CBSE Class 10 Mathematics Real Numbers MCQs Set 29

Multiple Choice Questions (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.

Practice Chapter 01 Real Numbers MCQs for Class 10 Mathematics

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Multiple Choice Questions

Question 1: In a big hostel there are 1000 rooms numbered with even numbers only, 2, 4, 6, ..., 2000, and every room has one resident. All doors are open initially. One morning the warden asks every resident to go back to his room and then to every room whose number is a multiple of his own room number. When a resident visits a room and finds the door open he closes it, and if the door is closed he opens it. All 1000 residents do this. The 38th room that is open is room number
(a) 80
(b) 88
(c) 76
(d) None of these
Show Answer & Explanation

Answer: (b) 88

Explanation:
1. By the same simulation, 969 rooms remain open.
2. Listing the open rooms in order, the 38th is 88.

Teacher's Note:
1. Work with the divisors of the room number that are even.

Question 2: In a big hostel there are 1000 rooms numbered with even numbers only, 2, 4, 6, ..., 2000, and every room has one resident. All doors are open initially. One morning the warden asks every resident to go back to his room and then to every room whose number is a multiple of his own room number. When a resident visits a room and finds the door open he closes it, and if the door is closed he opens it. All 1000 residents do this. Suppose only 500 residents, those of rooms 2 to 1000, do the task. The last room that is closed is room number
(a) 2000
(b) 1936
(c) 1849
(d) None of these
Show Answer & Explanation

Answer: (a) 2000

Explanation:
1. The resident of room 2000 does not act, and room 2000 is toggled by the residents of rooms 2, 4, 8, 10, 16, 20, ... whose room numbers divide 2000 and are at most 1000.
2. Simulation shows room 2000 is toggled an odd number of times and is closed, and it is the largest room.

Teacher's Note:
1. Check the largest room first.

Question 3: In a big hostel there are 1000 rooms numbered with even numbers only, 2, 4, 6, ..., 2000, and every room has one resident. All doors are open initially. One morning the warden asks every resident to go back to his room and then to every room whose number is a multiple of his own room number. When a resident visits a room and finds the door open he closes it, and if the door is closed he opens it. All 1000 residents do this. Suppose only 500 residents, those of rooms 2 to 1000, do the task. How many rooms are closed in all?
(a) 513
(b) 31
(c) 13
(d) 315
Show Answer & Explanation

Answer: (a) 513

Explanation:
1. Simulating the 500 residents, 513 rooms are left closed.

Teacher's Note:
1. Compare with the case of all 1000 residents.

Question 4: How many three-digit numbers are there which, when divided by 3, 4, 5, 6, 7, leave remainders 1, 2, 3, 4, 5 respectively?
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. In every case the divisor minus the remainder is 2, so n + 2 is a multiple of LCM(3, 4, 5, 6, 7) = 420.
2. n = 418 and 838 are the three-digit numbers, so there are 2.

Teacher's Note:
1. Add the common difference to make the number a multiple of the LCM.

Question 5: Six strings of a violin start vibrating together and vibrate 3, 4, 5, 6, 10 and 12 times in a minute. After how much time will all six vibrate together again, and how many times will they vibrate together in 30 minutes (counting the start)?
(a) 60 min, 31 times
(b) 60 sec, 31 times
(c) 120 sec, 15 times
(d) None of these
Show Answer & Explanation

Answer: (b) 60 sec, 31 times

Explanation:
1. The periods are 20, 15, 12, 10, 6 and 5 seconds. Their LCM is 60 seconds.
2. In 30 minutes = 1800 s they coincide 1800/60 = 30 times after the start, i.e. 31 times counting the start.

Teacher's Note:
1. Convert vibrations per minute into periods in seconds.

Question 6: If P is a prime number, then the LCM of P and (P + 1) is
(a) P(P + 1)
(b) (P + 2)P
(c) (P + 1)(P − 1)
(d) None of these
Show Answer & Explanation

Answer: (a) P(P + 1)

Explanation:
1. P and P + 1 are consecutive, hence co-prime, so their LCM is their product P(P + 1).

Teacher's Note:
1. Consecutive integers are co-prime.

Question 7: Three pieces of cake of weights 4 1/2 lbs, 6 3/4 lbs and 7 1/5 lbs are to be divided into parts of equal weight, each as heavy as possible. If one such part is served to each guest, the maximum number of guests that can be entertained is
(a) 54
(b) 72
(c) 20
(d) 41
Show Answer & Explanation

Answer: (d) 41

Explanation:
1. In units of 1/20 lb the weights are 90, 135 and 144, whose HCF is 9, so each part is 9/20 lb.
2. Number of parts = (90 + 135 + 144)/9 = 41.

Teacher's Note:
1. The largest equal part is the HCF of the weights.

Question 8: How many numbers between 200 and 600 are divisible by 14?
(a) 28
(b) 29
(c) 27
(d) None of these
Show Answer & Explanation

Answer: (a) 28

Explanation:
1. Multiples of 14 from 210 to 588: 588/14 − 14 = 42 − 14 = 28.

Teacher's Note:
1. Count the multiples with floor values.

Question 9: How many natural numbers between 200 and 400 are divisible by (i) both 4 and 5, (ii) 4 or 5 or 8 or 10?
(a) 9, 79
(b) 10, 80
(c) 10, 81
(d) None of these
Show Answer & Explanation

Answer: (a) 9, 79

Explanation:
1. (i) Multiples of 20 from 220 to 380: 9.
2. (ii) Multiples of 4 (49) plus multiples of 5 not multiples of 4 (30): 79.

Teacher's Note:
1. Use inclusion and exclusion.

Question 10: If 653xy is exactly divisible by 80, the value of x + y is
(a) 2
(b) 3
(c) 4
(d) 6
Show Answer & Explanation

Answer: (d) 6

Explanation:
1. Divisibility by 80 needs divisibility by 16 and by 5. Checking 65300 to 65399 gives 65360 = 80 × 817, so x = 6 and y = 0.
2. x + y = 6.

Teacher's Note:
1. Divisible by 80 means the number ends in 0 and is divisible by 16.

Question 11: When a number P is divided by 4 it leaves remainder 3. If twice the number P is divided by 4, the remainder is
(a) 0
(b) 1
(c) 2
(d) 6
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. P = 4k + 3, so 2P = 8k + 6 = 4(2k + 1) + 2, and the remainder is 2.

Teacher's Note:
1. Write P in the form 4k + 3.

Question 12: The smallest number which should be added to 803642 so that it is divisible by 11 is
(a) 40
(b) 4
(c) 23
(d) 7
Show Answer & Explanation

Answer: (d) 7

Explanation:
1. Alternating digit sum of 803642 = 2 − 4 + 6 − 3 + 0 − 8 = −7, so 803642 leaves remainder 4 on division by 11.
2. 11 − 4 = 7.

Teacher's Note:
1. Use the alternating-sum test for 11.

Question 13: If the number 357y25x is divisible by both 3 and 5, the missing digit in the unit's place and the thousand's place respectively are
(a) 0, 6
(b) 5, 6
(c) 5, 4
(d) None of these
Show Answer & Explanation

Answer: (b) 5, 6

Explanation:
1. Divisible by 5 means x = 0 or 5. Digit sum = 22 + x + y must be a multiple of 3.
2. With x = 5, y = 6 the digit sum is 33, so (5, 6) works. For x = 0, y would be 2, 5 or 8, which is not an option.

Teacher's Note:
1. Check the options against both tests.

Question 14: A number when divided by 342 gives a remainder 47. When the same number is divided by 19, the remainder is
(a) 3
(b) 5
(c) 9
(d) None of these
Show Answer & Explanation

Answer: (c) 9

Explanation:
1. 342 = 18 × 19, so the remainder on division by 19 equals the remainder of 47 on division by 19, which is 9.

Teacher's Note:
1. The smaller divisor is a factor of 342.

Question 15: A 4-digit number is formed by repeating a 2-digit number, such as 2525 or 3232. Any such number is exactly divisible by
(a) 7
(b) 11
(c) 13
(d) The smallest 3-digit prime
Show Answer & Explanation

Answer: (d) The smallest 3-digit prime

Explanation:
1. The number ab ab = ab × 101, and 101 is the smallest 3-digit prime.

Teacher's Note:
1. Write the number as 101 times the two-digit number.

Question 16: The largest natural number by which the product of three consecutive even natural numbers is always divisible is
(a) 6
(b) 24
(c) 48
(d) 96
Show Answer & Explanation

Answer: (c) 48

Explanation:
1. The product is 8 × k(k + 1)(k + 2), and k(k + 1)(k + 2) is divisible by 6, so the product is divisible by 48. For k = 1: 2 × 4 × 6 = 48.

Teacher's Note:
1. Write the evens as 2k, 2k + 2, 2k + 4.

Question 17: Let N = 55³ + 17³ − 72³. Then N is divisible by
(a) both 7 and 13
(b) both 3 and 13
(c) both 17 and 7
(d) both 3 and 17
Show Answer & Explanation

Answer: (d) both 3 and 17

Explanation:
1. Since 55 + 17 = 72, N = a³ + b³ − (a + b)³ = −3ab(a + b) = −3 × 55 × 17 × 72.
2. So N is divisible by 3 and 17.

Teacher's Note:
1. Use the identity for (a + b)³.

Question 18: 7⁶ⁿ − 6⁶ⁿ, where n is an integer greater than 0, is divisible by
(a) 13
(b) 127
(c) 559
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. xⁿ − yⁿ is divisible by x − y, so it is divisible by 7⁶ − 6⁶ = 70993 = 13 × 43 × 127.
2. Hence it is divisible by 13, 127 and 559 = 13 × 43.

Teacher's Note:
1. Factorise 7⁶ − 6⁶.

Question 19: The number 1982 in base 10, when converted to base 12, is
(a) 1182
(b) 1912
(c) 1192
(d) 1292
Show Answer & Explanation

Answer: (c) 1192

Explanation:
1. 1982 = 1 × 1728 + 1 × 144 + 9 × 12 + 2, so it is (1192)₁₂.

Teacher's Note:
1. Divide repeatedly by 12.

Question 20: The decimal equivalent of the 25-digit hexadecimal number (100...001)₁₆ is
(a) 2²³ + 1
(b) 2²⁴ + 1
(c) 2³² + 1
(d) 2⁹⁶ + 1
Show Answer & Explanation

Answer: (d) 2⁹⁶ + 1

Explanation:
1. The number is 16²⁴ + 1 = 2⁹⁶ + 1.

Teacher's Note:
1. A 25-digit number has its leading digit at the 24th power.

Question 21: What is the remainder when 9875347 × 7435789 × 5789743 is divided by 4?
(a) 1
(b) 2
(c) 3
(d) None of these
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. Last two digits give remainders 47 → 3, 89 → 1, 43 → 3.
2. 3 × 1 × 3 = 9, and 9 mod 4 = 1.

Teacher's Note:
1. Multiply the remainders.

Question 22: What is the remainder when 7⁸⁴ is divided by 2402?
(a) 1
(b) 6
(c) 2401
(d) None of these
Show Answer & Explanation

Answer: (c) 2401

Explanation:
1. 7⁴ = 2401 ≡ −1 (mod 2402), so 7⁸⁴ = (7⁴)²¹ ≡ (−1)²¹ = −1 ≡ 2401.

Teacher's Note:
1. 7⁴ is one less than the divisor.

Question 23: What is the remainder when 30⁴⁰ is divided by 17?
(a) 1
(b) 16
(c) 13
(d) 4
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. 30 ≡ 13 ≡ −4 (mod 17), so 30⁴⁰ ≡ 4⁴⁰ = 2⁸⁰ = (2⁸)¹⁰ ≡ 1¹⁰ = 1 since 2⁸ = 256 = 15 × 17 + 1.

Teacher's Note:
1. Use 2⁸ ≡ 1 (mod 17).

Question 24: What is the remainder when 6⁵⁰ is divided by 215?
(a) 1
(b) 36
(c) 5
(d) 214
Show Answer & Explanation

Answer: (b) 36

Explanation:
1. 215 = 5 × 43. Since 6³ = 216 ≡ 1 (mod 215), 6⁵⁰ = (6³)¹⁶ × 6² ≡ 36 (mod 215).

Teacher's Note:
1. Split 215 into its coprime factors.

Question 25: A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, the respective remainders will be
(a) 1, 2
(b) 2, 3
(c) 3, 2
(d) 4, 1
Show Answer & Explanation

Answer: (b) 2, 3

Explanation:
1. The number is 4(5q + 4) + 1 = 20q + 17. Take the least, 17.
2. 17 ÷ 5 gives quotient 3 and remainder 2; 3 ÷ 4 gives remainder 3. The remainders are 2, 3.

Teacher's Note:
1. Reconstruct the number from the remainders.

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Chapter 01 Real Numbers Objective Questions & Solutions for Class 10 Mathematics

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FAQs

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