CBSE Class 10 Mathematics Real Numbers MCQs Set 18

Answer: (a) 195 and 143

Explanation:
1. Let the numbers be 15x and 11x. Their H.C.F. is x, so x = 13.
2. The numbers are 15 × 13 = 195 and 11 × 13 = 143.

Teacher's Note:
1. The common multiplier of a ratio is the H.C.F.
2. Multiply each ratio term by the H.C.F.

Answer: (a) terminating

Explanation:
1. \( \frac{35}{50}=\frac{7}{10}=0.7 \), which ends after one digit.
2. So the decimal expansion is terminating.

Teacher's Note:
1. A fraction in lowest terms terminates if the denominator has only the prime factors 2 and 5.
2. Here the denominator 10 = 2 × 5.

Answer: (d) 0.21 lies between 0.2 and 0.3

Explanation:
1. 0.2 < 0.21 < 0.3, so (d) is true.
2. (a) is false: \( \frac{22}{7} \) is only an approximation of \( \pi \). (b) is false: irrational numbers are also real. (c) is false: a non-terminating, non-recurring decimal is not periodic.

Teacher's Note:
1. Check each statement one by one.
2. 22/7 is only a close value of pi.

Answer: (a) 111

Explanation:
1. Let the numbers be 37a and 37b with a, b co-prime. Then 1369ab = 4107, so ab = 3.
2. Co-prime pair: (1, 3), so the numbers are 37 and 111.

Teacher's Note:
1. Write numbers as HCF × co-prime factors.
2. Check that the product is 4107.

Answer: (b) non-terminating and repeating

Explanation:
1. \( \frac{21}{45}=\frac{7}{15} \), and 15 = 3 × 5 has the prime factor 3.
2. So the expansion is non-terminating and repeating.

Teacher's Note:
1. Reduce the fraction first.
2. Check the prime factors of the denominator.

Answer: (c) 1

Explanation:
1. Co-prime numbers have no common factor except 1, so HCF = 1.

Teacher's Note:
1. Co-prime means HCF 1.
2. Then LCM = pq.

Answer: (d) equal

Explanation:
1. HCF ≤ each number ≤ LCM. If HCF = LCM, both numbers equal that value.

Teacher's Note:
1. HCF is never more than the smaller number.
2. LCM is never less than the larger number.

Answer: (c) p³q²

Explanation:
1. LCM takes the highest power of each prime: p³ and q².
2. LCM = p³q².

Teacher's Note:
1. Highest powers for LCM, lowest for HCF.
2. Here HCF = pq.

Answer: (d) 138

Explanation:
1. 285 − 9 = 276 and 1249 − 7 = 1242.
2. HCF(276, 1242): 1242 = 4 × 276 + 138 and 276 = 2 × 138, so HCF = 138 (greater than both remainders).

Teacher's Note:
1. Subtract the remainders first.
2. Take the HCF of the results.

Answer: (c) 154

Explanation:
1. 2011 − 9 = 2002 and 2623 − 5 = 2618.
2. 2618 = 2002 + 616, 2002 = 3 × 616 + 154, 616 = 4 × 154, so HCF = 154.

Teacher's Note:
1. Subtract the remainders, then use Euclid's algorithm.
2. The divisor must exceed the remainders.

Answer: (a) 1

Explanation:
1. 2⁴ = 16 ≡ −1 (mod 17), so 2⁸ ≡ 1 (mod 17).
2. 2²⁵⁶ = (2⁸)³² ≡ 1, so the remainder is 1.

Teacher's Note:
1. Find a power of 2 that is close to a multiple of 17.
2. Then raise it to the needed power.

Answer: (a) 13

Explanation:
1. H.C.F. × L.C.M. = product of the numbers.
2. H.C.F. = 91 × 26 ÷ 182 = 13.

Teacher's Note:
1. Use HCF × LCM = a × b.
2. Check that 13 divides both 91 and 26.

Answer: (a) 13

Explanation:
1. 70 − 5 = 65 and 125 − 8 = 117.
2. HCF(65, 117) = 13, which is greater than both remainders.

Teacher's Note:
1. Subtract the remainders first.
2. The HCF must exceed the remainders.

Answer: (b) 16

Explanation:
1. 245 − 5 = 240 and 1029 − 5 = 1024.
2. HCF(240, 1024) = 16.

Teacher's Note:
1. Subtract the remainder from each number.
2. Take the HCF.

Answer: (d) \( \frac{23}{8} \)

Explanation:
1. 8 = 2³ has only the prime factor 2, so 23/8 terminates.
2. 210, 30 and 441 contain the factors 3 or 7, so the others do not terminate.

Teacher's Note:
1. Only factors 2 and 5 in the denominator allow termination.
2. Factorise each denominator.

Answer: (d) 4

Explanation:
1. H.C.F. × L.C.M. = x × 18, so 2 × 36 = 18x and x = 4.
2. Check: HCF(4, 18) = 2 and LCM(4, 18) = 36.

Teacher's Note:
1. Use HCF × LCM = product.
2. Check with both statements.

Answer: (a) 515

Explanation:
1. Each remainder is 10 less than the divisor, so N + 10 is a common multiple of 15, 25 and 35.
2. LCM = 525, so N = 525 − 10 = 515.

Teacher's Note:
1. Notice divisor − remainder = 10 each time.
2. Subtract 10 from the LCM.

Answer: (d) 24

Explanation:
1. Put n = 2k: n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2) = 4k(k + 1)(2k + 1).
2. k(k + 1) is even and the product of three consecutive numbers is divisible by 3, so the expression is divisible by 8 × 3 = 24 (n = 2 gives 24).

Teacher's Note:
1. Test n = 2: 2 × 3 × 4 = 24.
2. Prove divisibility by 8 and 3.

Answer: (d) 3600

Explanation:
1. LCM(16, 20, 24) = 240 = 2⁴ × 3 × 5.
2. A perfect square needs even powers, so multiply by 3 × 5 to get 2⁴ × 3² × 5² = 3600.

Teacher's Note:
1. Find the LCM first.
2. Make every prime power even.

Answer: (c) 576

Explanation:
1. LCM(64, 72): 64 = 2⁶ and 72 = 2³ × 3², so LCM = 2⁶ × 3² = 576.

Teacher's Note:
1. Use the LCM for the smallest common multiple.
2. Check that 576 ÷ 64 = 9 and 576 ÷ 72 = 8.

Answer: (d) 8

Explanation:
1. 64 = 2⁶ and 72 = 2³ × 3², so HCF = 2³ = 8.
2. Check: 8 × 576 = 4608 = 64 × 72.

Teacher's Note:
1. Take the lowest power of each common prime.
2. Verify with HCF × LCM.

Answer: (a) \( 2^3\times3^2 \)

Explanation:
1. 72 = 8 × 9 = 2³ × 3².

Teacher's Note:
1. Divide by 2 repeatedly, then by 3.
2. Check: 8 × 9 = 72.

Answer: (b) composite number

Explanation:
1. Take 19 common: 19 × (5 × 13 × 17 + 1) = 19 × 1106.
2. It has factors other than 1 and itself, so it is composite.

Teacher's Note:
1. Look for a common factor.
2. A number with a proper factor is composite.

Answer: (b) \( a^2b^2 \)

Explanation:
1. HCF takes the lowest power of each prime: a² and b².
2. HCF = a²b².

Teacher's Note:
1. Lowest powers give the HCF.
2. Highest powers give the LCM.

Answer: (b) 2

Explanation:
1. The LCM takes the highest power of 3, which must be 3².
2. The highest power among 3¹, 3¹ and 3ⁿ is 3², so n = 2.

Teacher's Note:
1. LCM uses the highest power of each prime.
2. Only c can supply 3².

Answer: (a) 9720

Explanation:
1. LCM(12, 15, 18, 27) = 540.
2. 9999 ÷ 540 = 18 remainder 279, so the number is 18 × 540 = 9720.

Teacher's Note:
1. Find the LCM first.
2. Subtract the remainder from 9999.

Answer: (b) \( 1.\overline{5} \)

Explanation:
1. x = 0.777... = 7/9.
2. 2x = 14/9 = 1.555... = \( 1.\overline{5} \).

Teacher's Note:
1. Convert a repeating decimal using 10x − x.
2. Then divide 14 by 9.

Answer: (b) 11

Explanation:
1. The numbers are 10x + y and 10y + x, with sum 11(x + y).
2. For a perfect square, x + y must be 11 (11 × 11 = 121); the digit sum cannot be larger than 17 and 11 is the only multiple of 11 in range.

Teacher's Note:
1. Add the two numbers.
2. 11 must divide the square, so 11² divides it.