Real Numbers — Class 10 Maths PYQs

131 previous-year board questions (2023–2025) 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.

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)$9$
  • (c)$10$
  • (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) $5^{10}$

Find lcm & hcf

1 Mark Questions
81 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
91 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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Sol. (a) $1:2$
101 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) $p^5q^4$
111 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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Sol. (D) $4$
121 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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Sol. (d) $180 a^3b^4$
131 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)$165$
  • (c)$1600$
  • (d)$1625$
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(A) $1650$
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 ↗
Given HCF $(2520, 6600) = 40$, LCM $(2520, 6600) = 252 \times k$, then the value of $k$ is:
  • (a)$1650$
  • (b)$1600$
  • (c)$165$
  • (d)$1625$
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(A) $1650$
171 Mark · March 2024 · Standardopen ↗
Given HCF $(2520, 6600) = 40$, LCM $(2520, 6600) = 252 \times k$, then the value of $k$ is:
  • (a)$1650$
  • (b)$165$
  • (c)$1600$
  • (d)$1625$
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(A) $1650$
181 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)$40$
  • (c)$13$
  • (d)$8$
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(B) $13$
191 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) $924$
201 Mark · March 2024 · Standardopen ↗
If the product of two co-prime numbers is $553$, then their HCF is :
  • (a)$1$
  • (b)$7$
  • (c)$553$
  • (d)$79$
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(A) $1$
211 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) $2^3 \times 3^2 \times 5$
221 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
231 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) $98$
241 Mark · March 2025 · Standardopen ↗
If HCF($98$, $28$) = $m$ and LCM($98$, $28$) = $n$, then the value of $n-7m$ is:
  • (a)$0$
  • (b)$28$
  • (c)$98$
  • (d)$198$
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(C) $98$
251 Mark · March 2025 · Standardopen ↗
If HCF($98$, $28$) = m and LCM($98$, $28$) = n, then the value of $n-7m$ is:
  • (a)$0$
  • (b)$28$
  • (c)$98$
  • (d)$198$
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(C) $98$
261 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) $10$
271 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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Sol. (A) $p = 35x$
281 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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Sol. (A) $p = 35x$
291 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) $ab(1 - ab)(1 + ab)$
301 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
311 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
321 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) $17$
331 Mark · March 2025 · Standardopen ↗
Directions: In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). 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.
341 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.
351 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).
361 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) $7$, $462$
2 Marks Questions
372 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 = $2520$
382 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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Sol. Let the numbers be $2x, 3x$
LCM = $6x = 180 \Rightarrow x = 30$
$\therefore$ Numbers are $60, 90$
HCF $(60, 90) = 30$
392 Marks · March 2023 · Standardopen ↗
Using prime factorisation, find HCF and LCM of $96$ and $120$.
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$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3$
$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$
$HCF = 24$
$LCM = 480$
402 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
412 Marks · March 2023 · Standardopen ↗
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=$360$
422 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$=360$
432 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, $6z - 9 = 255$
$\Rightarrow z = 44$
442 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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Sol. Let the two numbers be $4x$ and $5x$ where $x$ is common factor
Now HCF $= 11$
$\therefore x = 11$
Numbers are $44$ and $55$
$\operatorname{LCM}(44,55) = \frac{44 \times 55}{11} = 220$
452 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
3 Marks Questions
463 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 = $3 \times 5 \times 101 = 1515$
473 Marks · March 2023 · Standardopen ↗
Find the HCF and LCM of $26, 65$ and $117$, using prime factorisation.
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$26= 13 \times 2$
$65=13 \times 5$
$117=13 \times 3 \times 3$
$\therefore HCF = 13$
$LCM = 13 \times 2 \times 3 \times 5 \times 3 = 1170$
483 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 $(p, q, r) \neq p \times q \times r$

Unit digit 0

1 Mark Questions
491 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)$(5\times3)^n$
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(b) $(2 \times 5)^n$
501 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
511 Mark · March 2023 · Standardopen ↗
Questions number $19$ and $20$ are Assertion and Reason based questions carrying $1$ mark each. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). 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.
Assertion (A): The number $5^n$ cannot end with the digit $0$, where $n$ is a natural number.
Reason (R): Prime factorisation of $5$ has only two factors, $1$ and $5$.
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(c) Assertion (A) is true, but Reason (R) is false
521 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
531 Mark · March 2025 · Standardopen ↗
Directions:
In question number $19$ and $20$, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option :
(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.
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 $n$.
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(d) Assertion (A) is false but Reason (R) is true
541 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 $n$.
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(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
551 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
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(C) 0
2 Marks Questions
562 Marks · July 2023 · Standardopen ↗
Check whether $6^n$ can end with the digit $0$ for any natural number $n$.
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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 $0$.
572 Marks · March 2023 · Standardopen ↗
Prove that $4^n$ can never end with digit $0$, where $n$ is a natural number.
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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.
582 Marks · March 2023 · Standardopen ↗
Show that $6^n$ can not end with digit $0$ for any natural number 'n'.
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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.
592 Marks · July 2024 · Standardopen ↗
Check whether there is any natural number 'n' for which $(14)^n$ ends with the digit '0' or '5'.
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$(14)^n = 2^n7^n$
as it does not have $5$ as the prime factor
Hence, it cannot end with the digit $0$ or $5$
602 Marks · July 2025 · Standardopen ↗
Show that $14^n$ cannot end with the digit $0$ or $5$ for any natural number $n$.
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$14^n = 2^n \times 7^n$
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 $5$.

Prime number

2 Marks Questions
612 Marks · March 2024 · Standardopen ↗
Show that the number $5\times 11\times 17+3\times 11$ is a composite number.
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Sol. $5 \times 11 \times 17 + 3 \times 11 = 11 \times (5 \times 17 + 3)$
$= 11\times 88$ 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.
622 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 ($\frac{1}{2}$)
3 Marks Questions
633 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.
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(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.
643 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.
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$p \cdot q \cdot r + q = q(pr + 1)$. 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
651 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
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(D) Assertion (A) is false but Reason (R) is true
661 Mark · March 2023 · Standardopen ↗
If $p^2 = \frac{32}{50}$, then $p$ is a/an
  • (a)whole number
  • (b)rational number
  • (c)integer
  • (d)irrational number
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(C) rational
671 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)$(\sqrt{36}, \sqrt{2})$
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(C) $(\sqrt{3}, \sqrt{27})$
681 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)$5$
  • (c)$\sqrt{2}$
  • (d)$\sqrt{5}$
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Sol.
(D) $\sqrt{5}$
691 Mark · July 2025 · Standardopen ↗
$2.35$ is:
  • (a)an integer
  • (b)a rational number
  • (c)an irrational number
  • (d)a natural number
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(B) a rational number
701 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.
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(c) a positive irrational number
711 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.
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(c) a positive irrational number
721 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
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(A) positive rational number
731 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)$a^{1/8}$
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(A) $a$
741 Mark · March 2025 · Standardopen ↗
$\sqrt{0.4}$ is a/an
  • (a)natural number
  • (b)integer
  • (c)rational number
  • (d)irrational number
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(D) irrational number
2 Marks Questions
752 Marks · July 2023 · Standardopen ↗
If $\sqrt{2}$ is given as an irrational number, then prove that $(5-2\sqrt{2})$ is an irrational number.
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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.
762 Marks · March 2023 · Standardopen ↗
Prove that $2 + \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
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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
772 Marks · March 2023 · Standardopen ↗
Prove that $6 - \sqrt{7}$ is irrational number, given that $\sqrt{7}$ is an irrational number.
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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 ($\frac{1}{2}$)
782 Marks · March 2024 · Standardopen ↗
Prove that $7 - 3\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
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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.
792 Marks · July 2024 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3}= \frac{p}{q}$, where $q \neq 0$ and p & q are coprime.
$3q^2 = p^2 \Rightarrow p^2$ is divisible by $3 \Rightarrow p$ is divisible by $3$ ----- (i)
$\Rightarrow p = 3a$, where ‘a' is some integer
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $3$ ----- (ii)
(i) and (ii) leads to contradiction as ‘p' and ‘q' are coprime.
$\therefore \sqrt{3}$ is an irrational number.
802 Marks · July 2024 · Standardopen ↗
Find whether each of the following is an irrational number or a rational number.
(i) $(\sqrt{5}-3)^2$
(ii) $(5+\sqrt{3})(5-\sqrt{3})$
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Sol. (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.
812 Marks · March 2024 · Standardopen ↗
Prove that $5-2\sqrt{3}$ is an irrational number. It is given that $\sqrt{3}$ is an irrational number.
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Sol. 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.
822 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.
3 Marks Questions
833 Marks · July 2023 · Standardopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q\neq0$ and let $p$ & $q$ be co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5 \Rightarrow p$ is divisible by $5$
$\Rightarrow p = 5a$, 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 $5$
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{5}$ is an irrational number.
843 Marks · March 2023 · Standardopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q\neq0$ and let $p \& q$ be co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5 \Rightarrow p$ is divisible by $5$
$\Rightarrow p = 5a$, where ‘$a$' is some integer ----- (i)
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by $5 \Rightarrow q$ is divisible by $5$
$\Rightarrow q = 5b$, where '$b$' is some integer ----- (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{5}$ is an irrational number.
853 Marks · March 2023 · Standardopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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Sol. Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q\neq0$ and let $p \& q$ be co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5 \Rightarrow p$ is divisible by $5$
$\Rightarrow p = 5a$, where ‘a' is some integer
dots (i)
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by $5 \Rightarrow q$ is divisible by $5$
$\Rightarrow q = 5b$, where ‘b' is some integer
dots (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{5}$ is an irrational number.
863 Marks · March 2023 · 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.
873 Marks · March 2023 · Standardopen ↗
Prove that $\sqrt{2}$ is an irrational number.
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Let $\sqrt{2}$ be a rational number.
$\therefore \sqrt{2} = \frac{p}{q}$, where $q\neq0$ and let $p \& q$ be co-primes.
$2q^2 = p^2 \Rightarrow p^2$ is divisible by $2 \Rightarrow p$ is divisible by $2$
$\Rightarrow p = 2a$, where 'a' is some integer ----- (i)
$4a^2 = 2q^2 \Rightarrow q^2 = 2a^2 \Rightarrow q^2$ is divisible by $2 \Rightarrow q$ is divisible by $2$
$\Rightarrow q = 2b$, where 'b' is some integer ----- (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{2}$ is an irrational number.
883 Marks · July 2024 · Standardopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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Sol. Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{a}{b}$, where $a$, $b$ are coprime and $b \neq 0$.
$\Rightarrow a^2 = 5b^2 \Rightarrow a^2$ is divisible by $5$.
$\Rightarrow a$ is divisible by $5$. ----- (i)
Let $a = 5m$, where 'm' is any natural number.
$\Rightarrow b^2 = 5m^2 \Rightarrow b^2$ is divisible by $5$.
$\Rightarrow b$ is divisible by $5$. ----- (ii)
From (i) and (ii), $a$ and $b$ have common factors which is contrary to our assumption.
Hence, $\sqrt{5}$ is an irrational number.
893 Marks · March 2024 · Standardopen ↗
Prove that $\frac{2-\sqrt{3}}{5}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
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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 $$\begin{aligned}& \& q \neq 0 \\ & \sqrt{3} = \frac{2q-5p}{q} \\ & \text{Here RHS is rational but LHS is irrational.} \\ & \text{Therefore our assumption is wrong.} \\ & \text{Hence } \frac{2-\sqrt{3}}{5} \text{ is an irrational number.}\end{aligned}$$
903 Marks · March 2024 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3} = \frac{p}{q}$, where $q\neq 0$ and $p \& q$ are coprime.
$3q^2 = p^2 \Rightarrow p^2$ is divisible by $3$
$\Rightarrow p$ is divisible by $3----- (i)$
$\Rightarrow p = 3a$, where 'a' is a positive integer
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3$
$\Rightarrow q$ is divisible by $3 ----- (ii)$
(i) and (ii) leads to contradiction as 'p' and 'q' are coprime.
$\therefore \sqrt{3}$ is an irrational number.
913 Marks · March 2024 · Standardopen ↗
Prove that $(\sqrt{2} + \sqrt{3})^2$ is an irrational number, given that $\sqrt{6}$ is an irrational number.
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$(\sqrt{2}+\sqrt{3})^2 = 2+3+2\sqrt{6} = 5 + 2\sqrt{6}$
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
923 Marks · March 2024 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3} = \frac{p}{q}$, where $q \neq 0$ and $p \& q$ are coprime.
$3q^2 = p^2 \Rightarrow p^2$ is divisible by $3 \Rightarrow p$ is divisible by $$\begin{aligned}& 3 \dots (i) \\ & \Rightarrow p = 3a\end{aligned}$$, where '$a$' is some integer
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $$\begin{aligned}& 3 \dots (ii) \\ & (i)\end{aligned}$$ and $(ii)$ leads to contradiction as 'p' and 'q' are coprime.
$\therefore \sqrt{3}$ is an irrational number.
933 Marks · March 2025 · Standardopen ↗
Prove that $\frac{1}{\sqrt{5}}$ is an irrational number.
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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.
943 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.
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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
953 Marks · March 2025 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3} = \frac{p}{q}$, where $q \neq 0$ and let $p \& q$ be coprimes.
$\Rightarrow 3q^2 = p^2$
$p^2$ is divisible by $3$.
$\Rightarrow p$ is divisible by $3$. ----- (1)
Let $p = 3a$, where 'a' is some integer
$\therefore 9a^2 = 3q^2$
$\Rightarrow q^2 = 3a^2$
$\Rightarrow q^2$ is divisible by $3$
$\therefore q$ is divisible by $3$. ----- (2)
(1) and (2) leads to contradiction as $p$ and $q$ are coprimes.
Hence, $\sqrt{3}$ is an irrational number.
963 Marks · March 2025 · Standardopen ↗
Prove that $(5\sqrt{3} + \frac{2}{3})$ is an irrational number given that $\sqrt{3}$ is an irrational number.
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Sol. 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.
973 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.
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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.
983 Marks · March 2025 · Standardopen ↗
Prove that $\sqrt{2}$ is an irrational number.
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Let $\sqrt{2}$ be a rational number.
$\therefore \sqrt{2} = \frac{p}{q}$, where $q \neq 0$ and let $p$ & $q$ be co-primes.
$2q^2 = p^2 \Rightarrow p^2$ is divisible by $2 \Rightarrow p$ is divisible by $2$ -----(i)
$\Rightarrow p = 2a$, where 'a' is some integer
$4a^2 = 2 q^2 \Rightarrow q^2 = 2 a^2 \Rightarrow q^2$ is divisible by $2 \Rightarrow q$ is divisible by $2$ -----(ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{2}$ is an irrational number.
993 Marks · March 2025 · Standardopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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Let $\sqrt{5}$ be a rational number. $\therefore \sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and let $p$ & $q$ be the coprimes ($\frac{1}{2}$ mark). $\Rightarrow 5q^2 = p^2 \Rightarrow p^2$ is divisible by 5 $\Rightarrow p$ is divisible by 5 (1 mark). Let $p = 5a$, where 'a' is some integer $\therefore 25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by 5 $\Rightarrow q$ is divisible by 5 (1 mark). $\therefore 5$ divides both $p$ & $q$. ① and ② leads to contradiction as $p$ and $q$ are coprimes. Hence, $\sqrt{5}$ is an irrational number ($\frac{1}{2}$ mark).

Word Problem

1 Mark Questions
1001 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)$39$
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Ans. (c) $26$
1011 Mark · March 2023 · Standardopen ↗
The LCM of smallest $2$-digit number and smallest composite number is
  • (a)$12$
  • (b)$20$
  • (c)$4$
  • (d)$40$
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(C) $20$
1021 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)$69$
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(C) $138$
1031 Mark · March 2025 · Standardopen ↗
The LCM of the smallest prime number and the smallest odd composite number is:
  • (a)$10$
  • (b)$9$
  • (c)$6$
  • (d)$18$
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(D) $18$
1041 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
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(C) any odd number
1051 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)$1750$
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(A) $13$
1061 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)$2400$
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The correct option is not available in the given options. Full marks may be awarded to every attempt.
2 Marks Questions
1072 Marks · March 2023 · Standardopen ↗
Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.
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We have to find HCF of $85 – 1 = 84$ and $72 – 2 = 70$.
HCF of 84 and 70 = 14
1082 Marks · March 2023 · Standardopen ↗
Find the least number which when divided by 12, 16 and 24 leaves remainder 7 in each case
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LCM of 12, 16, 24 = 48
Required number is $48 + 7 = 55$.
1092 Marks · March 2023 · Standardopen ↗
Find the greatest $3$-digit number which is divisible by $18, 24$ and $36$.
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LCM of $18, 24, 36$ is $72$
Required greatest $3$-digit number $= 936$.
1102 Marks · July 2024 · Standardopen ↗
Find the smallest $4$-digit number exactly divisible by $15$, $24$ and $36$.
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Sol. 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 $15, 24 \& 36$.
1112 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?
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$$\begin{aligned}& 9 = 3^2 \\ & 12 = 2^2 \times 3 \\ & 15 = 3 \times 5 \\ & L.C.M = 2^2 \times 3^2 \times 5 = 180 \\ & \text{Three bells will toll together after } 180 \text{ min.}\end{aligned}$$
1122 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.
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$40 = 2^3 \times 5$
$48 = 2^4 \times 3$
L.C.M. $(40, 48) = 240$
Minimum number of books required in library is $240$.
1132 Marks · March 2024 · Standardopen ↗
Find the smallest number that is divisible by each of $8, 9$ and $10$.
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$$\begin{aligned}& 8 = 2^3 \\ & 9 = 3^2 \\ & 10 = 2 \times 5 \\ & LCM (8, 9, 10) = 2^3 \times 3^2 \times 5 = 360 \\ & \therefore\end{aligned}$$ smallest number divisible by each $8, 9$ and $10$ is $360$.
1142 Marks · March 2025 · Standardopen ↗
Find the smallest number which is divisible by both $644$ and $462$.
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Sol. $462 = 2 \times 3 \times 7 \times 11$
$644 = 2^2 \times 7 \times 23$
$\operatorname{LCM}(462, 644) = 2^2 \times 3 \times 7 \times 11 \times 23 = 21252$
$\therefore$ Smallest number which is divisible by both $462$ and $644$ is $21252$
3 Marks Questions
1153 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?
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LCM of $6, 12, 18 = 36$
So, all the three bells ring together after $36$ minutes at $6:36$ AM
1163 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?
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$LCM = 432$
i.e. $\frac{432}{60} = 7$ min $12$ sec.
$\Rightarrow$ traffic lights will change simultaneously again at $7 : 7 : 12$ a.m.
1173 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.
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HCF $(42, 56, 70) = 14$
Minimum number of tables required $= \frac{42}{14} + \frac{56}{14} + \frac{70}{14}$
$= 3 + 4 + 5 = 12$
1183 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.
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Sol. 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 = $\frac{48}{16} + \frac{80}{16} + \frac{144}{16} = 17$
1193 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?
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LCM $(15, 18, 20) = 180$
They will start new sweater again after $180$ days.
Total number of sweaters completed in $180$ days = $\frac{180}{15} + \frac{180}{18} + \frac{180}{20}$
$= 31$
1203 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.
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$144 = 2^4 \times 3^2$
$180 = 2^2 \times 3^2 \times 5$
$192 = 2^6 \times 3$
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)
4 Marks Questions
1214 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.
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(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
1224 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 ?
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Sol.
$173250= 2 \times 5^3 \times 3^2 \times 7 \times 11$
(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) $5$
1234 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.
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(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
1241 Mark · July 2023 · Standardopen ↗
Questions number $19$ and $20$ are Assertion and Reason based questions carrying $1$ mark each. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). 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).
(e) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
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.
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Sol. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
1251 Mark · March 2023 · Standardopen ↗
Assertion - Reason Based Questions : In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option out of the following :
(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.
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.
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(A)
1261 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.
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(A)
1271 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.
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(A)
3 Marks Questions
1283 Marks · March 2023 · Standardopen ↗
A natural number, when increased by $12$, equals $160$ times its reciprocal. Find the number.
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Let the natural number be $x$
ATQ, $x + 12 = \frac{160}{x}$
$x^2 + 12x = 160$
$x^2 + 12x - 160 = 0$
$(x + 20)(x - 8) = 0$
$x \neq -20, x= 8$
$\Rightarrow \text{Required natural number is } 8$
1293 Marks · March 2024 · Standardopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{p}{q}$, where $q \ne 0$ and let $p$ & $q$ be co-prime.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by $5 \Rightarrow p$ is divisible by $5$
dots (i)
$\Rightarrow p = 5a$, 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 $5$
dots (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-prime.
$\therefore \sqrt{5}$ is an irrational number.
1303 Marks · March 2025 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3} = \frac{p}{q}$, where $q \neq 0$ and let $p \& q$ be coprimes.
$\implies 3q^2 = p^2$
$\implies p^2$ is divisible by 3.
$\implies p$ is divisible by 3. ----- (1)
Let $p = 3a$, where 'a' is some integer
$\therefore 9a^2 = 3q^2$
$\implies q^2 = 3a^2$
$\implies q^2$ is divisible by 3
$\implies q$ is divisible by 3 ----- (2)
$\therefore 3$ divides both $p \& q$.
(1) and (2) leads to contradiction as $p$ and $q$ are coprimes.
Hence, $\sqrt{3}$ is an irrational number.
1313 Marks · March 2025 · Standardopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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Let $\sqrt{5}$ be a rational number. $\therefore \sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and let $p \& q$ be co-primes. $5q^2 = p^2 \implies p^2$ is divisible by $5 \implies p$ is divisible by $5$ ----- (i) $\implies p = 5a$, where '$a$' is some integer. $25a^2 = 5q^2 \implies q^2 = 5a^2 \implies q^2$ is divisible by $5 \implies q$ is divisible by $5$ ----- (ii) (i) and (ii) leads to contradiction as '$p$' and '$q$' are co-primes. $\therefore \sqrt{5}$ is an irrational number.