<\/span><\/h2>\nQuestion 1.<\/strong>
The exponent of 2 in the prime factorization of 144, is
(a) 4
(b) 5
(c) 6
(d) 3
Solution:<\/strong>
(a)<\/strong>
![\"a1\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a1.png\")
<\/p>\nQuestion 2.<\/strong>
The LCM of the two numbers is 1200. Which of the following cannot be their HCF?
(a) 600
(b) 500
(c) 400
(d) 200
Solution:<\/strong>
(b)<\/strong> LCM of two numbers = 1200
The HCF of these two numbers will be the factor of 1200
500 cannot be its HCF<\/p>\nQuestion 3.<\/strong>
If n = 23<\/sup> x 34<\/sup> x 44<\/sup> 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
Solution:<\/strong>
(c)<\/strong> Because it has four factors n = 23<\/sup> x 34<\/sup> x 44<\/sup> x 7
It has 4 zeroes<\/p>\nQuestion 4.<\/strong>
The sum of the exponents of the prime factors in the prime factorization of 196, is
(a) 1
(b) 2
(c) 4
(d) 6
Solution:<\/strong>
(c)<\/strong>
![\"a2\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a2.png\")
<\/p>\nQuestion 5.<\/strong>
The number of decimal places after which the decimal expansion of the rational number 2322\u00d75 will terminate is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:<\/strong><\/p>\n(b)<\/strong>
![\"a3\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a3.png\")
<\/p>\nQuestion 6.<\/strong>
![\"a4\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a4.png\")
(a) an even number
(b) an odd number
(c) An odd prime number
(d) a prime number
Solution:<\/strong>
(a)<\/strong>
![\"a5\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a5.png\")
<\/p>\nQuestion 7.<\/strong>
If two positive integers a and b are expressible in the form a = pq2<\/sup> and b = p2<\/sup>q ; p, q being prime numbers, then LCM (a, b) is
(a) pq
(b) p3<\/sup>q3<\/sup>
(c) p3<\/sup>q2<\/sup>
(d) p2<\/sup>q2<\/sup>
Solution:<\/strong>
(c)<\/strong> a and b are two positive integers and a =pq2<\/sup> and b = p3<\/sup>q, where p and q are prime numbers, then LCM=p3<\/sup>q2<\/sup><\/p>\nQuestion 8.<\/strong>
In Q. No. 7, HCF (a, b) is
(a) pq
(b) p3<\/sup>q3<\/sup>
(c) p3<\/sup>q2<\/sup>
(d) p2<\/sup>q2<\/sup>
Solution:<\/strong>
(a)<\/strong> a = pq2<\/sup> and b =p3<\/sup>q where a and b are positive integers and p, q are prime numbers, then HCF =pq<\/p>\nQuestion 9.<\/strong>
If two positive integers tn and n arc are expressible in the form m = pq3<\/sup> and n = p3<\/sup>q2<\/sup>, where p, q are prime numbers, then HCF (m, n) =
(a) pq
(b) pq2<\/sup>
(c) p3<\/sup>q3<\/sup>
(d) p2<\/sup>q3<\/sup>
Solution:<\/strong>
(b)<\/strong> m and n are two positive integers and m = pq3<\/sup> and n = pq2<\/sup>, where p and q are prime numbers, then HCF = pq2<\/sup><\/p>\nQuestion 10.<\/strong>
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
Solution:<\/strong>
(c)<\/strong> 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\u00d7218 = 4<\/p>\nQuestion 11.<\/strong>
The HCF of 95 and 152, is
(a) 57
(b) 1
(c) 19
(d) 38
Solution:<\/strong>
(c)<\/strong> HCF of 95 and 152 = 19
![\"a6\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a6.png\")
<\/p>\nQuestion 12.<\/strong>
If HCF (26, 169) = 13, then LCM (26, 169) =
(a) 26
(b) 52
(f) 338
(d) 13
Solution:<\/strong>
(c)<\/strong> HCF (26, 169) = 13
LCM (26, 169) = 26\u00d716913 = 338<\/p>\nQuestion 13.<\/strong>
![\"a7\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a7.png\")
(a) 1
(b) 2
(c) 3
(d) 4
Solution:<\/strong>
(b)<\/strong>
![\"a8\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a8.png\")
![\"a9\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a9.png\")
<\/p>\nQuestion 14.<\/strong>
The decimal expansion of the rational 145871250 number will terminate after
(a) one decimal place
(b) two decimal place
(c) three decimal place
(d) four decimal place
Solution:<\/strong>
(d)<\/strong>
![\"a10\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a10.png\")
<\/p>\nQuestion 15.<\/strong>
If p and q are co-prime numbers, then p2<\/sup> and q2<\/sup> are
(a) co-prime
(b) not co-prime
(c) even
(d) odd
Solution:<\/strong>
(a)<\/strong> p and q are co-prime, then
p2<\/sup> and q2<\/sup> will also be coprime<\/p>\nQuestion 16.<\/strong>
Which of the following rational numbers has a terminating decimal?
![\"a11\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a11.png\")
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i) and (iv)
Solution:<\/strong>
(d)<\/strong> We know that a rational number has a terminating decimal if the prime factors of its denominator are in the form of 2m<\/sup> x 5n<\/sup>
16225 and 7250 have terminating decimals<\/p>\nQuestion 17.<\/strong>
If 3 is the least prime factor of number a and 7 is the least prime factor of number b, then the least prime factor of a + b is
(a) 2
(b) 3
(c) 5
(d) 10
Solution:<\/strong>
(a)<\/strong> 3 is the least prime factor of a
7 is the least prime factor of b, then
The Sum of a and b will be divisible by 2
2 is the least prime factor of a + b<\/p>\nQuestion 18.<\/strong>
3.27\u00af is
(a) an integer
(b) a rational number
(c) a natural number
(d) an irrational number
Solution:<\/strong>
(b)<\/strong> 3.27\u00af is a rational number<\/p>\nQuestion 19.<\/strong>
The smallest number by which \u221a27 should be multiplied to get a rational number is
(a) \u221a27
(b) 3\u221a3
(c) \u221a3
(d) 3
Solution:<\/strong>
(c)<\/strong>
![\"a12\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a12.png\")
<\/p>\nQuestion 20.<\/strong>
The smallest rational number by which 13 should be multiplied so that its decimal expansion terminates after one place of decimal, is
(a) 310
(b) 110
(c) 3
(d) 3100
Solution:<\/strong>
(a)<\/strong> The smallest rational number which should be multiplied by 13 to get a terminating
![\"a13\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a13.png\")
<\/p>\nQuestion 21.<\/strong>
If n is a natural number, then 92n<\/sup> \u2013 42n<\/sup> is always divisible by
(a) 5
(b) 13
(c) Both 5 and 13
(d) None of these
Solution:<\/strong>
(c)<\/strong>
![\"a14\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a14.png\")
<\/p>\nQuestion 22.<\/strong>
If n is any natural number, then 6n<\/sup> \u2013 5n<\/sup> always ends with
(a) 1
(b) 3
(c) 5
(d) 7
Solution:<\/strong>
(a)<\/strong> n is any natural number and 6n<\/sup> \u2013 5n<\/sup>
We know that 6n<\/sup> ends with 6 and 5n<\/sup> end with 5
6n<\/sup> \u2013 5n<\/sup> will end with 6 \u2013 5 = 1<\/p>\nQuestion 23.<\/strong>
If the LCM and HCF of two rational numbers are equal, then the numbers must be
(a) prime
(b) co-prime
(c) composite
(d) equal
Solution:<\/strong>
(d)<\/strong> LCM and HCF of two rational numbers are equal Then those must be equal<\/p>\nQuestion 24.<\/strong>
If the sum of the LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of the two numbers is
(a) 203400
(b) 194400
(c) 198400
(d) 205400
Solution:<\/strong>
(b)<\/strong> Sum of LCM and HCF of two numbers = 1260
LCM = 900 more than HCF
LCM = 900 +HCF
But LCM = HCF = 1260
900 + HCF + HCF = 1260
=> 2HCF = 1260 \u2013 900 = 360
=> HCF = 180
and LCM = 1260 \u2013 180 = 1080
Product = LCM x HCF = 1080 x 180 = 194400
Product of numbers = 194400<\/p>\nQuestion 25.<\/strong>
The remainder when the square of any prime number greater than 3 is divided by 6, is
(a) 1
(b) 3
(c) 2
(d) 4
Solution:<\/strong>
(a)<\/strong>
![\"a15\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a15.png\")
<\/p>\nQuestion 26.<\/strong>
For some integer m, every even integer is of the form
(a) m
(b) m + 1
(c) 2m
(d) 2m + 1
Solution:<\/strong>
(c)<\/strong> We know that even integers are 2, 4, 6, \u2026
So, it can be written in the form of 2m Where, m = Integer = Z
[Since the integer is represented by Z]
or m = \u2026, -1, 0, 1, 2, 3, \u2026
2m = \u2026, -2, 0, 2, 4, 6, \u2026
Alternate Method<\/strong>
Let \u2018a\u2019 be a positive integer.
On dividing \u2018a\u2019 by 2, let m be the quotient and r be the remainder.
Then, by Euclid\u2019s division algorithm, we have
a \u2013 2m + r, where a \u2264 r < 2 i.e., r = 0 and r = 1
=> a = 2 m or a = 2m + 1
When, a = 2m for some integer m, then clearly a is even.<\/p>\nQuestion 27.<\/strong>
For some integer q, every odd integer is of the form
(a) q
(b) q + 1
(c) 2q
(d) 2q + 1
Solution:<\/strong>
(d)<\/strong> We know those odd integers are 1, 3, 5,\u2026
So, it can be written in the form of 2q + 1 Where q = integer = Z
or q = \u2026, -1, 0, 1, 2, 3, \u2026
2q + 1 = \u2026, -3, -1, 1, 3, 5, \u2026
Alternate Method<\/strong>
Let \u2018a\u2019 be given a positive integer.
On dividing \u2018a\u2019 by 2, let q be the quotient and r be the remainder.
Then, by Euclid\u2019s division algorithm, we have
a = 2q + r, where 0 \u2264 r < 2
=> a = 2q + r, where r = 0 or r = 1
=> a = 2q or 2q + 1
When a = 2q + 1 for some integer q, then clearly a is odd.<\/p>\nQuestion 28.<\/strong>
n2<\/sup> \u2013 1 is divisible by 8, if n is
(a) an integer
(b) a natural number
(c) an odd integer
(d) an even integer
Solution:<\/strong>
(c)<\/strong>
![\"aa\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/aa.png\")
![\"aa1\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/aa1.png\")
At k = -1, a = 4(-1)(-1 + 1) = 0 which is divisible by 8.
At k = 0, a = 4(0)(0 + 1) = 4 which is divisible by 8.
At k = 1, a = 4(1)(1 + 1) = 8 which is divisible by 8.
Hence, we can conclude from the above two cases, if n is odd, then n2<\/sup> \u2013 1 is divisible by 8.<\/p>\nQuestion 29.<\/strong>
The decimal expansion of the rational number 3322\u00d75 will terminate after
(a) one decimal place
(b) two decimal places.
(c) three decimal places
(d) more than 3 decimal places
Solution:<\/strong>
(b)<\/strong>
![\"aa2\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/aa2.png\")
<\/p>\nQuestion 30.<\/strong>
If two positive integers a and b are written as a = x3<\/sup>y2<\/sup> and b = xy3<\/sup> ; x, y are prime numbers, then HCF (a, b) is
(a) xy
(b) xy2<\/sup>
(c) x3<\/sup>y3<\/sup>
(d) x2<\/sup>y2<\/sup>
Solution:<\/strong>
(b)<\/strong>
![\"aa3\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/aa3.png\")
[Since HCF is the product of the smallest power of each common prime factor involved in the numbers]<\/p>\nQuestion 31.<\/strong>
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(a) 10
(b) 100
(c) 504
(d) 2520
Solution:<\/strong>
(d)<\/strong> Factors of 1 to 10 numbers
1 = 1
2 = 1 x 2
3 = 1 x 3
4 = 1 x 2 x 2
5 = 1 x 5
6 = 1 x 2 x 3
7 = 1 x 7
8 = 1 x 2 x 2 x 2
9 = 1 x 3 x 3
10 = 1 x 2 x 5
LCM of numbers 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 1 x 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520<\/p>\nQuestion 32.<\/strong>
The largest number which divides 70 and 125, leaving the remaining 5 and 8, respectively, is
(a) 13
(b) 65
(c) 875
(d) 1750
Solution:<\/strong>
(a)<\/strong> Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70 \u2013 5), 117 = (125 \u2013 8), which is divisible by the required number.
Now, the required number = HCF of 65, 117
[For the largest number]
For this, 117 = 65 x 1 + 52 [Dividend = divisor x quotient + remainder]
=> 65 = 52 x 1 + 13
=> 52 = 13 x 4 + 0
HCF = 13
Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.<\/p>\nQuestion 33.<\/strong>
If the HCF of 65 and 117 is expressible in form 65m \u2013 117, then the value of m is
(a) 4
(b) 2
(c) 1
(d) 3
Solution:<\/strong>
(b)<\/strong> By Euclid\u2019s division algorithm,
b = aq + r, 0 \u2264 r < a [dividend = divisor x quotient + remainder]
=> 117 = 65 x 1 + 52
=> 65 = 52 x 1 + 13
=> 52 = 13 x 4 + 0
HCF (65, 117)= 13 \u2026(i)
Also, given that HCF (65, 117) = 65m \u2013 117 \u2026..(ii)
From equations (i) and (ii),
65m \u2013 117 = 13
=> 65m = 130
=> m = 2<\/p>\nQuestion 34.<\/strong>
The decimal expansion of the rational number 145871250 will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Solution:<\/strong>
(d)<\/strong>
![\"aa4\"](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/aa4.png\")
Hence, the given rational number will terminate after four decimal places.<\/p>\nQuestion 35.<\/strong>
Euclid\u2019s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy
(a) 1 < r < b
(b) 0 < r \u2264 b
(c) 0 \u2264 r < b
(d) 0 < r < b
Solution:<\/strong>
(c)<\/strong> According to Euclid\u2019s Division lemma, for a positive pair of integers there exists unique integers q and r, such that
a = bq + r, where 0 \u2264 r < b.<\/p>\n
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-MCQs.jpg\")
<\/p>\n