Real Numbers – Assertion and Reason Questions

Q 1 :

Assertion (A): $6^{n}$ never ends with the digit zero, where n is natural number.
Reason (R): Any number ends with digit zero, if its prime factor is of the form $2^{m} \times 5^{n}$ , where m, n are natural numbers.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true but Reason (R) is false.
Assertion (A) is false but Reason (R) is true.

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(1)

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

$6^{n} = {(2 \times 3)}^{n} = 2^{n} \times 3^{n}$ , Its prime factors do not contain 5 i.e., of the form $2^{m} \times 5^{n}$ ,

where m, n are natural numbers. Hence, $6^{n}$ never ends with the digit zero.

Q 2 :

Assertion (A): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340
Reason (R) : HCF is always a factor of LCM

Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
Assertion (A) is true but reason (R) is false.
Assertion (A) is false but reason (R) is true.

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(2)

Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

Q 3 :

Assertion (A): The HCF of two numbers is 16 and their product is 3072. Then their LCM = 162.
Reason (R): If a and b are two positive integers, then their HCF × LCM = a × b.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false.
A is false, but R is true.

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(4)

As we know that HCF(a, b) × LCM(a, b) = a × b.
But 16 × 162 = 2592, i.e., 2592 ≠ 3072

Q 4 :

Assertion (A): If LCM(p, q) = 30 and HCF(p, q) = 5, then p × q = 150.
Reason (R): LCM of (a, b) × HCF of (a, b) = a × b.

Both A and R are true, and R is the correct explanation of A
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false.
A is false, but R is true

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(1)

We know that LCM(p, q) × HCF(p, q) = p × q.
⇒ 30 × 5 = p × q ⇒ 150 = p × q.

Q 5 :

Assertion (A): HCF(306, 657) = 9, and LCM(306, 657) = 22338.

Reason (R): If a and b are two positive integers and HCF(a, b) = 9, then LCM(a, b) = (a + b)/2.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A
A is true, but R is false.
A is false, but R is true.

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(3)

We have, HCF(306, 657) = 9
⇒ LCM(306, 657) = (306 × 657) / 9 = 22338.

Q 6 :

Assertion (A): $6^{n}$ can end with the digit 0 for any natural number n.

Reason (R): Any positive integer ending with the digits 0 or 5 is divisible by 5 and so its prime factorisation must contain the prime 5.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A
A is true, but R is false.
A is false, but R is true.

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(4)

As we know that any number whose prime factors consist of 2? × 5?, where m and n are natural numbers, can end with the digit 0, but $6^{2}$ has only prime factors as 2 and 3. So $6^{2}$ cannot end with the digit 0.

Q 7 :

Assertion (A): The HCF of two numbers is 4 and their product is 38784. Then their LCM is 9696.

Reason (R): LCM(a, b) × HCF(a, b) = a × b.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false
A is false, but R is true

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(1)

For Assertion (A): Given, HCF of two numbers = 4 and product of these numbers = 38784.
We know that HCF × LCM = Product of numbers.
⇒ LCM = 38784 ÷ 4 = 9696.

Q 8 :

Assertion (A): 2 is a rational number.

Reason (R): The square root of all positive integers are irrationals.

Both A and R are true, and R is the correct explanation of A
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false.
A is false, but R is true

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(3)

Clearly, 2 is a rational number but square root of all positive integers are not always irrational.
It may be rational or irrational number.
For example, √2 is an irrational number but √4 is not an irrational number because √4 = 2 which is rational.
So, Assertion (A) is true but Reason (R) is false

Q 9 :

Assertion (A): √5 is an irrational number.

Reason (R): If m is an odd number greater than 1, then √m is irrational.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false.
A is false, but R is true

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(3)

Clearly, √5 is irrational so Assertion (A) is true.
Let m = 9 ⇒ √m = √9 = 3 which is rational.
So, for m is an odd number greater than 1, √m need not be irrational always.
∴ Assertion (A) is true but Reason (R) is false.

Q 10 :

Assertion (A): 15/1600 is a terminating decimal fraction.
Reason (R): If x = p/q is a rational number, such that the prime factorisation of q is of the form $2^{n} 5^{m},$
where n, m are non-negative integers, then x = p/q is a terminating decimal fraction.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false.
A is false, but R is true

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(1)

For Assertion (A)
We have, 15/1600 = 15 / (2 × 2 × 2 × 2 × 2 × 5 × 5) = 15 / (2? × 5²)
⇒ The prime factorisation of denominator 1600 of 15/1600 is of the form 2?5?, where n, m are non-negative integers.
So, 15/1600 is a terminating decimal fraction.
Hence, Assertion (A) is true.
Also Reason (R) is true because it is the statement of a known theorem which says
“If x = p/q is a rational number, such that the prime factorisation of q is of the form 2?5?, where n, m are non-negative integers,
then x = p/q is a terminating decimal fraction.”
Also, it helps to prove Assertion (A) and so Reason (R) is the correct explanation of Assertion (A).

Topic Question Set

Q 1 :

Assertion (A): $6^{n}$ never ends with the digit zero, where n is natural number.
Reason (R): Any number ends with digit zero, if its prime factor is of the form $2^{m} \times 5^{n}$ , where m, n are natural numbers.

Q 2 :

Assertion (A): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340
Reason (R) : HCF is always a factor of LCM

Q 3 :

Assertion (A): The HCF of two numbers is 16 and their product is 3072. Then their LCM = 162.
Reason (R): If a and b are two positive integers, then their HCF × LCM = a × b.

Q 4 :

Assertion (A): If LCM(p, q) = 30 and HCF(p, q) = 5, then p × q = 150.
Reason (R): LCM of (a, b) × HCF of (a, b) = a × b.

Q 5 :

Assertion (A): HCF(306, 657) = 9, and LCM(306, 657) = 22338.

Reason (R): If a and b are two positive integers and HCF(a, b) = 9, then LCM(a, b) = (a + b)/2.

Q 6 :

Assertion (A): $6^{n}$ can end with the digit 0 for any natural number n.

Reason (R): Any positive integer ending with the digits 0 or 5 is divisible by 5 and so its prime factorisation must contain the prime 5.

Q 7 :

Assertion (A): The HCF of two numbers is 4 and their product is 38784. Then their LCM is 9696.

Reason (R): LCM(a, b) × HCF(a, b) = a × b.

Q 8 :

Assertion (A): 2 is a rational number.

Reason (R): The square root of all positive integers are irrationals.

Q 9 :

Assertion (A): √5 is an irrational number.

Reason (R): If m is an odd number greater than 1, then √m is irrational.

Q 10 :

Assertion (A): 15/1600 is a terminating decimal fraction.
Reason (R): If x = p/q is a rational number, such that the prime factorisation of q is of the form $2^{n} 5^{m},$
where n, m are non-negative integers, then x = p/q is a terminating decimal fraction.

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