# CBSE Class 10th Math 1 – Real Numbers MCQs

#### The maximum number of students among them 91 pens and 26 pencils can be distributed in such away that each student gets same number of pens, and pencils is ____.

Given, 91 pens and 26 pencils. The HCF of 91 and 26 gives the maximum number of students. 91 = 13 × 7 + 0 26 = 13 × 2 + 0 ⇒ HCF (91, 26) = 13 Hence, the required number of students is 13.

#### The total number of prime numbers between 120 and 140 is ____.

Numbers between 120 and 140 are: 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138 and 139. Among the above numbers only 127, 131, 137 and 139 are primes. Hence, the number of prime numbers between 120 and 140 is 4.

#### The HCF of 135 and 225 is _____.

Given, 135 and 225. 225 = 135 × 1 + 90 135 = 90 × 1 + 45 90 = 45 × 2 + 0 at this step the remainder has become zero. The divisor at this stage i.e., 45 is the HCF of 135 and 225. Hence, the HCF of 135 and 225 is 45.

#### The counting numbers 1, 2, 3, 4, are called whole numbers.

The counting numbers 1, 2, 3, 4, are called natural numbers.

#### The HCF of 6, 72 and 120 is ____.

Given, 6, 72 and 120. 6 = 6 × 1 + 0 72 = 6 × 12 + 0 120 = 6 × 20 + 0 6 is the highest common factor of 6, 72 and 120. Hence, the HCF of 6, 72 and 120 is 6.

#### The LCM of the numbers 6 and 20 is ____.

Given numbers are 6 and 20. We have, 6 = 3 ×2 20 = 2 × 2 × 5 =2^{2} × 5 LCM (6, 20) = Product of the greatest power of each prime factor, involved in the numbers =2^{2} × 3 × 5 =60 Hence, the LCM of 6 and 20 is 60.

#### 156 as the product of primes is ____.

156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13 ∴ 156 = 2 × 2 × 3 × 13

#### If 'a' is any positive integer and b = 2, then by Euclid's Division Lemma ____.

Euclid's division lemma: Given positive integers a and b, there exist unique integer’s q and r satisfying a = bq + r, 0 ≤ r < b given integers are 'a' and '2'. ⇒ a = 2q + r, 0 ≤ r < 2

#### The HCF of 852 and 1491 is 213, then their LCM is ____.

Given the HCF of 852 and 1491 is 213. We know that, HCF (a, b) × LCM (a, b) = a × b ⇒ 213 × LCM = 852 × 1491 ⇒ LCM = 852 × 1491 /213 ⇒ LCM = 5964 Hence, the LCM of 852 and 1491 is 5964.