NCERT Solutions›Class 10 Mathematics›Chapter 1

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Chapter 1 · Class 10 Mathematics

Real Numbers

3 exercises15 questions solved

Exercise 1.1 Exercise 1.2 Exercise 1.3

Exercise 1.1Euclid's Division Algorithm

Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Solution

(i) HCF of 135 and 225: Step 1: 225 = 1 × 135 + 90 Step 2: 135 = 1 × 90 + 45 Step 3: 90 = 2 × 45 + 0 Since remainder = 0, HCF(135, 225) = 45 (ii) HCF of 196 and 38220: Step 1: 38220 = 195 × 196 + 0 Since remainder = 0, HCF(196, 38220) = 196 (iii) HCF of 867 and 255: Step 1: 867 = 3 × 255 + 102 Step 2: 255 = 2 × 102 + 51 Step 3: 102 = 2 × 51 + 0 Since remainder = 0, HCF(867, 255) = 51

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution

Let a be any positive integer and b = 6. By Euclid's division algorithm: a = 6q + r, where 0 ≤ r < 6. So r can be 0, 1, 2, 3, 4, or 5. This gives: a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. • 6q = 2(3q) → even • 6q+1 → odd ✓ • 6q+2 = 2(3q+1) → even • 6q+3 = 3(2q+1) → odd ✓ • 6q+4 = 2(3q+2) → even • 6q+5 = 6q+5 → odd ✓ Therefore, any positive odd integer is of the form 6q+1, 6q+3, or 6q+5.

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution

The maximum number of columns = HCF(616, 32) Using Euclid's division algorithm: Step 1: 616 = 19 × 32 + 8 Step 2: 32 = 4 × 8 + 0 Since the remainder is 0, HCF(616, 32) = 8 The maximum number of columns in which they can march is 8.

Use Euclid's Lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [Hint: Let x = 3q + r, where 0 ≤ r < 3]

Solution

Let x be any positive integer. By Euclid's division algorithm: x = 3q + r, where 0 ≤ r < 3, so r = 0, 1, or 2. Case 1: r = 0 → x = 3q x² = 9q² = 3(3q²) = 3m, where m = 3q² Case 2: r = 1 → x = 3q + 1 x² = (3q+1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1, where m = 3q² + 2q Case 3: r = 2 → x = 3q + 2 x² = (3q+2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1, where m = 3q² + 4q + 1 In all cases, x² is either of the form 3m or 3m + 1.

Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution

Let x be any positive integer. By Euclid's division algorithm: x = 3q + r, where r = 0, 1, or 2. Case 1: r = 0 → x = 3q x³ = 27q³ = 9(3q³) = 9m, where m = 3q³ Case 2: r = 1 → x = 3q + 1 x³ = (3q+1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1, where m = 3q³ + 3q² + q Case 3: r = 2 → x = 3q + 2 x³ = (3q+2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8, where m = 3q³ + 6q² + 4q Therefore, the cube of any positive integer is of the form 9m, 9m+1, or 9m+8.

Exercise 1.2Fundamental Theorem of Arithmetic

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

Solution

(i) 140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7 = 2² × 5 × 7 (ii) 156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13 = 2² × 3 × 13 (iii) 3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17 (iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13 (v) 7429 = 17 × 437 = 17 × 19 × 23

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: 26 = 2 × 13; 91 = 7 × 13 HCF = 13; LCM = 2 × 7 × 13 = 182 Verification: 182 × 13 = 2366 = 26 × 91 ✓ (ii) 510 and 92: 510 = 2 × 3 × 5 × 17; 92 = 2² × 23 HCF = 2; LCM = 2² × 3 × 5 × 17 × 23 = 23460 Verification: 23460 × 2 = 46920 = 510 × 92 ✓ (iii) 336 and 54: 336 = 2⁴ × 3 × 7; 54 = 2 × 3³ HCF = 2 × 3 = 6; LCM = 2⁴ × 3³ × 7 = 3024 Verification: 3024 × 6 = 18144 = 336 × 54 ✓

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: 12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7 HCF = 3 (common factor) LCM = 2² × 3 × 5 × 7 = 420 (ii) 17, 23 and 29: All three are prime numbers. HCF = 1 (no common factor) LCM = 17 × 23 × 29 = 11339 (iii) 8, 9 and 25: 8 = 2³; 9 = 3²; 25 = 5² HCF = 1 (no common factor) LCM = 2³ × 3² × 5² = 1800

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

Solution

We know that for any two positive integers a and b: LCM × HCF = a × b LCM(306, 657) × HCF(306, 657) = 306 × 657 LCM(306, 657) × 9 = 306 × 657 LCM(306, 657) = (306 × 657) / 9 LCM(306, 657) = 201042 / 9 LCM(306, 657) = 22338

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

Solution

For 6ⁿ to end with 0, it must be divisible by 10 = 2 × 5. So 6ⁿ must have 5 as a prime factor. Prime factorisation of 6 = 2 × 3 Therefore, 6ⁿ = 2ⁿ × 3ⁿ The only prime factors of 6ⁿ are 2 and 3. Since 5 is not a factor of 6ⁿ, 6ⁿ cannot be divisible by 5. By the Fundamental Theorem of Arithmetic, the prime factorisation of any number is unique. Therefore, 6ⁿ can never end with the digit 0 for any natural number n.

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

Solution

(i) 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 × 2 × 39 = 13 × 2 × 3 × 13 This number has factors 2, 3, and 13 (other than 1 and itself), so it is a COMPOSITE number. ✓ (ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009 This number has factors 5 and 1009 (other than 1 and itself), so it is a COMPOSITE number. ✓

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 at the starting point after a time equal to the LCM of 18 and 12 minutes. Prime factorisation: 18 = 2 × 3² 12 = 2² × 3 LCM(18, 12) = 2² × 3² = 4 × 9 = 36 Sonia and Ravi will meet again at the starting point after 36 minutes. (Check: In 36 minutes, Sonia completes 36/18 = 2 rounds and Ravi completes 36/12 = 3 rounds — both back at start. ✓)

Exercise 1.3Irrational Numbers

Prove that √5 is irrational.

Solution

Proof by contradiction: Assume that √5 is rational. Then it can be written as p/q, where p and q are integers with HCF(p, q) = 1 (i.e., p/q is in lowest terms). √5 = p/q Squaring both sides: 5 = p²/q² ⟹ p² = 5q² ... (1) This means 5 divides p², and since 5 is prime, 5 divides p. Let p = 5m for some integer m. Substituting in (1): (5m)² = 5q² ⟹ 25m² = 5q² ⟹ q² = 5m² This means 5 divides q², and since 5 is prime, 5 divides q. But we now have 5 dividing both p and q, which contradicts HCF(p, q) = 1. Our assumption was wrong. Therefore, √5 is irrational.

Prove that 3 + 2√5 is irrational.

Solution

Given that √5 is irrational (proved in Q1). Assume that 3 + 2√5 is rational. Then: 3 + 2√5 = p/q, where p, q are integers, q ≠ 0, HCF(p, q) = 1 2√5 = p/q − 3 2√5 = (p − 3q)/q √5 = (p − 3q)/(2q) Since p, q are integers, (p − 3q)/(2q) is rational. This means √5 is rational. But this contradicts the fact that √5 is irrational. Therefore, our assumption is wrong. 3 + 2√5 is irrational.

Prove that the following are irrational: (i) 1/√2 (ii) 7√5 (iii) 6 + √2

Solution

(i) Prove 1/√2 is irrational: Assume 1/√2 is rational = p/q (lowest terms). Then √2 = q/p, which is rational. Contradiction (√2 is irrational). Therefore, 1/√2 is irrational. (ii) Prove 7√5 is irrational: Assume 7√5 = p/q (rational, lowest terms). Then √5 = p/(7q), which is rational. Contradiction (√5 is irrational). Therefore, 7√5 is irrational. (iii) Prove 6 + √2 is irrational: Assume 6 + √2 = p/q (rational). Then √2 = p/q − 6 = (p − 6q)/q, which is rational. Contradiction (√2 is irrational). Therefore, 6 + √2 is irrational.

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