GSEB Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.2

Question 1. Express each number as a product of its prime factors.
1. 140
2. 156
3. 3825
4. 5005
5. 7429
Answer:
1. Prime factorisation of \( 140 = 2 \times 2 \times 5 \times 7 \)
2. Prime factorisation of \( 156 = 2 \times 2 \times 3 \times 13 \)
3. Prime factorisation of \( 3825 = 3 \times 3 \times 5 \times 5 \times 17 \)
4. Prime factorisation of \( 5005 = 5 \times 7 \times 11 \times 13 \)
5. Prime factorisation of \( 7429 = 17 \times 19 \times 23 \)
In simple words: To find the prime factors, you break down each number into smaller numbers that can only be divided by 1 and themselves. These are called prime numbers. You keep breaking them down until all the parts are prime.

Exam Tip: Remember to use only prime numbers (2, 3, 5, 7, 11, etc.) when doing prime factorisation. Start with the smallest prime number and continue dividing until the number becomes 1.

Question 2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the two numbers.
1. 26 and 91
2. 510 and 92
3. 336 and 54
Answer:
1. For 26 and 91:
Prime factorisation of \( 26 = 2 \times 13 \)
Prime factorisation of \( 91 = 7 \times 13 \)
HCF of 26 and 91 is 13.
LCM of 26 and 91 is \( 2 \times 7 \times 13 = 182 \).
Now, let's verify:
Product of the two numbers \( = 26 \times 91 = 2366 \).
LCM \( \times \) HCF \( = 182 \times 13 = 2366 \).
Since both results are 2366, the verification is complete.
2. For 510 and 92:
Prime factorisation of \( 510 = 2 \times 3 \times 5 \times 17 \)
Prime factorisation of \( 92 = 2 \times 2 \times 23 \)
HCF of 510 and 92 is 2.
LCM of 510 and 92 is \( 2 \times 2 \times 3 \times 5 \times 17 \times 23 = 23460 \).
Now, let's verify:
Product of the two numbers \( = 510 \times 92 = 46920 \).
LCM \( \times \) HCF \( = 23460 \times 2 = 46920 \).
Since both results are 46920, the verification is complete.
3. For 336 and 54:
Prime factorisation of \( 336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 \)
Prime factorisation of \( 54 = 2 \times 3 \times 3 \times 3 \)
HCF of 336 and 54 is \( 2 \times 3 = 6 \).
LCM of 336 and 54 is \( 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 = 3024 \).
Now, let's verify:
Product of the two numbers \( = 336 \times 54 = 18144 \).
LCM \( \times \) HCF \( = 3024 \times 6 = 18144 \).
Since both results are 18144, the verification is complete.
In simple words: First, break down each pair of numbers into their prime factors. Then, find their HCF (highest common factor) and LCM (least common multiple). Finally, multiply the original numbers together and separately multiply their HCF and LCM. These two products should always match.

Exam Tip: To find the HCF, take the common prime factors with the lowest power. To find the LCM, take all prime factors (common and uncommon) with the highest power. Always remember to check the relationship LCM \( \times \) HCF = Product of the two numbers.

Question 3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
1. 12, 15 and 21
2. 17, 23 and 29
3. 8, 9 and 25
Answer:
1. For 12, 15 and 21:
Prime factorisation of \( 12 = 2 \times 2 \times 3 \)
Prime factorisation of \( 15 = 3 \times 5 \)
Prime factorisation of \( 21 = 3 \times 7 \)
HCF of 12, 15 and 21 is 3 (the only common prime factor).
LCM of 12, 15 and 21 is \( 2 \times 2 \times 3 \times 5 \times 7 = 420 \).
2. For 17, 23 and 29:
Prime factorisation of \( 17 = 17 \times 1 \)
Prime factorisation of \( 23 = 23 \times 1 \)
Prime factorisation of \( 29 = 29 \times 1 \)
HCF of 17, 23 and 29 is 1 (since they are all prime numbers and 1 is the only common factor).
LCM of 17, 23 and 29 is \( 17 \times 23 \times 29 = 11339 \).
3. For 8, 9 and 25:
Prime factorisation of \( 8 = 2 \times 2 \times 2 \)
Prime factorisation of \( 9 = 3 \times 3 \)
Prime factorisation of \( 25 = 5 \times 5 \)
HCF of 8, 9 and 25 is 1 (since they have no common prime factors other than 1).
LCM of 8, 9 and 25 is \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 = 1800 \).
In simple words: First, break down each number into its prime factors. For HCF, find the prime numbers that appear in all lists, using the smallest power. For LCM, list all distinct prime factors that appear in any of the lists, using the largest power. If numbers are prime or have no common factors, their HCF is 1 and their LCM is their product.

Exam Tip: When dealing with more than two numbers, the relationship LCM \( \times \) HCF = Product of the numbers does NOT always hold. For numbers with no common prime factors (like 17, 23, 29 or 8, 9, 25), the HCF is always 1 and the LCM is simply the product of those numbers.

Question 4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer: We know that for any two positive integers a and b,
LCM (a, b) \( \times \) HCF (a, b) = a \( \times \) b
Given a = 306, b = 657, and HCF (306, 657) = 9.
So, LCM (306, 657) \( \times 9 = 306 \times 657 \)
LCM (306, 657) \( = \frac { 306 \times 657 }{ 9 } \)
LCM (306, 657) \( = 34 \times 657 \)
LCM (306, 657) \( = 22338 \).
In simple words: We use the rule that the product of two numbers is equal to the product of their HCF and LCM. Since we already know the HCF and the two numbers, we can simply divide the product of the numbers by the HCF to get the LCM.

Exam Tip: This formula (LCM \( \times \) HCF = product of two numbers) is very important for solving problems involving HCF and LCM of exactly two integers. Make sure to apply it correctly when one of the values is missing.

Question 5. Check whether 6\(^n\) can end with the digit 0 for any natural number n.(CBSE)
Answer: A number that ends with the digit 0 must have 5 as a prime factor. This is because any number ending in 0 is a multiple of 10, and \( 10 = 2 \times 5 \).
The prime factorisation of 6 is \( 2 \times 3 \).
So, 6\(^n\) can be written as \( (2 \times 3)^n = 2^n \times 3^n \).
In the prime factorisation of 6\(^n\), the only prime factors present are 2 and 3. There is no factor of 5.
According to the Fundamental Theorem of Arithmetic, the prime factorisation of a natural number is unique.
Therefore, for 6\(^n\) to end with the digit 0, it must have 5 as a prime factor, which it does not.
Hence, 6\(^n\) cannot end with the digit 0 for any natural number n.
In simple words: For a number to finish with a zero, it must be divisible by both 2 and 5. The number 6 has prime factors 2 and 3, but not 5. Because prime factorisation is unique, 6 raised to any power will only have 2 and 3 as its prime factors, never 5. So, 6\(^n\) can never end with a zero.

Exam Tip: To determine if a number \( x^n \) can end with the digit 0, always check its prime factors. If 5 is not a prime factor of x, then \( x^n \) will never end with 0. This concept is based on the Fundamental Theorem of Arithmetic.

Question 6. 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.
Answer: We need to show that both numbers have more than two factors (1 and the number itself) to prove they are composite.
Consider the first number: \( 7 \times 11 \times 13 + 13 \)
We can take 13 as a common factor:
\( = 13 \times (7 \times 11 \times 1 + 1) \)
\( = 13 \times (77 + 1) \)
\( = 13 \times 78 \)
Since \( 78 = 2 \times 3 \times 13 \), the expression becomes \( 13 \times 2 \times 3 \times 13 \).
This number clearly has factors 1, 2, 3, 13, 78, 169, and itself. Because it has more than two factors, it is a composite number.
Consider the second number: \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \)
We can take 5 as a common factor:
\( = 5 \times (7 \times 6 \times 1 \times 4 \times 3 \times 2 \times 1 + 1) \)
\( = 5 \times (1008 + 1) \)
\( = 5 \times 1009 \)
Since 1009 is a prime number, the factors of this expression are 1, 5, 1009, and \( 5 \times 1009 \). Because it has more than two factors (1, 5, 1009, 5045), it is also a composite number.
In simple words: A composite number is one that can be divided evenly by more than just 1 and itself. By taking out common factors from each expression, we can show that both numbers can be written as a product of several smaller numbers. This means they have factors other than 1 and themselves, proving they are composite.

Exam Tip: To show that a number is composite, you just need to prove that it has at least one factor other than 1 and itself. Often, factoring out a common term is the easiest way to demonstrate this.

Question 7. 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?
Answer: Time taken by Sonia to complete one round = 18 minutes.
Time taken by Ravi to complete one round = 12 minutes.
To find when they will meet again at the starting point, we need to find the Least Common Multiple (LCM) of 18 and 12.
Prime factorisation of \( 18 = 2 \times 3 \times 3 \)
Prime factorisation of \( 12 = 2 \times 2 \times 3 \)
LCM of 12 and 18 is \( 2 \times 2 \times 3 \times 3 = 36 \).
Therefore, they will meet again at the starting point after 36 minutes.
In simple words: To find when Sonia and Ravi will next meet at the starting line, we need to find the smallest number that both 18 and 12 can divide into evenly. This number is called the LCM. By finding the prime factors of 18 and 12, we can work out their LCM, which tells us the time they will meet.

Exam Tip: Problems involving two or more events happening at different intervals and asking when they will next happen together usually require finding the LCM. Make sure to clearly state the prime factors and show the calculation for the LCM.