CBSE Class 10 Mathematics Real Numbers MCQs Set 23

Practice MCQs for Class 10 Mathematics Chapter 01 Real Numbers

Access targeted multiple-choice questions for Chapter 01 Real Numbers designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: Any positive odd integer is of the form (q is an integer):
(a) \( 6q+1,\ 6q+3\ \text{or}\ 6q+5 \)
(b) \( 6q,\ 6q+2\ \text{or}\ 6q+4 \)
(c) \( 6q+1\ \text{or}\ 6q+5\ \text{only} \)
(d) \( 6q+3\ \text{only} \)
Show Answer & Explanation

Answer: (a) \( 6q+1,\ 6q+3\ \text{or}\ 6q+5 \)

Explanation:
1. Writing a = 6q + r with r = 0, 1, 2, 3, 4, 5, the forms 6q, 6q + 2 and 6q + 4 are even. So an odd integer is 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 2: The cube of any positive integer is of the form (m is an integer):
(a) \( 9m,\ 9m+2\ \text{or}\ 9m+7 \)
(b) \( 9m+1,\ 9m+4\ \text{or}\ 9m+7 \)
(c) \( 9m,\ 9m+3\ \text{or}\ 9m+6 \)
(d) \( 9m,\ 9m+1\ \text{or}\ 9m+8 \)
Show Answer & Explanation

Answer: (d) \( 9m,\ 9m+1\ \text{or}\ 9m+8 \)

Explanation:
1. Any integer is 3q, 3q + 1 or 3q + 2. (3q)³ = 9(3q³); (3q + 1)³ = 9q(3q² + 3q + 1) + 1; (3q + 2)³ = 9q(3q² + 6q + 4) + 8. So the cube is 9m, 9m + 1 or 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 3: The square of a positive integer of the form 5q + 1 is of the form:
(a) \( 5m+2 \)
(b) \( 5m+4 \)
(c) \( 5m+1 \)
(d) \( 5m \)
Show Answer & Explanation

Answer: (c) \( 5m+1 \)

Explanation:
1. (5q + 1)² = 25q² + 10q + 1 = 5(5q² + 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 4: Using Euclid's division algorithm, the HCF of 196 and 38318 is:
(a) 196
(b) 98
(c) 49
(d) 2
Show Answer & Explanation

Answer: (b) 98

Explanation:
1. 38318 = 196 x 195 + 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 5: In the first step of Euclid's algorithm applied to 38318 and 196, the quotient and remainder are:
(a) 196 and 98
(b) 195 and 98
(c) 195 and 196
(d) 194 and 294
Show Answer & Explanation

Answer: (b) 195 and 98

Explanation:
1. 38318 = 196 x 195 + 98, so the quotient is 195 and the 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 6: The HCF of 657 and 963 is expressible in the form 657x + 963 x (-15). The value of x is:
(a) 22
(b) 15
(c) 18
(d) 20
Show Answer & Explanation

Answer: (a) 22

Explanation:
1. By Euclid's algorithm 963 = 657 x 1 + 306, 657 = 306 x 2 + 45, 306 = 45 x 6 + 36, 45 = 36 x 1 + 9, 36 = 9 x 4 + 0, so the HCF is 9. Then 657x - 14445 = 9, so 657x = 14454 and 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 7: The HCF of 657 and 963 is:
(a) 3
(b) 18
(c) 27
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. 963 = 657 + 306, 657 = 2 x 306 + 45, 306 = 6 x 45 + 36, 45 = 36 + 9, 36 = 4 x 9. The last non-zero remainder is 9.

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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 8: The largest number that divides 626, 3127 and 15628 leaving remainders 1, 2 and 3 respectively is:
(a) 313
(b) 125
(c) 625
(d) 3125
Show Answer & Explanation

Answer: (c) 625

Explanation:
1. The number divides 626 - 1 = 625, 3127 - 2 = 3125 and 15628 - 3 = 15625. HCF(625, 3125) = 625 and HCF(625, 15625) = 625. So the number 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 9: 144 cartons of coke cans and 90 cartons of Pepsi cans are stacked in a canteen, with every stack of the same height and containing cartons of the same drink. The greatest number of cartons in each stack is:
(a) 9
(b) 36
(c) 18
(d) 6
Show Answer & Explanation

Answer: (c) 18

Explanation:
1. The number of cartons in a stack must divide both 144 and 90. HCF(144, 90) = 18 (144 = 90 + 54, 90 = 54 + 36, 54 = 36 + 18, 36 = 2 x 18). So the greatest number 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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 10: The prime factorisation of 45470971 is:
(a) \( 7^2\times11^2\times17^2\times19 \)
(b) \( 7^2\times13^2\times17^2\times19 \)
(c) \( 7\times13^2\times17^2\times19^2 \)
(d) \( 7^2\times13\times17^2\times19^2 \)
Show Answer & Explanation

Answer: (b) \( 7^2\times13^2\times17^2\times19 \)

Explanation:
1. Multiplying 7² x 13² x 17² x 19 = 49 x 169 x 289 x 19 = 45470971. By the Fundamental Theorem of Arithmetic this factorisation is unique.

Teacher's Note:
1. Every composite number is a product of primes in a unique way, apart from the order of the factors.
2. HCF is the product of the smallest powers of the common primes and LCM is the product of the greatest powers of all primes.

Question 11: For any natural number n, the number 6^n ends with the digit 0:
(a) never, because 5 is not a prime factor of 6^n
(b) always, because 6 is even
(c) only when n is even
(d) only when n is a multiple of 5
Show Answer & Explanation

Answer: (a) never, because 5 is not a prime factor of 6^n

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

Teacher's Note:
1. Every composite number is a product of primes in a unique way, apart from the order of the factors.
2. HCF is the product of the smallest powers of the common primes and LCM is the product of the greatest powers of all primes.

Question 12: The HCF and LCM of 84, 90 and 120, found by prime factorisation, are respectively:
(a) 6 and 5040
(b) 12 and 2520
(c) 3 and 1260
(d) 6 and 2520
Show Answer & Explanation

Answer: (d) 6 and 2520

Explanation:
1. 84 = 2² x 3 x 7, 90 = 2 x 3² x 5, 120 = 2³ x 3 x 5. HCF = 2 x 3 = 6. LCM = 2³ x 3² x 5 x 7 = 2520.

Teacher's Note:
1. Every composite number is a product of primes in a unique way, apart from the order of the factors.
2. HCF is the product of the smallest powers of the common primes and LCM is the product of the greatest powers of all primes.

Question 13: Three persons start walking together with steps of 80 cm, 85 cm and 90 cm. The minimum distance each should walk so that each can cover it in complete steps is:
(a) 12 m 24 cm
(b) 61 m 20 cm
(c) 244 m 80 cm
(d) 122 m 40 cm
Show Answer & Explanation

Answer: (d) 122 m 40 cm

Explanation:
1. The distance is the LCM of 80, 85 and 90. 80 = 2⁴ x 5, 85 = 5 x 17, 90 = 2 x 3² x 5. LCM = 16 x 9 x 5 x 17 = 12240 cm = 122 m 40 cm.

Teacher's Note:
1. Every composite number is a product of primes in a unique way, apart from the order of the factors.
2. HCF is the product of the smallest powers of the common primes and LCM is the product of the greatest powers of all primes.

Question 14: In the proof that √2 is irrational, assuming √2 = a/b with a and b co-prime leads to a² = 2b². The contradiction is that:
(a) a is odd and even at once
(b) b is zero
(c) a and b both turn out to be multiples of 2, but they are co-prime
(d) √2 is an integer
Show Answer & Explanation

Answer: (c) a and b both turn out to be multiples of 2, but they are co-prime

Explanation:
1. From a² = 2b², a² is even so a is even, a = 2c. Then 2b² = 4c² gives b² = 2c², so b is also even. So 2 is a common factor of a and b, contradicting that they are co-prime.

Teacher's Note:
1. Every composite number is a product of primes in a unique way, apart from the order of the factors.
2. HCF is the product of the smallest powers of the common primes and LCM is the product of the greatest powers of all primes.

Question 15: If a prime p divides a², where a is a positive integer, then:
(a) p does not divide a
(b) p divides a
(c) p divides a + 1
(d) p divides a² only
Show Answer & Explanation

Answer: (b) p divides a

Explanation:
1. This is a result of the Fundamental Theorem of Arithmetic: the primes of a² are exactly the primes of a, so p divides a.

Teacher's Note:
1. Every composite number is a product of primes in a unique way, apart from the order of the factors.
2. HCF is the product of the smallest powers of the common primes and LCM is the product of the greatest powers of all primes.

Question 16: To prove that 3 - √5 is irrational, assume 3 - √5 = a/b (a, b co-prime). This gives √5 = (3b - a)/b. The contradiction is that:
(a) a rational number would equal the irrational number √5
(b) √5 equals 3
(c) b equals zero
(d) a and b are not integers
Show Answer & Explanation

Answer: (a) a rational number would equal the irrational number √5

Explanation:
1. 3b - a and b are integers, so (3b - a)/b is rational. That would make √5 rational, contradicting the fact that √5 is irrational.

Teacher's Note:
1. Every composite number is a product of primes in a unique way, apart from the order of the factors.
2. HCF is the product of the smallest powers of the common primes and LCM is the product of the greatest powers of all primes.

Question 17: Without long division, 13/3125 has a decimal expansion that is:
(a) terminating
(b) non-terminating and repeating
(c) non-terminating and non-repeating
(d) recurring with period 5
Show Answer & Explanation

Answer: (a) terminating

Explanation:
1. 3125 = 5⁵ = 2⁰ x 5⁵, which has only the prime factors 2 and 5. So the expansion terminates.

Teacher's Note:
1. A rational number p/q (in lowest terms) has a terminating decimal expansion exactly when q = 2^m x 5^n.
2. Otherwise the expansion is non-terminating and repeating.

Question 18: The rational number 43.123456789 has a terminating decimal expansion. The prime factorisation of the denominator (in lowest terms) is of the form:
(a) \( 3^m\times7^n \)
(b) \( 2^m\times3^n \)
(c) \( 5^m\times11^n \)
(d) \( 2^m\times5^n \)
Show Answer & Explanation

Answer: (d) \( 2^m\times5^n \)

Explanation:
1. A rational number with a terminating decimal expansion has a denominator whose only prime factors are 2 and 5.

Teacher's Note:
1. A rational number p/q (in lowest terms) has a terminating decimal expansion exactly when q = 2^m x 5^n.
2. Otherwise the expansion is non-terminating and repeating.

Question 19: The number 43.̅123456789 (the block 123456789 repeating) is rational with a non-terminating repeating expansion. The denominator in lowest terms:
(a) is of the form 2^m x 5^n
(b) is 1
(c) has prime factors other than 2 and 5
(d) is a power of 2
Show Answer & Explanation

Answer: (c) has prime factors other than 2 and 5

Explanation:
1. A non-terminating repeating decimal has a denominator with a prime factor other than 2 or 5.

Teacher's Note:
1. A rational number p/q (in lowest terms) has a terminating decimal expansion exactly when q = 2^m x 5^n.
2. Otherwise the expansion is non-terminating and repeating.

Question 20: Which of the following is the statement of Euclid's division lemma for positive integers a and b?
(a) \( a=bq+r,\ 0(b) \( a=bq+r,\ 0\le r\lt b\ \text{for unique integers}\ q,r \)
(c) \( a=b+q+r,\ 0\le r\lt b \)
(d) \( a=bq,\ \text{for every}\ a,b \)
Show Answer & Explanation

Answer: (b) \( a=bq+r,\ 0\le r\lt b\ \text{for unique integers}\ q,r \)

Explanation:
1. For given positive integers a and b there exist unique integers q and r such that a = bq + r with 0 ≤ r < b. If r = 0 then b divides a.

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.
2. The HCF of a and b is found by repeating the lemma on (b, r) until the remainder is 0.

Question 21: A rational number x has a terminating decimal expansion. Then x = p/q in lowest terms, where the prime factorisation of q is of the form:
(a) \( 3^m\times5^n \)
(b) \( 2^m\times5^n \)
(c) \( 2^m\times7^n \)
(d) \( any prime power \)
Show Answer & Explanation

Answer: (b) \( 2^m\times5^n \)

Explanation:
1. A number with a terminating expansion can be written with a denominator 10^k = 2^k x 5^k, so q 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 q = 2^m x 5^n.
2. Otherwise the expansion is non-terminating and repeating.

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