NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

We’ll find the prime factorization of both numbers to determine HCF, then use the relationship to find LCM.

We’ll use prime factorization for HCF. Since numbers are larger, we can also use Euclidean algorithm as an alternative.

We’ll find prime factorization for HCF. Note that these numbers have common factors beyond just small primes.

First, we find the prime factors of each number by dividing them by prime numbers starting from 2.

LCM is the product of the highest powers of all prime factors present in any of the numbers. The prime factors involved are 2, 3, 5, and 7.

HCF is the product of the lowest powers of common prime factors. The only common prime factor among all three numbers is 3.

These numbers are all prime numbers themselves, so their prime factorisation is straightforward.

Since all numbers are distinct primes with no common factors, LCM is simply the product of all numbers.

HCF requires common prime factors. Since these are distinct primes with no common factors, HCF is 1.

Find prime factors by dividing each number by primes. Note that these are composite numbers with no common primes.

LCM takes the highest power of each prime factor from all numbers. The primes are distinct:2,3,and5.

HCF requires common prime factors.Since these numbers have no common primes,HCF is1.

A number ending with digit 0 is divisible by 10. For example: 10, 20, 30, 100, etc. all end with 0 and are divisible by 10. Mathematically, if a number ends with digit 0, it can be written as \(10 \times k\) where \(k\) is some integer.

To understand divisibility by 10, we need to examine the prime factors of 10. According to the Fundamental Theorem of Arithmetic, every composite number can be expressed as a product of primes in a unique way (up to the order of factors).

For a number to be divisible by 10, it must contain both prime factors 2 and 5 in its prime factorization. This is because if a number is divisible by 10, it must be divisible by both 2 and 5 (since 10 = 2 × 5).

Now let’s examine the prime factors of the base number 6. We need to understand what happens when we raise 6 to any power \(n\).

When we raise a number to a power, we multiply all its prime factors by that exponent. According to the laws of exponents: \((a \times b)^n = a^n \times b^n\).

Now we examine the prime factorization of \(6^n\). From step 5, we see that \(6^n = 2^n \times 3^n\). Notice that this expression contains only prime factors 2 and 3. There is no factor of 5 in this expression.

Since \(6^n\) does not contain the prime factor 5 in its prime factorization, it cannot be divisible by 5. Remember from step 3: for \(6^n\) to be divisible by 10, it must be divisible by both 2 and 5.

Since \(6^n\) is not divisible by 5 (it lacks the prime factor 5), it cannot be divisible by 10. Therefore, \(6^n\) cannot end with digit 0.

We start with the expression \(7 \times 11 \times 13 + 13\). Notice that \(13\) is a common factor in both terms. By factoring out \(13\), we can rewrite the expression in a product form, which will help identify its factors.

Now, \(13 \times 78\) is clearly a product of two integers greater than 1. To confirm it’s composite, we factorize \(78\) into its prime factors, showing that the entire expression has multiple factors.

Now consider \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\). Here, \(5\) is a common factor. Factor out \(5\) to simplify the expression.

To determine if \(5 \times 1009\) is composite, we need to check whether \(1009\) is prime or composite. If \(1009\) is prime, then \(5 \times 1009\) has exactly two prime factors, making it composite because it has more than two divisors. We test divisibility by primes up to \(\sqrt{1009}\).

Since \(5\) and \(1009\) are both primes greater than \(1\), their product \(5 \times 1009\) has divisors: \(1, 5, 1009,\) and \(5045\). This satisfies the definition of a composite number.

First, note the time each person takes to complete one full round of the circular path. These times are given directly in the problem.

For Sonia and Ravi to meet again at the starting point, both must have completed an integer number of laps. Let Sonia complete \( n_s \) laps and Ravi complete \( n_r \) laps in the meeting time \( T \). Since they start together, the time taken is the same for both. The distance covered in terms of laps is related to time by their speeds. Alternatively, we can think in terms of time: \( T \) must be a multiple of both 18 and 12 minutes.

To find the earliest time they meet, we need the smallest positive common multiple, i.e., the LCM of 18 and 12. We can find LCM using prime factorization or division method.

The LCM, 36 minutes, is the smallest time that is a multiple of both 18 and 12. At this time, Sonia completes exactly \( \frac{36}{18} = 2 \) laps and Ravi completes exactly \( \frac{36}{12} = 3 \) laps. Both return to the starting point simultaneously.

Check that at 36 minutes, both are at the starting point. Sonia takes 18 minutes per lap, so in 36 minutes she completes \( 36 / 18 = 2 \) laps, ending at start. Ravi takes 12 minutes per lap, so in 36 minutes he completes \( 36 / 12 = 3 \) laps, also ending at start. Thus, they meet.