EXERCISE 1.1

1. Product of its prime factors:  (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Prime Factorization:

(i) 140 = 2 × 2 × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

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

LCM and HCF:

(i) 26 = 2 × 13, 91 = 7 × 13 HCF(26, 91) = 13 LCM(26, 91) = 2 × 7 × 13 = 182 Verify: 13 × 182 = 26 × 91

(ii) 510 = 2 × 3 × 5 × 17, 92 = 2 × 2 × 23 HCF(510, 92) = 2 LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460 Verify: 2 × 23460 = 510 × 92

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7, 54 = 2 × 3 × 3 × 3 HCF(336, 54) = 6 LCM(336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024 Verify: 6 × 3024 = 336 × 54

3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

LCM and HCF by Prime Factorization:

(i) 12 = 2 × 2 × 3, 15 = 3 × 5, 21 = 3 × 7 HCF(12, 15, 21) = 3 LCM(12, 15, 21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23, 29 are prime numbers. HCF(17, 23, 29) = 1 LCM(17, 23, 29) = 17 × 23 × 29 = 11339

(iii) 8 = 2 × 2 × 2, 9 = 3 × 3, 25 = 5 × 5 HCF(8, 9, 25) = 1 LCM(8, 9, 25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

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

LCM and HCF Relationship:

HCF × LCM = Product of the numbers 9 × LCM(306, 657) = 306 × 657 LCM(306, 657) = (306 × 657) / 9 = 22338

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

6^n Ending with 0:

For a number to end with 0, it must be divisible by 10. This means it must have both 2 and 5 as factors. However, 6^n only has 2 and 3 as prime factors, so it can never end with 0.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Composite Numbers:

• 7 × 11 × 13 + 13 = 13(7 × 11 + 1) is divisible by 13.
• 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1) is divisible by 5. Both numbers have factors other than 1 and themselves, making them composite.

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 minuteswill they meet again at the starting point?

To find the time they meet again, we need to find the LCM of their round times. LCM(18, 12) = 36 So, they will meet again after 36 minutes.

EXERCISE 1.2

1. Prove that √5 is irrational.

Solution :

Let's assume that √5 is rational. This means it can be expressed as a fraction p/q, where p and q are integers with no common factors other than 1, and q ≠ 0.

So, √5 = p/q

Squaring both sides:

5 = p²/q²

=> 5q² = p²

This implies that 5 divides p². If 5 divides p², then 5 must also divide p.

Let p = 5k, where k is an integer.

Substituting this in the equation:

5q² = (5k)²

=> q² = 5k²

This implies that 5 divides q². If 5 divides q², then 5 must also divide q.

But this contradicts our assumption that p and q have no common factors other than 1.

Hence, our assumption that √5 is rational is¹ incorrect. Therefore, √5 is irrational.

2. Prove that 3+2√5 is irrational.

Solution :

Let's assume that 3+2√5 is rational. This means it can be expressed as a fraction p/q, where p and q are integers with no common factors other than 1, and q ≠ 0.

So, 3+2√5 = p/q

Rearranging:

2√5 = (p/q) - 3

√5 = [(p/q) - 3]/2

The right-hand side is a rational number (difference of two rational numbers divided by a rational number). However, we know that √5 is irrational.

This contradiction proves our assumption wrong. Therefore, 3+2√5 is irrational.

3. Prove that the following are irrationals :

(i) 1 /√2

(i) 1/√2: If 1/√2 is rational, then its reciprocal, √2, must also be rational. But we know that √2 is irrational. Therefore, 1/√2 is irrational.

(ii) 7√5

If 7√5 is rational, then (7√5)/7 = √5 must also be rational. But we know that √5 is irrational. Therefore, 7√5 is irrational.

(iii) 6+√2 

If 6+√2 is rational, then (6+√2) - 6 = √2 must also be rational. But we know that √2 is irrational. Therefore, 6+√2 is irrational.

Test Your Knowledge

1. Which of the following is an irrational number?

• a) 22/7
• b) √16
• c) √3
• d) 3.14 

2. The decimal expansion of a rational number is:

• a) always terminating
• b) always non-terminating
• c) either terminating or non-terminating repeating  
• d) non-terminating non-repeating 

3. The HCF of 144 and 198 is:

• a) 6
• b) 9
• c) 18
• d) 36 

4. The LCM of two numbers is 180 and their HCF is 3. If one of the numbers is 45, the other number is:

• a) 12
• b) 24
• c) 36
• d) 72

5. If a = 2³ × 3² × 5 and b = 2² × 3³ × 7, then HCF(a, b) is:

• a) 2³ × 3²
• b) 2² × 3³
• c) 2² × 3²
• d) 2³ × 3³ 

6. Euclid's division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where:  

• a) 0 < r ≤ b
• b) 0 ≤ r < b
• c) 0 < r < b
• d) 0 ≤ r ≤ b 

7. The decimal expansion of the rational number 43/2⁴ × 5³ will terminate after:

• a) 3 decimal places
• b) 4 decimal places
• c) 5 decimal places
• d) 6 decimal places 

8. Which of the following is a pair of co-prime numbers?

• a) 17, 51
• b) 18, 25
• c) 21, 35
• d) 23, 92 

9. The product of two irrational numbers is:

• a) always an irrational number
• b) always a rational number
• c) may be rational or irrational
• d) none of these 

10. The sum of a rational number and an irrational number is:

• a) always an irrational number
• b) always a rational number
• c) may be rational or irrational
• d) none of these 

Answers : 1 (c), 2 (c),3 (c),4 (b),5 (c),6 (b),7 (c),8 (b),9 (c),10(a).