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 :
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.
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.
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 $55$ $\operatorname{LCM}(44,55) = \frac{44 \times 55}{11} = 220$
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 $(p, q, r) \neq p \times q \times r$
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
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
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
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 $n$.
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 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.
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).
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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 $$\begin{aligned}& \& q \neq 0 \\ & \sqrt{3} = \frac{2q-5p}{q} \\ & \text{Here RHS is rational but LHS is irrational.} \\ & \text{Therefore our assumption is wrong.} \\ & \text{Hence } \frac{2-\sqrt{3}}{5} \text{ is an irrational number.}\end{aligned}$$
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.
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
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.
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
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.
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.
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.
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.
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).
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 $15, 24 \& 36$.
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 $240$.
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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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 = $\frac{48}{16} + \frac{80}{16} + \frac{144}{16} = 17$
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)
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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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) $5$
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.
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).
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.
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.
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.
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.
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.
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.