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RD Chapter 1 Real Numbers Ex MCQS Solutions

Question - 1 : -
The exponent of 2 in the prime factorisation of 144, is
(a) 4
(b) 5
(c) 6
(d) 3

Answer - 1 : - (a)

Question - 2 : -
The LCM of two numbersls 1200. Which of the following cannot be their HCF ?
(a) 600
(b) 500
(c) 400
(d) 200

Answer - 2 : -

(b) LCM of two number = 1200
Their HCF of these two numbers will be the factor of 1200
500 cannot be its HCF

Question - 3 : -
If n = 23 x 34 x 44 x 7, then the number of consecutive zeroes in n, where n is a natural number, is
(a) 2
(b) 3
(c) 4
(d) 7

Answer - 3 : -

(c) Because it has four factors n = 23 x34 x 44 x 7
It has 4 zeroes

Question - 4 : -
The sum of the exponents of the prime factors in the prime factorisation of 196, is
(a) 1
(b) 2
(c) 4
(d) 6

Answer - 4 : - (c)

Question - 5 : -
The number of decimal places after which the decimal expansion of the rational number 23/22×5 will terminate, is
(a) 1
(b) 2
(c) 3
(d) 4

Answer - 5 : - (b)

Question - 6 : -
(a) an even number
(b) an odd number
(c) an odd prime number
(d) a prime number

Answer - 6 : - (a)

Question - 7 : - If two positive integers a and b are expressible inthe form a = pq2 and b = p2q ; p, q being primenumbers, then LCM (a, b) is
(a) pq
(b) p3q3
(c) p3q2
(d) p2q2

Answer - 7 : -

(c) a and b are two positive integers anda =pq2 and b = p3q, where p and q are prime numbers,then LCM=p3q2

Question - 8 : - In Q. No. 7, HCF (a, b) is
(a) pq
(b) p3q3
(c) p3q2
(d) p2q2

Answer - 8 : -

(a) a = pq2 and b =p3qwhere a and b are positive integers and p, q are prime numbers, then HCF =pq

Question - 9 : - If two positive integers tn and n arc expressible inthe form m = pq3 and n = p3q2, where p, qare prime numbers, then HCF (m, n) =
(a) pq
(b) pq2
(c) p3q3
(d) p2q3

Answer - 9 : -

(b) m and n are two positive integers andm = pq3 and n = pq2, where p and q are primenumbers, then HCF = pq2

Question - 10 : -
If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
(a) 2
(b) 3
(c) 4
(d) 1

Answer - 10 : -

(c) LCM of a and 18 = 36
and HCF of a and 18 = 2
Product of LCM and HCF = product of numbers
36 x 2 = a x 18
a = 36×2/18 = 4

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