CBSE Class 10 Mathematics - Chapter - 1 - Real Number Exercise 1.1 solution

Q1.

Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Ans.

(i) $140 = 2 \times 2 \times 5 \times 7$.  So, the prime factorization of $140$ is $2^2 \times 5 \times 7$. (ii) $156 = 2 \times 2 \times 3 \times 13$.  So, the prime factorization of $156$ is $2^2 \times 3 \times 13$. (iii) $3825 = 3 \times 5 \times 5 \times 7 \times 11$.  So, the prime factorization of $3825$ is $3 \times 5^2 \times 7 \times 11$. (iv) $5005 = 5 \times 7 \times 11 \times 13$.  So, the prime factorization of $5005$ is $5 \times 7 \times 11 \times 13$. (v) $7429$  is a prime number, so its prime factorization is itself:  $7429 = 7429$.

Q2.

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 (ii) 510 and 92 (iii) 336 and 54

Ans.

(i) $26 \text{ and } 91$: Prime factors of  $26 = 2 \times 13$ Prime factors of $91 = 7 \times 13$ HCF = Highest common factor = $13$ (since it's the only common prime factor) LCM = Least common multiple $= 2 \times 7 \times 13 = 182$ Now, let's verify if $\text {LCM} \times \text{HCF}$ equals the product of the two numbers: $\text{LCM} \times \text{HCF} = 182 \times 13 = 2366$ $\text{Product of } 26 \text{ and } 91 = 26 \times 91 = 2366$ $\text{So, LCM} \times \text{HCF} = \text{Product of the two numbers}$. It’s verified. (ii) $510 \text{ and } 92$:  Prime factors of  $510 = 2 × 3 × 5 × 17$ Prime factors of  $92 = 2 × 2 × 23$ HCF = Highest common factor = $2$ (since it's the only common prime factor) LCM = Least common multiple = $2 × 2 × 3 × 5 × 17 × 23 = 23460$ Now, let's verify if $\text {LCM} × \text {HCF}$ equals the product of the two numbers: $\text{LCM} \times \text{HCF} = 23460 × 2 = 46920$ Product of $510$ and $92 = 510 × 92 = 46920$ So, $\text{LCM} \times \text{HCF} = \text{Product of the two numbers}$. It’s verified. (iii) $336 \text{ and } 54$: Prime factors of $336 = 2 × 2 × 2 × 2 × 3 × 7 = 2^4 × 3 × 7$ Prime factors of $54 = 2 × 3 × 3 × 3 = 2 × 3^3$ HCF = Highest common factor = $2 × 3 = 6$ LCM = Least common multiple = $2^4 × 3^3 × 7 = 3024$ Now, let's verify if $\text {LCM} × \text{HCF}$ equals the product of the two numbers: $\text {LCM} × \text {HCF} = 3024 × 6 = 18144$ Product of $336$ and $54 = 336 × 54 = 18144$ So, $\text {LCM} × \text {HCF} = \text {Product of the two numbers.}$ It’s verified.

Q3.

Find the LCM and HCF of the following integers by applying the prime factorization method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Ans.

(i) 12, 15, and 21: Prime factors of $12 = 2 × 2 × 3 = 2^2 × 3$ Prime factors of $15 = 3 × 5 = 3 × 5$ Prime factors of $21 = 3 × 7 = 3 × 7$ HCF = Highest common factor = Common prime factors with the lowest exponent = $3$ LCM = Least common multiple = All prime factors with the highest exponent $= 2^2 × 3 × 5 × 7 = 420$ (ii) 17, 23, and 29: Prime factors of $17 = 17$ Prime factors of $23 = 23$ Prime factors of $29 = 29$ HCF = Highest common factor = $1$ (since there are no common prime factors) LCM = Least common multiple = Product of all the numbers $= 17 × 23 × 29 = 11389$ (iii) 8, 9, and 25: Prime factors of $8 = 2 × 2 × 2 = 2^3$ Prime factors of $9 = 3 × 3 = 3^2$ Prime factors of $25 = 5 × 5 = 2^2$ HCF = Highest common factor = Common prime factors with the lowest exponent = $1$ (since there are no common prime factors) LCM = Least common multiple = All prime factors with the highest exponent $= 2^3 × 3^2 × 5^2 = 3600$

Q4.

Given that HCF (306, 657) = 9, find LCM (306, 657).

Ans.

To find the LCM of 306 and 657 when their HCF is known, we can use the formula as given below: $$\bf \text {LCM (a, b)} = \frac {\left(a \times b\right)}{\text {HCF (a, b)}}$$ So, $\text{HCF} (306, 657) = 9$      (Given) Now, we can calculate the LCM as follows: $\text {LCM}(306, 657) = \frac {(306 \times 657)}{9}$ $\text {LCM}(306, 657) = \frac {200,562}{9}$ $\text {LCM}(306, 657) = 22,284$ So, the LCM of 306 and 657 is 22,284.

Q5.

Check whether $6n$ can end with the digit $0$ for any natural number n.

Ans.

Yes, $6n$ can end with the digit $0$ for certain natural numbers $n$. In fact, any natural number $n$ that is a multiple of $5$ will result in $6n$ ending with the digit $0$. Here's why: $6n = 6 \times n$ When $n$ is a multiple of $5$, it can be expressed as $n = 5k$, where k is a natural number. Now, substituting this value into $6n$: $6n = 6 \times (5k) = 30k$ As we can see, $6n$ is equal to $30k$, which ends with the digit $0$. So, for any natural number n that is a multiple of $5$, $6n$ will end with the digit $0$. It’s verified.

Q6.

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Ans.

(i) \(7 \times 11 \times 13 + 13\): We can be simplified this expression as follows: $7 \times 11 \times 13 + 13 = 1001+13 = 1014$ $1014$ has factors other than 1 and itself which are 2, 3, 6, 169, 338 and 507. Because it has more then two factors, It is a composite numbers. (ii) \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\): We can be simplified this expression as follows: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 =  5040 + 5 = 5045$ $5045$ has factors other than 1 and itself which are 5 and 1009. Because it has more then two factors, It is a composite numbers.

Q7.

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?

Ans.

To find out when Sonia and Ravi will meet again at the starting point, we must calculate the LCM of their individual times to complete one round of the field. Given : Sonia takes $18$ minutes to complete one round Ravi takes $12$ minutes to complete one round The LCM of $18$ and $12$ will give us the time it takes for them to meet again at the starting point. So, firstly we find LCM. Prime factors of $18: 2 × 3 × 3$ (by prime factorization) Prime factors of $12: 2 × 2 × 3$ (by prime factorization) Now, calculate the LCM by taking the highest power of each prime factor: $\text {LCM}(18, 12) = 2^2 \times 3^2  = 4 \times 9 = 36$ Sonia and Ravi will meet again at the starting point after $36$ minutes.