CBSE 10th Maths Exam 2021: Important MCQs from Chapter 1 Real Numbers with Detailed Solutions
Class 10 Maths important MCQs for Chapter 1- Real Numbers are provided here with answers and detailed solutions.
Important Multiple Choice Questions (MCQs) from Real Numbers are provided here for students who are preparing for the CBSE Class 10 Board Exam 2021. Detailed solutions are also provided for all questions. These MCQs will help students clear all the fundamental concepts and prepare effectively for Class 10 Maths exam.
Students must note that Euclid's division lemma has been excluded in the revised CBSE syllabus of Class10 Maths. So, students should prepare according to the new syllabus only.
Check below the solved MCQs from Class 10 Maths Chapter 1 Real Numbers:
(A) one decimal place
(B) two decimal places
(C) three decimal places
(D) more than 3 decimal places
Explanation: The termination of any rational number depends upon the power of 2 in the prime factorization of denominator.
2. For some integer m, every odd integer is of the form
(B) m + 1
(D) 2m + 1
Explanation: As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.
3. If two positive integers a and b are written as a = p3q2 and b = pq3; p, q are prime numbers, then HCF (a, b) is:
Explanation: Since a = p × p × p × q × q,
b = p × q × q × q
Therefore H.C.F of a and b = pq2
4. The product of a non-zero number and an irrational number is:
(A) always irrational
(B) always rational
(C) rational or irrational
Explanation: Product of a non-zero rational and an irrational number is always irrational i.e.,
5. If the HCF of 65 and 117 is expressible in the form 65 m – 117, then the value of m is
Explanation: By Euclid’s division algorithm,
6. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is
Explanation: Since 5 and 8 are the remainders of 70 and 125, respectively. Thus after subtracting these remainders from the numbers, we have the numbers
65 = (70 − 5), 117 = (125 − 8) which is divisible by the required number.
Now required number = H.C.F of (65,117)
7. If two positive integers p and q can be expressed as p = ab2 and q = a3b; a, b being prime numbers, then LCM (p, q) is
p = a × b × b
q = a × a × a × b
Since L.C.M is the product of the greatest power of each prime factor involved in the numbers
Therefore, L.C.M of p and q = a3b2
8. The values of the remainder r, when a positive integer a is divided by 3 are:
(A) 0, 1, 2, 3
(B) 0, 1
(C) 0, 1, 2
(D) 2, 3, 4
According to Euclid’s division lemma,
a = 3q + r, where 0 r < 3
As the number is divided by 3.So the remainder cannot be greater than divisor 3 also r is an integer. Therefore, the values of r can be 0, 1 or 2.
(A) Terminating decimal expansion
(B) Non-Terminating Non repeating decimal expansion
(C) Non-Terminating repeating decimal expansion
(D) None of these
Explanation: After simplification,
As the denominator has factor 53 × 22 and which is of the type 5m × 2n, So this is a terminating decimal expansion.
10. A rational number in its decimal expansion is 327.7081. What would be the prime factors of q when the number is expressed in the p/q form?
(A) 2 and 3
(B) 3 and 5
(C) 2, 3 and 5
(D) 2 and 5
Explanation: This can be explained as,
11. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
Explanation: Factors of 1 to 10 numbers
L.C.M of numbers from 1 to 10 is =
12. n2 – 1 is divisible by 8, if n is
(A) an integer
(B) a natural number
(C) an odd integer greater than 1
(D) an even integer
Explanation: n can be even or odd
Case 1: If n is even
Case 2: If n is odd
Which is divisible by 8.
Similarly we can check for any integer.
13. If n is a rational number, then 52n − 22n is divisible by
(C) Both 3 and 7
(D) None of these
52n −22n is of the form a2n − b2n which is divisible by both (a + b) and (a – b).
So, 52n − 22n is divisible by both 7, 3.
14. The H.C.F of 441, 567 and 693 is
693 = 3×3×7×7
567 = 3×3×3×3×7
441 = 3×3×7×11
Therefore H.C.F of 693, 567 and 441 is 63.
15. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
Explanation: We need to find the L.C.M of 40, 42 and 45 cm to get the required minimum distance.
40 = 2×2×2×5
42 = 2×3×7
45 = 3×3×5
L.C.M. = 2×3×5×2×2×3×7 = 2520
Important Articles for Maths Exam Preparation
We have prepared some articles which will help the class 10 students in preparations of their Maths exam by summarising all important resources in one place. Students can check the following links to explore the important articles to prepare in effective manner and perform well in the exam: