Kaliyuga Ekalavya

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 1(ii) Question: Express 156 as a product of its prime factors. Given: The number 156. To Find: The prime factorization of 156. Formula: Prime factorization is done by dividing the number by prime numbers successively until all factors are prime. Solution: Step 1: Divide 156 by 2 (the smallest prime number) because 156 is even. $$156 \div 2 = 78$$ Step 2: Divide 78 by 2 again, since 78 is also even. $$78 \div 2 = 39$$ Step 3: Check 39 for divisibility by 3. The sum of its digits (3 + 9 = 12) is divisible by 3, so divide by 3. $$39 \div 3 = 13$$ Step 4: 13 is a prime number. Divide 13 by itself. $$13 \div 13 = 1$$ Step 5: Collect all the prime factors found. $$156 = 2 \times 2 \times 3 \times 13$$ Step 6: Express repeated factors using exponents. $$156 = 2^2 \times 3 \times 13$$ ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 1(iii) Question: Express 3825 as a product of its prime factors. Given: The number 3825. To Find: Prime factorization of 3825. Formula: Successively divide the given number by prime numbers to express it as a product of its prime factors. Solution: Step 1: Check divisibility by 2. 3825 is odd, so it is not divisible by 2. Step 2: Check divisibility by 3. Sum of digits: 3 + 8 + 2 + 5 = 18, which is divisible by 3. Divide by 3: $$ 3825 \div 3 = 1275 $$ Step 3: Check divisibility of 1275 by 3. Sum of digits: 1 + 2 + 7 + 5 = 15, which is divisible by 3. Divide by 3: $$ 1275 \div 3 = 425 $$ Step 4: Check divisibility of 425 by 5. 425 ends with 5, so it is divisible by 5. Divide by 5: $$ 425 \div 5 = 85 $$ Step 5: Check divisibility of 85 by 5. 85 ends with 5, so it is divisible by 5. Divide by 5: $$ 85 \div 5 = 17 $$ ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 1(iv) Question: Express 5005 as a product of its prime factors. Given: The number to be factorized is \( 5005 \). To Find: Express \( 5005 \) as a product of its prime factors. Formula: To write a number as a product of its prime factors, divide it successively by prime numbers starting from the smallest, until the quotient is 1. Solution: Step 1: Check divisibility of \( 5005 \) by 2. \( 5005 \) is odd, so it is not divisible by 2. Step 2: Check divisibility by 3. Sum of the digits: \( 5 + 0 + 0 + 5 = 10 \). Since 10 is not divisible by 3, \( 5005 \) is not divisible by 3. Step 3: Check divisibility by 5. \( 5005 \) ends in 5, so it is divisible by 5. $$ 5005 \div 5 = 1001 $$ So, $$ 5005 = 5 \times 1001 $$ Step 4: Factorize \( 1001 \). Check divisibility by 2, 3, and 5. ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 1(v) Question: Express 7429 as a product of its prime factors. Given: The number given is 7429. To Find: Express 7429 as a product of its prime factors. Formula: The prime factorization of a number \( n \) is written as: $$ n = p_1 \times p_2 \times \cdots \times p_k $$ where each \( p_i \) is a prime number. Solution: Step 1: Start dividing 7429 by the smallest prime numbers to find its prime factors. 7429 is not divisible by 2, 3, 5, 7, 11, or 13 (checked by divisibility tests). Step 2: Check divisibility by 17. $$ 7429 \div 17 = 437 $$ So, 17 is a prime factor. We write: $$ 7429 = 17 \times 437 $$ Step 3: Factorize 437 by dividing by primes. 437 is not divisible by 2, 3, 5, 7, 11, 13, or 17. Check divisibility by 19: $$ 437 \div 19 = 23 ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 2(ii) Question: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers: 510 and 92. Given: The two numbers are 510 and 92. To Find: 1. The Highest Common Factor (HCF) of 510 and 92. 2. The Least Common Multiple (LCM) of 510 and 92. 3. Verify that $$\text{LCM} \times \text{HCF} = 510 \times 92$$. Formula: $$ \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b $$ Solution: Step 1: Find the prime factorization of 510. $$ 510 \div 2 = 255 \\ 255 \div 3 = 85 \\ 85 \div 5 = 17 \\ 17 \text{ is prime} $$ So, $$510 = 2 \times 3 \times 5 \times 17$$ Step 2: Find the prime factorization of 92. $$ 92 \div 2 = 46 \\ 46 \div 2 = 23 \\ 23 \text{ is prime} $$ So, $$92 = 2^2 \times 23$$ Step 3: Find the HCF (Highest Common Factor). List the prime factors: 510: \(2^1 \times 3^1 \times 5^1 \times 17^1\...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 2(iii) Question: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF equals the product of the two numbers. 336 and 54 Given: Two integers: 336 and 54 To Find: The LCM and HCF of 336 and 54. Verify that LCM × HCF = product of the two numbers. Formula: Prime factorization is used to find LCM and HCF. $$ \text{LCM} \times \text{HCF} = \text{Product of the two numbers} $$ Solution: Step 1: Find the prime factorization of 336. $$ 336 \div 2 = 168 $$ $$ 168 \div 2 = 84 $$ $$ 84 \div 2 = 42 $$ $$ 42 \div 2 = 21 $$ $$ 21 \div 3 = 7 $$ Thus, $$ 336 = 2^4 \times 3^1 \times 7^1 $$ Step 2: Find the prime factorization of 54. $$ 54 \div 2 = 27 $$ $$ 27 \div 3 = 9 $$ $$ 9 \div 3 = 3 $$ $$ 3 \div 3 = 1 $$ Thus, $$ 54 = 2^1 \times 3^3 $$ Step 3: Find the HCF (Highest Common Factor) of 336 and 54. Take the ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 2(i) Question: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (26 and 91) Given: Integers: 26 and 91 To Find: LCM of 26 and 91 HCF of 26 and 91 Verify that LCM × HCF = 26 × 91 Formula: For any two positive integers \( a \) and \( b \): $$ \mathrm{LCM}(a, b) \times \mathrm{HCF}(a, b) = a \times b $$ Solution: Step 1: Write the prime factorization of each number. $$ 26 = 2 \times 13 \\ 91 = 7 \times 13 $$ Step 2: Express the prime factorization using exponents. $$ 26 = 2^1 \times 13^1 $$ $$ 91 = 7^1 \times 13^1 $$ Step 3: Find the HCF by taking the common prime factors with the lowest exponent. The common prime factor is 13. $$ \mathrm{HCF}(26, 91) = 13^1 = 13 $$ Step 4: Find the LCM by taki...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 3(ii) Question: Find the LCM and HCF of the following integers by applying the prime factorisation method: 17, 23, and 29. Given: The integers are 17, 23, and 29. To Find: Find the LCM and HCF of 17, 23, and 29 using the prime factorisation method. Formula: For any set of numbers: HCF is the product of the smallest powers of all common prime factors. LCM is the product of the highest powers of all prime factors present in any of the numbers. For distinct prime numbers \(p, q, r\): $$\text{HCF}(p, q, r) = 1$$ $$\text{LCM}(p, q, r) = p \times q \times r$$ Solution: Step 1: Write the prime factorisation of each number. 17 is a prime number: $$17 = 17^1$$ 23 is a prime number: $$23 = 23^1$$ 29 is a prime number: $$29 = 29^1$$ Step 2: Find the HCF (Highest Common Factor). ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 3(iii) Question: Find the LCM and HCF of the following integers by applying the prime factorisation method: 8, 9, and 25 Given: Numbers: 8, 9, 25 To Find: LCM and HCF of 8, 9, and 25 Formula: Prime Factorisation: Express each number as a product of its prime factors. HCF: Product of the lowest powers of all common prime factors. LCM: Product of the highest powers of all prime factors present in any number. Solution: Step 1: Write each number as a product of its prime factors. $$ 8 = 2 \times 2 \times 2 = 2^3 $$ $$ 9 = 3 \times 3 = 3^2 $$ $$ 25 = 5 \times 5 = 5^2 $$ Step 2: Find the HCF (Highest Common Factor). Check for common prime factors in all three numbers. There is no prime factor common to 8, 9, and 25. $$ \gcd(8,\,9,\,25) = 1 $$ Step 3: Find the LCM (Lowest Common Multiple). List...

NCERT Class X Chapter 1: Real Numbers Question: Find the LCM and HCF of 12, 15, and 21 by applying the prime factorization method. Given: Numbers: 12, 15, 21 To Find: LCM and HCF of 12, 15, and 21 Formula: HCF (Highest Common Factor): The product of the smallest powers of all common prime factors in the numbers. LCM (Least Common Multiple): The product of the greatest powers of all prime numbers present in any of the numbers. Solution: Step 1: Write the prime factorization of each number. $$ 12 = 2 \times 2 \times 3 = 2^2 \times 3 $$ $$ 15 = 3 \times 5 $$ $$ 21 = 3 \times 7 $$ Step 2: Express each number using exponents for the prime factors. $$ 12 = 2^2 \times 3^1 $$ $$ 15 = 3^1 \times 5^1 $$ $$ 21 = 3^1 \times 7^1 $$ Step 3: Find the HCF by taking the lowest power of each common prime factor. The only common prime factor is 3. The lowe...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 6 Question: 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. Given: The two expressions: \(7 \times 11 \times 13 + 13\) \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\) To Find: Show, with step-by-step reasoning, that both expressions are composite numbers . Formula: A composite number is a whole number greater than 1 that has more than two distinct positive divisors (i.e., it is not prime). To show a number is composite, we factorize it or show it has divisors other than 1 and itself. Solution: Step 1: Consider the first expression: \(7 \times 11 \times 13 + 13\). $$ 7 \times 11 \times 13 + 13 $$ Step 2: Factor out 13 from both terms. $$ = (7 \times 11 \times 13) + 13 \\ = 13 \times (7 \times 11) + 13 \times 1 \\ = ...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 4 Question: Given that HCF (306, 657) = 9, find LCM (306, 657). Given: HCF (306, 657) = 9 To Find: LCM (306, 657) Formula: For any two positive integers \(a\) and \(b\): $$ \gcd(a, b) \times \mathrm{LCM}(a, b) = a \times b $$ Solution: Step 1: Let \(a = 306\) and \(b = 657\). The HCF is given as 9. $$ \gcd(306, 657) = 9 $$ Step 2: Write the HCF-LCM relationship for two numbers. $$ \gcd(306, 657) \times \mathrm{LCM}(306, 657) = 306 \times 657 $$ Step 3: Substitute the known value of HCF into the formula. $$ 9 \times \mathrm{LCM}(306, 657) = 306 \times 657 $$ Step 4: Rearrange to solve for the LCM. $$ \mathrm{LCM}(306, 657) = \frac{306 \times 657}{9} $$ Step 5: Calculate the product \(306 \times 657\). $$ 306 \times 657 = 201{,}042 $$ Step 6: Substitute the product into the LCM formula. $$ \mathrm{LCM}(306, 657) = \frac{201{,}...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 5 Question: Check whether \( 6^n \) can end with the digit 0 for any natural number \( n \). Given: The expression \( 6^n \), where \( n \) is a natural number. To Find: Whether \( 6^n \) can end with the digit 0 for any natural number \( n \). Formula: To end with 0, a number must be divisible by 10. \( 10 = 2 \times 5 \) Solution: Step 1: For a number to end with 0, it must be divisible by 10. The prime factorization of 10 is \( 10 = 2 \times 5 \). $$ 10 = 2 \times 5 $$ Step 2: Write 6 as a product of its prime factors. $$ 6 = 2 \times 3 $$ Step 3: Raise both sides to the power \( n \). $$ 6^n = (2 \times 3)^n $$ Step 4: Apply the exponent rule: \( (a \times b)^n = a^n \times b^n \). $$ 6^n = 2^n \times 3^n $$ Step 5: Observe the prime factors of \( 6^n \). The only p...

NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 7 Question: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? Given: Sonia takes 18 minutes to complete one round. Ravi takes 12 minutes to complete one round. To Find: The time after which both Sonia and Ravi will meet again at the starting point. Formula: The required time is the Least Common Multiple (LCM) of the times taken by Sonia and Ravi to complete one round. If Sonia's time is \( a \) and Ravi's time is \( b \), then $$ \text{Time to meet again} = \operatorname{LCM}(a, b) $$ Definitions: Least Common Multiple (LCM): The smallest positive integer that is exactly divisible b...

NCERT Class X Chapter 1: Real Numbers Example 5 Question: Prove that \( \sqrt{3} \) is irrational. Given: We need to prove that \( \sqrt{3} \) is an irrational number. To Find: Show that \( \sqrt{3} \) cannot be expressed as a ratio of two integers. Formula: No specific formula is required. We use proof by contradiction and properties of rational and irrational numbers. Solution: Step 1: Assume, for contradiction, that \( \sqrt{3} \) is rational. That is, $$ \sqrt{3} = \frac{a}{b} $$ where \( a \) and \( b \) are coprime integers and \( b \neq 0 \). Step 2: Square both sides to remove the square root: $$ (\sqrt{3})^2 = \left( \frac{a}{b} \right)^2 $$ So, $$ 3 = \frac{a^2}{b^2} $$ Step 3: Multiply both sides by \( b^2 \) to clear the denominator: $$ 3b^2 = a^2 $$ Step 4: Since 3 divides \( a^2 \), 3 must also divide \( a \) (because 3 is prime). Let \( a = 3k \) for some integer \( k \). Step 5: ...

NCERT Class X Chapter 1: Real Numbers Example 6 Question: Show that \(5 - \sqrt{3}\) is irrational. Given: Assume, for contradiction, that \(5 - \sqrt{3}\) is rational. To Find: Prove that \(5 - \sqrt{3}\) is irrational. Formula: Proof by contradiction using properties of rational and irrational numbers. Definitions: Rational number: A number that can be expressed as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\). Irrational number: A number that cannot be expressed as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\). Solution: Step 1: Assume that \(5 - \sqrt{3}\) is rational. Let \(5 - \sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are integers, \(b \ne 0\). Step 2: Rearranging the equation to isolate \(\sqrt{3}\). $$ 5 - \sqrt{3} = \frac{a}{b} $$ Add \(\sqrt{3}\) to both sides: $$ 5 = \frac{a}{b} + \sqrt{3} $$ ...

NCERT Class X Chapter 1: Real Numbers Example 7 Question: Show that \(3\sqrt{2}\) is irrational. Given: The number \(3\sqrt{2}\). To Find: To prove that \(3\sqrt{2}\) is irrational. Formula: We use proof by contradiction and the property that the product of a non-zero rational number and an irrational number is irrational. Solution: Step 1: Assume, to the contrary, that \(3\sqrt{2}\) is rational. $$ 3\sqrt{2} = \frac{a}{b} $$ where \(a\) and \(b\) are coprime integers, \(b \neq 0\). Step 2: Divide both sides by 3 to isolate \(\sqrt{2}\). $$ \sqrt{2} = \frac{a}{3b} $$ Step 3: Note that \(\frac{a}{3b}\) is a rational number, since \(a\) and \(b\) are integers and \(3b \neq 0\). Step 4: But it is a well-known fact that \(\sqrt{2}\) is irrational. Step 5: This leads to a contradiction, since \(\sqrt{2}\) cannot be rational. Step 6: Therefore, our assumption i...

NCERT Class X Chapter 1: Real Numbers Example 4 Question: Find the HCF and LCM of 6, 72, and 120 using the prime factorisation method. Given: The numbers are 6, 72, and 120. To Find: The Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of 6, 72, and 120 using the prime factorisation method. Formula: HCF (by prime factorisation): Multiply the lowest powers of all common prime factors present in each number. LCM (by prime factorisation): Multiply the highest powers of all prime factors present in any of the numbers. Solution: Step 1: Write the prime factorisation of each number. $$6 = 2 \times 3$$ $$72 = 2 \times 2 \times 2 \times 3 \times 3$$ $$120 = 2 \times 2 \times 2 \times 3 \times 5$$ Step 2: Express each number in exponential form. $$6 = 2^1 \times 3^1$$ $$72 = 2^3 \times 3^2$$ $$120 = 2^3 \times 3^1 \times 5^1$$ Step 3: Find the HCF by taking the lowe...