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 :
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.
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.
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?
Assertion (A) : H.C.F. ($36 \text{ m}^2$, 18 m) = 18 m, where m is a prime number. Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) : For any two natural numbers a and b, the HCF of a and b is a factor of the LCM of a and b. Reason (R) : HCF of any two natural numbers divides both the numbers.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(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.
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:
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 :
Two numbers are in the ratio $4: 5$ and their HCF is $11$. Find the LCM of these numbers.
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Let the two numbers be $4x$ and $5x$ where $x$ is common factor Now HCF $= 11$ $\therefore x = 11$ Numbers are $44$ and $55$ $\operatorname{LCM}(44,55) = \frac{44 \times 55}{11} = 220$
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 $(p, q, r) \neq p \times q \times r$
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?
Three measuring rods are of lengths $120 \text{ cm}, 100 \text{ cm}$ and $150 \text{ cm}$. Find the least length of a fence that can be measured an exact number of times, using any of the rods. How many times each rod will be used to measure the length of the fence?
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$120 = 2^3 \times 3 \times 5, 100 = 2^2 \times 5^2, 150 = 2 \times 3 \times 5^2$ ($1\frac{1}{2}$ marks) $LCM(120, 100, 150) = 2^3 \times 3 \times 5^2 = 600$ (1 mark) $\therefore \text{Least length of the fence is } 600 \text{ cm.}$ $\text{Each rod is used 5, 6 and 4 times respectively}$ ($\frac{1}{2}$ mark)
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
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
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).
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).
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 $0$.
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.
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.
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 $5$.
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 $0$.
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$45^n = (3 \times 3)^n \times 5^n$ [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]
Check whether $15^n \times 2^n$, n being a natural number, ends with the digit zero.
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$15^n \times 2^n = 5^n \times 3^n \times 2^n$ $\Rightarrow 2$ and 5 both are the factors of the given number $\therefore$ The given number ends with the digit zero
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?
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$6^n = 2^n \times 3^n$ 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 $5$.
Show that the number $5\times 11\times 17+3\times 11$ is a composite number.
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$5 \times 11 \times 17 + 3 \times 11 = 11 \times (5 \times 17 + 3)$ $= 11\times 88$ 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.
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.
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).
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.
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(C) Assertion (A) is true, but Reason (R) is false.
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.
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(D) Assertion (A) is false but Reason (R) is true.
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.
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
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 ($\frac{1}{2}$)
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.
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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(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.
Prove that $5-2\sqrt{3}$ is an irrational number. It is given that $\sqrt{3}$ is an irrational number.
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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.
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.
Prove that $2 + 3\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational number.
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Let $2 + 3\sqrt{5}$ be a rational number. $\therefore 2 + 3\sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and $p$ and $q$ are integers. (I) ($\frac{1}{2}$ Mark) $\Rightarrow \sqrt{5} = \frac{p-2q}{3q}$ (II) ($\frac{1}{2}$ Mark) As $\frac{p-2q}{3q}$ is a rational number, so $\sqrt{5}$ is rational. But we know that $\sqrt{5}$ is irrational. $\therefore$ Our assumption is wrong. Hence, $2 + 3\sqrt{5}$ is an irrational number. (III) (1 Mark)
Prove that $2 - 5\sqrt{3}$ is an irrational number given that $\sqrt{3}$ is irrational.
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Let $2 - 5\sqrt{3}$ be a rational number. $2 - 5\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (I) (1/2) $ \sqrt{3} = \frac{2b-a}{5b}$ (II) (1/2) RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $2 - 5\sqrt{3}$ is an irrational number. (III) (1)
Prove that $4 - 2\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational.
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Let $4 - 2\sqrt{5}$ be a rational number. (I Mark) $\therefore 4 - 2\sqrt{5} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (II Mark) $\sqrt{5} = \frac{4b - a}{2b}$ RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $4 - 2\sqrt{5}$ is an irrational number. (III Mark)
Prove that $14 - 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.
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Let $14 - 2\sqrt{3}$ be a rational number. (1/2 Mark) $14 - 2\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. (1/2 Mark) $\sqrt{3} = \frac{14b - a}{2b}$ (1 Mark) RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $14 - 2\sqrt{3}$ is an irrational number.
3 Marks Questions
1503 Marks · 🔁 March 2023 & July 2023 & March 2024 & July 2024 & March 2025 & March 2026 · Basicopen ↗
Prove that $\sqrt{5}$ is an irrational number.
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(a) Let $\sqrt{5}$ be a rational number such that $\sqrt{5} = \frac{p}{q}$ ($p$ and $q$ are co-prime numbers, $q \neq 0$) [1/2 mark] $\sqrt{5}q = p \Rightarrow 5q^2 = p^2$ $5$ divides $p^2 \Rightarrow 5$ divides $p$ as well [1 mark] $p = 5m$ (for some integer $m$) $5q^2 = 25m^2 \Rightarrow q^2 = 5m^2$ $5$ divides $q^2 \Rightarrow 5$ divides $q$ as well [1 mark] $p$ and $q$ have a common factor $5$ which is a contradiction as $p$ and $q$ are co-prime. $\therefore \text{our assumption is wrong}$ Hence, $\sqrt{5}$ is an irrational number [1/2 mark]
1513 Marks · 🔁 March 2023 & March 2024 & July 2024 & March 2025 & March 2026 · Standardopen ↗
Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number. $\therefore \sqrt{3} = \frac{p}{q}$, let $p \& q$ be co-primes and $q \neq 0$ $3q^2 = p^2 \Rightarrow p^2$ is divisible by $3 \Rightarrow p$ is divisible by $3$ $\Rightarrow p = 3a$, where 'a' is some integer quad ----- (i) $9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $3$ $\Rightarrow q = 3b$, where 'b' is some integer quad ----- (ii) (i) and (ii) leads to contradiction as 'p' and 'q' are co-primes. $\therefore \sqrt{3}$ is an irrational number.
1523 Marks · 🔁 March 2023 & March 2025 & March 2026 · Basicopen ↗
Prove that $\sqrt{2}$ is an irrational number.
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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]
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 $$\begin{aligned}& \& q \neq 0 \\ & \sqrt{3} = \frac{2q-5p}{q} \\ & \text{Here RHS is rational but LHS is irrational.} \\ & \text{Therefore our assumption is wrong.} \\ & \text{Hence } \frac{2-\sqrt{3}}{5} \text{ is an irrational number.}\end{aligned}$$
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
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.
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
Prove that $(5\sqrt{3} + \frac{2}{3})$ is an irrational number given that $\sqrt{3}$ is an irrational number.
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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.
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.
Prove that $\frac{2 + 3\sqrt{5}}{7}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
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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.
State the "Fundamental Theorem of Arithmetic" and use it to find LCM of $36$ and $54$.
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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]
Given that $\sqrt{5}$ is an irrational number, prove that $2 + 3\sqrt{5}$ is an irrational number.
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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
Prove that $\frac{\sqrt{3}+5}{2}$ is an irrational number, where it is given that $\sqrt{3}$ is irrational.
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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
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.
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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)
Given that $\sqrt{2}$ is an irrational number, prove that $5-2\sqrt{2}$ is also an irrational number.
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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.
Given that $\sqrt{3}$ is an irrational number, prove that $2-5\sqrt{3}$ is also an irrational number.
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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)
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.
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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)
Find the smallest $4$-digit number exactly divisible by $15$, $24$ and $36$.
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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 $15, 24 \& 36$.
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?
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 $240$.
Find the length of the plank that can be used to measure the lengths $4$ m $20$ cm and $5$ m $4$ cm exactly, in the least time.
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$4$ m $20$ cm $= 420$ cm and $5$ m $4$ cm $= 504$ cm (1/2 Mark) Size of plank should be maximum, so we will find HCF $(420, 504)$ $420 = 2^2 \times 3 \times 5 \times 7$ and $504 = 2^3 \times 3^2 \times 7$ (1 Mark) HCF $(420, 504) = 84$ (1/2 Mark) $\therefore$ the required length of the plank is $84$ cm
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.
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.
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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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 = $\frac{48}{16} + \frac{80}{16} + \frac{144}{16} = 17$
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 = $\frac{180}{15} + \frac{180}{18} + \frac{180}{20}$ $= 31$
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)
A trader has three different types of oils of volume $870$ l, $812$ l and $638$ l. Find the least number of containers of equal size required to store all the oil without getting mixed.
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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)
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.
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$156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$ (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)
Three friends plan to go for a morning walk. They step off together and their steps measures $48$ cm, $52$ cm and $56$ cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ten times?
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$48 = 2^4 \times 3, 52 = 2^2 \times 13, 56 = 2^3 \times 7$ LCM $= 2^4 \times 3 \times 13 \times 7 = 4368$ $\Rightarrow$ Minimum distance each walked in complete steps ten times $= 43680$ cm
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
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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$173250= 2 \times 5^3 \times 3^2 \times 7 \times 11$ (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) $5$
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.
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.
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(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
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).