156 as the product of primes is ____.

Correct! Wrong!

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

The product of the LCM and HCF of 16 and 36 is ____.

Correct! Wrong!

Given numbers are 16 and 36. We know that, HCF (a, b) × LCM (a, b) = a × b ⇒ HCF × LCM = 16 × 36 = 576 Hence, the product of the LCM and HCF of 16 and 36 is 576.

The LCM of the numbers 6 and 20 is ____.

Correct! Wrong!

Given numbers are 6 and 20. We have, 6 = 3 ×2 20 = 2 × 2 × 5 =22 × 5 LCM (6, 20) = Product of the greatest power of each prime factor, involved in the numbers =22 × 3 × 5 =60 Hence, the LCM of 6 and 20 is 60.

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

Correct! Wrong!

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

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

Correct! Wrong!

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.

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

Correct! Wrong!

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

Correct! Wrong!

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 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 ____.

Correct! Wrong!

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 HCF of 135 and 225 is _____.

Correct! Wrong!

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 HCF of 6, 72 and 120 is ____.

Correct! Wrong!

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.

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CBSE Class 10th Math 1 - Real Numbers MCQs
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