CBSE Class 10 Mathematics Real Numbers MCQs Set 27

Answer: (a) 6q + 1, 6q + 3 or 6q + 5

Explanation:
1. Dividing by 6 the remainder r is 0, 1, 2, 3, 4 or 5. The forms 6q, 6q + 2 and 6q + 4 are even, so an odd integer must be 6q + 1, 6q + 3 or 6q + 5.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (d) 9m, 9m + 1 or 9m + 8

Explanation:
1. Write x = 3q, 3q + 1 or 3q + 2. Then x^3 = 27q^3 = 9(3q^3), (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9m + 1 and (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9m + 8.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (c) 5m + 1

Explanation:
1. (5q + 1)^2 = 25q^2 + 10q + 1 = 5(5q^2 + 2q) + 1, which has the form 5m + 1.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (b) 98

Explanation:
1. 38318 = 195 x 196 + 98 and 196 = 98 x 2 + 0. The last non-zero remainder is 98.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (b) 22

Explanation:
1. The HCF is 9 (963 = 657 + 306, 657 = 2 x 306 + 45, 306 = 6 x 45 + 36, 45 = 36 + 9, 36 = 4 x 9). Then 657x = 9 + 14445 = 14454, so x = 22.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (a) 625

Explanation:
1. The required number is the HCF of 626 - 1 = 625, 3127 - 2 = 3125 and 15628 - 3 = 15625. Since 3125 = 5 x 625 and 15625 = 25 x 625, the HCF is 625.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (d) 18

Explanation:
1. This is HCF(144, 90). 144 = 90 + 54, 90 = 54 + 36, 54 = 36 + 18, 36 = 2 x 18. The HCF is 18.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (c) 7^2 x 13^2 x 17^2 x 19

Explanation:
1. By successive division, 45470971 = 49 x 169 x 289 x 19 = 7^2 x 13^2 x 17^2 x 19.

Teacher's Note:
1. Fundamental theorem of arithmetic: every composite number is a product of primes in exactly one way (apart from order). HCF takes the least exponent of each common prime and LCM the greatest exponent of every prime.

Answer: (c) for no natural number n

Explanation:
1. A number ending in 0 is divisible by 5, so 5 would appear in its prime factorisation. But 6^n = 2^n x 3^n has only the primes 2 and 3, so 6^n never ends in 0.

Teacher's Note:
1. Fundamental theorem of arithmetic: every composite number is a product of primes in exactly one way (apart from order). HCF takes the least exponent of each common prime and LCM the greatest exponent of every prime.

Answer: (b) 6

Explanation:
1. 84 = 2^2 x 3 x 7, 90 = 2 x 3^2 x 5, 120 = 2^3 x 3 x 5. The common primes are 2 and 3 with least exponents 1 and 1, so the HCF is 6.

Teacher's Note:
1. Fundamental theorem of arithmetic: every composite number is a product of primes in exactly one way (apart from order). HCF takes the least exponent of each common prime and LCM the greatest exponent of every prime.

Answer: (a) 2520

Explanation:
1. Taking the greatest exponents: 2^3 x 3^2 x 5 x 7 = 8 x 9 x 5 x 7 = 2520.

Teacher's Note:
1. Fundamental theorem of arithmetic: every composite number is a product of primes in exactly one way (apart from order). HCF takes the least exponent of each common prime and LCM the greatest exponent of every prime.

Answer: (d) 122 m 40 cm

Explanation:
1. It is the LCM of 80, 85 and 90: 80 = 2^4 x 5, 85 = 5 x 17, 90 = 2 x 3^2 x 5, so LCM = 16 x 9 x 5 x 17 = 12240 cm = 122 m 40 cm.

Teacher's Note:
1. Fundamental theorem of arithmetic: every composite number is a product of primes in exactly one way (apart from order). HCF takes the least exponent of each common prime and LCM the greatest exponent of every prime.

Answer: (d) 2^m x 5^n

Explanation:
1. 43.123456789 terminates, so it equals p/10^9 and the denominator of its lowest-terms form has only the primes 2 and 5.

Teacher's Note:
1. A rational number p/q in lowest terms has a terminating decimal expansion exactly when the prime factors of q are only 2 and 5; otherwise the expansion is non-terminating and repeating.

Answer: (c) with at least one prime factor other than 2 and 5

Explanation:
1. Terminating decimals correspond exactly to denominators of the form 2^m x 5^n, so a repeating non-terminating expansion needs another prime factor in the denominator.

Teacher's Note:
1. A rational number p/q in lowest terms has a terminating decimal expansion exactly when the prime factors of q are only 2 and 5; otherwise the expansion is non-terminating and repeating.

Answer: (b) 2

Explanation:
1. 814 = 14 x 56 + 30, 56 = 1 x 30 + 26, 30 = 1 x 26 + 4, 26 = 6 x 4 + 2, 4 = 2 x 2. The HCF is 2.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (a) 179

Explanation:
1. 76254 = 12 x 6265 + 1074, 6265 = 5 x 1074 + 895, 1074 = 1 x 895 + 179, 895 = 5 x 179. The HCF is 179.

Teacher's Note:
1. Euclid's division lemma: for positive integers a and b there are unique integers q and r with a = bq + r and 0 <= r < b. Euclid's algorithm repeats this on (b, r) until the remainder is 0; the last divisor is the HCF.

Answer: (a) 6 and 40896

Explanation:
1. 426 = 2 x 3 x 71 and 576 = 2^6 x 3^2. HCF = 2 x 3 = 6 and LCM = 2^6 x 3^2 x 71 = 40896.

Teacher's Note:
1. Fundamental theorem of arithmetic: every composite number is a product of primes in exactly one way (apart from order). HCF takes the least exponent of each common prime and LCM the greatest exponent of every prime.

Answer: (d) 125 and 95625

Explanation:
1. 625 = 5^4, 1125 = 3^2 x 5^3, 2125 = 5^3 x 17. HCF = 5^3 = 125 and LCM = 3^2 x 5^4 x 17 = 95625.

Teacher's Note:
1. Fundamental theorem of arithmetic: every composite number is a product of primes in exactly one way (apart from order). HCF takes the least exponent of each common prime and LCM the greatest exponent of every prime.

Answer: (c) \( 5+\sqrt2 \)

Explanation:
1. sqrt2 is irrational and a rational number added to an irrational number is irrational, so 5 + sqrt2 is irrational. sqrt25 = 5, sqrt2 x sqrt8 = 4 and 0.333... = 1/3 are rational.

Teacher's Note:
1. If a prime p divides a^2 then p divides a. Using this one shows that square roots of non-square integers are irrational. A rational plus or times a non-zero irrational number is irrational.

Answer: (b) 3 divides both a and b, contradicting that a/b is in lowest terms

Explanation:
1. From a^2 = 3b^2, 3 divides a^2 so 3 divides a. Writing a = 3c gives b^2 = 3c^2, so 3 divides b. Then a and b have the common factor 3, a contradiction.

Teacher's Note:
1. If a prime p divides a^2 then p divides a. Using this one shows that square roots of non-square integers are irrational. A rational plus or times a non-zero irrational number is irrational.