CBSE Class 10 Mathematics Real Numbers MCQs Set 35

Answer: (a) 2

Explanation:
1. The least prime factor of a is 3, so a has no factor 2 and is odd. Similarly b is odd.
2. The sum of two odd numbers is even, so a + b is divisible by 2.
3. The least prime factor of a + b is 2.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (c) 0 ≤ r < b

Explanation:
1. The remainder can be zero (when b divides a) but must be smaller than the divisor b.
2. So 0 ≤ r < b.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (c) any odd positive integer

Explanation:
1. If n is odd, write n = 2k + 1, then n² − 1 = 4k(k + 1).
2. k(k + 1) is a product of two consecutive integers and is even, so 4k(k + 1) is divisible by 8.
3. For even n, n² − 1 is odd and cannot be divisible by 8.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (b) 2

Explanation:
1. n = 9k + 7, so 3n − 1 = 27k + 21 − 1 = 27k + 20.
2. 27k is divisible by 9, and 20 = 2 × 9 + 2.
3. The remainder is 2.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (b) 2

Explanation:
1. Euclid's algorithm: 153 = 85 × 1 + 68, 85 = 68 × 1 + 17, 68 = 17 × 4 + 0, so HCF = 17.
2. 85n − 153 = 17 gives 85n = 170, so n = 2.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (c) 10000

Explanation:
1. HCF × LCM = product of the two numbers = 125 × 80 = 10000.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (a) 12

Explanation:
1. HCF takes the smallest power of each common prime: 2² and 3¹.
2. HCF = 4 × 3 = 12.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (b) 40

Explanation:
1. HCF × LCM = 253 × 440, so 11 × 253 × R = 253 × 440.
2. R = 440/11 = 40.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (b) 4

Explanation:
1. 5005 = 5 × 1001 = 5 × 7 × 11 × 13.
2. There are 4 prime factors.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (d) 8

Explanation:
1. The smallest prime is 2 and the smallest composite number is 4.
2. HCF(2, 4) = 2 and LCM(2, 4) = 4, so the product is 8 (which equals 2 × 4).

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (c) 1/7

Explanation:
1. 35/14 = 5/2 = 2.5 terminates, 14/35 = 2/5 = 0.4 terminates and 7/8 = 0.875 terminates.
2. 1/7 has denominator 7, which is not of the form 2ᵐ5ⁿ, so it is a non-terminating repeating decimal.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (a) terminating

Explanation:
1. 400 = 2⁴ × 5², which has no prime factors other than 2 and 5.
2. So 27/400 = 0.0675 is a terminating decimal.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (d) infinite

Explanation:
1. Between any two distinct rational numbers a and b the mean (a + b)/2 is rational and lies between them, and the process can be repeated.
2. So there are infinitely many.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (d) prime

Explanation:
1. If A and B were both odd, A + B would be even and larger than 2, so not prime; A = 2 is impossible since A − B must be positive. So A is odd and B = 2.
2. Then A − 2, A, A + 2 are three primes; among three numbers differing by 2 one is divisible by 3, so they must be 3, 5, 7 and A = 5.
3. The primes are 5, 2, 3, 7 and the sum is 17, which is prime.

Teacher's Note:
1. Parity and divisibility by 3 force the only possible set 3, 5, 7.
2. Test the case A = 5, B = 2 directly.

Answer: (d) 0

Explanation:
1. A + k = 41! + (k + 1) for k = 1, ..., 40, so the numbers are 41! + 2, 41! + 3, ..., 41! + 41.
2. 41! + m is divisible by m for each m from 2 to 41, and is larger than m, so it is composite.
3. Hence none of them is prime.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (b) 7

Explanation:
1. (a − b)(a + b) = 13, and 13 is prime, so a − b = 1 and a + b = 13.
2. a = 7 and b = 6.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (c) 8

Explanation:
1. Write a = 6x and b = 6y with gcd(x, y) = 1 and x + y = 15.
2. x must be coprime to 15 and between 1 and 14: x = 1, 2, 4, 7, 8, 11, 13, 14, which gives 8 values.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (c) 6

Explanation:
1. Divisible by 2 needs d to be even; divisible by 9 needs the digit sum 2d + 15 to be a multiple of 9.
2. 2d + 15 = 27 for d = 6, which is a multiple of 9, and 6 is even, so 64566 is divisible by 18.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (d) 114345

Explanation:
1. A number divisible by 99 is divisible by 9 and 11.
2. 114345: digit sum 18 (divisible by 9) and alternating sum 1 − 1 + 4 − 3 + 4 − 5 = 0 (divisible by 11).
3. The others fail one of the tests.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (d) 9

Explanation:
1. A number leaves the same remainder as its digit sum when divided by 9.
2. So the difference is divisible by 9.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (c) 6

Explanation:
1. x divides 2x + 12 exactly when x divides 12.
2. The positive divisors of 12 are 1, 2, 3, 4, 6, 12, so there are 6 values.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (b) 3455

Explanation:
1. In each case the divisor minus the remainder is 10, so the number is 10 less than a common multiple.
2. LCM(35, 45, 55) = 3465.
3. The least number is 3465 − 10 = 3455.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (a) 1

Explanation:
1. The number is 14k + 8 = 7(2k + 1) + 1.
2. The remainder on division by 7 is 1.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.

Answer: (d) 83

Explanation:
1. Let the number be 10a + b. Then a + b = 11 and (10a + b) − (10b + a) = 9(a − b) = 45, so a − b = 5.
2. a = 8 and b = 3, so the number is 83.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers and prime factorisation wherever possible.
2. Check your answer by testing it on the original numbers.