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)$3$
Show SolutionHide Solution
(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)$7$
Show SolutionHide Solution
(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)$4$
Show SolutionHide Solution
(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)$1.857142$
Show SolutionHide Solution
(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)$2$
Show SolutionHide Solution
(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)$0$
Show SolutionHide Solution
(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)$5^{12}$
Show SolutionHide Solution
(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.
Show SolutionHide Solution
(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
Show SolutionHide Solution
(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)$1:3$
Show SolutionHide Solution
(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)$p^5q^2+ p^3q^4$
Show SolutionHide Solution
(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)$4$
Show SolutionHide Solution
(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)$180 a^3b^4$
Show SolutionHide Solution
(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)$1625$
Show SolutionHide Solution
(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)$0$
Show SolutionHide Solution
(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)$8$
Show SolutionHide Solution
(B) $13$
171 Mark · March 2024 · Standardopen ↗
The LCM of three numbers $28, 44, 132$ is:
  • (a)$258$
  • (b)$231$
  • (c)$462$
  • (d)$924$
Show SolutionHide Solution
(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)$79$
Show SolutionHide Solution
(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)$2^3 \times 3^3 \times 5$
Show SolutionHide Solution
(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
Show SolutionHide Solution
(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)$198$
Show SolutionHide Solution
(C) $98$
221 Mark · March 2025 · Standardopen ↗
The HCF of $40$, $110$ and $360$ is :
  • (a)$40$
  • (b)$110$
  • (c)$360$
  • (d)$10$
Show SolutionHide Solution
(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)$p = 16y$
Show SolutionHide Solution
(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)$ab (1-ab) (1 + ab)$
Show SolutionHide Solution
(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
Show SolutionHide Solution
(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
Show SolutionHide Solution
(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)$14$
Show SolutionHide Solution
(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.
Show SolutionHide Solution
(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.
Show SolutionHide Solution
(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 $HCF(x, y)$.
Show SolutionHide Solution
(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)$21$, $77$
Show SolutionHide Solution
(C) $7$, $462$
321 Mark · March 2026 · Standardopen ↗
The HCF of $960$ and $432$ is:
  • (a)$48$
  • (b)$54$
  • (c)$72$
  • (d)$36$
Show SolutionHide Solution
(a) $48$ (1 Mark)
331 Mark · March 2026 · Standardopen ↗
The LCM of $960$ and $240$ is:
  • (a)$960$
  • (b)$240$
  • (c)$60$
  • (d)$15$
Show SolutionHide Solution
(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)$272$
Show SolutionHide Solution
(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.
Show SolutionHide Solution
(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)$3$
Show SolutionHide Solution
(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)$ab$
Show SolutionHide Solution
(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.
Show SolutionHide Solution
(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)$-14$
Show SolutionHide Solution
(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)$2^3 \times 3^3 \times 5$
Show SolutionHide Solution
(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)$10$
Show SolutionHide Solution
(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)$150$
Show SolutionHide Solution
(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)$2^3.3^3$
Show SolutionHide Solution
(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)$2^5.3^5$
Show SolutionHide Solution
(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)$3^7 \cdot 7^7$
Show SolutionHide Solution
(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)$1,1$
Show SolutionHide Solution
(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
Show SolutionHide Solution
(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)$2,1$
Show SolutionHide Solution
(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)$2,2$
Show SolutionHide Solution
(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)$2^2.3^3$
Show SolutionHide Solution
(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)$2p + 1$
Show SolutionHide Solution
(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)$1$
Show SolutionHide Solution
(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)$3^5 \cdot 5^3$
Show SolutionHide Solution
$3^0 \cdot 5^0$
541 Mark · March 2026 · Basicopen ↗
If HCF $(850, 325)$ is $25$, then LCM $(850, 325)$ is :
  • (a)$442$
  • (b)$11050$
  • (c)$8450$
  • (d)$2210$
Show SolutionHide Solution
(B) $11050$
551 Mark · March 2026 · Basicopen ↗
HCF of two consecutive natural numbers is :
  • (a)$2$
  • (b)$1$
  • (c)$0$
  • (d)smaller number
Show SolutionHide Solution
(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)$20$
Show SolutionHide Solution
(B) $1000$
2 Marks Questions
572 Marks · July 2023 · Standardopen ↗
Find the HCF and LCM of $84, 90$ and $120$ by prime factorization method.
Show SolutionHide Solution
$84 = 2^2 \times 3\times7$
$90 = 2 \times 3^2 \times 5$
$120 = 2^3 \times 3\times5$
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 ?
Show SolutionHide Solution
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 $120$.
Show SolutionHide Solution
$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3$
$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$
$HCF = 24$
$LCM = 480$
602 Marks · March 2023 · Standardopen ↗
Find the HCF and LCM of $72$ and $120$.
Show SolutionHide Solution
$72=2^3 \times 3^2$
$120=2^3 \times 3 \times 5$
HCF = 24
LCM=360
612 Marks · March 2023 · Standardopen ↗
OR
Find the LCM and HCF of $72$ and $120$
Show SolutionHide Solution
$72=2^3 \times 3^2$
$120=2^3 \times 3 \times 5$
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 $z$.
Show SolutionHide Solution
$51 = 3 \times 17$, $85 = 5 \times 17$
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.
Show SolutionHide Solution
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.
Show SolutionHide Solution
$35 = 5 \times 7$
$55 = 5 \times 11$
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 $m$.
Show SolutionHide Solution
$210 = 2 \times 3 \times 5 \times 7$
$55 = 5 \times 11$
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 $65$.
Show SolutionHide Solution
$48 = 2^4 \times 3$, $60 = 2^2 \times 3 \times 5$, $65 = 5 \times 13$ (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.
Show SolutionHide Solution
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.
Show SolutionHide Solution
$180 = 2^2 \times 3^2 \times 5, 140 = 2^2 \times 5 \times 7, 210 = 2 \times 3 \times 5 \times 7$ ($1\frac{1}{2}$ 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 $192$.
Show SolutionHide Solution
$144 = 2^4 \times 3^2$, $180 = 2^2 \times 3^2 \times 5$, $192 = 2^6 \times 3$
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 $2040$.
Show SolutionHide Solution
$1530 = 2 \times 3^2 \times 5 \times 17$;
$2040 = 2^3 \times 3 \times 5 \times 17$ (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 $x$.
Show SolutionHide Solution
$135x^2 = 3^3 \times 5 \times x^2$, $189x^3 = 3^3 \times 7 \times x^3$ (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 $312$.
Show SolutionHide Solution
$408 = 2^3 \times 3 \times 17$ (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.
Show SolutionHide Solution
$18180 = 2^2 \times 3^2 \times 5 \times 101$
$7575 = 3 \times 5^2 \times 101$
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.
Show SolutionHide Solution
$26= 13 \times 2$
$65=13 \times 5$
$117=13 \times 3 \times 3$
$\therefore HCF = 13$
$LCM = 13 \times 2 \times 3 \times 5 \times 3 = 1170$
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.
Show SolutionHide Solution
$p = x^2y^3, q = xy^4, r = x^5y^2$
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?
Show SolutionHide Solution
$45 = 3^2 \times 5, 75 = 3 \times 5^2, 60 = 2^2 \times 3 \times 5$
$LCM (45, 75, 60) = 2^2 \times 3^2 \times 5^2 = 900$
$900$ 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?
Show SolutionHide Solution
$120 = 2^3 \times 3 \times 5, 100 = 2^2 \times 5^2, 150 = 2 \times 3 \times 5^2$ ($1\frac{1}{2}$ 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 $5$.
Show SolutionHide Solution
(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.
Show SolutionHide Solution
$15^n = (5 \times 3)^n$
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 · Standardopen ↗
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 SolutionHide Solution
(d) Assertion (A) is false but Reason (R) is true
821 Mark · March 2025 · Standardopen ↗
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 $n$.
Show SolutionHide Solution
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
831 Mark · March 2025 · Standardopen ↗
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 SolutionHide Solution
(C) 0
841 Mark · March 2026 · Standardopen ↗
For any natural number $n$, $6^n$ ends with the digit :
  • (a)$0$
  • (b)$6$
  • (c)$3$
  • (d)$2$
Show SolutionHide Solution
(b) $6$ (1 Mark)
851 Mark · March 2026 · Standardopen ↗
For any natural number $n$, $5^n$ ends with the digit:
  • (a)$0$
  • (b)$5$
  • (c)$3$
  • (d)$2$
Show SolutionHide Solution
(B) $5$ (1 Mark)
861 Mark · March 2025 · Basicopen ↗
The number $2^n$, where $n$ is a natural number, cannot end with the digit :
  • (a)$4$
  • (b)$6$
  • (c)$2$
  • (d)$0$
Show SolutionHide Solution
(D) $0$
871 Mark · March 2025 · Basicopen ↗
The number $3^n$, where $n$ is a natural number, cannot end with the digit :
  • (a)$1$
  • (b)$3$
  • (c)$5$
  • (d)$7$
Show SolutionHide Solution
(C) $5$
881 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)$8$
Show SolutionHide Solution
Answer (C) $6$
891 Mark · March 2026 · Basicopen ↗
Which of the following numbers will not end with 0 for any natural
number $n$ ?
  • (a)$4^n$
  • (b)$4^n$
  • (c)$3^n+1$
  • (d)$10^{n+1}$
Show SolutionHide Solution
(b) $4^n$
901 Mark · March 2026 · Basicopen ↗
Assertion (A): $4^n$ can not end with the digit zero.
Reason (R): Prime factorisation of $4^n$ is unique.
  • (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
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A)
2 Marks Questions
912 Marks · July 2023 · Standardopen ↗
Check whether $6^n$ can end with the digit $0$ for any natural number $n$.
Show SolutionHide Solution
If the number $6^n$ ends with the digit $0$, then it should be divisible by $2$ and $5$.
But prime factorisation of $6^n$ is $(2 \times 3)^n$.
$\therefore$ Prime factorisation of $6^n$ does not contain prime number $5$.
Hence, $6^n$ can't end with the digit $0$.
922 Marks · March 2023 · Standardopen ↗
Prove that $4^n$ can never end with digit $0$, where $n$ is a natural number.
Show SolutionHide Solution
If the number $4^n$, for any $n$, were to end with digit zero, it would be divisible by $5$. So, the prime factorization of $4^n$ should contain the prime factor $5$.
But in $4^n = (2 \times 2)^n = 2^{2n}$, the only prime factor is $2$.
$\therefore$ By fundamental theorem of arithmetic, there is no natural number $n$ for which $4^n$ ends with digit zero.
932 Marks · March 2023 · Standardopen ↗
Show that $6^n$ can not end with digit $0$ for any natural number 'n'.
Show SolutionHide Solution
If $6^n$ ends with digit $0$, it would be divisible by $5$. So, prime factorization of $6^n$ would contain $5$. But $6^n = (2 \times 3)^n$, the only prime factorization of $6^n$ are $2$ and $3$ as per fundamental theorem of Arithmetic . There is no other prime in the factorization of $6^n$. So, there is no natural number $n$ for which $6^n$ ends with digit zero.
942 Marks · July 2024 · Standardopen ↗
Check whether there is any natural number 'n' for which $(14)^n$ ends with the digit '0' or '5'.
Show SolutionHide Solution
$(14)^n = 2^n7^n$
as it does not have $5$ as the prime factor
Hence, it cannot end with the digit $0$ or $5$
952 Marks · July 2025 · Standardopen ↗
Show that $14^n$ cannot end with the digit $0$ or $5$ for any natural number $n$.
Show SolutionHide Solution
$14^n = 2^n \times 7^n$
To end with a digit $0$ or $5$, $14^n$ must have at least one prime factor $5$, which is not there.
$\therefore 14^n$ can not end with digit $0$ or $5$.
962 Marks · March 2025 · Basicopen ↗
Show that $45^n$ can not end with the digit $0$, $n$ being a natural number. Write the prime number '$a$' which on multiplying with $45^n$ makes the product end with the digit $0$.
Show SolutionHide Solution
$45^n = (3 \times 3)^n \times 5^n$ [1 mark]
To end with digit $0$, $45^n$ should have prime factors $2$ and $5$ both. So it cannot end with digit $0$. [1/2 mark]
$45^n$ should be multiplied by $2 \Rightarrow a = 2$ [1/2 mark]
972 Marks · March 2025 · Basicopen ↗
Check whether $15^n \times 2^n$, n being a natural number, ends with the digit zero.
Show SolutionHide Solution
$15^n \times 2^n = 5^n \times 3^n \times 2^n$
$\Rightarrow 2$ and 5 both are the factors of the given number
$\therefore$ The given number ends with the digit zero
982 Marks · March 2025 · Basicopen ↗
Prove that, for a natural number $n$, $6^n$ can not end with the digit $0$. Which prime number must be multiplied with $6^n$ so that the resultant ends with the digit zero?
Show SolutionHide Solution
$6^n = 2^n \times 3^n$
To end with the digit $0$, $6^n$ should have $2$ and $5$ both as prime factors.
$\therefore 6^n$ cannot end with the digit $0$.
To end with digit $0$, $6^n$ should be multiplied by the prime number $5$.

Prime number

1 Mark Questions
991 Mark · March 2026 · Standardopen ↗
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 SolutionHide Solution
(a) a prime number (1 Mark)
1001 Mark · March 2026 · Standardopen ↗
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 SolutionHide Solution
(D) neither prime nor composite (1 Mark)
1011 Mark · March 2026 · Standardopen ↗
$(3 \times 11 \times 13 + 3)$ is :
  • (a)a prime number
  • (b)divisible by $13$
  • (c)a composite number
  • (d)an odd number
Show SolutionHide Solution
(C) a composite number
1021 Mark · March 2026 · Standardopen ↗
$(3 \times 11 \times 13 + 3)$ is :
  • (a)a prime number
  • (b)divisible by $13$
  • (c)a composite number
  • (d)an odd number
Show SolutionHide Solution
(C) a composite number
1031 Mark · March 2026 · Standardopen ↗
$(3 \times 11 \times 13 + 3)$ is :
  • (a)a prime number
  • (b)divisible by $13$
  • (c)a composite number
  • (d)an odd number
Show SolutionHide Solution
(C) a composite number (1 Mark)
1041 Mark · March 2025 · Basicopen ↗
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 SolutionHide Solution
(B) exactly one prime factor
1051 Mark · March 2025 · Basicopen ↗
The sum of first five prime numbers is
  • (a)$18$
  • (b)$26$
  • (c)$28$
  • (d)$39$
Show SolutionHide Solution
(C) $28$
1061 Mark · March 2026 · Basicopen ↗
Assertion (A): $7 \times 2+3$ is a composite number.
Reason (R): A composite number has more than two factors.
Show SolutionHide Solution
(d) Assertion (A) is false, but Reason (R) is true.
1071 Mark · March 2026 · Basicopen ↗
$7 \times 29 \times 23 + 1$ is:
  • (a)a prime number.
  • (b)divisible by 23.
  • (c)an odd number.
  • (d)a composite number.
Show SolutionHide Solution
(D) a composite number
1081 Mark · March 2026 · Basicopen ↗
$7 \times 29 \times 23 + 1$ is:
  • (a)a prime number.
  • (b)divisible by $23$.
  • (c)an odd number.
  • (d)a composite number.
Show SolutionHide Solution
(D) a composite number
1091 Mark · March 2026 · Basicopen ↗
$7 \times 29 \times 23 + 1$ is:
  • (a)a prime number.
  • (b)divisible by $23$.
  • (c)an odd number.
  • (d)a composite number.
Show SolutionHide Solution
(D) a composite number
1101 Mark · March 2026 · Basicopen ↗
$7 x 11 x 13 + 5$ is
  • (a)a prime number.
  • (b)an odd number.
  • (c)a composite number.
  • (d)a multiple of 5.
Show SolutionHide Solution
Answer (C) a composite number.
1111 Mark · March 2026 · Basicopen ↗
$7 \times 11 \times 13 + 5$ is
  • (a)a prime number.
  • (b)an odd number.
  • (c)a composite number.
  • (d)a multiple of $5$.
Show SolutionHide Solution
Answer (C) a composite number.
1121 Mark · March 2026 · Basicopen ↗
$17 \times 11 \times 13 + 11$ is
  • (a)a prime number.
  • (b)multiple of $17$.
  • (c)a composite number.
  • (d)an odd number.
Show SolutionHide Solution
Answer (C) a composite number.
2 Marks Questions
1132 Marks · March 2024 · Standardopen ↗
Show that the number $5\times 11\times 17+3\times 11$ is a composite number.
Show SolutionHide Solution
$5 \times 11 \times 17 + 3 \times 11 = 11 \times (5 \times 17 + 3)$
$= 11\times 88$ or $11 \times 11 \times 2^3$
It means the number can be expressed as a product of two factors other than $1$, therefore the given number is a composite number.
1142 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.
Show SolutionHide Solution
$7 \times 11 \times 13 + 13 = 13 \times 78$ 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 ($\frac{1}{2}$)
3 Marks Questions
1153 Marks · March 2025 · Standardopen ↗
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.
Show SolutionHide Solution
(i) True, $\because 2 \times 3 \times 5 \times 7 + 7 = 7 \times (2 \times 3 \times 5 + 1)$ has more than two factors.
(ii) False, $\because 2 \times 3 \times 5 \times 7 + 1 = 211$ has only two factors.
1163 Marks · March 2025 · Standardopen ↗
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.
Show SolutionHide Solution
$p \cdot q \cdot r + q = q(pr + 1)$. Thus, the given number has more than 2 factors. Hence it is composite ($\frac{1}{2} + \frac{1}{2}$ marks). (i) Taking $p=3, q=5$ and $r=7$, $pqr + 1 = 3 \cdot 5 \cdot 7 + 1 = 106$ is a composite number (1 mark). (ii) Taking $p=2, q=3$ and $r=5$, $pqr + 1 = 2 \cdot 3 \cdot 5 + 1 = 31$ is a prime number (1 mark).

Irrational

1 Mark Questions
1171 Mark · March 2023 · Standardopen ↗
Assertion (A): The perimeter of $\triangle ABC$ is a rational number. Reason (R): The sum of the squares of two rational numbers is always rational.
figure for this question
Show SolutionHide Solution
(D) Assertion (A) is false but Reason (R) is true
1181 Mark · March 2023 · Standardopen ↗
If $p^2 = \frac{32}{50}$, then $p$ is a/an
  • (a)whole number
  • (b)integer
  • (c)rational number
  • (d)irrational number
Show SolutionHide Solution
(C) rational
1191 Mark · March 2024 · Standardopen ↗
A pair of irrational numbers whose product is a rational number is :
  • (a)$(\sqrt{16}, \sqrt{4})$
  • (b)$(\sqrt{5}, \sqrt{2})$
  • (c)$(\sqrt{3}, \sqrt{27})$
  • (d)$(\sqrt{36}, \sqrt{2})$
Show SolutionHide Solution
(C) $(\sqrt{3}, \sqrt{27})$
1201 Mark · March 2024 · Standardopen ↗
The smallest irrational number by which $\sqrt{20}$ should be multiplied so as to get a rational number, is:
  • (a)$\sqrt{20}$
  • (b)$\sqrt{2}$
  • (c)$5$
  • (d)$\sqrt{5}$
Show SolutionHide Solution
(D) $\sqrt{5}$
1211 Mark · July 2025 · Standardopen ↗
$2.35$ is:
  • (a)an integer
  • (b)a rational number
  • (c)an irrational number
  • (d)a natural number
Show SolutionHide Solution
(B) a rational number
1221 Mark · March 2025 · Standardopen ↗
$(1+\sqrt{3})^2 -(1-\sqrt{3})^2$ is:
  • (a)a positive rational number.
  • (b)a negative integer.
  • (c)a positive irrational number.
  • (d)a negative irrational number.
Show SolutionHide Solution
(c) a positive irrational number
1231 Mark · March 2025 · Standardopen ↗
$(1 + \sqrt{3})^2 - (1 - \sqrt{3})^2$ is :
  • (a)a positive rational number.
  • (b)a negative integer.
  • (c)a positive irrational number.
  • (d)a negative irrational number.
Show SolutionHide Solution
(c) a positive irrational number
1241 Mark · March 2025 · Standardopen ↗
$(\sqrt{3}+2)^2+(\sqrt{3}-2)^2$ is a/an
  • (a)positive rational number
  • (b)negative rational number
  • (c)positive irrational number
  • (d)negative irrational number
Show SolutionHide Solution
(A) positive rational number
1251 Mark · March 2025 · Standardopen ↗
For any prime number $p$, if $p$ divides $a^2$, where $a$ is any real number then $p$ also divides
  • (a)$a$
  • (b)$a^{1/2}$
  • (c)$a^{3/2}$
  • (d)$a^{1/8}$
Show SolutionHide Solution
(A) $a$
1261 Mark · March 2025 · Standardopen ↗
$\sqrt{0.4}$ is a/an
  • (a)natural number
  • (b)integer
  • (c)rational number
  • (d)irrational number
Show SolutionHide Solution
(D) irrational number
1271 Mark · March 2026 · Standardopen ↗
Assertion (A): $(\sqrt{3} + \sqrt{5})$ is an irrational number. Reason (R) : Sum of the any two irrational numbers is always irrational.
Show SolutionHide Solution
(C) Assertion (A) is true, but Reason (R) is false.
1281 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
1291 Mark · March 2025 · Basicopen ↗
$\frac{\sqrt{3} - 3}{\sqrt{3}}$ is :
  • (a)a rational number
  • (b)an irrational number
  • (c)an integer
  • (d)a natural number
Show SolutionHide Solution
(B) an irrational number
1301 Mark · March 2025 · Basicopen ↗
$(2 + \sqrt{2})^2$ is :
  • (a)a rational number
  • (b)an irrational number
  • (c)an integer
  • (d)a natural number
Show SolutionHide Solution
(B) an irrational number
1311 Mark · March 2025 · Basicopen ↗
$(2 - 5\sqrt{3})^2$ is
  • (a)a negative integer
  • (b)an irrational number
  • (c)a rational number
  • (d)a positive integer
Show SolutionHide Solution
(B) an irrational number
1321 Mark · March 2025 · Basicopen ↗
$(2 - 5\sqrt{3})^2$ is
  • (a)a negative integer
  • (b)an irrational number
  • (c)a rational number
  • (d)a positive integer
Show SolutionHide Solution
(B) an irrational number
1331 Mark · March 2025 · Basicopen ↗
$\sqrt{2}(\sqrt{2} - 1)$ is
  • (a)an integer
  • (b)a rational number
  • (c)an irrational number
  • (d)equal to 1
Show SolutionHide Solution
(C) an irrational number
1341 Mark · March 2025 · Basicopen ↗
Assertion (A) : $(a + \sqrt{b}) \cdot (a - \sqrt{b})$ 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 SolutionHide Solution
(C) Assertion (A) is true, but Reason (R) is false.
1351 Mark · March 2026 · Basicopen ↗
Assertion (A): $(\sqrt{2} + \sqrt{3})$ is an irrational number.
Reason (R): The sum of two irrational numbers is always an irrational number.
  • (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
(C) Assertion (A) is true, Reason (R) is false.
1361 Mark · March 2026 · Basicopen ↗
Assertion (A): $(\sqrt{3} + 1)^2$ is a rational number.
Reason (R): $(\sqrt{3})^2 = 3$ is a rational number.
Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.
(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 SolutionHide Solution
(D) Assertion (A) is false but Reason (R) is true.
1371 Mark · March 2026 · Basicopen ↗
Assertion (A) : $(2 - \sqrt{5})^2$ is a rational number.
Reason (R): $(\sqrt{5})^2$ is a rational 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 SolutionHide Solution
(D) Assertion (A) is false but Reason (R) is true.
1381 Mark · March 2026 · Basicopen ↗
Assertion (A): $(3 + 2\sqrt{2})^2$ is a rational number.
Reason (R): $\sqrt{2}$ is an irrational 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 SolutionHide Solution
Assertion (A) is false but Reason (R) is true.
2 Marks Questions
1392 Marks · July 2023 · Standardopen ↗
If $\sqrt{2}$ is given as an irrational number, then prove that $(5-2\sqrt{2})$ is an irrational number.
Show SolutionHide Solution
Let us assume that $5 - 2\sqrt{2}$ be a rational number.
$\therefore 5 - 2\sqrt{2} = \frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
$\Rightarrow \sqrt{2} = \frac{5q-p}{2q}$
RHS is a rational number. So, LHS is also a rational number which contradict the given fact that $\sqrt{2}$ is an irrational number.
So, our assumption is wrong.
Hence, $5 - 2\sqrt{2}$ is an irrational number.
1402 Marks · March 2023 · Standardopen ↗
Prove that $2 + \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
Show SolutionHide Solution
Let us assume that $2 + \sqrt{3}$ is rational
Let $2 + \sqrt{3} = \frac{P}{q}$ ; $q \neq 0$ and p, q are integers
$\Rightarrow \sqrt{3} = \frac{p-2q}{q}$
p and q are integers, $\therefore p - 2q$ is an integer
$\Rightarrow \frac{p - 2q}{q}$ is a rational number
$\Rightarrow \sqrt{3}$ is a rational number which contradicts our assumption that $\sqrt{3}$ is an irrational number.
$\Rightarrow 2 + \sqrt{3}$ is an irrational number
1412 Marks · March 2023 · Standardopen ↗
Prove that $6 - \sqrt{7}$ is irrational number, given that $\sqrt{7}$ is an irrational number.
Show SolutionHide Solution
Let us assume that $6 - \sqrt{7}$ is rational
$\therefore 6 - \sqrt{7} = \frac{p}{q}$ ; $q \neq 0$ and $p, q$ are integers ($\frac{1}{2}$)
$\Rightarrow \sqrt{7} = \frac{6q - p}{q}$ ($\frac{1}{2}$)
p, q are integers, $\therefore 6q - p$ is an integer
$\Rightarrow \frac{6q - p}{q}$ is a rational number ($\frac{1}{2}$)
$\Rightarrow \sqrt{7}$ is rational number which contradicts our assumption that $\sqrt{7}$ is an irrational number
$\Rightarrow 6-\sqrt{7}$ is an irrational number ($\frac{1}{2}$)
1422 Marks · March 2024 · Standardopen ↗
Prove that $7 - 3\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
Show SolutionHide Solution
Assuming $7-3\sqrt{5}$ to be a rational number.
Let $7 - 3\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are integers & $b \neq 0$
$\Rightarrow \sqrt{5} = \frac{7b-a}{3b}$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, $7 - 3\sqrt{5}$ is an irrational number.
1432 Marks · July 2024 · Standardopen ↗
Find whether each of the following is an irrational number or a rational number.
(i) $(\sqrt{5}-3)^2$
(ii) $(5+\sqrt{3})(5-\sqrt{3})$
Show SolutionHide Solution
(i) $(\sqrt{5}-3)^2 = 8-2\sqrt{15}$
So, $(\sqrt{5}-3)^2$ is an irrational number.
(ii) $(5+\sqrt{3})(5-\sqrt{3}) = 25 - 3 = 22$
So, $(5+\sqrt{3})(5-\sqrt{3})$ is a rational number.
1442 Marks · March 2024 · Standardopen ↗
Prove that $5-2\sqrt{3}$ is an irrational number. It is given that $\sqrt{3}$ is an irrational number.
Show SolutionHide Solution
Assuming $5 - 2\sqrt{3}$ to be a rational number.
Let $5 - 2\sqrt{3}= \frac{a}{b}$ where $a$ and $b$ are integers & $b\neq 0$
$\Rightarrow \sqrt{3} = \frac{5b-a}{2b}$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, $5 - 2\sqrt{3}$ is an irrational number.
1452 Marks · July 2025 · Standardopen ↗
If $\sqrt{7}$ is an irrational number, then prove that $2\sqrt{7}$ is also an irrational number.
Show SolutionHide Solution
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.
1462 Marks · March 2026 · Standardopen ↗
Prove that $2 + 3\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational number.
Show SolutionHide Solution
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)
1472 Marks · March 2026 · Standardopen ↗
Prove that $2 - 5\sqrt{3}$ is an irrational number given that $\sqrt{3}$ is irrational.
Show SolutionHide Solution
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)
1482 Marks · March 2026 · Standardopen ↗
Prove that $4 - 2\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational.
Show SolutionHide Solution
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)
1492 Marks · March 2026 · Standardopen ↗
Prove that $14 - 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.
Show SolutionHide Solution
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
1503 Marks · 🔁 March 2023 & July 2023 & March 2024 & July 2024 & March 2025 & March 2026 · Basicopen ↗
Prove that $\sqrt{5}$ is an irrational number.
Show SolutionHide Solution
(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]
1513 Marks · 🔁 March 2023 & March 2024 & July 2024 & March 2025 & March 2026 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
Show SolutionHide Solution
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.
1523 Marks · 🔁 March 2023 & March 2025 & March 2026 · Basicopen ↗
Prove that $\sqrt{2}$ is an irrational number.
Show SolutionHide Solution
Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{p}{q}$ [$\frac{1}{2}$ mark]
($p$ and $q$ are co-prime numbers, $q \neq 0$)
$\sqrt{2}q = p \Rightarrow 2q^2 = p^2$
$2$ divides $p^2 \Rightarrow 2$ divides $p$ as well [$1$ mark]
$p = 2m$ (for some integer $m$)
$2q^2 = 4m^2 \Rightarrow q^2 = 2m^2$
$2$ divides $q^2 \Rightarrow 2$ divides $q$ as well
$p$ and $q$ have a common factor $2$ which is a contradiction as $p$ and $q$ are co-prime. [$1$ mark]
$\therefore$ our assumption is wrong
Hence, $\sqrt{2}$ is an irrational number [$\frac{1}{2}$ mark]
1533 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 $$\begin{aligned}& \& q \neq 0 \\ & \sqrt{3} = \frac{2q-5p}{q} \\ & \text{Here RHS is rational but LHS is irrational.} \\ & \text{Therefore our assumption is wrong.} \\ & \text{Hence } \frac{2-\sqrt{3}}{5} \text{ is an irrational number.}\end{aligned}$$
1543 Marks · March 2024 · Standardopen ↗
Prove that $(\sqrt{2} + \sqrt{3})^2$ is an irrational number, given that $\sqrt{6}$ is an irrational number.
Show SolutionHide Solution
$(\sqrt{2}+\sqrt{3})^2 = 2+3+2\sqrt{6} = 5 + 2\sqrt{6}$
Let us assume, to the contrary, that $5 + 2\sqrt{6}$ is rational
$\therefore 5 + 2\sqrt{6} = \frac{a}{b}$; $a, b$ are integers, $b \neq 0$
$\sqrt{6} = \frac{a-5b}{2b}$
RHS is a rational number, whereas LHS is an irrational number.
$\therefore$ Our assumption is wrong.
$\Rightarrow 5 + 2\sqrt{6} = (\sqrt{2} + \sqrt{3})^2$ is an irrational number
1553 Marks · March 2025 · Standardopen ↗
Prove that $\frac{1}{\sqrt{5}}$ is an irrational number.
Show SolutionHide Solution
Let $\frac{1}{\sqrt{5}}$ be a rational number. $\therefore \frac{1}{\sqrt{5}} = \frac{p}{q}$, where $q \neq 0$ and $p, q$ are co-primes. $5p^2 = q^2 \implies q^2$ is divisible by $5 \implies q$ is divisible by $5$. Let $q = 5a$. $25a^2 = 5p^2 \implies p^2 = 5a^2 \implies p^2$ is divisible by $5 \implies p$ is divisible by $5$. This leads to contradiction as $p, q$ are co-primes. $\therefore \frac{1}{\sqrt{5}}$ is an irrational number.
1563 Marks · March 2025 · Standardopen ↗
This section has $6$ Short Answer (SA) type questions carrying $3$ marks each.
Prove that $\sqrt{5}$ is an irrational number.
Show SolutionHide Solution
Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and let p & q are co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5$
$\Rightarrow p$ is divisible by $5$----- (i)
$\Rightarrow$let $p = 5a$, where 'a' is some integer
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by $5$.
$\Rightarrow q$ is divisible by $5$. ----- (ii)
(i) and (ii) leads to contradiction as p and q are coprimes.
$\therefore \sqrt{5}$ is an irrational number
1573 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.
1583 Marks · March 2025 · Standardopen ↗
Prove that $\left(4\sqrt{2} + \frac{5}{3}\right)$ is an irrational number given that $\sqrt{2}$ is an irrational number.
Show SolutionHide Solution
Let $4\sqrt{2} + \frac{5}{3}$ be a rational number.
$\therefore 4\sqrt{2} + \frac{5}{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $$\begin{aligned}& b \neq 0 \\ & 4\sqrt{2} = \frac{a}{b} - \frac{5}{3} \\ & sqrt{2} = \frac{3a - 5b}{12b} \\ & 3a - 5b\end{aligned}$$ and $12b$ are integers.
$\therefore$ RHS is rational.
But LHS = $\sqrt{2}$ is an irrational number which is contradiction to our supposition.
Hence $4\sqrt{2} + \frac{5}{3}$ is an irrational number.
1593 Marks · March 2026 · Standardopen ↗
Prove that $\frac{2 + 3\sqrt{5}}{7}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
Show SolutionHide Solution
Let $\frac{2 + 3\sqrt{5}}{7}$ be a rational number. ($\frac{1}{2}$ Mark)
Then $\frac{2 + 3\sqrt{5}}{7} = \frac{p}{q}$, where $q \neq 0$ and $p$ and $q$ are integers. ($\frac{1}{2}$ Mark)
$\Rightarrow 2 + 3\sqrt{5} = \frac{7p}{q}$ (1 Mark)
$\Rightarrow \sqrt{5} = \frac{7p - 2q}{3q}$ ($\frac{1}{2}$ Mark)
Since '$p$' and '$q$' are integers.
$\therefore \frac{7p - 2q}{3q}$ is rational. ($\frac{1}{2}$ Mark)
But this contradicts the fact that $\sqrt{5}$ is irrational.
Hence, $\frac{2 + 3\sqrt{5}}{7}$ is an irrational number.
1603 Marks · March 2025 · Basicopen ↗
State the "Fundamental Theorem of Arithmetic" and use it to find LCM of $36$ and $54$.
Show SolutionHide Solution
Statement: "Every composite number can be factorized as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur." [1 mark]
$36 = 2^2 \times 3^2$ [1/2 mark]
$54 = 2 \times 3^3$ [1/2 mark]
$\text{LCM}(36, 54) = 2^2 \times 3^3$ or $108$ [1 mark]
1613 Marks · March 2025 · Basicopen ↗
Find which among the following numbers $a, b$ and $c$ is/are composite numbers.
$a = 7 \times 11 \times 13 + 13$
$b = 6 \times 5 \times 4 + 4$
$c = 7 \times 13 + 6$
Show SolutionHide Solution
$a$ and $b$ are only composite numbers. [$3$ marks]
1623 Marks · March 2025 · Basicopen ↗
Given that $\sqrt{5}$ is an irrational number, prove that $2 + 3\sqrt{5}$ is an irrational number.
Show SolutionHide Solution
Let $2 + 3\sqrt{5} = a$, where '$a$' is a rational number
$\Rightarrow \sqrt{5} = \frac{a - 2}{3}$
Here L.H.S. is an irrational number but R.H.S. is a rational number
So, our assumption is wrong
Hence, $2 + 3\sqrt{5}$ is an irrational number
1633 Marks · March 2025 · Basicopen ↗
Prove that $\frac{\sqrt{3}+5}{2}$ is an irrational number, where it is given that $\sqrt{3}$ is irrational.
Show SolutionHide Solution
Let $\frac{\sqrt{3}+5}{2} = a$, where '$a$' is a rational number
$\implies \sqrt{3} = 2a - 5$
Here LHS is an irrational number but RHS is a rational number
So, our assumption is wrong
Hence, $\frac{\sqrt{3}+5}{2}$ is an irrational number
1643 Marks · March 2026 · Basicopen ↗
This section comprises $6$ Short Answer (SA) type questions of $3$ marks each.
Given that $\sqrt{5}$ is an irrational number, prove that $3+2\sqrt{5}$ is also an irrational number.
Show SolutionHide Solution
Let $3 + 2\sqrt{5} = r$ be a rational number (1/2 Mark)
So $\sqrt{5} = \frac{r-3}{2}$ (1 Mark)
RHS is a rational number
So LHS is a rational number which is a contradiction (1 Mark)
Hence $3 + 2\sqrt{5}$ is an irrational number. (1/2 Mark)
1653 Marks · March 2026 · Basicopen ↗
Given that $\sqrt{2}$ is an irrational number, prove that $5-2\sqrt{2}$ is also an
irrational number.
Show SolutionHide Solution
Let $5 - 2\sqrt{2} = r$ be a rational number (1/2 Mark)
So $\sqrt{2} = \frac{5-r}{2}$ (1 Mark)
RHS is a rational number (1 Mark)
So LHS is a rational number which is a contradiction (1/2 Mark)
Hence, $5 - 2\sqrt{2}$ is an irrational number.
1663 Marks · March 2026 · Basicopen ↗
Given that $\sqrt{3}$ is an irrational number, prove that $2-5\sqrt{3}$ is also an irrational number.
Show SolutionHide Solution
Let $2 - 5\sqrt{3} = r$ be a rational number (1/2 Mark)
So, $\sqrt{3} = \frac{2-r}{5}$ (1 Mark)
RHS is a rational number
So LHS is a rational number which is a contradiction (1 Mark)
Hence, $2 - 5\sqrt{3}$ is an irrational number. (1/2 Mark)
1673 Marks · March 2026 · Basicopen ↗
Neha claimed that there does not exist any irrational number between 1 and 2. Raunak claimed that $\sqrt{2}$ lies between 1 and 2 and $\sqrt{2}$ is an irrational number. Who do you think is correct? Justify by proving either $\sqrt{2}$ as an irrational number or otherwise.
Show SolutionHide Solution
Raunak is correct (1/2 Mark)
Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are coprime and $b \neq 0$ (1/2 Mark)
$\sqrt{2}b = a$
$2b^2 = a^2$
2 divides $a^2$
2 divides $a$ as well. (1/2 Mark)
Let $a = 2p$ (for some integer $p$)
$a^2 = 4p^2$
$2b^2 = 4p^2$
$b^2 = 2p^2$
2 divides $b^2$
2 divides $b$ as well (1 Mark)
$\therefore$ 2 is a common factor of $a$ and $b$ which is a contradiction as $a$ and $b$ are coprime.
$\therefore$ Our assumption is wrong. Hence, $\sqrt{2}$ is an irrational number. (1/2 Mark)

Word Problem

1 Mark Questions
1681 Mark · July 2023 · Standardopen ↗
The greatest number which divides both $83$ and $138$, leaving remainders $5$ and $8$ respectively, is :
  • (a)$13$
  • (b)$65$
  • (c)$26$
  • (d)$39$
Show SolutionHide Solution
(c) $26$
1691 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) $20$
1701 Mark · March 2024 · Standardopen ↗
The greatest number which divides $281$ and $1249$, leaving remainder $5$ and $7$ respectively, is :
  • (a)$23$
  • (b)$276$
  • (c)$138$
  • (d)$69$
Show SolutionHide Solution
(C) $138$
1711 Mark · March 2025 · Standardopen ↗
The LCM of the smallest prime number and the smallest odd composite number is:
  • (a)$10$
  • (b)$6$
  • (c)$9$
  • (d)$18$
Show SolutionHide Solution
(D) $18$
1721 Mark · March 2025 · Standardopen ↗
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 SolutionHide Solution
(C) any odd number
1731 Mark · March 2025 · Standardopen ↗
The greatest number which divides $70$ and $125$, leaving remainders $5$ and $8$ respectively, is:
  • (a)$13$
  • (b)$65$
  • (c)$875$
  • (d)$1750$
Show SolutionHide Solution
(A) $13$
1741 Mark · March 2025 · Standardopen ↗
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 SolutionHide Solution
The correct option is not available in the given options. Full marks may be awarded to every attempt.
2 Marks Questions
1752 Marks · March 2023 · Standardopen ↗
Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.
Show SolutionHide Solution
We have to find HCF of $85 – 1 = 84$ and $72 – 2 = 70$.
HCF of 84 and 70 = 14
1762 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 $48 + 7 = 55$.
1772 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 $= 936$.
1782 Marks · July 2024 · Standardopen ↗
Find the smallest $4$-digit number exactly divisible by $15$, $24$ and $36$.
Show SolutionHide Solution
LCM $(15, 24, 36) = 360$
Therefore, the smallest $4$-digit number which is a multiple of $360$ is $360 \times 3 = 1080$ which is divisible by $15, 24 \& 36$.
1792 Marks · March 2024 · Standardopen ↗
Three bells toll at intervals of $9$, $12$ and $15$ minutes respectively. If they start tolling together, after what time will they next toll together?
Show SolutionHide Solution
$$\begin{aligned}& 9 = 3^2 \\ & 12 = 2^2 \times 3 \\ & 15 = 3 \times 5 \\ & L.C.M = 2^2 \times 3^2 \times 5 = 180 \\ & \text{Three bells will toll together after } 180 \text{ min.}\end{aligned}$$
1802 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 $240$.
1812 Marks · March 2024 · Standardopen ↗
Find the smallest number that is divisible by each of $8, 9$ and $10$.
Show SolutionHide Solution
$$\begin{aligned}& 8 = 2^3 \\ & 9 = 3^2 \\ & 10 = 2 \times 5 \\ & LCM (8, 9, 10) = 2^3 \times 3^2 \times 5 = 360\ \therefore\end{aligned}$$ smallest number divisible by each $8, 9$ and $10$ is $360$.
1822 Marks · March 2025 · Standardopen ↗
Find the smallest number which is divisible by both $644$ and $462$.
Show SolutionHide Solution
$462 = 2 \times 3 \times 7 \times 11$
$644 = 2^2 \times 7 \times 23$
$\operatorname{LCM}(462, 644) = 2^2 \times 3 \times 7 \times 11 \times 23 = 21252$
$\therefore$ Smallest number which is divisible by both $462$ and $644$ is $21252$
1832 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.
Show SolutionHide Solution
$4$ 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
1842 Marks · March 2025 · Basicopen ↗
Find the greatest number which divides $285$ and $1249$ leaving remainders $9$ and $7$ respectively.
Show SolutionHide Solution
$285 - 9 = 276$ , $1249 - 7 = 1242$
$HCF(276, 1242) = 138$
Greatest number which divides $276$ and $1242$ is $138$
3 Marks Questions
1853 Marks · March 2023 · Standardopen ↗
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?
Show SolutionHide Solution
LCM of $6, 12, 18 = 36$
So, all the three bells ring together after $36$ minutes at $6:36$ AM
1863 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.
1873 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 $= \frac{42}{14} + \frac{56}{14} + \frac{70}{14}$
$= 3 + 4 + 5 = 12$
1883 Marks · March 2024 · Standardopen ↗
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.
Show SolutionHide Solution
Minimum number of rooms required means there should be maximum number of teachers in a room. We have to find HCF of $48, 80$ and $144$.
$48 = 2^4 \times 3$
$80 = 2^4 \times 5$
$144 = 2^4 \times 3^2$
HCF $(48, 80, 144) = 2^4 = 16$
Therefore, total number of rooms required = $\frac{48}{16} + \frac{80}{16} + \frac{144}{16} = 17$
1893 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 = $\frac{180}{15} + \frac{180}{18} + \frac{180}{20}$
$= 31$
1903 Marks · March 2025 · Standardopen ↗
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.
Show SolutionHide Solution
$144 = 2^4 \times 3^2$
$180 = 2^2 \times 3^2 \times 5$
$192 = 2^6 \times 3$
HCF = $2^2 \times 3 = 12$ (1/2)
Number of stacks of Physics Books = $\frac{144}{12} = 12$ (1/2)
Number of stacks of Chemistry Books = $\frac{180}{12} = 15$ (1)
Number of stacks of Mathematics Books = $\frac{192}{12} = 16$ (1)
1913 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)
1923 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 $108$.
1933 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)
1943 Marks · March 2025 · Basicopen ↗
Find the smallest number which when increased by $20$, is exactly divisible by $72, 90$ and $150$.
Show SolutionHide Solution
$72 = 2^3 \times 3^2, 90 = 3^2 \times 2 \times 5, 150 = 5^2 \times 2 \times 3$
$LCM (72, 90, 150) = 2^3 \times 3^2 \times 5^2 = 1800$
Required smallest number is $1800 - 20 = 1780$
1953 Marks · March 2025 · Basicopen ↗
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?
Show SolutionHide Solution
$48 = 2^4 \times 3, 52 = 2^2 \times 13, 56 = 2^3 \times 7$
LCM $= 2^4 \times 3 \times 13 \times 7 = 4368$
$\Rightarrow$ Minimum distance each walked in complete steps ten times $= 43680$ cm
4 Marks Questions
1964 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
1974 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) $5$
1984 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
1991 Mark · 🔁 March 2023 & July 2023 · Standardopen ↗
Assertion (A): A fair die is thrown once. The probability of getting a prime number is $\frac{1}{2}$.
Reason (R): A natural number is a prime number if it has only two factors.
  • (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 false, but Reason (R) is true.
  • (d)Assertion (A) is false, but Reason (R) is true.
Show SolutionHide Solution
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2001 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
2013 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, $x + 12 = \frac{160}{x}$
$x^2 + 12x = 160$
$x^2 + 12x - 160 = 0$
$(x + 20)(x - 8) = 0$
$x \neq -20, x= 8$
$\Rightarrow \text{Required natural number is } 8$