Rbse Solutions for Class 10 Maths Chapter 1 Exercise 1.1 | Real Numbers

Get step-by-step solutions for NCERT Class 10 Maths Chapter 1 Exercise 1.1. Master the Fundamental Theorem of Arithmetic by expressing numbers as a product of prime factors (Q.1). Learn to find the LCM (Least Common Multiple) and HCF (Highest Common Factor) using the prime factorization method (Q.2-3). Verify the relationship: LCM × HCF = Product of two numbers. Solutions include checking if a number ends with zero (Q.5) and explaining why numbers are composite (Q.6-7). Essential practice for Board Exams.

This exercise focuses on the Fundamental Theorem of Arithmetic (prime factorization) and its applications in finding LCM (Least Common Multiple) and HCF (Highest Common Factor).

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1. Express each number as a product of its prime factors:

Prime factorization is the process of finding which prime numbers multiply together to make the original number.

Question	Factorization	Product of Prime Factors
(i) 140	$140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7$	$\mathbf{2^2 \times 5 \times 7}$
(ii) 156	$156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$	$\mathbf{2^2 \times 3 \times 13}$
(iii) 3825	$3825 = 5 \times 765 = 5 \times 5 \times 153 = 5 \times 5 \times 3 \times 51 = 5 \times 5 \times 3 \times 3 \times 17$	$\mathbf{3^2 \times 5^2 \times 17}$
(iv) 5005	$5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$	$\mathbf{5 \times 7 \times 11 \times 13}$
(v) 7429	$7429 = 17 \times 437 = 17 \times 19 \times 23$	$\mathbf{17 \times 19 \times 23}$

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2. Find the LCM and HCF and verify $\text{LCM} \times \text{HCF} = \text{Product of numbers}$.

(i) 26 and 91

1. Prime Factors:$$26 = 2 \times 13$$$$91 = 7 \times 13$$
2. HCF: Product of the smallest power of each common prime factor.$$\text{HCF}(26, 91) = \mathbf{13}$$
3. LCM: Product of the greatest power of all prime factors.$$\text{LCM}(26, 91) = 2^1 \times 7^1 \times 13^1 = 182$$
4. Verification:$$\text{LCM} \times \text{HCF} = 182 \times 13 = 2366$$$$\text{Product of numbers} = 26 \times 91 = 2366$$Since $2366 = 2366$, the verification holds.

(ii) 510 and 92

1. Prime Factors:$$510 = 2 \times 3 \times 5 \times 17$$$$92 = 2 \times 2 \times 23 = 2^2 \times 23$$
2. HCF:$$\text{HCF}(510, 92) = 2^1 = \mathbf{2}$$
3. LCM:$$\text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 4 \times 3 \times 5 \times 17 \times 23 = \mathbf{23460}$$
4. Verification:$$\text{LCM} \times \text{HCF} = 23460 \times 2 = 46920$$$$\text{Product of numbers} = 510 \times 92 = 46920$$Since $46920 = 46920$, the verification holds.

(iii) 336 and 54

1. Prime Factors:$$336 = 2^4 \times 3 \times 7$$$$54 = 2 \times 3^3$$
2. HCF: Product of the smallest power of common prime factors (2 and 3).$$\text{HCF}(336, 54) = 2^1 \times 3^1 = \mathbf{6}$$
3. LCM: Product of the greatest power of all prime factors (2, 3, and 7).$$\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = \mathbf{3024}$$
4. Verification:$$\text{LCM} \times \text{HCF} = 3024 \times 6 = 18144$$$$\text{Product of numbers} = 336 \times 54 = 18144$$Since $18144 = 18144$, the verification holds.

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3. Find the LCM and HCF using the prime factorisation method.

(i) 12, 15 and 21

1. Prime Factors:$$12 = 2^2 \times 3$$$$15 = 3 \times 5$$$$21 = 3 \times 7$$
2. HCF: Smallest power of the common prime factor (3).$$\text{HCF}(12, 15, 21) = 3^1 = \mathbf{3}$$
3. LCM: Greatest power of all prime factors (2, 3, 5, 7).$$\text{LCM}(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = \mathbf{420}$$

(ii) 17, 23 and 29

1. Prime Factors: (All are prime numbers)$$17 = 17$$$$23 = 23$$$$29 = 29$$
2. HCF: The only common factor is 1.$$\text{HCF}(17, 23, 29) = \mathbf{1}$$
3. LCM: Product of the numbers.$$\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = \mathbf{11339}$$

(iii) 8, 9 and 25

1. Prime Factors:$$8 = 2^3$$$$9 = 3^2$$$$25 = 5^2$$
2. HCF: There are no common prime factors.$$\text{HCF}(8, 9, 25) = \mathbf{1}$$
3. LCM: Product of the numbers.$$\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = \mathbf{1800}$$

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4. Given $\text{HCF}(306, 657) = 9$, find $\text{LCM}(306, 657)$.

We use the fundamental relationship between LCM and HCF for two positive integers $a$ and $b$:

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

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

$$\text{LCM}(306, 657) = \frac{306 \times 657}{9}$$

$$\text{LCM}(306, 657) = 34 \times 657$$

$$\text{LCM}(306, 657) = \mathbf{22338}$$

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

A number ends with the digit 0 if and only if its prime factorization includes factors of $\mathbf{2}$ and $\mathbf{5}$ (i.e., it is divisible by $2 \times 5 = 10$).

The prime factorization of the base $6$ is:

$$6 = 2 \times 3$$

The prime factorization of $6^n$ is:

$$6^n = (2 \times 3)^n = 2^n \times 3^n$$

According to the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique. The factorization of $6^n$ contains only the primes 2 and 3, but lacks the prime factor 5.

Therefore, $6^n$ is not divisible by 5, and hence $6^n$ cannot end with the digit 0 for any natural number $n$.

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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.

A composite number is a natural number greater than 1 that is not a prime number; that is, it has factors other than 1 and itself.

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

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 the number can be expressed as a product of two factors, 13 and 78, other than 1 and the number itself, it is a composite number.

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

We can take 5 as a common factor:

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

$$= 5 \times (1008 + 1)$$

$$= 5 \times 1009$$

Since the number can be expressed as a product of two factors, 5 and 1009, other than 1 and the number itself, it is a composite number.

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7. Circular Path Problem

To find the time when Sonia and Ravi will meet again at the starting point, we need to find the Least Common Multiple (LCM) of the time taken by each.

• Sonia’s time: 18 minutes
• Ravi’s time: 12 minutes

1. Prime Factorization:$$18 = 2 \times 3^2$$$$12 = 2^2 \times 3$$
2. LCM: Product of the greatest power of all prime factors (2 and 3).$$\text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$$

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