# Assertion Reason Questions for Class 10 Maths Chapter 1 Real Numbers

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## Assertion Reason Questions for Class 10 Maths Chapter 1 Real Numbers

Here we are providing Assertion Reason Questions for Class 10 Maths Chapter 1 Real Numbers. Students can practice these questions for better understanding of the topics. This will also help to perform better in the examination.

Direction: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of
Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Q.1. Assertion: The H.C.F. of two numbers is 16 and their product is 3072. Then their L.C.M. = 162.
Reason: If a and b are two positive integers, then H.C.F. × L.C.M. = a × b.

Answer Answer: (d) Assetion (A) is false but Reason (R) is true.
Explanation: Since HCF × LCM = a × b
⇒ 3072 = 16 × 162
⇒ 3072 ≠ 2592

Q.2. Assertion: Denominator of 34.12345. When expressed in the form p/q, q ≠ 0, is of the form
2m × 5n, where m and n are non-negative integers.
Reason: 34.12345 is a terminating decimal fraction.

Q.3. Assertion: 13/3125 is a terminating decimal fraction.
Reason: If q = 2n.5m where n and m are non-negative integers, then p/q is a terminating decimal fraction.

Q.4. Assertion: When a positive integer a is divided by 3, the values of remainder can be 0, 1 or 2.

Reason: According to Euclid’s Division Lemma a = bq + r, where 0 ≤ r < b and r is an integer.

Given positive integers A and B, there exists unique integers Q and R satisfying a = bq + r, where 0 ≤ r < b.
This is known as Euclid’s Division Lemma.

Q.5. Assertion: A number N when divided by 15 gives the remainder 2. Then the remainder is same when N is divided by 5.
Reason: √3 is an irrational number.

Answer Answer: Explanation: Let we take three numbers which are divisible by 5 and 15 both, are 30, 45, 60.
Now, we add the remainder 2, we get 32, 47, 62
Therefore, we can see that as one numbers are divisible
by 5 & 15 but remainder is same as 2.

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