NCERT Solutions For Class 10 Maths: Real Numbers

EXERCISE 1.1

Question 1: Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Solution-

(i) 140
140 = 2 × 70
140 = 2 × 2 × 35
140 = 2 × 2 × 5 × 7
Therefore, $140 = 2^2 \times 5 \times 7$.

(ii) 156
156 = 2 × 78
156 = 2 × 2 × 39
156 = 2 × 2 × 3 × 13
Therefore, $156 = 2^2 \times 3 \times 13$.

(iii) 3825
3825 = 5 × 765
3825 = 5 × 5 × 153
3825 = 5 × 5 × 3 × 51
3825 = 5 × 5 × 3 × 3 × 17
Therefore, $3825 = 3^2 \times 5^2 \times 17$.

(iv) 5005
5005 = 5 × 1001
5005 = 5 × 7 × 143
5005 = 5 × 7 × 11 × 13
Therefore, $5005 = 5 \times 7 \times 11 \times 13$.

(v) 7429
7429 = 17 × 437
7429 = 17 × 19 × 23
Therefore, $7429 = 17 \times 19 \times 23$.

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

Solution-

(i) 26 and 91

Prime factorisation of $26 = 2 \times 13$.
Prime factorisation of $91 = 7 \times 13$.

HCF is the product of the smallest power of each common prime factor.
HCF$(26, 91) = 13^1 = 13$.

LCM is the product of the greatest power of each prime factor involved in the numbers.
LCM$(26, 91) = 2^1 \times 7^1 \times 13^1 = 182$.

Verification:
Product of the two numbers $= 26 \times 91 = 2366$.
Product of HCF and LCM $= 13 \times 182 = 2366$.
Since LCM $\times$ HCF $=$ product of the two numbers, the verification holds true.

(ii) 510 and 92

Prime factorisation of $510 = 2 \times 3 \times 5 \times 17$.
Prime factorisation of $92 = 2 \times 2 \times 23 = 2^2 \times 23$.

HCF is the product of the smallest power of each common prime factor in the numbers.
HCF$(510, 92) = 2^1 = 2$.

LCM is the product of the greatest power of each prime factor, involved in the numbers.
LCM$(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$.

Verification:
Product of the two numbers $= 510 \times 92 = 46920$.
Product of HCF and LCM $= 2 \times 23460 = 46920$.
Since LCM $\times$ HCF $=$ product of the two numbers, the verification holds true.

(iii) 336 and 54

Prime factorisation of $336 = 2 \times 168 = 2^4 \times 3 \times 7$.
Prime factorisation of $54 = 2 \times 27 = 2 \times 3^3$.

HCF is the product of the smallest power of each common prime factor in the numbers.
The common factors are 2 and 3. Smallest powers are $2^1$ and $3^1$.
HCF$(336, 54) = 2^1 \times 3^1 = 6$.

LCM is the product of the greatest power of each prime factor, involved in the numbers.
Greatest powers are $2^4$, $3^3$, and $7^1$.
LCM$(336, 54) = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$.

Verification:
Product of the two numbers $= 336 \times 54 = 18144$.
Product of HCF and LCM $= 6 \times 3024 = 18144$.
Since LCM $\times$ HCF $=$ product of the two numbers, the verification holds true.

Question 3: Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25

Solution-

(i) 12, 15 and 21

Prime factorisation:
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$

HCF is the product of the smallest power of each common prime factor.
The only common prime factor is 3.
HCF$(12, 15, 21) = 3^1 = 3$.

LCM is the product of the greatest power of each prime factor involved.
LCM$(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$.

(ii) 17, 23 and 29

Prime factorisation:
$17 = 17^1$
$23 = 23^1$
$29 = 29^1$

HCF is the product of the smallest power of each common prime factor. Since 17, 23, and 29 are all prime numbers, they share no common prime factors other than 1.
HCF$(17, 23, 29) = 1$.

LCM is the product of the greatest power of each prime factor involved.
LCM$(17, 23, 29) = 17 \times 23 \times 29 = 11339$.

(iii) 8, 9 and 25

Prime factorisation:
$8 = 2 \times 2 \times 2 = 2^3$
$9 = 3 \times 3 = 3^2$
$25 = 5 \times 5 = 5^2$

HCF is the product of the smallest power of each common prime factor. Since they share no common prime factors, the HCF is 1.
HCF$(8, 9, 25) = 1$.

LCM is the product of the greatest power of each prime factor involved.
LCM$(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$.

Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution-

We know that for any two positive integers $a$ and $b$:
HCF$(a, b) \times$ LCM$(a, b) = a \times b$.

Here, $a = 306$, $b = 657$, and HCF$(306, 657) = 9$.

$9 \times$ LCM$(306, 657) = 306 \times 657$
LCM$(306, 657) = \frac{306 \times 657}{9}$
LCM$(306, 657) = 34 \times 657$
LCM$(306, 657) = 22338$.

Question 5: Check whether $6^n$ can end with the digit 0 for any natural number $n$.

Solution-

If the number $6^n$, for any natural number $n$, were to end with the digit zero, then it would be divisible by 5.

If a number is divisible by 5, then the prime factorisation of that number must contain the prime 5.

The prime factorisation of $6^n$:
$6^n = (2 \times 3)^n = 2^n \times 3^n$.

The only primes in the factorisation of $6^n$ are 2 and 3. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of $6^n$. Since the prime factorisation of $6^n$ does not contain the prime 5, it is not divisible by 5.

Therefore, there is no natural number $n$ for which $6^n$ ends with the digit zero.

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.

Solution-

A composite number is any natural number that can be expressed (factorised) as a product of primes.

First Number: $7 \times 11 \times 13 + 13$

We can take 13 as a common factor:
$7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1)$
$13 \times (77 + 1)$
$13 \times 78$

Since $78 = 2 \times 39 = 2 \times 3 \times 13$, we can write:
$13 \times (2 \times 3 \times 13)$
$13^2 \times 2 \times 3$

Since the number is expressed as a product of primes (13, 2, and 3), it is a composite number.

Second Number: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

We can take 5 as a common factor:
$5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$
$5 \times (1008 + 1)$
$5 \times 1009$

Since 1009 is a prime number (this must be verified, similar to checking 3803 and 3607 in Example 8), the number is expressed as a product of primes (5 and 1009).

Since both given expressions can be factorized as a product of primes, they are composite numbers.

Question 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?

Solution-

They will meet again at the starting point when the time elapsed is a common multiple of both 18 minutes and 12 minutes. Since they start at the same time and go in the same direction, the moment they meet again for the first time will be the Least Common Multiple (LCM) of 18 and 12.

Find the prime factorisation of 18 and 12:
$18 = 2 \times 9 = 2 \times 3^2$
$12 = 2 \times 6 = 2^2 \times 3$

LCM is the product of the greatest power of each prime factor involved in the numbers.
LCM$(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$.

They will meet again at the starting point after 36 minutes.

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EXERCISE 1.2

Question 1: Prove that $\sqrt{5}$ is irrational.

Solution-

We use the method of proof by contradiction.

Assume, to the contrary, that $\sqrt{5}$ is rational.
If $\sqrt{5}$ is rational, we can find integers $a$ and $b$ ($b \ne 0$) such that $\sqrt{5} = \frac{a}{b}$.

Suppose $a$ and $b$ have a common factor other than 1. We divide by this common factor and assume that $a$ and $b$ are coprime.

So, $\sqrt{5}b = a$.

Squaring on both sides, we get $5b^2 = a^2$.
This means that $a^2$ is divisible by 5.
By Theorem 1.2 (If $p$ divides $a^2$, then $p$ divides $a$, where $p$ is a prime), it follows that $a$ is also divisible by 5.

We can write $a = 5c$ for some integer $c$.

Substituting $a=5c$ into the equation $5b^2 = a^2$:
$5b^2 = (5c)^2$
$5b^2 = 25c^2$
$b^2 = 5c^2$.

This means that $b^2$ is divisible by 5, and by Theorem 1.2 (with $p=5$), it follows that $b$ is also divisible by 5.

Therefore, $a$ and $b$ have at least 5 as a common factor.
But this contradicts the fact that $a$ and $b$ are coprime (have no common factors other than 1).

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.
We conclude that $\sqrt{5}$ is irrational.

Question 2: Prove that $3 + 2\sqrt{5}$ is irrational.

Solution-

We use the method of proof by contradiction.

Assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.
That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that $3 + 2\sqrt{5} = \frac{a}{b}$.

Rearranging the equation to isolate $\sqrt{5}$:
$2\sqrt{5} = \frac{a}{b} - 3$
$2\sqrt{5} = \frac{a - 3b}{b}$
$\sqrt{5} = \frac{a - 3b}{2b}$

Since $a$ and $b$ are integers, $a-3b$ and $2b$ are also integers. Therefore, $\frac{a - 3b}{2b}$ is a rational number.

This implies that $\sqrt{5}$ is rational.
But this contradicts the fact that $\sqrt{5}$ is irrational (as proven in Question 1).

This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.
We conclude that $3 + 2\sqrt{5}$ is irrational.

Question 3: Prove that the following are irrationals:
(i) $\frac{1}{\sqrt{2}}$
(ii) $7\sqrt{5}$
(iii) $6 + \sqrt{2}$

Solution-

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

Assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational.
We can write $\frac{1}{\sqrt{2}} = \frac{a}{b}$, where $a$ and $b$ are coprime integers ($b \ne 0$).

Rearranging the equation gives $\sqrt{2} = \frac{b}{a}$.

Since $a$ and $b$ are integers, $\frac{b}{a}$ is rational.
This implies that $\sqrt{2}$ is rational.

But this contradicts the known fact that $\sqrt{2}$ is irrational.
This contradiction proves that $\frac{1}{\sqrt{2}}$ is irrational.

(ii) $7\sqrt{5}$

Assume, to the contrary, that $7\sqrt{5}$ is rational.
We can write $7\sqrt{5} = \frac{a}{b}$, where $a$ and $b$ are coprime integers ($b \ne 0$).

Rearranging the equation to isolate $\sqrt{5}$:
$\sqrt{5} = \frac{a}{7b}$

Since $a, 7$, and $b$ are integers, $\frac{a}{7b}$ is a rational number.
This implies that $\sqrt{5}$ is rational.

But this contradicts the fact that $\sqrt{5}$ is irrational.
We conclude that $7\sqrt{5}$ is irrational. (This follows the principle that the product and quotient of a non-zero rational and an irrational number is irrational).

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

Assume, to the contrary, that $6 + \sqrt{2}$ is rational.
We can find coprime integers $a$ and $b$ ($b \ne 0$) such that $6 + \sqrt{2} = \frac{a}{b}$.

Rearranging the equation to isolate $\sqrt{2}$:
$\sqrt{2} = \frac{a}{b} - 6$
$\sqrt{2} = \frac{a - 6b}{b}$

Since $a$ and $b$ are integers, $\frac{a - 6b}{b}$ is a rational number.
This implies that $\sqrt{2}$ is rational.

But this contradicts the known fact that $\sqrt{2}$ is irrational.
This contradiction proves that $6 + \sqrt{2}$ is irrational. (This follows the principle that the sum or difference of a rational and an irrational number is irrational).