46. If A = 2n + 13, B = n + 7, where n is a
natural number, then HCF of A and B is:
(a) 2
(b) 1
(c) 3
(d) 4Answer/ Explanation
Answer: b
Explaination:
Taking different values of n we find
that A and B are coprime.
∴ HCF = 1
47. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:
(a) 22
(b) 16
(c) 36
(d) 21Answer/ Explanation
Answer: b
Explaination:
HCF of 576 and 448 = 64
∴ Number of sections = =57664+44864
= 9 + 7 = 16
48. The HCF of 2472, 1284 and a third number N is 12. If their LCM is 23 x 32 x 5 x 103 x 107, then the number Nis :
(a) 22 x 32 x 7
(b) 22 x 33 x 103
(c) 22 x 32 x 5
(d) 24 x 32 x llAnswer/ Explanation
Answer: c
Explaination:
2472 = 23 × 3 × 103
1284 = 2² × 3 × 107
∵ LCM = 23 × 3² × 5 × 103 × 107
∴ N = 2² × 3² × 5 = 180
49. Two natural numbers whose difference is 66 and the least common multiple is 360, are:
(a) 120 and 54
(b) 90 and 24
(c) 180 and 114
(d) 130 and 64Answer/ Explanation
Answer: b
Explaination:
Difference of 90 and 24 = 66
and LCM of 90 and 24 = 360
∴ Numbers are 90 and 24.
50. HCF of 52 x 32 and 35 x 53 is:
(a) 53 x 35
(b) 5 x 33
(c) 53 x 32
(d) 52 x 32Answer/ Explanation
Answer: d
Explaination:
HCF of 5² × 3² and 35 × 53
= 5² × 3²
51. A number 10x + y is multiplied by another number 10a + b and the result comes as 100p + 10q + r, where r = 2y, q = 2(x + y) and p = 2x; y < 5, q ≠ 0. The value of 10a + b may be _______.Answer/ Explanation
Answer:
Explaination:
(10x + y)(10a + b) = 100p + 10q + r
⇒ (10x + y)(10a + b) = 100 × 2x + 10 × 2(x + y) + 2y
⇒ (10x + y)(lOa + b) = 200x + 20(x + y) + 2y
⇒ (10x + y)(10a + b) = 220x + 22y
⇒ (10x + y)(\0a + b) = 22(10x + y)
⇒ 10a + b = 22
52. If the HCF of 55 and 99 is expressible in the form 55m – 99, then the value of m is _______.Answer/ Explanation
Answer:
Explaination:
55 = 5 × 11, 99 = 9 × 11
∴ HCF(55, 99) = 11
ATQ, 55m – 99 = 11
55 × 2-99 = 11
m = 2
53. Euclid’s division lemma states for any two positive integers a and b, there exists integers q and r such that a = bq + r. If a = 5, b = 8, then write the value of q and r.Answer/ Explanation
Answer:
Explaination:
Using Euclid’s division lemma, we get
a = bq + r
5 = 8 × 0 + 5
q = 0 and r = 5
54. If a and b are two positive integers such that a = 14b. Find the HCF of a and b.Answer/ Explanation
Answer:
Explaination:
We can write
a = 14b + 0
∵ remainder is 0
∴ HCF is b.
55. Find HCF of 1001 and 385.Answer/ Explanation
Answer:
Explaination:
1001 =385 × 2 + 231
385 =231 × l + 154
231 =154 × 1 + 77
154 = 77 × 2 + 0
∴ HCF =77
56. 4 Bells toll together at 9.00 am. They toll after 7, 8, 11 and 12 seconds respectively. How many times will they toll together again in the next 3 hours?
(a) 3
(b) 4
(c) 5
(d) 6Answer/ Explanation
Answer: c
Explaination:
LCM of 7, 8, 11, 12 = 1848
∴ Bells will toll together after every 1848 sec.
∴ In next 3 hrs, number of times the bells will toll together
=3×36001848 = 5.84
⇒ 5 times.
57. The HCF and LCM of two numbers are 33 and 264 respectively. When the first number is completely divided by 2 the quotient is 33. The other number is ________ .Answer/ Explanation
Answer:
Explaination:First number = 2 × 33 = 66
58. Given that LCM (91, 26) = 182, then HCF (91, 26) is _______.Answer/ Explanation
Answer:
Explaination:
LCM (91, 26) × HCF (91, 26) = 91 × 26
182 × HCF (91, 26) = 91 × 26
⇒ HCF (91, 26) = =91×26182
⇒ HCF (91, 26) = 13
59. Find the LCM of smallest prime and smallest odd composite natural number.Answer/ Explanation
Answer:
Explaination:
Smallest prime number = 2
Smallest composite odd number = 9
LCM of 2 and 9 = 2 × 9 = 18
60. Decompose 32760 into prime factors.Answer/ Explanation
Answer:
Explaination:
32760 =2 × 2 × 2 × 3 × 3 × 5 × 7 × 13
= 23 × 32 × 5 × 7 × 13
61. Write the sum of exponents of prime factors in the prime factorisation of 250.Answer/ Explanation
Answer:
Explaination:
250 = 2 × 53
∴ Sum of exponents =1 + 3 = 4
62. Complete the missing entries in the following factor tree:Answer/ Explanation
Answer:
Explaination:
63. What is the HCF of smallest prime number and the smallest composite number? [CBSE 2018]Answer/ Explanation
Answer:
Explaination:
The smallest prime number = 2
The smallest composite number = 4
∴ HCF of 2 and 4 = 2
64. If the prime factorisation of a natural number N is 24 × 34 × 53 × 7, write the number of consecutive zeroes in N.Answer/ Explanation
Answer:
Explaination:
Number of consecutive zeroes = zeroes in 23 × 53 = zeroes in (10)3 = 3
65. If product of two numbers is 3691 and their LCM is 3691, find their HCF.Answer/ Explanation
Answer:
Explaination:
66. If p and q are two coprime numbers, then find the HCF and LCM of p and q.Answer/ Explanation
Answer:
Explaination:
∵ p and q are co-prime numbers
∴ Common factor of p and q = 1
⇒ HCF of p and q = 1
67. The HCF of 45 and 105 is 15. Write their LCM.Answer/ Explanation
Answer: b
Explaination:
68. The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, then find the other number.Answer/ Explanation
Answer:
Explaination:
69. Can two numbers have 18 as their HCF and 380 as their LCM? Give reason. [NCERT Exemplar Problems]Answer/ Explanation
Answer:
Explaination:
No, because HCF is always a factor of LCM. But 18 is not a factor of 380.
70. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.Answer/ Explanation
Answer:
Explaination:
Required number is HCF of 2053 – 5 and 967 – 7 = HCF of 2048 and 960 = 64
71. The decimal expansion of the rational number 14587/1250 will terminate after
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal placesAnswer/ Explanation
Answer:
Explaination:
72. Which of the following rational numbers have a terminating decimal expansion?Answer/ Explanation
Answer:
Explaination:
The denominator 26 × 52 is 0f the form 2m × 5n where m and n are non-negative integers. Hence, it is a terminating decimal expansion.
73. The decimal expansion of number 44122×53×7 has _______ decimal representation.Answer/ Explanation
Answer:
Explaination:
The denominator 26 × 52 is 0f the form 2m × 5n where m and n are non-negative integers.
74. From the following, the rational number whose decimal expansion terminating is:Answer/ Explanation
Answer:
Explaination:
The denominator 25 × 51 is 0f the form 2m × 5n where m and n are non-negative integers. Hence, it is a terminating decimal expansion.
75. Without actually performing the long division, state whether the following rational numbers will have terminating decimal expansion or non-terminating repeating decimal expansion.Answer/ Explanation
Answer:
Explaination:
76. Write whether 245√+320√25√ on simplification gives an irrational or a rational number. [CBSE 2018 (C)]Answer/ Explanation
Answer:
Explaination:
77. If p/q is a rational number (q ≠ 0), what is condition of q so that the decimal representation of p/q is terminating?Answer/ Explanation
Answer:
Explaination:
For any rational number pq with terminating decimal representation, the prime factorisation of q is of the form 2n.5m, where n and m are non-negative integers.
78. Find a rational number between √2 and √3 [Delhi 2019]Answer/ Explanation
Answer:
Explaination:
Let p be rational number between √2 and √3
∴ √2 < p < √3
On squaring throughout, we have 2 < p² < 3
The perfect squares which lie between 2 and 3 are 2.25, 2.56, 2.89.
We have, 2 < 2.25 < 2.56 < 2.89 < 3
Taking square root throughout √2 < 1.5 < 1.6 < 1.7 < √3
The rational numbers between √2 and √3 are 1.5, 1.6 1.7 and more.
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