Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the two numbers: (i) 45 and 120 (ii) 84 and 126 (iii) 360 and 225

(i) 45 and 120: HCF = 15, LCM = 360. Product = 5400 (verified). (ii) 84 and 126: HCF = 42, LCM = 252. Product = 10584 (verified). (iii) 360 and 225: HCF = 45, LCM = 1800. Product = 81000 (verified).

Find the LCM and HCF of the following numbers: (i) 18, 24 and 30 (ii) 11, 17 and 31 (iii) 12, 20 and 45

(i) HCF = 6, LCM = 360 (ii) HCF = 1, LCM = 5797 (iii) HCF = 1, LCM = 180

Explain why the following numbers are composite: (i) 11 × 13 × 17 + 17 (ii) 8 × 7 × 6 × 5 × 4 × 3 + 6

(i) 11 × 13 × 17 + 17 = 17(11 × 13 + 1). Since 17 is a factor, it is composite. (ii) 8 × 7 × 6 × 5 × 4 × 3 + 6 = 6(8 × 7 × 5 × 4 × 3 + 1). Since 6 is a factor, it is composite.

Class 10 Maths Chapter 1 Exercise 1.1 NCERT Solutions

Q: Question 1

(i) 168 = 2^3 × 3 × 7 (ii) 252 = 2^2 × 3^2 × 7 (iii) 4095 = 3^2 × 5 × 7 × 13 (iv) 6006 = 2 × 3 × 7 × 11 × 13 (v) 7425 = 3^3 × 5^2 × 11

Q: Question 4

Using formula LCM × HCF = Product of numbers, LCM = 420 / 6 = 70.

Q: Question 5

Prime factorisation of 9^n = (3^2)^n = 3^(2n). A number ends with 0 only if it contains both prime factors 2 and 5. Since 9^n has neither 2 nor 5, it can never end with 0.

Q: Question 7

They will meet again after LCM(20, 30) = 60 minutes.

Question 1

Express each of the following numbers as a product of their prime factors:

(i) $168$ (ii) $252$ (iii) $4095$ (iv) $6006$ (v) $7425$

(i) Prime Factorisation of 168

$$\begin{aligned} 168 &\div 2 = 84 \\ 84 &\div 2 = 42 \\ 42 &\div 2 = 21 \\ 21 &\div 3 = 7 \\ 7 &\div 7 = 1 \end{aligned}$$ $$\therefore 168 = 2 \times 2 \times 2 \times 3 \times 7 = 2^3 \times 3 \times 7$$

(ii) Prime Factorisation of 252

$$\begin{aligned} 252 &\div 2 = 126 \\ 126 &\div 2 = 63 \\ 63 &\div 3 = 21 \\ 21 &\div 3 = 7 \\ 7 &\div 7 = 1 \end{aligned}$$ $$\therefore 252 = 2^2 \times 3^2 \times 7$$

(iii) Prime Factorisation of 4095

$$\begin{aligned} 4095 &\div 3 = 1365 \\ 1365 &\div 3 = 455 \\ 455 &\div 5 = 91 \\ 91 &\div 7 = 13 \\ 13 &\div 13 = 1 \end{aligned}$$ $$\therefore 4095 = 3^2 \times 5 \times 7 \times 13$$

(iv) Prime Factorisation of 6006

$$\begin{aligned} 6006 &\div 2 = 3003 \\ 3003 &\div 3 = 1001 \\ 1001 &\div 7 = 143 \\ 143 &\div 11 = 13 \\ 13 &\div 13 = 1 \end{aligned}$$ $$\therefore 6006 = 2 \times 3 \times 7 \times 11 \times 13$$

(v) Prime Factorisation of 7425

$$\begin{aligned} 7425 &\div 3 = 2475 \\ 2475 &\div 3 = 825 \\ 825 &\div 3 = 275 \\ 275 &\div 5 = 55 \\ 55 &\div 5 = 11 \\ 11 &\div 11 = 1 \end{aligned}$$ $$\therefore 7425 = 3^3 \times 5^2 \times 11$$

Fundamental Theorem of Arithmetic: Every composite number can be expressed uniquely as a product of prime powers.

Question 2

Find the LCM and HCF of the following pairs of integers and verify that:

$$\text{LCM} \times \text{HCF} = \text{Product of the two numbers}$$

(i) 45 and 120 (ii) 84 and 126 (iii) 360 and 225

(i) 45 and 120

Prime factorisation:
$$45 = 3^2 \times 5$$ $$120 = 2^3 \times 3 \times 5$$ $$\text{HCF} = 3 \times 5 = 15$$ $$\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$$
Verification:
$$\text{LCM} \times \text{HCF} = 360 \times 15 = 5400$$ $$\text{Product of numbers} = 45 \times 120 = 5400$$ $$\text{LCM} \times \text{HCF} = \text{Product of numbers} \quad \checkmark$$

(ii) 84 and 126

Prime factorisation:
$$84 = 2^2 \times 3 \times 7$$ $$126 = 2 \times 3^2 \times 7$$ $$\text{HCF} = 2 \times 3 \times 7 = 42$$ $$\text{LCM} = 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 252$$
Verification:
$$\text{LCM} \times \text{HCF} = 252 \times 42 = 10584$$ $$\text{Product of numbers} = 84 \times 126 = 10584$$ $$\text{LCM} \times \text{HCF} = \text{Product of numbers} \quad \checkmark$$

(iii) 360 and 225

Prime factorisation:
$$360 = 2^3 \times 3^2 \times 5$$ $$225 = 3^2 \times 5^2$$ $$\text{HCF} = 3^2 \times 5 = 9 \times 5 = 45$$ $$\text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$$
Verification:
$$\text{LCM} \times \text{HCF} = 1800 \times 45 = 81000$$ $$\text{Product of numbers} = 360 \times 225 = 81000$$ $$\text{LCM} \times \text{HCF} = \text{Product of numbers} \quad \checkmark$$

Verified: In all pairs, $\text{LCM} \times \text{HCF} = \text{Product of the two numbers}$.

Question 3

Find the LCM and HCF of the following numbers by applying the prime factorisation method:

(i) 18, 24 and 30 (ii) 11, 17 and 31 (iii) 12, 20 and 45

(i) 18, 24 and 30

$$18 = 2 \times 3^2$$ $$24 = 2^3 \times 3$$ $$30 = 2 \times 3 \times 5$$ $$\text{HCF} = 2 \times 3 = 6$$ $$\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$$

(ii) 11, 17 and 31

11, 17, and 31 are all prime numbers with no common factor other than 1.
$$\text{HCF} = 1$$ $$\text{LCM} = 11 \times 17 \times 31 = 5797$$

(iii) 12, 20 and 45

$$12 = 2^2 \times 3$$ $$20 = 2^2 \times 5$$ $$45 = 3^2 \times 5$$ $$\text{HCF} = 1 \quad \text{(no prime factor common to all three)}$$ $$\text{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$$

Note: For three or more numbers, $\text{LCM} \times \text{HCF} \neq \text{Product of numbers}$.

Question 4

If the HCF of two numbers is 6 and their product is 420, find their LCM.

Using Product Formula

We know the relation between LCM and HCF of two numbers: $$\text{LCM} \times \text{HCF} = \text{Product of the two numbers}$$ Substitute the given values: $$\text{LCM} \times 6 = 420$$ $$\text{LCM} = \frac{420}{6} = 70$$

$$\therefore \text{LCM} = 70$$

Question 5

Check whether $9^n$ can end with the digit 0 for any natural number $n \in \mathbb{N}$.

Divisibility Condition for Digit 0

A number ends with the digit 0 if and only if it is divisible by 10.
Since $10 = 2 \times 5$, its prime factorisation must contain both prime factors 2 and 5.

Prime Factorisation of $9^n$

$$9 = 3 \times 3 = 3^2$$ $$9^n = (3^2)^n = 3^{2n}$$ The only prime factor of $9^n$ is 3.
By the uniqueness of the Fundamental Theorem of Arithmetic, no prime factor other than 3 exists in $9^n$.

$\therefore$ Since $9^n$ has neither 2 nor 5 as prime factors, $9^n$ cannot end with the digit 0 for any natural number $n$.

Question 6

Explain why the following numbers are composite:

(i) $11 \times 13 \times 17 + 17$ (ii) $8 \times 7 \times 6 \times 5 \times 4 \times 3 + 6$

(i) Evaluating $11 \times 13 \times 17 + 17$

Take common factor 17: $$\begin{aligned} 11 \times 13 \times 17 + 17 &= 17 \times (11 \times 13 + 1) \\ &= 17 \times (143 + 1) \\ &= 17 \times 144 \end{aligned}$$ Since the number has factors other than 1 and itself (specifically 17 and 144), it is a composite number.

(ii) Evaluating $8 \times 7 \times 6 \times 5 \times 4 \times 3 + 6$

Take common factor 6: $$\begin{aligned} 8 \times 7 \times 6 \times 5 \times 4 \times 3 + 6 &= 6 \times (8 \times 7 \times 5 \times 4 \times 3 + 1) \\ &= 6 \times (3360 + 1) \\ &= 6 \times 3361 \end{aligned}$$ Since the number has factors other than 1 and itself (specifically 6 and 3361), it is a composite number.

By definition, any number having factors other than 1 and itself is a composite number. Both expressions are composite.

Question 7

Two students take part in a circular race. The first student completes one round in 20 minutes and the second student completes one round in 30 minutes. If both start at the same time from the same point and in the same direction, after how many minutes will they meet again at the starting point?

Understanding the Problem

Time taken by first student to complete one round = 20 minutes.
Time taken by second student to complete one round = 30 minutes.
They will meet again at the starting point after a time which is the Least Common Multiple (LCM) of 20 and 30 minutes.

Finding $\text{LCM}(20, 30)$

Prime factorisation: $$20 = 2^2 \times 5$$ $$30 = 2 \times 3 \times 5$$ $$\text{LCM}(20, 30) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$$

$\therefore$ They will meet again at the starting point after 60 minutes (1 hour).

Real Numbers - Exercise 1.1

Overview