Class 10-NCERT Solutions-Chapter-01-Real Numbers

Solution:

We use the division method or factor tree to find the prime factors.

(i) 140

\[ 140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7 \]

\[ \Rightarrow 140 = 2^2 \times 5 \times 7 \]

(ii) 156

\[ 156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13 \]

\[ \Rightarrow 156 = 2^2 \times 3 \times 13 \]

(iii) 3825

\[ 3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3 \times 3 \times 5 \times 85 = 3 \times 3 \times 5 \times 5 \times 17 \]

\[ \Rightarrow 3825 = 3^2 \times 5^2 \times 17 \]

(iv) 5005

\[ 5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13 \]

\[ \Rightarrow 5005 = 5 \times 7 \times 11 \times 13 \]

(v) 7429

\[ 7429 = 17 \times 437 = 17 \times 19 \times 23 \]

\[ \Rightarrow 7429 = 17 \times 19 \times 23 \]

Solution:

(i) 26 and 91

Prime factors:

\[ 26 = 2 \times 13 \]

\[ 91 = 7 \times 13 \]

\[ \text{HCF} = 13 \]

\[ \text{LCM} = 2 \times 7 \times 13 = 182 \]

Verification:

\[ \text{LCM} \times \text{HCF} = 182 \times 13 = 2366 \]

\[ \text{Product of numbers} = 26 \times 91 = 2366 \]

Hence verified.

(ii) 510 and 92

Prime factors:

\[ 510 = 2 \times 3 \times 5 \times 17 \]

\[ 92 = 2 \times 2 \times 23 = 2^2 \times 23 \]

\[ \text{HCF} = 2 \]

\[ \text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \]

Verification:

\[ \text{LCM} \times \text{HCF} = 23460 \times 2 = 46920 \]

\[ \text{Product of numbers} = 510 \times 92 = 46920 \]

Hence verified.

(iii) 336 and 54

Prime factors:

\[ 336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^4 \times 3 \times 7 \]

\[ 54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3 \]

\[ \text{HCF} = 2 \times 3 = 6 \]

\[ \text{LCM} = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024 \]

Verification:

\[ \text{LCM} \times \text{HCF} = 3024 \times 6 = 18144 \]

\[ \text{Product of numbers} = 336 \times 54 = 18144 \]

Hence verified.

Solution:

(i) 12, 15 and 21

\[ 12 = 2^2 \times 3 \]

\[ 15 = 3 \times 5 \]

\[ 21 = 3 \times 7 \]

\[ \text{HCF} = 3 \]

\[ \text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420 \]

(ii) 17, 23 and 29

These numbers are prime numbers.

\[ 17 = 17 \times 1 \]

\[ 23 = 23 \times 1 \]

\[ 29 = 29 \times 1 \]

\[ \text{HCF} = 1 \]

\[ \text{LCM} = 17 \times 23 \times 29 = 11339 \]

(iii) 8, 9 and 25

\[ 8 = 2^3 \]

\[ 9 = 3^2 \]

\[ 25 = 5^2 \]

There is no common factor other than 1.

\[ \text{HCF} = 1 \]

\[ \text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \]

Solution:

We know that for any two positive integers \( a \) and \( b \):

\[ \text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b \]

Given \( a = 306, b = 657, \text{HCF} = 9 \).

\[ \text{LCM}(306, 657) \times 9 = 306 \times 657 \]

\[ \text{LCM} = \frac{306 \times 657}{9} \]

\[ \text{LCM} = 34 \times 657 = 22338 \]

Answer: 22338

Solution:

If any number ends with the digit 0, it should be divisible by 10, which means it must be divisible by both 2 and 5.

The prime factorisation of \( 6^n \) is:

\[ 6^n = (2 \times 3)^n = 2^n \times 3^n \]

It can be observed that 5 is not in the prime factorisation of \( 6^n \). By the Fundamental Theorem of Arithmetic, this factorisation is unique.

Therefore, for any natural number \( n \), \( 6^n \) is not divisible by 5.

Consequently, \( 6^n \) cannot end with the digit 0 for any natural number \( n \).

Solution:

A number is called a composite number if it has factors other than 1 and itself.

First Number:

\[ 7 \times 11 \times 13 + 13 \]

\[ = 13 \times (7 \times 11 + 1) \]

\[ = 13 \times (77 + 1) \]

\[ = 13 \times 78 \]

Since this number has factors 13 and 78 (other than 1 and itself), it is a composite number.

Second Number:

\[ 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \]

\[ = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \]

\[ = 5 \times (1008 + 1) \]

\[ = 5 \times 1009 \]

Since this number has factors 5 and 1009, it is a composite number.

Solution:

They will meet again at the starting point at a time that is a common multiple of the time taken by both to complete one round. To find the minimum time, we need to calculate the LCM of 18 and 12.

\[ 18 = 2 \times 3^2 \]

\[ 12 = 2^2 \times 3 \]

\[ \text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36 \]

Therefore, Ravi and Sonia will meet again at the starting point after 36 minutes.

Solution:

Let us assume, to the contrary, that \( \sqrt{5} \) is rational.

Then we can find coprime integers \( a \) and \( b \) (\( b \neq 0 \)) such that:

\[ \sqrt{5} = \frac{a}{b} \]

\[ \Rightarrow \sqrt{5}b = a \]

Squaring on both sides:

\[ 5b^2 = a^2 \quad \dots(1) \]

Therefore, 5 divides \( a^2 \). By Theorem 1.2, it follows that 5 divides \( a \).

So, we can write \( a = 5c \) for some integer \( c \).

Substituting for \( a \) in equation (1):

\[ 5b^2 = (5c)^2 \]

\[ 5b^2 = 25c^2 \]

\[ b^2 = 5c^2 \]

This means that 5 divides \( b^2 \), and so 5 divides \( b \).

Therefore, \( a \) and \( b \) have at least 5 as a common factor. But this contradicts the fact that \( a \) and \( b \) are coprime.

This contradiction has arisen because of our incorrect assumption that \( \sqrt{5} \) is rational. So, we conclude that \( \sqrt{5} \) is irrational.

Solution:

Let us assume, to the contrary, that \( 3 + 2\sqrt{5} \) is rational.

That is, we can find coprime integers \( a \) and \( b \) (\( b \neq 0 \)) such that:

\[ 3 + 2\sqrt{5} = \frac{a}{b} \]

Rearranging the equation:

\[ 2\sqrt{5} = \frac{a}{b} - 3 \]

\[ 2\sqrt{5} = \frac{a - 3b}{b} \]

\[ \sqrt{5} = \frac{a - 3b}{2b} \]

Since \( a \) and \( b \) are integers, \( \frac{a - 3b}{2b} \) is rational. Thus, \( \sqrt{5} \) is rational.

But this contradicts the fact that \( \sqrt{5} \) is irrational.

This contradiction has arisen because of our incorrect assumption that \( 3 + 2\sqrt{5} \) is rational.

So, we conclude that \( 3 + 2\sqrt{5} \) is irrational.

Solution:

(i) \( \frac{1}{\sqrt{2}} \)

Assume \( \frac{1}{\sqrt{2}} \) is rational. Then,

\[ \frac{1}{\sqrt{2}} = \frac{a}{b} \quad \text{(where a, b are coprime integers, } b \neq 0 \text{)} \]

\[ \Rightarrow \sqrt{2} = \frac{b}{a} \]

Since \( a, b \) are integers, \( \frac{b}{a} \) is rational. This implies \( \sqrt{2} \) is rational.

This contradicts the fact that \( \sqrt{2} \) is irrational. Hence, \( \frac{1}{\sqrt{2}} \) is irrational.

(ii) \( 7\sqrt{5} \)

Assume \( 7\sqrt{5} \) is rational. Then,

\[ 7\sqrt{5} = \frac{a}{b} \quad \text{(where a, b are coprime integers, } b \neq 0 \text{)} \]

\[ \Rightarrow \sqrt{5} = \frac{a}{7b} \]

Since \( a, b \) are integers, \( \frac{a}{7b} \) is rational. This implies \( \sqrt{5} \) is rational.

This contradicts the fact that \( \sqrt{5} \) is irrational. Hence, \( 7\sqrt{5} \) is irrational.

(iii) \( 6 + \sqrt{2} \)

Assume \( 6 + \sqrt{2} \) is rational. Then,

\[ 6 + \sqrt{2} = \frac{a}{b} \quad \text{(where a, b are coprime integers, } b \neq 0 \text{)} \]

\[ \Rightarrow \sqrt{2} = \frac{a}{b} - 6 \]

\[ \Rightarrow \sqrt{2} = \frac{a - 6b}{b} \]

Since \( a, b \) are integers, \( \frac{a - 6b}{b} \) is rational. This implies \( \sqrt{2} \) is rational.

This contradicts the fact that \( \sqrt{2} \) is irrational. Hence, \( 6 + \sqrt{2} \) is irrational.