NCERT Solutions for Class 10 Maths Exercise 1.1 Chapter 1 Real Numbers
1. Express each number as a product of its prime factors:
(i) 140
Step 1: Start with the smallest prime number that divides 140, which is 2.
140 ÷ 2 = 70
Step 2: Continue dividing by 2.
70 ÷ 2 = 35
Step 3: The number 35 is not divisible by 2, so move to the next prime number, which is 5.
35 ÷ 5 = 7
Step 4: The number 7 is a prime number, so the factorization is complete.
140 = 2 × 2 × 5 × 7
(ii) 156
Step 1: Start with the smallest prime number that divides 156, which is 2.
156 ÷ 2 = 78
Step 2: Continue dividing by 2.
78 ÷ 2 = 39
Step 3: The number 39 is not divisible by 2, so move to the next prime number, which is 3.
39 ÷ 3 = 13
Step 4: The number 13 is a prime number, so the factorization is complete.
156 = 2 × 2 × 3 × 13
(iii) 3825
Step 1: Start with the smallest prime number that divides 3825, which is 3.
3825 ÷ 3 = 1275
Step 2: Continue dividing by 3.
1275 ÷ 3 = 425
Step 3: The number 425 is not divisible by 3, so move to the next prime number, which is 5.
425 ÷ 5 = 85
Step 4: Continue dividing by 5.
85 ÷ 5 = 17
Step 5: The number 17 is a prime number, so the factorization is complete.
3825 = 3 × 3 × 5 × 5 × 17
(iv) 5005
Step 1: Start with the smallest prime number that divides 5005, which is 5.
5005 ÷ 5 = 1001
Step 2: The number 1001 is not divisible by 5, so move to the next prime number, which is 7.
1001 ÷ 7 = 143
Step 3: The number 143 is not divisible by 7, so move to the next prime number, which is 11.
143 ÷ 11 = 13
Step 4: The number 13 is a prime number, so the factorization is complete.
5005 = 5 × 7 × 11 × 13
(v) 7429
Step 1: Start with the smallest prime number that divides 7429, which is 17.
7429 ÷ 17 = 437
Step 2: The number 437 is not divisible by 17, so move to the next prime number, which is 19.
437 ÷ 19 = 23
Step 3: The number 23 is a prime number, so the factorization is complete.
7429 = 17 × 19 × 23
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
Step 1: Prime factorize both numbers.
26 = 2 × 13
91 = 7 × 13
Step 2: HCF (Highest Common Factor) is the product of the smallest power of common prime factors in the numbers.
HCF of 26 and 91 = 13
Step 3: LCM (Least Common Multiple) is the product of the highest power of all prime factors in the numbers.
LCM of 26 and 91 = 2 × 7 × 13 = 182
Step 4: Verify that LCM × HCF = product of the two numbers.
LCM × HCF = 182 × 13 = 2366
Product of the two numbers = 26 × 91 = 2366
Since LCM × HCF = Product of the two numbers, the verification is correct.
(ii) 510 and 92
Step 1: Prime factorize both numbers.
510 = 2 × 3 × 5 × 17
92 = 2² × 23
Step 2: HCF is the product of the smallest power of common prime factors in the numbers.
HCF of 510 and 92 = 2
Step 3: LCM is the product of the highest power of all prime factors in the numbers.
LCM of 510 and 92 = 2² × 3 × 5 × 17 × 23 = 2 × 3 × 5 × 17 × 46 = 23460
Step 4: Verify that LCM × HCF = product of the two numbers.
LCM × HCF = 23460 × 2 = 46920
Product of the two numbers = 510 × 92 = 46920
Since LCM × HCF = Product of the two numbers, the verification is correct.
(iii) 336 and 54
Step 1: Prime factorize both numbers.
336 = 2⁴ × 3 × 7
54 = 2 × 3³
Step 2: HCF is the product of the smallest power of common prime factors in the numbers.
HCF of 336 and 54 = 2 × 3 = 6
Step 3: LCM is the product of the highest power of all prime factors in the numbers.
LCM of 336 and 54 = 2⁴ × 3³ × 7 = 16 × 27 × 7 = 3024
Step 4: Verify that LCM × HCF = product of the two numbers.
LCM × HCF = 3024 × 6 = 18144
Product of the two numbers = 336 × 54 = 18144
Since LCM × HCF = Product of the two numbers, the verification is correct.
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Step 1: Prime factorize all numbers.
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
Step 2: HCF is the product of the smallest power of common prime factors in the numbers.
HCF of 12, 15, and 21 = 3
Step 3: LCM is the product of the highest power of all prime factors in the numbers.
LCM of 12, 15, and 21 = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
Step 1: Since all numbers are prime, the HCF is 1.
HCF of 17, 23, and 29 = 1
Step 2: LCM is the product of the numbers as all are prime and do not share any common factors.
LCM of 17, 23, and 29 = 17 × 23 × 29
(iii) 8, 9 and 25
Step 1: Prime factorize all numbers.
8 = 2³
9 = 3²
25 = 5²
Step 2: HCF is the product of the smallest power of common prime factors in the numbers, and since there are no common prime factors, the HCF is 1.
HCF of 8, 9, and 25 = 1
Step 3: LCM is the product of the highest power of all prime factors in the numbers.
LCM of 8, 9, and 25 = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Step 1: Use the relationship between HCF and LCM of two numbers, which states that the product of the HCF and LCM of two numbers is equal to the product of the numbers themselves.
Product of numbers = HCF × LCM
Step 2: Calculate the product of the two numbers.
Product of 306 and 657 = 306 × 657
Step 3: Divide the product by the HCF to find the LCM.
LCM = (Product of numbers) / HCF
LCM = (306 × 657) / 9
Step 4: Perform the division to find the LCM.
LCM = (306 × 657) / 9
= (306 × 73)
= 22338
The LCM of 306 and 657 is 22338.
5. Check whether 6n can end with the digit 0 for any natural number n.
Step 1: For a number to end with the digit 0, it must be a multiple of 10.
Step 2: A multiple of 10 is always a multiple of 5 and 2.
Step 3: Since 6n is a multiple of 6 (which is 2 × 3), it will always contain the factor 2.
Step 4: For 6n to end with a digit 0, n must provide the factor 5 to the product.
Step 5: There is no restriction on n, and it can be any natural number, including multiples of 5.
Conclusion: 6n can end with the digit 0 if n is a multiple of 5.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Step 1: Look at the structure of each number.
Step 2: For the number 7 × 11 × 13 + 13:
Notice that 13 is a common factor.
It can be factored as 13(7 × 11 + 1).
Step 3: This shows that the number is divisible by 13, and since there is another factor (7 × 11 + 1), the number is not prime.
Step 4: For the number 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5:
Factor out 5, resulting in 5(7 × 6 × 4 × 3 × 2 × 1 + 1).
Step 5: This shows that the number is divisible by 5, and since there is another factor (7 × 6 × 4 × 3 × 2 × 1 + 1), the number is not prime.
Conclusion: Both numbers have more than two factors and therefore are composite numbers.
7. 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?
Step 1: Calculate the LCM of the times taken by Sonia and Ravi to find the time after which they will meet at the starting point.
Time taken by Sonia = 18 minutes
Time taken by Ravi = 12 minutes
Step 2: Find the LCM of 18 and 12 to determine after how many minutes they will meet.
LCM of 18 and 12 = 2 × 3² × 2 = 36
Conclusion: Sonia and Ravi will meet again at the starting point after every 36 minutes.