbers is $553$, then their HCF is :
  • (a)$1$
  • (b)$553$
  • (c)$7$
  • (d)$79$
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(A) $1$
191 Mark · March 2024 · Standardopen ↗
The LCM of $24, 36$ and $60$ in terms of their prime factors is :
  • (a)$2^2 \times 3 \times 5$
  • (b)$2^3 \times 3^2$
  • (c)$2^3 \times 3^2 \times 5$
  • (d)$2^3 \times 3^3 \times 5$
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(C) $2^3 \times 3^2 \times 5$
201 Mark · July 2025 · Standardopen ↗
The ratio of the HCF to the LCM of $7, 21$ and $28$ is :
  • (a)1:4
  • (b)3:4
  • (c)1:8
  • (d)1:12
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(D) 1 : 12
211 Mark · March 2025 · Standardopen ↗
If $\text{HCF}(98, 28) = m$ and $\text{LCM}(98, 28) = n$, then the value of $n - 7m$ is:
  • (a)$0$
  • (b)$28$
  • (c)$98$
  • (d)$198$
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(C) $98$
221 Mark · March 2025 · Standardopen ↗
The HCF of $40$, $110$ and $360$ is :
  • (a)$40$
  • (b)$110$
  • (c)$360$
  • (d)$10$
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(D) $10$
231 Mark · March 2025 · Standardopen ↗
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)$p = 35x$
  • (b)$p = 4y$
  • (c)$p = 8x$
  • (d)$p = 16y$
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(A) $p = 35x$
241 Mark · March 2025 · Standardopen ↗
If $x = ab^3$ and $y = a^3b$, where $a$ and $b$ are prime numbers, then [HCF $(x, y)$ – LCM $(x, y)$] is equal to :
  • (a)$1-ab^3$
  • (b)$ab (1-ab)$
  • (c)$ab-a^3b^3$
  • (d)$ab (1-ab) (1 + ab)$
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(d) $ab(1 - ab)(1 + ab)$
251 Mark · March 2025 · Standardopen ↗
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
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(B) 7
261 Mark · March 2025 · Standardopen ↗
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 $3n - 2m$ is
  • (a)-1
  • (b)1
  • (c)3
  • (d)-3
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(B) 1
271 Mark · March 2025 · Standardopen ↗
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 $(2m + 3n)$ is
  • (a)$18$
  • (b)$17$
  • (c)$15$
  • (d)$14$
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(B) $17$
281 Mark · March 2025 · Standardopen ↗
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.
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(D) Assertion (A) is false, but Reason (R) is true.
291 Mark · March 2025 · Standardopen ↗
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.
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(D) Assertion (A) is false, but Reason (R) is true.
301 Mark · March 2025 · Standardopen ↗
Assertion (A) : For two odd prime numbers $x$ and $y$, $(x \neq y)$, $LCM(2x, 4y) = 4xy$. Reason (R) : $LCM(x, y)$ is a multiple of $HCF(x, y)$.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation of Assertion (A).
311 Mark · March 2025 · Standardopen ↗
The HCF and the LCM of $14$, $21$ and $77$ respectively are
  • (a)$7$, $77$
  • (b)$14$, $462$
  • (c)$7$, $462$
  • (d)$21$, $77$
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(C) $7$, $462$
321 Mark · March 2026 · Standardopen ↗
The HCF of $960$ and $432$ is:
  • (a)$48$
  • (b)$54$
  • (c)$72$
  • (d)$36$
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(a) $48$ (1 Mark)
331 Mark · March 2026 · Standardopen ↗
The LCM of $960$ and $240$ is:
  • (a)$960$
  • (b)$240$
  • (c)$60$
  • (d)$15$
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(A) $960$
341 Mark · March 2026 · Standardopen ↗
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)$272$
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(C) $420$ (1 Mark)
351 Mark · March 2026 · Standardopen ↗
Assertion (A) : H.C.F. ($36 \text{ m}^2$, 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.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
361 Mark · March 2026 · Standardopen ↗
If HCF of $66$ and $99$ is expressible in the form of $55m - 132$, then the value of $m$ is :
  • (a)$4$
  • (b)$2$
  • (c)$1$
  • (d)$3$
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(D) $3$
371 Mark · March 2025 · Basicopen ↗
If the HCF of two positive integers a and b is $1$, then their LCM is :
  • (a)$a + b$
  • (b)$a$
  • (c)$b$
  • (d)$ab$
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(D) $ab$
381 Mark · March 2025 · Basicopen ↗
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.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
391 Mark · March 2025 · Basicopen ↗
The value of $(\text{HCF} - \text{LCM})$ for the two numbers $3$ and $5$ is :
  • (a)$2$
  • (b)$4$
  • (c)$14$
  • (d)$-14$
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(D) $-14$
401 Mark · March 2025 · Basicopen ↗
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)$2^3 \times 3^3 \times 5$
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(D) $2^3 \times 3^3 \times 5$
411 Mark · March 2025 · Basicopen ↗
If $HCF(x, 20) = 2$ and $LCM(x, 20) = 60$, then value of $x$ is :
  • (a)$3$
  • (b)$6$
  • (c)$20$
  • (d)$10$
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(b) $6$
421 Mark · March 2025 · Basicopen ↗
The LCM of two numbers is $3600$. Which of the following can not be their HCF ?
  • (a)$600$
  • (b)$400$
  • (c)$500$
  • (d)$150$
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(c) $500$
431 Mark · March 2026 · Basicopen ↗
The HCF of $2^2.3^3$ and $3^2.2^3$ is :
  • (a)$1$
  • (b)$2.3$
  • (c)$2^2.3^2$
  • (d)$2^3.3^3$
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(c) $2^2.3^2$
441 Mark · March 2026 · Basicopen ↗
The LCM of $2^2.3^3$ and $3^2.2^3$ is:
  • (a)$1$
  • (b)$2^1.3^1$
  • (c)$2^3.3^3$
  • (d)$2^5.3^5$
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(c) $2^3.3^3$
451 Mark · March 2026 · Basicopen ↗
The HCF of $3^7 \cdot 7^3$ and $3^3 \cdot 7^7$ is:
  • (a)$1$
  • (b)$3 \cdot 7$
  • (c)$3^3 \cdot 7^3$
  • (d)$3^7 \cdot 7^7$
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(c) $3^3 \cdot 7^3$
461 Mark · March 2026 · Basicopen ↗
The HCF of the smallest prime number and the smallest 3-digit number is $2^m 5^n$. The respective values of $m$ and $n$ are :
  • (a)$0,0$
  • (b)$1,0$
  • (c)$0,1$
  • (d)$1,1$
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(B) $1,0$
471 Mark · March 2026 · Basicopen ↗
(i) Which of the following statements is true for HCF and LCM of two distinct natural numbers $a$ and $b$?
HCF is always greater than LCM.
(ii) HCF is a factor of LCM.
(iii) LCM is a factor of HCF.
  • (a)(i) only
  • (b)(i) and (iii)
  • (c)(i) and (ii)
  • (d)(ii) only
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(D) (ii) only
481 Mark · March 2026 · Basicopen ↗
If the HCF of the smallest odd prime number and the greatest $2$-digit number is expressed as $3^m \cdot 11^n$, then the values of $m$ and $n$ respectively are:
  • (a)$0,0$
  • (b)$1,0$
  • (c)$1,1$
  • (d)$2,1$
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(B) $1,0$
491 Mark · March 2026 · Basicopen ↗
The LCM of the smallest $2$-digit number and the smallest $3$-digit number is expressed in the form $2^P \cdot 5^q$. The respective values of $p$ and $q$ are :
  • (a)$1,2$
  • (b)$1,1$
  • (c)$2,1$
  • (d)$2,2$
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(D) $2,2$
501 Mark · March 2026 · Basicopen ↗
The HCF of $2^3$ and $3^2$ is:
  • (a)$2^0.3^0$
  • (b)$2^1.3^1$
  • (c)$2^3.3^2$
  • (d)$2^2.3^3$
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(A) $2^0.3^0$
511 Mark · March 2026 · Basicopen ↗
The LCM of two consecutive natural numbers $p$ and $p + 1$ is:
  • (a)$p$
  • (b)$p^2 + p$
  • (c)$1$
  • (d)$2p + 1$
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(B) $p^2 + p$
521 Mark · March 2026 · Basicopen ↗
The HCF of $4^3$ and $6^2$ is :
  • (a)$2^2$
  • (b)$2^3$
  • (c)$2^2 \cdot 3^2$
  • (d)$1$
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(A) $2^2$
531 Mark · March 2026 · Basicopen ↗
The HCF of $3^5$ and $5^3$ is :
  • (a)$3^0 \cdot 5^0$
  • (b)$3^3 \cdot 5^3$
  • (c)$3^1 \cdot 5^1$
  • (d)$3^5 \cdot 5^3$
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$3^0 \cdot 5^0$
541 Mark · March 2026 · Basicopen ↗
If HCF $(850, 325)$ is $25$, then LCM $(850, 325)$ is :
  • (a)$442$
  • (b)$11050$
  • (c)$8450$
  • (d)$2210$
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(B) $11050$
551 Mark · March 2026 · Basicopen ↗
HCF of two consecutive natural numbers is :
  • (a)$2$
  • (b)$1$
  • (c)$0$
  • (d)smaller number
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(B) $1$
561 Mark · March 2026 · Basicopen ↗
The product of the H.C.F. and L.C.M. of two numbers $50$ and $20$ is
  • (a)$100$
  • (b)$1000$
  • (c)$50$
  • (d)$20$
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(B) $1000$
2 Marks Questions
572 Marks · July 2023 · Standardopen ↗
Find the HCF and LCM of $84, 90$ and $120$ by prime factorization method.
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$84 = 2^2 \times 3\times7$
$90 = 2 \times 3^2 \times 5$
$120 = 2^3 \times 3\times5$
HCF = $6$
LCM = $2520$
582 Marks · March 2023 · Standardopen ↗
Two numbers are in the ratio $2 : 3$ and their LCM is $180$. What is the HCF of these numbers ?
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Let the numbers be $2x, 3x$
LCM = $6x = 180 \Rightarrow x = 30$
$\therefore$ Numbers are $60, 90$
HCF $(60, 90) = 30$
592 Marks · March 2023 · Standardopen ↗
Using prime factorisation, find HCF and LCM of $96$ and $120$.
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$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3$
$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$
$HCF = 24$
$LCM = 480$
602 Marks · March 2023 · Standardopen ↗
Find the HCF and LCM of $72$ and $120$.
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$72=2^3 \times 3^2$
$120=2^3 \times 3 \times 5$
HCF = 24
LCM=360
612 Marks · March 2023 · Standardopen ↗
OR
Find the LCM and HCF of $72$ and $120$
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$72=2^3 \times 3^2$
$120=2^3 \times 3 \times 5$
HCF $= 24$
LCM$=360$
622 Marks · July 2025 · Standardopen ↗
If LCM of $51$ and $85$ can be expressed in the form of $6z - 9$, then find the value of $z$.
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$51 = 3 \times 17$, $85 = 5 \times 17$
LCM $(51, 85) = 3 \times 5 \times 17 = 255$
Now, $6z - 9 = 255$
$\Rightarrow z = 44$
632 Marks · March 2025 · Standardopen ↗
Two numbers are in the ratio $4: 5$ and their HCF is $11$. Find the LCM of these numbers.
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Let the two numbers be $4x$ and $5x$ where $x$ is common factor
Now HCF $= 11$
$\therefore x = 11$
Numbers are $44$ and $55$
$\operatorname{LCM}(44,55) = \frac{44 \times 55}{11} = 220$
642 Marks · March 2025 · Standardopen ↗
Find HCF and LCM of $35$ and $55$ and verify your answer.
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$35 = 5 \times 7$
$55 = 5 \times 11$
HCF = $5$
LCM = $5 \times 7 \times 11 = 385$
HCF $\times$ LCM = $5 \times 385 = 1925$
Product of two numbers = $35 \times 55 = 1925$
Hence HCF $\times$ LCM = Product of two numbers
652 Marks · March 2026 · Standardopen ↗
If the HCF of $210$ and $55$ is expressed as $210 \times 5 + 55m$, then find the value of $m$.
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$210 = 2 \times 3 \times 5 \times 7$
$55 = 5 \times 11$
H.C.F. $(210, 55) = 5$ (I) (1 Mark)
$\therefore 5 = 210 \times 5 + 55m$ (II) ($\frac{1}{2}$ Mark)
$\Rightarrow m = -19$ (III) ($\frac{1}{2}$ Mark)
662 Marks · March 2026 · Standardopen ↗
Find the greatest number less than $10,000$ which is exactly divisible by $48, 60$ and $65$.
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$48 = 2^4 \times 3$, $60 = 2^2 \times 3 \times 5$, $65 = 5 \times 13$ (1/2 Mark)
L.C.M. $(48,60,65) = 2^4 \times 3 \times 5 \times 13 = 3120$ (1 Mark)
$\therefore$ Highest multiple of $3120$, less than $10,000 = 3120 \times 3 = 9360$ (1/2 Mark)
672 Marks · March 2026 · Standardopen ↗
Two numbers are in the ratio $3 : 5$ and their LCM is $180$. Find the HCF of these two numbers.
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Let the numbers be $3x$ and $5x$. HCF = $x$ ($\frac{1}{2}$ Mark)
HCF $\times$ LCM = Product of two numbers ($\frac{1}{2}$ Mark)
$x \times 180 = 3x \times 5x$ (1 Mark)
$x = 12$ ($\frac{1}{2}$ Mark)
HCF of numbers is $12$. ($\frac{1}{2}$ Mark)
682 Marks · March 2025 · Basicopen ↗
Using prime factorisation, find the HCF of 180, 140 and 210.
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$180 = 2^2 \times 3^2 \times 5, 140 = 2^2 \times 5 \times 7, 210 = 2 \times 3 \times 5 \times 7$ ($1\frac{1}{2}$ marks)
$HCF(180, 140, 210) = 2 \times 5 = 10$ ($\frac{1}{2}$ mark)
692 Marks · March 2025 · Basicopen ↗
Using prime factorisation, find the HCF of $144, 180$ and $192$.
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$144 = 2^4 \times 3^2$, $180 = 2^2 \times 3^2 \times 5$, $192 = 2^6 \times 3$
HCF $(144, 180, 192) = 2^2 \times 3 = 12$
702 Marks · March 2026 · Basicopen ↗
Find the H.C.F. and L.C.M. of $1530$ and $2040$.
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$1530 = 2 \times 3^2 \times 5 \times 17$;
$2040 = 2^3 \times 3 \times 5 \times 17$ (½ + ½ Mark)
$\text{HCF}(1530, 2040) = 2 \times 3 \times 5 \times 17 = 510$ (½ Mark)
$\text{LCM } (1530, 2040) = 2^3 \times 3^2 \times 5 \times 17 = 6120$ (½ Mark)
712 Marks · March 2026 · Basicopen ↗
If H.C.F. of $135x^2$ and $189x^3$ is $108$, then find the value of $x$.
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$135x^2 = 3^3 \times 5 \times x^2$, $189x^3 = 3^3 \times 7 \times x^3$ (1 Mark)
HCF($135x^2$, $189x^3$) = $27x^2$
$27x^2 = 108 \Rightarrow x = 2$ (1 Mark)
722 Marks · March 2026 · Basicopen ↗
Find the H.C.F. and L.C.M. of $408$ and $312$.
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$408 = 2^3 \times 3 \times 17$ (1/2 Mark)
$312 = 2^3 \times 3 \times 13$ (1/2 Mark)
HCF = $24$ (1/2 Mark)
LCM = $5304$ (1/2 Mark)
3 Marks Questions
733 Marks · March 2023 · Standardopen ↗
Find by prime factorisation the LCM of the numbers $18180$ and $7575$. Also, find the HCF of the two numbers.
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$18180 = 2^2 \times 3^2 \times 5 \times 101$
$7575 = 3 \times 5^2 \times 101$
LCM = $2^2 \times 3^2 \times 5^2 \times 101 = 90900$
HCF = $3 \times 5 \times 101 = 1515$
743 Marks · March 2023 · Standardopen ↗
Find the HCF and LCM of $26, 65$ and $117$, using prime factorisation.
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$26= 13 \times 2$
$65=13 \times 5$
$117=13 \times 3 \times 3$
$\therefore HCF = 13$
$LCM = 13 \times 2 \times 3 \times 5 \times 3 = 1170$
753 Marks · March 2025 · Standardopen ↗
Let $x$ and $y$ be two distinct prime numbers and $p = x^2 y^3$, $q = xy^4$, $r = x^5 y^2$. 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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$p = x^2y^3, q = xy^4, r = x^5y^2$
HCF $(p,q,r) = xy^2$
LCM $(p,q,r) = x^5y^4$
HCF $\times$ LCM $= x^6y^6$
$p \times q \times r = x^8y^9$
$\Rightarrow$ HCF $(p, q, r) \times$ LCM $(p, q, r) \neq p \times q \times r$
763 Marks · March 2025 · Basicopen ↗
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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$45 = 3^2 \times 5, 75 = 3 \times 5^2, 60 = 2^2 \times 3 \times 5$
$LCM (45, 75, 60) = 2^2 \times 3^2 \times 5^2 = 900$
$900$ seconds $= 15$ minutes
Lights will change together at $5:15$ a.m. again
773 Marks · March 2025 · Basicopen ↗
Three measuring rods are of lengths $120 \text{ cm}, 100 \text{ cm}$ and $150 \text{ cm}$. 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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$120 = 2^3 \times 3 \times 5, 100 = 2^2 \times 5^2, 150 = 2 \times 3 \times 5^2$ ($1\frac{1}{2}$ marks)
$LCM(120, 100, 150) = 2^3 \times 3 \times 5^2 = 600$ (1 mark)
$\therefore \text{Least length of the fence is } 600 \text{ cm.}$
$\text{Each rod is used 5, 6 and 4 times respectively}$ ($\frac{1}{2}$ mark)

Unit digit 0

1 Mark Questions
781 Mark · March 2023 · Standardopen ↗
If 'n' is a natural number, then which of the following numbers end with zero?
  • (a)$(3\times2)^n$
  • (b)$(2\times5)^n$
  • (c)$(6\times2)^n$
  • (d)$(5\times3)^n$
Show SolutionHide Solution
(b) $(2 \times 5)^n$
791 Mark · March 2023 · Standardopen ↗
Assertion (A): The number $5^{\text{n}}$ cannot end with the digit $0$, where $n$ is a natural number.
Reason (R): Prime factorisation of $5$ has only two factors, $1$ and $5$.
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(c) Assertion (A) is true, but Reason (R) is false
801 Mark · March 2024 · Standardopen ↗
Can the number $(15)^n$, $n$ being a natural number, end with the digit $0$? Give reasons.
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$15^n = (5 \times 3)^n$
A number ends with zero if it has two prime factors $2$ and $5$ both. Since $15^n$ does not have $2$ as a prime factor, so it can't end with zero
811 Mark · March 2025 · Basicopen ↗
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)$8$
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Answer (C) $6$

Prime number

2 Marks Questions
822 Marks · March 2024 · Standardopen ↗
Explain why $7 \times 11 \times 13 + 13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+ 5$ are composite numbers.
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$7 \times 11 \times 13 + 13 = 13 \times 78$ or $2 \times 3 \times 13^2$ (1)
and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times 1009$ ($\frac{1}{2}$)
Both numbers have two factors other than $1$
$\therefore$ both the numbers are composite ($\frac{1}{2}$)

Irrational

1 Mark Questions
831 Mark · March 2025 · Basicopen ↗
The number $3 + \sqrt{2}$ is :
  • (a)a rational number
  • (b)an irrational number
  • (c)an integer
  • (d)a natural number
Show SolutionHide Solution
(B) an irrational number
2 Marks Questions
842 Marks · July 2025 · Standardopen ↗
If $\sqrt{7}$ is an irrational number, then prove that $2\sqrt{7}$ is also an irrational number.
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Let $2\sqrt{7}$ be a rational number.
Let $2\sqrt{7}=\frac{a}{b}$ where a & b are co-prime.
$\Rightarrow \sqrt{7}=\frac{a}{2b}$
RHS is rational which contradicts the fact that $\sqrt{7}$ is irrational.
Therefore $2\sqrt{7}$ is an irrational number.
852 Marks · March 2026 · Standardopen ↗
Prove that $2 + 3\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational number.
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Let $2 + 3\sqrt{5}$ be a rational number.
$\therefore 2 + 3\sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and $p$ and $q$ are integers. (I) ($\frac{1}{2}$ Mark)
$\Rightarrow \sqrt{5} = \frac{p-2q}{3q}$ (II) ($\frac{1}{2}$ Mark)
As $\frac{p-2q}{3q}$ is a rational number, so $\sqrt{5}$ is rational.
But we know that $\sqrt{5}$ is irrational.
$\therefore$ Our assumption is wrong. Hence, $2 + 3\sqrt{5}$ is an irrational number. (III) (1 Mark)
862 Marks · March 2026 · Standardopen ↗
Prove that $2 - 5\sqrt{3}$ is an irrational number given that $\sqrt{3}$ is irrational.
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Let $2 - 5\sqrt{3}$ be a rational number.
$2 - 5\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (I) (1/2)
$ \sqrt{3} = \frac{2b-a}{5b}$ (II) (1/2)
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $2 - 5\sqrt{3}$ is an irrational number. (III) (1)
872 Marks · March 2026 · Standardopen ↗
Prove that $4 - 2\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational.
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Let $4 - 2\sqrt{5}$ be a rational number. (I Mark)
$\therefore 4 - 2\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (II Mark)
$\sqrt{5} = \frac{4b - a}{2b}$
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $4 - 2\sqrt{5}$ is an irrational number. (III Mark)
882 Marks · March 2026 · Standardopen ↗
Prove that $14 - 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.
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Let $14 - 2\sqrt{3}$ be a rational number. (1/2 Mark)
$14 - 2\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (1/2 Mark)
$\sqrt{3} = \frac{14b - a}{2b}$ (1 Mark)
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $14 - 2\sqrt{3}$ is an irrational number.
3 Marks Questions
893 Marks · 🔁 March 2023 & July 2023 & March 2024 & July 2024 & March 2025 & March 2026 · Basicopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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(a) Let $\sqrt{5}$ be a rational number such that $\sqrt{5} = \frac{p}{q}$ ($p$ and $q$ are co-prime numbers, $q \neq 0$) [1/2 mark]
$\sqrt{5}q = p \Rightarrow 5q^2 = p^2$
$5$ divides $p^2 \Rightarrow 5$ divides $p$ as well [1 mark]
$p = 5m$ (for some integer $m$)
$5q^2 = 25m^2 \Rightarrow q^2 = 5m^2$
$5$ divides $q^2 \Rightarrow 5$ divides $q$ as well [1 mark]
$p$ and $q$ have a common factor $5$ which is a contradiction as $p$ and $q$ are co-prime.
$\therefore \text{our assumption is wrong}$
Hence, $\sqrt{5}$ is an irrational number [1/2 mark]
903 Marks · 🔁 March 2023 & March 2024 & July 2024 & March 2025 & March 2026 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3} = \frac{p}{q}$, let $p \& q$ be co-primes and $q \neq 0$
$3q^2 = p^2 \Rightarrow p^2$ is divisible by $3 \Rightarrow p$ is divisible by $3$
$\Rightarrow p = 3a$, where 'a' is some integer
quad ----- (i)
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $3$
$\Rightarrow q = 3b$, where 'b' is some integer
quad ----- (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{3}$ is an irrational number.
913 Marks · March 2024 · Standardopen ↗
Prove that $\frac{2-\sqrt{3}}{5}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
Show SolutionHide Solution
Assuming $\frac{2-\sqrt{3}}{5}$ to be a rational number.
$\frac{2-\sqrt{3}}{5} = \frac{p}{q}$, where $p$ and $q$ are integers $$\begin{aligned}& \& q \neq 0 \\ & \sqrt{3} = \frac{2q-5p}{q} \\ & \text{Here RHS is rational but LHS is irrational.} \\ & \text{Therefore our assumption is wrong.} \\ & \text{Hence } \frac{2-\sqrt{3}}{5} \text{ is an irrational number.}\end{aligned}$$
923 Marks · March 2025 · Standardopen ↗
Prove that $(5\sqrt{3} + \frac{2}{3})$ is an irrational number given that $\sqrt{3}$ is an irrational number.
Show SolutionHide Solution
Let $5\sqrt{3} + \frac{2}{3}$ be a rational number.
$\therefore 5\sqrt{3} + \frac{2}{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$.
$5\sqrt{3} = \frac{a}{b} - \frac{2}{3}$
$\sqrt{3} = \frac{3a - 2b}{15b}$
$3a - 2b$ and $15b$ are integers.
$\therefore$ RHS is rational.
But LHS $= \sqrt{3}$ is an irrational number which is contradiction to our supposition.
Hence $5\sqrt{3} + \frac{2}{3}$ is an irrational number.

Word Problem

1 Mark Questions
931 Mark · March 2023 · Standardopen ↗
The LCM of smallest $2$-digit number and smallest composite number is
  • (a)$12$
  • (b)$4$
  • (c)$20$
  • (d)$40$
Show SolutionHide Solution
(C) $20$
2 Marks Questions
942 Marks · March 2023 · Standardopen ↗
Find the least number which when divided by 12, 16 and 24 leaves remainder 7 in each case
Show SolutionHide Solution
LCM of 12, 16, 24 = 48
Required number is $48 + 7 = 55$.
952 Marks · March 2023 · Standardopen ↗
Find the greatest $3$-digit number which is divisible by $18, 24$ and $36$.
Show SolutionHide Solution
LCM of $18, 24, 36$ is $72$
Required greatest $3$-digit number $= 936$.
962 Marks · March 2024 · Standardopen ↗
In a school, there are two sections of class X. There are $40$ students in the first section and $48$ students in the second section. Determine the minimum number of books required for their class library so that they can be distributed equally among students of both sections.
Show SolutionHide Solution
$40 = 2^3 \times 5$
$48 = 2^4 \times 3$
L.C.M. $(40, 48) = 240$
Minimum number of books required in library is $240$.
972 Marks · March 2026 · Standardopen ↗
Find the length of the plank that can be used to measure the lengths $4$ m $20$ cm and $5$ m $4$ cm exactly, in the least time.
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$4$ m $20$ cm $= 420$ cm and $5$ m $4$ cm $= 504$ cm (1/2 Mark)
Size of plank should be maximum, so we will find HCF $(420, 504)$
$420 = 2^2 \times 3 \times 5 \times 7$ and $504 = 2^3 \times 3^2 \times 7$ (1 Mark)
HCF $(420, 504) = 84$ (1/2 Mark)
$\therefore$ the required length of the plank is $84$ cm
3 Marks Questions
983 Marks · March 2023 · Standardopen ↗
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?
Show SolutionHide Solution
$LCM = 432$
i.e. $\frac{432}{60} = 7$ min $12$ sec.
$\Rightarrow$ traffic lights will change simultaneously again at $7 : 7 : 12$ a.m.
993 Marks · March 2024 · Standardopen ↗
A school has invited $42$ Mathematics teachers, $56$ Physics teachers and $70$ Chemistry teachers to attend a Science workshop. Find the minimum number of tables required, if the same number of teachers are to sit at a table and each table is occupied by teachers of the same subject.
Show SolutionHide Solution
HCF $(42, 56, 70) = 14$
Minimum number of tables required $= \frac{42}{14} + \frac{56}{14} + \frac{70}{14}$
$= 3 + 4 + 5 = 12$
1003 Marks · July 2025 · Standardopen ↗
Ranjita, Neha and Salma start weaving sweaters at the same time for the children of an orphan home. They need $15, 18$ and $20$ days, respectively, to complete a sweater. After how many days will all of them start making a new sweater again? By that time how many sweaters will have been competed by them?
Show SolutionHide Solution
LCM $(15, 18, 20) = 180$
They will start new sweater again after $180$ days.
Total number of sweaters completed in $180$ days = $\frac{180}{15} + \frac{180}{18} + \frac{180}{20}$
$= 31$
1013 Marks · March 2026 · Standardopen ↗
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.
Show SolutionHide Solution
Least number of containers means maximum volume in each container. (I) ($\frac{1}{2}$ Mark)
$870 = 2 \times 3 \times 5 \times 29$ (II) ($\frac{1}{2}$ Mark)
$812 = 2^2 \times 7 \times 29$ (III) ($\frac{1}{2}$ Mark)
$638 = 2 \times 11 \times 29$
$\therefore$ H.C.F. $(870, 812, 638) = 2 \times 29 = 58$ (IV) ($\frac{1}{2}$ Mark)
Number of containers of different types required $= \frac{870}{58} + \frac{812}{58} + \frac{638}{58}$ (V) ($\frac{1}{2}$ Mark)
$= 15 + 14 + 11$
$= 40$ (VI) ($\frac{1}{2}$ Mark)
1023 Marks · March 2026 · Standardopen ↗
Find the greatest number which divides $764$ and $1198$, leaving remainders $8$ and $10$ respectively.
Show SolutionHide Solution
$764-8=756$ and $1198-10 = 1188$ (I) (1 Mark)
$756 = 2^2 \times 3^3 \times 7$ (II) (1/2 Mark)
$1188 = 2^2 \times 3^3 \times 11$ (III) (1/2 Mark)
H.C.F. $(756, 1188) = 2^2 \times 3^3 = 108$ (IV) (1 Mark)
$\therefore$ Required greatest number is $108$.
1033 Marks · March 2026 · Standardopen ↗
The dimensions of a window are $156$ cm $\times 216$ 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.
Show SolutionHide Solution
$156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$ (I Mark)
$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3$ (II Mark)
$\therefore$ Required side length of the square = HCF($156,216$) = $12$ cm (III Mark)
Number of squares formed = $\frac{156 \times 216}{12 \times 12} = 234$ (IV Mark)
4 Marks Questions
1044 Marks · July 2023 · Standardopen ↗
February $14$ is celebrated as International Book Giving Day and many countries in the world celebrate this day. Some people in India also started celebrating this day and donated the following number of books of various subjects to a public library :
History = $96$, Science = $240$, Mathematics = $336$.
These books have to be arranged in minimum number of stacks such that each stack contains books of only one subject and the number of books on each stack is the same.
Based on the above information, answer the following questions :
(i) How many books are arranged in each stack?
(ii) How many stacks are used to arrange all the Mathematics books?
(iii) (a) Determine the total number of stacks that will be used for arranging all the books.
OR
(iii) (b) If the thickness of each book of History, Science and Mathematics is $1.8$ cm, $2.2$ cm and $2.5$ cm respectively, then find the height of each stack of History, Science and Mathematics books.
Show SolutionHide Solution
(i) HCF $(96, 240, 336) = 48$
(ii) Number of stacks $= \frac{336}{48} = 7$
(iii) (a) Total number of stacks $= \frac{96}{48} + \frac{240}{48} + \frac{336}{48}$
$= 14$
OR
(b) Height of each stack of History $= 48 \times 1.8 = 86.4$ cm
Height of each stack of Science $= 48 \times 2.2 = 105.6$ cm
Height of each stack of Mathematics $= 48 \times 2.5 = 120$ cm
1054 Marks · March 2024 · Standardopen ↗
Teaching Mathematics through activities is a powerful approach that enhances students' understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class $5$ students. She announces the number $2$ in her class and asked the first student to multiply it by a prime number and then pass it to second student. Second student also multiplied it by a prime number and passed it to third student. In this way by multiplying to a prime number, the last student got $173250$.
Now, Mukta asked some questions as given below to the students :
(i) What is the least prime number used by students?
(ii) (a) How many students are in the class ?
OR
(b) What is the highest prime number used by students?
(iii) Which prime number has been used maximum times ?
Show SolutionHide Solution
$173250= 2 \times 5^3 \times 3^2 \times 7 \times 11$
(i) $3$
(ii) (a) $173250= 2 \times 5^3 \times 3^2 \times 7 \times 11$
Number of students in the class = $3+2+1+1=7$
OR
(ii) (b) $173250= 2 \times 5^3 \times 3^2 \times 7 \times 11$
Highest prime number used by students = $11$
(iii) $5$
1064 Marks · July 2025 · Standardopen ↗
This section has $3$ case study based questions carrying $4$ marks each.
Case Study – 1
The science department of a college is conducting an international seminar in which the number of participants in Physics, Chemistry and Biology are $65, 91$ and $117$ respectively. The coordinator has made the arrangement such that in each room, the same number of participants are to be seated with all of them being in the same subject.
Based on the information given above, answer the following questions :
(i) Find the HCF of $65, 91$ and $117$.
(ii) Find the LCM of $65, 91$ and $117$.
(iii) (a) Find the minimum number of rooms required based on the above conditions.
OR
(iii) (b) Find the minimum number of participants to be accommodated in each of the rooms.
Show SolutionHide Solution
(i) $65 = 5 \times 13$; $91 = 7 \times 13$ ; $117 = 3^2 \times 13$
HCF $(65, 91 \& 117) = 13$
(ii) LCM $(65, 91 \& 117) = 3^2 \times 5 \times 7 \times 13 = 4095$
(iii) (a) minimum number of rooms = $\frac{65}{13} + \frac{91}{13} + \frac{117}{13}$
$= 21$
OR
(b) $1$

General

1 Mark Questions
1071 Mark · March 2023 · Standardopen ↗
Statement A (Assertion) : If $5 + \sqrt{7}$ is a root of a quadratic equation with rational co-efficients, then its other root is $5 - \sqrt{7}$.
Statement R (Reason) : Surd roots of a quadratic equation with rational co-efficients occur in conjugate pairs.
  • (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 SolutionHide Solution
(A)
3 Marks Questions
1083 Marks · March 2023 · Standardopen ↗
A natural number, when increased by $12$, equals $160$ times its reciprocal. Find the number.
Show SolutionHide Solution
Let the natural number be $x$
ATQ, $x + 12 = \frac{160}{x}$
$x^2 + 12x = 160$
$x^2 + 12x - 160 = 0$
$(x + 20)(x - 8) = 0$
$x \neq -20, x= 8$
$\Rightarrow \text{Required natural number is } 8$