CBSE Class 10 Mathematics Real Numbers MCQs Set 22

Answer: (b) No, because 3ⁿ is never divisible by 5

Explanation:
1. If 3ⁿ ended in 0 it would be divisible by 10, so by 5.
2. The prime factorisation of 3ⁿ is 3 × 3 × ... × 3, which contains no 5.
3. By the Fundamental Theorem of Arithmetic this cannot change, so 3ⁿ never ends in 0.

Teacher's Note:
1. A number ending in 0 must have both 2 and 5 among its prime factors.
2. By unique prime factorisation, 3ⁿ has no prime factor other than 3.

Answer: (a) 6 and 72

Explanation:
1. 18 = 2 × 3² and 24 = 2³ × 3.
2. HCF = 2¹ × 3¹ = 6.
3. LCM = 2³ × 3² = 72.

Teacher's Note:
1. HCF takes the smallest power of each common prime.
2. LCM takes the greatest power of every prime that appears.

Answer: (d) 18

Explanation:
1. 144 = 2⁴ × 3².
2. 198 = 2 × 3² × 11.
3. Common part with least powers: 2 × 3² = 18.

Teacher's Note:
1. Choose the least power of each common prime factor.
2. 11 occurs only in 198, so it is not in the HCF.

Answer: (c) 540

Explanation:
1. 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5, 27 = 3³.
2. Highest powers: 2², 3³ and 5¹.
3. LCM = 4 × 27 × 5 = 540.

Teacher's Note:
1. Take the highest power of every prime that occurs.
2. The highest power of 3 comes from 27.

Answer: (b) 14976

Explanation:
1. HCF × LCM = 1152 × 1664.
2. LCM = (1152 × 1664) ÷ 128.
3. 1152 ÷ 128 = 9, so LCM = 9 × 1664 = 14976.

Teacher's Note:
1. For two numbers, HCF × LCM = a × b.
2. Divide 1152 by 128 first to keep the arithmetic small.

Answer: (a) 660

Explanation:
1. 5 × 11 = 55.
2. 3 × 55 = 165.
3. 2 × 165 = 330.
4. 2 × 330 = 660.

Teacher's Note:
1. Work from the bottom of the tree upwards, multiplying at each step.
2. The top number equals 2 × 2 × 3 × 5 × 11.

Answer: (d) m = 2, n = −1

Explanation:
1. 117 = 65 × 1 + 52.
2. 65 = 52 × 1 + 13, so HCF = 13.
3. 13 = 65 − 52 = 65 − (117 − 65) = 2 × 65 − 1 × 117.
4. So m = 2 and n = −1.

Teacher's Note:
1. Apply Euclid's algorithm and then work backwards.
2. Check by substituting: 2 × 65 − 117 = 13.

Answer: (c) 6

Explanation:
1. 18 = 2 × 3², 24 = 2³ × 3, 36 = 2² × 3², 48 = 2⁴ × 3.
2. Least powers of the common primes: 2¹ and 3¹.
3. HCF = 6.

Teacher's Note:
1. The HCF of several numbers divides every one of them.
2. 18 = 2 × 3², 24 = 2³ × 3: common part 2 × 3.

Answer: (b) 30

Explanation:
1. The number is divisible by 2, 3 and 5.
2. The least such number is LCM(2, 3, 5) = 30.

Teacher's Note:
1. "No coconut left" means the number is divisible by each group size.
2. Look for the least common multiple.

Answer: (a) −19

Explanation:
1. 210 = 55 × 3 + 45, 55 = 45 × 1 + 10, 45 = 10 × 4 + 5, 10 = 5 × 2.
2. HCF = 5.
3. 210 × 5 + 55y = 5 gives 1050 + 55y = 5.
4. 55y = −1045, so y = −19.

Teacher's Note:
1. First find the HCF by Euclid's algorithm.
2. Then solve the linear equation for y.

Answer: (d) 64

Explanation:
1. 2053 − 5 = 2048 and 967 − 7 = 960.
2. 2048 = 2¹¹ and 960 = 2⁶ × 15.
3. HCF = 2⁶ = 64, which is larger than both remainders.

Teacher's Note:
1. Subtract the remainders first.
2. The required number is the HCF of the results.

Answer: (c) 170 litres

Explanation:
1. 850 = 680 × 1 + 170.
2. 680 = 170 × 4 + 0.
3. HCF = 170, so the container holds 170 litres.

Teacher's Note:
1. Largest equal measure means HCF.
2. 850 ÷ 170 = 5 and 680 ÷ 170 = 4.

Answer: (b) 24 inches and 20 tiles

Explanation:
1. 10 ft = 120 inches and 8 ft = 96 inches.
2. HCF(120, 96) = 24 inches.
3. Number of tiles = (120 ÷ 24) × (96 ÷ 24) = 5 × 4 = 20.

Teacher's Note:
1. Convert feet to inches before taking the HCF.
2. The tile side is the HCF of the two dimensions.

Answer: (a) 75 cm

Explanation:
1. 825 = 3 × 5² × 11, 675 = 3³ × 5², 450 = 2 × 3² × 5².
2. Common part with least powers: 3 × 5² = 75.
3. The longest rod is 75 cm.

Teacher's Note:
1. Convert all dimensions to centimetres.
2. The longest exact measure is the HCF.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Assertion: this is Euclid's division lemma for integers and is true.
2. Reason: dividing by 3 leaves remainder 1 or 2 when a is not a multiple of 3, so it is also true.
3. The Reason is only the special case b = 3 of the lemma; it does not explain why the lemma holds for every b.
4. So both are true but R is not the correct explanation of A.

Teacher's Note:
1. The division lemma is the foundation of Euclid's algorithm.
2. The remainder forms 3n + 1 and 3n + 2 are special cases of it.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Assertion: 9 and 4608 are integers and 4608 ≠ 0, so 9/4608 is rational. True.
2. Reason: a decimal that terminates or repeats can be written as p/q, so it is rational. True.
3. The decimal expansion of 9/4608 either terminates or repeats, and the Reason explains why this makes the number rational.
4. So R correctly explains A.

Teacher's Note:
1. A rational number is any number of the form p/q with q ≠ 0.
2. Its decimal expansion is always terminating or non-terminating recurring.

Answer: (d) A-Q, B-S, C-P, D-R

Explanation:
1. −10 is a negative integer, π is irrational, 3 is a natural number, and 5/2 is a fraction that is not an integer.

Teacher's Note:
1. Check each number against all four descriptions.
2. A negative integer is never a natural number.

Answer: (c) xy > yz

Explanation:
1. xz > yz (reversed), so (a) is false.
2. x/z > y/z (reversed), so (b) is false.
3. 1/x > 1/y and z < 0, so z/x < z/y, so (c) is false.
4. xy is positive and yz is negative, so xy > yz is true.

Teacher's Note:
1. An inequality reverses when both sides are multiplied or divided by a negative number.
2. A positive number is always greater than a negative number.

Answer: (b) (18, 25)

Explanation:
1. 14 = 2 × 7 and 35 = 5 × 7: HCF 7.
2. 18 = 2 × 3² and 25 = 5²: HCF 1.
3. 93 = 3 × 31, so HCF(31, 93) = 31.
4. HCF(32, 62) = 2.

Teacher's Note:
1. Coprime numbers have HCF 1.
2. Check the common factors of each pair.

Answer: (a) 60

Explanation:
1. Powers of 2: 3, 3, 2, so least is 2².
2. Powers of 3: 2, 3, 1, so least is 3¹.
3. Powers of 5: 1, 2, 3, so least is 5¹.
4. HCF = 4 × 3 × 5 = 60.

Teacher's Note:
1. Take the smallest exponent of every common prime.
2. 7 is not common to all three.

Answer: (d) 36 and 60

Explanation:
1. Let the numbers be 12a and 12b with HCF(a, b) = 1.
2. 144ab = 2160, so ab = 15.
3. Coprime pairs with product 15: (3, 5) or (1, 15). (1, 15) gives 12 and 180, not two-digit.
4. So the numbers are 36 and 60.

Teacher's Note:
1. Use the fact that the numbers are multiples of the HCF.
2. Both 36 and 60 are two-digit numbers.

Answer: (c) 80

Explanation:
1. 15 × HCF = 600, so HCF = 40 and LCM = 560.
2. Other number = (HCF × LCM) ÷ 280 = 22400 ÷ 280 = 80.

Teacher's Note:
1. Use HCF × LCM = product of the numbers.
2. Find the HCF from the sum first.

Answer: (b) 48

Explanation:
1. Numbers = 3 × 4 = 12 and 4 × 4 = 16.
2. LCM(12, 16) = 48.

Teacher's Note:
1. Numbers with ratio 3 : 4 and HCF 4 are 3 × 4 and 4 × 4.
2. LCM of 12 and 16 = 2⁴ × 3.

Answer: (a) −2

Explanation:
1. LCM = xy = 161 = 7 × 23.
2. x > y, so x = 23 and y = 7.
3. 3y − x = 21 − 23 = −2.

Teacher's Note:
1. The LCM of two distinct primes is their product.
2. Factorise 161 by trying small primes.