CBSE Class 10 Mathematics Real Numbers MCQs Set 32

Answer: (c) both 5 and 13

Explanation:
1. \( 9^{2n}-4^{2n} = 81^n-16^n \) is divisible by \( 81-16 = 65 = 5\times13 \).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (c) either 0 or 1

Explanation:
1. Write the number as 3k, 3k+1 or 3k+2. Squares give \( 9k^2 \), \( 9k^2+6k+1 \), \( 9k^2+12k+4 \), with remainders 0, 1, 1.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) 500

Explanation:
1. The HCF must divide the LCM. \( 1200/500 \) is not an integer, so 500 is impossible.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (d) None of these

Explanation:
1. \( \sqrt2 \) has rationalising factors \( \sqrt2 \), \( 2\sqrt2 \) and more, so (a) is false. \( \sqrt2+(-\sqrt2)=0 \) is rational but \( \sqrt2+\sqrt3 \) is not, so (b) is not always true. \( \sqrt2\times\sqrt2 = 2 \) is rational, so (c) is false.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (c) either 0 or 1

Explanation:
1. Even numbers give \( 4k^2 \) (remainder 0). Odd numbers give \( 4k^2+4k+1 \) (remainder 1).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 13

Explanation:
1. \( 3^{2n}+2^{2n} = 9^n+4^n \). For odd n this is divisible by \( 9+4 = 13 \).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (d) equal

Explanation:
1. HCF divides both numbers and LCM is a multiple of both. If they are equal, both numbers equal that common value.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 1

Explanation:
1. Such a prime is \( 6k\pm1 \). Its square is \( 36k^2\pm12k+1 \), which leaves remainder 1.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 1

Explanation:
1. \( 72 = 2^3\times3^2 \). Let \( q = 2^a3^b \). Then \( \min(3,a) = 2 \) gives a = 2, and \( \min(2,b) = 1 \) gives b = 1.
2. Only q = 12 works, so there is 1 value.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (c) 0

Explanation:
1. The product is \( 2^{35}\times35! \), which contains many factors of 10 (at least 7 factors of 5 and more factors of 2), so it ends in at least two zeros. The tens digit is 0.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (c) 2

Explanation:
1. LCM of fractions \( = \frac{\text{LCM of numerators}}{\text{HCF of denominators}} = \frac{2}{1} = 2 \).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) 5

Explanation:
1. The product contains the factor 5 and no even factor, so it is odd and a multiple of 5. It ends in 5.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) an even number

Explanation:
1. If the first number is k, the sum is \( 20k+190 \), which is even.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (d) 41 and 43

Explanation:
1. Twin primes are primes that differ by 2. 41 and 43 are both prime and differ by 2.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 1

Explanation:
1. The units digit of powers of 3 cycles 3, 9, 7, 1. Since \( 84 = 4\times21 \), the digit is 1.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) 1

Explanation:
1. The list contains consecutive numbers, which are co-prime, so the HCF is 1.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 7

Explanation:
1. \( 4^{44} \) ends in 6 (even power of 4). \( 5^{55} \) ends in 5. \( 8^{88} \) ends in 6 because \( 88 = 4\times22 \).
2. \( 6+5+6 = 17 \), so the units digit is 7.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (d) 6

Explanation:
1. Any power of a number ending in 6 ends in 6.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 60

Explanation:
1. LCM of numerators 5, 6, 3, 4 is 60. HCF of denominators 12, 5, 2, 17 is 1.
2. LCM \( = \frac{60}{1} = 60 \).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) 1

Explanation:
1. The numbers are p, p+2, p+4. One of any three such numbers is a multiple of 3, so one must equal 3. Only p = 3 works, giving (3, 5, 7).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) 13

Explanation:
1. Divisibility by 5 gives q = 0 or 5. Digit sum is \( 20+p+q \), which must be a multiple of 3, so \( p+q \equiv 1 \pmod 3 \).
2. Largest value not exceeding 14 with this property (q = 5, p = 8) is 13.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) 16

Explanation:
1. Values of 6k+1 for k = 1 to 25 are 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121, 127, 133, 139, 145, 151.
2. Composites among them: 25, 49, 55, 85, 91, 115, 121, 133, 145 (9 numbers), so primes number 25 - 9 = 16.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 17

Explanation:
1. Group size is HCF(72, 64) = 8.
2. Groups \( = \frac{72}{8}+\frac{64}{8} = 9+8 = 17 \).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (b) 432

Explanation:
1. 1 to 9 use 9 presses, 10 to 99 use \( 90\times2 = 180 \), 100 to 180 use \( 81\times3 = 243 \).
2. Total \( = 9+180+243 = 432 \).

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 2

Explanation:
1. Last digit 5 gives divisibility by 5. Digit sum \( = 34+X \) must be a multiple of 3, so X = 2, 5 or 8. Least is 2.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.

Answer: (a) 2

Explanation:
1. Every even number greater than 2 has 2 as a factor, so is composite. Hence 2 is the only even prime.

Teacher's Note:
1. Break the numbers into prime factors or use remainders.