CBSE Class 10 Mathematics Real Numbers MCQs Set 30

Download CBSE 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.

Chapter-wise Objective Questions: Chapter 01 Real Numbers

Navigate directly to the 50 objective questions for Chapter 01 Real Numbers using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?
(a) 0
(b) 9
(c) 3
(d) 6
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. 1421 ≡ 5, 1423 ≡ 7, 1425 ≡ 9 (mod 12). \( 5 \times 7 = 35 \) ≡ 11, and \( 11 \times 9 = 99 \) ≡ 3.

Teacher's Note:
1. Reduce each factor modulo 12.

Question 2: If n = 67, the unit digit of 3ⁿ + 2ⁿ is
(a) 1
(b) 10
(c) 5
(d) None
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. 3⁶⁷: 67 mod \( 4 = 3 \), unit digit 7. 2⁶⁷: unit digit 8.
2. \( 7 + 8 = 15 \), so the unit digit is 5.

Teacher's Note:
1. Use the cycle length of four.

Question 3: The last digit of (1³ + 2³ + 3³ + ... + 10³)⁶⁴ is
(a) 2
(b) 5
(c) 9
(d) 0
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. The sum is \( (10 \times 11/2)^{2} = 3025 \), which ends in 5, and any power of a number ending in 5 ends in 5.

Teacher's Note:
1. Sum of cubes = (sum of numbers)².

Question 4: The sum of the unit digits of the numbers 2²⁰⁰⁶ and 5²⁰⁰⁷ is
(a) 6
(b) 9
(c) 5
(d) 1
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. 2²⁰⁰⁶: 2006 mod \( 4 = 2 \), unit digit 4. 5²⁰⁰⁷ ends in 5.
2. Sum = 9.

Teacher's Note:
1. Powers of 5 always end in 5.

Question 5: Find A + B + C + D such that AB × CB = DDD, where AB and CB are two-digit numbers and DDD is a three-digit number.
(a) 21
(b) 19
(c) 17
(d) 18
Show Answer & Explanation

Answer: (a) 21

Explanation:
1. DDD = \( 111 \times D = 3 \times 37 \times D \). With both factors ending in the same digit, \( 27 \times 37 = 999 \) works.
2. A \( + B + C + D = 2 + 7 + 3 + 9 = 21 \).

Teacher's Note:
1. Factorise \( 111D \).

Question 6: A 3-digit number 4a3 is added to 984 to give the four-digit number 13b7, which is divisible by 11. Then (a + b) is
(a) 10
(b) 11
(c) 12
(d) 15
Show Answer & Explanation

Answer: (a) 10

Explanation:
1. \( 413 + 984 = 1397 \) and \( 1397 = 11 \times 127 \), so \( a = 1 \) and \( b = 9 \).
2. \( a + b = 10 \).

Teacher's Note:
1. Units digit gives \( 3 + 4 = 7 \); then fit the tens digit.

Question 7: The sum of the digits of a two-digit number is 11. If the digits are reversed the number decreases by 45. The number is
(a) 38
(b) 65
(c) 74
(d) 83
Show Answer & Explanation

Answer: (d) 83

Explanation:
1. \( 10a + b - (10b + a) = 9(a - b) = 45 \), so \( a - b = 5 \), and \( a + b = 11 \) gives \( a = 8, b = 3 \).
2. The number is 83.

Teacher's Note:
1. Use the difference of the number and its reverse.

Question 8: If log_x y = 100 and log₂ x = 10, then the value of y is
(a) 2¹⁰⁰⁰
(b) 2¹⁰⁰
(c) 2²⁰⁰⁰
(d) 2¹⁰⁰⁰⁰
Show Answer & Explanation

Answer: (a) 2¹⁰⁰⁰

Explanation:
1. \( x = 2^{10} \) and \( y = x^{100} = 2^{1000} \).

Teacher's Note:
1. Convert each logarithm to exponent form.

Question 9: log₁₀(1 + 1/2) + log₁₀(1 + 1/3) + log₁₀(1 + 1/4) + ... + log₁₀(1 + 1/1999), when simplified, has the value
(a) 1
(b) 3
(c) 10
(d) 100
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. The product \( (3/2)(4/3)...(2000/1999) = 1000 \), so the sum = \( \log_{10} 1000 = 3 \).

Teacher's Note:
1. The product telescopes.

Question 10: If log₁₀ N = 2.5, the total number of digits in N is
(a) 3
(b) 4
(c) 5
(d) cannot be determined
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. \( N = 10^{2.5} \) lies between 10² and 10³, so \( N \) has 3 digits.

Teacher's Note:
1. Digits = integer part of the \( \log \) plus one.

Question 11: If log x = n, then 2n is equal to
(a) log (x²)
(b) (log x)²
(c) log (x + 2)
(d) log 2x
Show Answer & Explanation

Answer: (a) log (x²)

Explanation:
1. \( 2n = 2 \log x = \log x^{2} \).

Teacher's Note:
1. Apply the power rule.

Question 12: The value of [log₁₀(5 log₁₀ 100)]² is
(a) 0
(b) 1
(c) 2
(d) 10
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. \( \log_{10} 100 = 2, 5 \times 2 = 10, \log_{10} 10 = 1 \), and \( 1^{2} = 1 \).

Teacher's Note:
1. Evaluate from the inside out.

Question 13: If log₁₀[log₁₀(log₁₀ x)] = 0, then
(a) x = 10³
(b) x = 10¹⁰
(c) x = 15⁵
(d) None
Show Answer & Explanation

Answer: (b) x = 10¹⁰

Explanation:
1. \( \log_{10}(\log_{10} x) = 1 \), so \( \log_{10} x = 10 \) and \( x = 10^{10} \).

Teacher's Note:
1. Undo each logarithm in turn.

Question 14: If x satisfies log₁₀ 1250 + log₁₀ 80 = x, then x is
(a) 5
(b) 4
(c) 8
(d) 7
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. \( x = \log_{10}(1250 \times 80) = \log_{10} 100000 = 5 \).

Teacher's Note:
1. Add logarithms by multiplying the numbers.

Question 15: Which of the following surds is the greatest in magnitude: the sixth root of 17, √2, the twelfth root of 25, the cube root of 4?
(a) sixth root of 17
(b) twelfth root of 25
(c) cube root of 4
(d) √2
Show Answer & Explanation

Answer: (a) sixth root of 17

Explanation:
1. Raise each to the 12th power: \( 17^{2} = 289, 2^{6} = 64, 25, 4^{4} = 256 \).
2. The largest is 289, for the sixth root of 17.

Teacher's Note:
1. Bring the surds to a common index.

Question 16: How many numbers between 400 and 600 begin or end with a digit 5?
(a) 40
(b) 100
(c) 110
(d) 120
Show Answer & Explanation

Answer: (c) 110

Explanation:
1. Numbers 500 to 599 begin with 5: 100. Numbers ending in 5 from 405 to 495: 10 more (those in 505 to 595 are already counted).
2. Total = 110.

Teacher's Note:
1. Avoid double counting.

Question 17: Anita had to do a multiplication. Instead of taking 35 as one of the multipliers she took 53. As a result the product went up by 540. The new product is
(a) 1050
(b) 540
(c) 1440
(d) 1590
Show Answer & Explanation

Answer: (d) 1590

Explanation:
1. (53 − 35) × other = 540, so the other multiplier = 30.
2. New product = \( 53 \times 30 = 1590 \).

Teacher's Note:
1. The change in product is the change in the multiplier times the other factor.

Question 18: Three friends stopped to eat. Sita took 1/3 of the mints in a bowl but returned four. Fatima then took 1/4 of what was left but returned three. Eswari then took half of the remainder but threw two back. The bowl had 17 mints left at the end. How many mints were originally in the bowl?
(a) 38
(b) 31
(c) 41
(d) 48
Show Answer & Explanation

Answer: (d) 48

Explanation:
1. Work backwards: after Eswari \( 17 = L/2 + 2 \), so \( L = 30 \) before her.
2. \( 30 = 3M/4 + 3 \) where \( M \) is what was in the bowl before Fatima, so \( M = 36 \); and \( 36 = 2N/3 + 4 \), so \( N = 48 \).

Teacher's Note:
1. Work backwards from the final count.

Question 19: The value of 1/(2² − 1) + 1/(4² − 1) + 1/(6² − 1) + ... + 1/(20² − 1) is
(a) 9/19
(b) 10/19
(c) 10/21
(d) 11/21
Show Answer & Explanation

Answer: (c) 10/21

Explanation:
1. \( 1/((2k)^{2} - 1) = (1/2)(1/(2k - 1) - 1/(2k + 1)) \), and the sum telescopes to \( (1/2)(1 - 1/21) = 10/21 \).

Teacher's Note:
1. Use partial fractions.

Question 20: (1 + 1/2)(1 + 1/3)(1 + 1/4)(1 + 1/5)(1 + 1/6)(1 + 1/7)(1 + 1/8) is equal to
(a) 9
(b) 8
(c) 4.5
(d) None of these
Show Answer & Explanation

Answer: (c) 4.5

Explanation:
1. The product telescopes: \( (3/2)(4/3)...(9/8) = 9/2 = 4.5 \).

Teacher's Note:
1. Write each factor as a single fraction.

Question 21: A rich merchant collected many gold coins. He said, 'If I divide the coins into two unequal numbers, then 48 times the difference between the two numbers equals the difference between the squares of the two numbers.' How many gold coins does the merchant have?
(a) 96
(b) 53
(c) 43
(d) 48
Show Answer & Explanation

Answer: (d) 48

Explanation:
1. \( 48(x - y) = x^{2} - y^{2} = (x - y)(x + y) \), and \( x \ne y \), so \( x + y = 48 \).

Teacher's Note:
1. Cancel the common factor of \( (x - y) \).

Question 22: P, Q and R are natural numbers such that P and Q are distinct primes and Q divides PR. Which of the following is correct?
(a) Q divides R
(b) P divides R
(c) Q divides PR only
(d) P divides QR
Show Answer & Explanation

Answer: (a) Q divides R

Explanation:
1. \( Q \) is a prime that does not divide \( P \), so \( Q \) must divide \( R \).

Teacher's Note:
1. Use Euclid's lemma.

Question 23: When expanded, the number of zeros in 1000¹⁰ is
(a) 13
(b) 30
(c) 4
(d) 10
Show Answer & Explanation

Answer: (b) 30

Explanation:
1. \( 1000^{10} = 10^{30} \), which has 30 zeros.

Teacher's Note:
1. Write 1000 as 10³.

Question 24: Let N = 28. The sum of all distinct factors of N is
(a) 27
(b) 28
(c) 55
(d) 56
Show Answer & Explanation

Answer: (d) 56

Explanation:
1. Factors: 1, 2, 4, 7, 14, 28, whose sum is 56.

Teacher's Note:
1. List the factors in pairs.

Question 25: The units digit of (1 + 9 + 9² + 9³ + ... + 9²⁰⁰⁹) is
(a) 0
(b) 1
(c) 9
(d) 3
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. Powers of 9 end in 1, 9, 1, 9, ... There are 2010 terms, forming 1005 pairs each ending in 0, so the units digit is 0.

Teacher's Note:
1. Group the terms in pairs.

Free study material for Mathematics

Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 01 Real Numbers

Chapter MCQs with Answers for Class 10 Mathematics

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FAQs

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Yes, our CBSE Class 10 Mathematics Real Numbers MCQs Set 30 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

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