Real Numbers – Exercise 1.1

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) 168
168 ÷ 2 = 84 84 ÷ 2 = 42 42 ÷ 2 = 21 21 ÷ 3 = 7 7 ÷ 7 = 1 ⇒ $168 = 2^3 × 3 × 7$

(ii) 252
252 ÷ 2 = 126 126 ÷ 2 = 63 63 ÷ 3 = 21 21 ÷ 3 = 7 7 ÷ 7 = 1 ⇒ $252 = 2^2 × 3^2 × 7$

(iii) 4095
4095 ÷ 3 = 1365 1365 ÷ 3 = 455 455 ÷ 5 = 91 91 ÷ 7 = 13 13 ÷ 13 = 1 ⇒ $4095 = 3^2 × 5 × 7 × 13$

(iv) 6006
6006 ÷ 2 = 3003 3003 ÷ 3 = 1001 1001 ÷ 7 = 143 143 ÷ 11 = 13 13 ÷ 13 = 1 ⇒ $6006 = 2 × 3 × 7 × 11 × 13$

(v) 7425
7425 ÷ 3 = 2475 2475 ÷ 3 = 825 825 ÷ 3 = 275 275 ÷ 5 = 55 55 ÷ 5 = 11 11 ÷ 11 = 1 ⇒ $7425 = 3^3 × 5^2 × 11$

Question 2

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
45 = $3^2 × 5$
120 = $2^3 × 3 × 5$
HCF = 3 × 5 = 15
LCM = $2^3 × 3^2 × 5 = 360$
Check: 15 × 360 = 5400 ✔

(ii) 84 and 126
84 = $2^2 × 3 × 7$
126 = $2 × 3^2 × 7$
HCF = 2 × 3 × 7 = 42
LCM = $2^2 × 3^2 × 7 = 252$
Check: 42 × 252 = 10584 ✔

(iii) 360 and 225
360 = $2^3 × 3^2 × 5$
225 = $3^2 × 5^2$
HCF = $3^2 × 5 = 45$
LCM = $2^3 × 3^2 × 5^2 = 1800$
Check: 45 × 1800 = 81000 ✔

Question 3

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)
18 = $2 × 3^2$
24 = $2^3 × 3$
30 = $2 × 3 × 5$
HCF = 2 × 3 = 6
LCM = $2^3 × 3^2 × 5 = 360$

(ii)
11, 17 and 31 are all prime numbers
HCF = 1
LCM = 11 × 17 × 31 = 5797

(iii)
12 = $2^2 × 3$
20 = $2^2 × 5$
45 = $3^2 × 5$
HCF = 1
LCM = $2^2 × 3^2 × 5 = 180$

Question 4

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

Using formula:
LCM × HCF = Product of numbers

LCM × 6 = 420
LCM = 420 ÷ 6
LCM = 70

Question 5

Check whether $9^n$ can end with the digit 0 for any natural number n.

9 = 3 × 3 So, $9^n = 3^{2n}$ Prime factorization of $9^n$ contains only factor 3. A number ends with 0 only if it has both factors 2 and 5. Since $9^n$ has neither 2 nor 5, it can never end with 0. ✅ Therefore, $9^n$ cannot end with 0 for any natural number n.

Question 6

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, the number is composite.

(ii)
$8×7×6×5×4×3 + 6$ = $6(8×7×5×4×3 + 1)$ Since 6 is a factor, the number is 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, after how many minutes will they meet again at the starting point?

Time taken by first student = 20 minutes Time taken by second student = 30 minutes They will meet again after the LCM of 20 and 30. Prime factorisation: 20 = $2^2 × 5$ 30 = $2 × 3 × 5$ LCM = $2^2 × 3 × 5 = 60$ ✅ Therefore, they will meet again after **60 minutes**.

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