CBSE Class 10 Mathematics Real Numbers MCQs Set 34

Answer: (b) irrational

Explanation:
1. Multiply by the conjugate: (5 − √5)²/(25 − 5) = (30 − 10√5)/20 = (3 − √5)/2.
2. This contains √5, so it is irrational.

Teacher's Note:
1. Always rationalise before deciding.
2. A rational plus a non-zero irrational multiple of √5 is irrational.

Answer: (a) \( 0.\overline{45} \)

Explanation:
1. \( 0.\overline{23} = \frac{23}{99} \) and \( 0.\overline{22} = \frac{22}{99} \).
2. Sum = 45/99 = \( 0.\overline{45} \).

Teacher's Note:
1. Add as fractions with denominator 99.
2. 45/99 = 5/11 = 0.4545…

Answer: (d) \( \frac{18}{999} \)

Explanation:
1. All three digits 018 repeat, so the fraction is 018/999 = 18/999.

Teacher's Note:
1. Three repeating digits mean three 9s in the denominator.
2. 18/999 = 2/111.

Answer: (a) \( \frac{53}{3} \)

Explanation:
1. \( 17.\overline{6} = 17 + \frac{6}{9} = 17 + \frac{2}{3} = \frac{53}{3} \).

Teacher's Note:
1. Split the whole part from the repeating part.
2. 6/9 = 2/3.

Answer: (a) terminating decimal

Explanation:
1. 625 = 5⁴, so the denominator has only the prime factor 5.
2. A fraction in lowest terms whose denominator has only 2s and 5s terminates (961 = 31² has no factor 5).

Teacher's Note:
1. Check the prime factors of the denominator of the reduced fraction.
2. 961/625 = 1.5376.

Answer: (b) \( \frac{2}{n} \)

Explanation:
1. Each factor is (k − 1)/k: (2/3)(3/4)(4/5)…((n − 1)/n).
2. The numerators and denominators cancel in a chain, leaving 2/n.

Teacher's Note:
1. This is a telescoping product.
2. Check with n = 3: the product is 2/3.

Answer: (c) 176

Explanation:
1. 99 = 3² × 11 has (2 + 1)(1 + 1) = 6 divisors.
2. 101 is prime: 2 divisors.
3. 176 = 2⁴ × 11 has (4 + 1)(1 + 1) = 10 divisors.
4. 182 = 2 × 7 × 13 has 8 divisors. So 176 has the most.

Teacher's Note:
1. Number of divisors = product of (power + 1) over prime factors.
2. Factorise each number first.

Answer: (a) 6

Explanation:
1. Divisors of 6 other than 6: 1, 2, 3, and 1 + 2 + 3 = 6.
2. 9: 1 + 3 = 4; 15: 1 + 3 + 5 = 9; 21: 1 + 3 + 7 = 11, none equal the number.

Teacher's Note:
1. The first two perfect numbers are 6 and 28.
2. Check by adding proper divisors.

Answer: (a) \( 2^2\times 3^2\times 5 \)

Explanation:
1. HCF takes the lowest power of each common prime.
2. 2: lowest power 2²; 3: lowest 3²; 5: lowest 5¹; 7 and 11 are not common to all three.
3. HCF = 2² × 3² × 5.

Teacher's Note:
1. HCF uses minimum exponents; LCM uses maximum exponents.
2. A prime missing from any number is left out of the HCF.

Answer: (b) (18, 25)

Explanation:
1. gcd(16, 62) = 2, gcd(21, 35) = 7 and gcd(23, 92) = 23.
2. gcd(18, 25) = 1, so (18, 25) are co-prime.

Teacher's Note:
1. Co-primes have no common factor other than 1.
2. They need not be prime themselves.

Answer: (d) \( 2^5\times 3^4\times 5^3\times 7^2\times 11 \)

Explanation:
1. LCM takes the highest power of every prime that appears.
2. 2⁵, 3⁴, 5³, 7², 11¹.

Teacher's Note:
1. Include every prime that appears in any of the numbers.
2. Take the maximum exponent of each.

Answer: (c) 23

Explanation:
1. LCM(5, 6, 4, 3) = 60.
2. 2497 ÷ 60 leaves remainder 37 (2460 + 37).
3. Number to add = 60 − 37 = 23.

Teacher's Note:
1. Add (LCM − remainder).
2. Check: 2520 = 60 × 42.

Answer: (b) 3600

Explanation:
1. LCM(16, 20, 24) = 240 = 2⁴ × 3 × 5.
2. For a perfect square every prime must have an even power: multiply by 3 × 5 to get 2⁴ × 3² × 5² = 3600.

Teacher's Note:
1. Make every exponent in the LCM even.
2. √3600 = 60.

Answer: (b) 1015

Explanation:
1. LCM(12, 16, 18, 21, 28) = 1008.
2. The number minus 7 must be a multiple of 1008; the smallest such positive number is 1008 + 7 = 1015.

Teacher's Note:
1. Smallest number = LCM + 7.
2. LCM: 2⁴ × 3² × 7 = 1008.

Answer: (b) 859

Explanation:
1. LCM(24, 32, 36, 54) = 2⁵ × 3³ = 864.
2. The number = 864 − 5 = 859.

Teacher's Note:
1. Increased by 5 means the number is 5 less than a multiple of the LCM.
2. 864 − 5 = 859.

Answer: (d) 548

Explanation:
1. LCM(12, 15, 20, 54) = 540.
2. Required number = 540 + 8 = 548.

Teacher's Note:
1. Same remainder in each case: LCM + remainder.
2. LCM = 2² × 3³ × 5 = 540.

Answer: (b) 9831

Explanation:
1. LCM(4, 7, 13) = 364.
2. 9999 ÷ 364 leaves remainder 171 (364 × 27 = 9828), so the largest multiple of 364 within four digits is 9828.
3. Required number = 9828 + 3 = 9831.

Teacher's Note:
1. Largest multiple of the LCM within the limit, plus the remainder.
2. 9831 − 3 = 9828 = 364 × 27.

Answer: (d) 364

Explanation:
1. LCM(6, 9, 15, 18) = 90, so numbers leaving remainder 4 are 94, 184, 274, 364, …
2. The first of these that is a multiple of 7: 94 = 7 × 13 + 3 no, 184 = 7 × 26 + 2 no, 274 = 7 × 39 + 1 no, 364 = 7 × 52 yes.

Teacher's Note:
1. Write the general form 90k + 4 and test successive values.
2. 364 = 7 × 52.

Answer: (b) 15110

Explanation:
1. Each divisor minus its remainder = 10 (48 − 38 = 60 − 50 = 72 − 62 = 108 − 98 = 140 − 130 = 10).
2. LCM(48, 60, 72, 108, 140) = 15120.
3. Required number = 15120 − 10 = 15110.

Teacher's Note:
1. When the gap (divisor − remainder) is constant, answer = LCM − gap.
2. LCM = 2⁴ × 3³ × 5 × 7.