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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Using prime factorization, find the HCF and LCM of (i) 36, 84 (ii) 23, 31 (iii) 96, 404 (iv) 144, 198 (v) 396, 1080 (vi) 1152, 1664 In each case verify that HCF × LCM = product of given numbers. |
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Answer» (i) Prime factorization: 36 = 22 × 3 84 = 22 × 3 × 7 HCF = product of smallest power of each common prime factor in the numbers = 22 × 3 = 12 LCM = product of greatest power of each prime factor involved in the numbers = 22 × 32 × 7 = 252 (ii) Prime factorization: 23 = 23 31 = 31 HCF = product of smallest power of each common prime factor in the numbers = 1 LCM = product of greatest power of each prime factor involved in the numbers = 23 × 31 = 713 (iii) Prime factorization: 96 = 25 × 3 404 = 22 × 101 HCF = product of smallest power of each common prime factor in the numbers = 22 = 4 LCM = product of greatest power of each prime factor involved in the numbers = 25 × 3 × 101 = 9696 (iv) Prime factorization: 144 = 24 × 32 198 = 2 × 32 × 11 HCF = product of smallest power of each common prime factor in the numbers = 2 × 32 = 18 LCM = product of greatest power of each prime factor involved in the numbers = 24 × 32 × 11 = 1584 (v) Prime factorization: 396 = 22 × 32 × 11 1080 = 23 × 33 × 5 HCF = product of smallest power of each common prime factor in the numbers = 22 × 32 = 36 LCM = product of greatest power of each prime factor involved in the numbers = 23 × 33 × 5 × 11 = 11880 (vi) Prime factorization: 1152 = 27 × 32 1664 = 27 × 13 HCF = product of smallest power of each common prime factor in the numbers = 27 = 128 LCM = product of greatest power of each prime factor involved in the numbers = 27× 32 × 13 = 14976 |
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| 102. |
Show that every positive even integer is of the form 4m and that every positive odd integer is of the form 4m + 1 for some integer m. |
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Answer» Let n be any arbitrary positive odd integer. On dividing n by 4, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have n = 4m + r, where 0 ≤ r ˂ 4. As 0 ≤ r ˂ 4 and r is an integer, r can take values 0, 1, 2, 3. ⇒ n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3 But n ≠ 4m or n ≠ 4m + 2 ( ∵ 4m, 4m + 2 are multiples of 2, so an even integer whereas n is an odd integer) ⇒ n = 4m + 1 or n = 4m + 3 Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer. |
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| 103. |
Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer. |
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Answer» Let s be any positive integer. On dividing s by 4, let m be the quotient and r be the remainder. By Euclid’s division lemma, s = 4m + r, where 0 ≤ r ˂ 4 So we have, s = 4m or s = 4m + 1 or s = 4m + 2 or s = 4m + 3. Here, 4m, 4m + 2 are multiples of 2, which revert even values to s. Again, s = 4m + 1 or s = 4m + 3 are odd values of s. Thus, any positive odd integer is of the form (4m + 1) or (4m + 3) where s is any odd integer. |
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| 104. |
For some integer q, every odd integer is of the form (A) q (B) q +1 (C) 2g (D) 2q +1 |
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Answer» (D) 2q+1 Explanation: Odd integers are those integers which are not divisible by 2. Hence, we can say that every integer which is a multiple of 2 must be an even integer, while 1 added to every integer which is multiplied by 2 is an odd integer. Therefore, let us conclude that, for an integer ‘q’, every odd integer must be of the form (2 × q)+1 = 2q+1. Hence, option (D) is the correct answer. |
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| 105. |
Show that every positive integer is either even or odd? |
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Answer» Let us assume that there exist a smallest positive integer that is neither odd nor even, say n. Since n is least positive integer which is neither even nor odd, n – 1 must be either odd or even. Case 1: If n – 1 is even, n – 1 = 2k for some k. But this implies n = 2k + 1 this implies n is odd. Case 2: If n – 1 is odd, n – 1 = 2k + 1 for some k. But this implies n = 2k + 2 (k+1) this implies n is even. In both ways we have a contradiction. Thus, every positive integer is either even or odd. |
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| 106. |
Show that every positive even integer is of the form 4m and that every positive odd integer is of the form 4m + 1 for some integer m. |
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Answer» Let n be any arbitrary positive odd integer. On dividing n by 4, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have n = 4m + r, where 0 ≤ r ˂ 4. As 0 ≤ r ˂ 4 and r is an integer, r can take values 0, 1, 2, 3. ⇒ n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3 But n ≠ 4m or n ≠ 4m + 2 ( ∵ 4m, 4m + 2 are multiples of 2, so an even integer whereas n is an odd integer) ⇒ n = 4m + 1 or n = 4m + 3 Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer. |
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| 107. |
Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer. |
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Answer» Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r, for some integer q≥ 0, and r = 0 or r = 1, because 0≤ r < 2. So, a = 2q or 2q + 1. If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1. |
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| 108. |
Show that every positive integer is either even or odd? |
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Answer» Let us assume that there exist a smallest positive integer that is neither odd nor even, say n. Since n is least positive integer which is neither even nor odd, n – 1 must be either odd or even. Case 1: If n – 1 is even, n – 1 = 2k for some k. But this implies n = 2k + 1 this implies n is odd. Case 2: If n – 1 is odd, n – 1 = 2k + 1 for some k. But this implies n = 2k + 2 (k+1) this implies n is even. In both ways we have a contradiction. Thus, every positive integer is either even or odd. |
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| 109. |
Using Euclid’s algorithm, find the HCF of 960 and 1575. |
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Answer» Step 1: Choose bigger number: 1575 > 960 On dividing 1575 by 960, we have Quotient = 1, remainder = 615 1575 = 960 x 1 + 615 Step 2: On dividing 960 by 615, we have Quotient = 1 and Remainder = 345 960 = 615 × 1 + 345 Step 3: On dividing 615 by 345 Quotient = 1 and Remainder = 270 615 = 345 × 1 + 270 Step 4: On dividing 345 by 270, we have Quotient = 1 and Remainder = 75 345 = 270 × 1 + 75 Step 5: Dividing 270 by 75, we get Quotient = 3, remainder =45 270 = 75 × 3 + 45 Step 6: Dividing 75 by 45, we get Quotient = 1, remainder = 30 75 = 45 × 1 + 30 Step 7: Dividing 45 by 30, we get Quotient = 1 and Remainder = 15 45 = 30 × 1 + 15 Step 8: Dividing 30 by 15, we get Quotient = 2 and Remainder = 0 Since remainder is zero, stop the process and write your answer. Therefore, H.C.F. of 1575 and 960 is 15. |
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| 110. |
Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers. |
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Answer» So, basically there are two types of numbers i.e., prime numbers and composite numbers. Understanding that, Prime numbers are those numbers having 1 and the number itself as factors. And, Composite numbers are those numbers having factors other than 1 and itself. It’s seen that, 7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) [taking 13 out as common] = 13 x (77 + 1) = 13 x 78 = 13 x 13 x 6 So, the given expression has 6 and 13 as its factors. Therefore, we can conclude that it is a composite number. Similarly, 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1) [taking 5 out- common] = 5 x (1008 + 1) = 5 x 1009 Since, 1009 is a prime number the given expression has 5 and 1009 as its factors other than 1 and the number itself. Hence, it is also a composite number. |
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| 111. |
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. |
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Answer» (i) 7 × 11 × 13 + 13 = 13 × [7 × 11 + 1] = 13 × 78 This is composite number. (ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × {7 × 6 × 5 × 4 × 3 × 2 + 1} = 5 × 1009. This is composite number. |
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| 112. |
Is 7 × 5 × 3 × 2 + 3 a composite number? Justify your answer. |
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Answer» Given: 7 × 5 × 3 × 2 + 3 = 3 (7 × 5 × 2 + 1) = 3 × (70 + 1) = 3 × 71 ∴ The given number has two factors namely 3 and 71. Hence it is a composite number. |
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| 113. |
How will you show that (17 × 11 × 2) + (17 × 11 × 5) is a composite number? Explain. |
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Answer» (17 × 11 × 2) + (17 × 11 × 5) = (17 × 11) (2 + 5) = (17 × 11) (7) = 187 × 7 Now the given expression is written as a product of two integers and hence it is a composite number. |
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| 114. |
What is a composite number? |
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Answer» A composite number is a positive integer which is not prime (i.e. which has factors other than 1 and itself). |
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| 115. |
Explain why 3 × 5 ×7 +7 is a composite number. |
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Answer» 3 × 5 ×7 + 7= 7 (3 × 5 +1) = 7 × 16, which has more than two factors. |
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| 116. |
What is the HCF of the smallest composite number and the smallest prime number? |
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Answer» Smallest composite number = 4 = 2 × 2 Smallest prime number = 2 We know that HCF = Product of the smallest power of each common prime factor in the numbers. HCF (4, 2) = 2 The HCF of the smallest composite number and the smallest prime number is 2. |
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| 117. |
Without actually performing the long division, state whether the rational number 27/8 have terminating or non-terminating repeating (recurring) decimal expansion. |
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Answer» Given rational number is 27/8 p/q is terminating if a) p and q are co-prime & b) q is of the form of 2n 5m where n and m are non-negative integers. Firstly, we check co-prime 27 = 3 × 3 × 3 8 = 2 × 2 × 2 27 and 8 have no common factors Therefore, 27 and 8 are co-prime. Now, we have to check that q is in the form of 2n5m 8 = 23 = 1 × 23 = 50 × 23 So, denominator is of the form 2n5m where n = 3 and m = 0 Thus, 27/8 is a terminating decimal. |
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| 118. |
Without actually performing the long division, state whether the rational number 17/8 have terminating or non-terminating repeating (recurring) decimal expansion. |
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Answer» Given rational number is 17/8 p/q is terminating if a) p and q are co-prime & b) q is of the form of 2n 5m where n and m are non-negative integers. Firstly, we check co-prime 17 = 17 × 1 8 = 2 × 2 × 2 17 and 8 have no common factors Therefore, 17 and 8 are co-prime. Now, we have to check that q is in the form of 2n5m 8 = 23 = 1 × 23 = 50 × 23 So, denominator is of the form 2n5m where n = 3 and m = 0 Thus, 17/8 is a terminating decimal. |
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| 119. |
Show that the following numbers are irrational.(i) \(\frac{1}{\sqrt2}\)(ii) 7\(\sqrt5\)(iii) 6 + \(\sqrt2\)(iv) 3 - \(\sqrt5\) |
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Answer» (i) Let assume that \(\frac{1}{\sqrt2}\) is rational Therefore it can be expressed in the form of \(\frac{p}{q}\), where p and q are integers and q ≠ 0 Therefore we can write \(\frac{1}{\sqrt2}\) = \(\frac{p}{q}\) √2 = \(\frac{q}{p}\) \(\frac{q}{p}\) is a rational number as p and q are integers. Therefore √2 is rational which contradicts the fact that √2 is irrational. Hence, our assumption is false and we can say that \(\frac{1}{\sqrt2}\) is irrational. (ii) Let assume that 7√5 is rational therefore it can be written in the form of \(\frac{p}{q}\) where p and q are integers and q ≠ 0. Moreover, let p and q have no common divisor > 1. 7√5 = \(\frac{p}{q}\) for some integers p and q Therefore √5 = \(\frac{p}{7q}\) \(\frac{p}{7q}\)is rational as p and q are integers, therefore √5 should be rational. This contradicts the fact that √5 is irrational. Therefore our assumption that 7√5 is rational is false. Hence 7√5 is irrational. (iii) Let assume that 6 + √2 is rational therefore it can be written in the form of \(\frac{p}{q}\) where p and q are integers and q ≠ 0. Moreover, let p and q have no common divisor > 1. 6 + √2 = \(\frac{p}{q}\) for some integers p and q Therefore √2 = \(\frac{p}{q}\) - 6 Since p and q are integers therefore \(\frac{p}{q}\) - 6 is rational, hence √2 should be rational. This contradicts the fact that √2 is irrational. Therefore our assumption is false. Hence, 6 + √2 is irrational. (iv) Let us assume that 3 - √5 is rational Therefore 3 - √5 can be written in the form of \(\frac{p}{q}\) where p and q are integers and q ≠ 0 3 - √5 = \(\frac{p}{q}\) ⇒ \(\frac{p}{q}\) - 3 = √5 ⇒ \(\frac{p-3q}{q}\) = √5 Since p, q and 3 are integers therefore \(\frac{p-3q}{q}\) is rational number But we know √5 is irrational number, Therefore it is a contradiction. Hence 3 - √5 is irrational |
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| 120. |
In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps? |
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Answer» To calculate minimum distance Minimum distance we should calculate LCM Prime factors of 80 = 2×2×2 × 2 ×5 Prime factors of 85 = 5 ×17 Prime factors of 90 = 2×3×3×5 LCM of 80, 85 and 90 ⇒ 2 ×2 × 2 ×2 × 3×3 ×5 × 17 = 12240 Minimum distance is 12240 cm, As 1 m = 100 cm So 12240 = 120 m 40 cm Therefore the minimum distance each should walk so that he can cover the distance in complete steps is 122 m 40 cm |
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| 121. |
Find the HCF of the pairs of numbers 240 and 6552. |
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Answer» By applying Euclid’s Division lemma on 6552 and 240 we get, 6552 = 240 x 27 + 72. Since remainder ≠ 0, apply division lemma on divisor 240 and remainder 72 240 = 72 x 3+ 24. Since remainder ≠ 0, apply division lemma on divisor 72 and remainder 24 72 = 24 x 3 + 0. Therefore, H.C.F. of 240 and 6552 is 24. |
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| 122. |
Find the HCF of the pairs of numbers 75 and 243. |
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Answer» By applying Euclid’s Division lemma on 243 and 75 243 = 75 x 3 + 18. Since remainder ≠ 0, apply division lemma on 75 and remainder 18 75 = 18 x 4 + 3. Since remainder ≠ 0, apply division lemma on divisor 18 and remainder 3 18 = 3 x 6+ 0. Therefore, H.C.F. of 75 and 243 is 3. |
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| 123. |
Define HCF of two positive integers and find the HCF of the following pairs of numbers:(i) 32 and 54(ii) 18 and 24(iii) 70 and 30(iv) 56 and 88(v) 475 and 495 |
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Answer» Definition of HCF (Highest Common Factor): The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF) or Greatest Common Divisor or Greatest Common Factor (GCF). (i) Prime factorization of 32 and 54 are: 32 = 2 × 2 × 2 × 2 × 2 54 = 2 × 3 × 3 × 3 From above prime factorization we got that the highest common factor of 32 and 54 is 2 (ii) Prime factorization of 18 and 24 are: 18 = 2 × 3 × 3 24 = 2 × 2 × 2 × 3 From above prime factorization we got that the highest common factor of 18 and 24 is 3×2 ⇒ 6 (iii) Prime factorization of 30 and 70 are: 30 = 2 × 3 × 5 70 = 2 × 5 × 7 From above prime factorization we got that the highest common factor of 30 and 70 is 2×5 ⇒ 10 (iv) Prime factorization of 56 and 88 are: 56 = 2 × 2 × 2 × 7 88 = 2 × 2 × 2 × 11 From above prime factorization we got that the highest common factor of 56 and 88 is 2×2×2 ⇒ 8 (v) Prime factorization of 475 and 495 are: 475 = 5 × 5 × 19 495 = 3 × 3 × 5 × 11 From above prime factorization we got that the highest common factor of 475 and 495 is 5 |
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| 124. |
Define HCF of two positive integers and find the HCF of the following pairs of numbers:(i) 75 and 243(ii)240 and 6552(iii)155 and 1385(iv) 100 and 190(v) 105 and 120 |
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Answer» Definition of HCF (Highest Common Factor): The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF) or Greatest Common Divisor or Greatest Common Factor (GCF). (i) Prime factorization of 75 and 243 are: 75 = 3 × 5 × 5 243 = 3 × 3 x 3 × 3 × 3 From above prime factorization we got that the highest common factor of 75 and 243 is 3 (ii) Prime factorization of 240 and 6552 are: 240 = 2 × 2 ×2 × 2 × 3 × 5 6552 = 2 × 2 × 2 × 3 x 3 × 7 × 13 From above prime factorization we got that the highest common factor of 240 and 6552 is 2×2×2×3 ⇒ 24 (iii) Prime factorization of 155 and 1385 are: 155 = 5 × 31 1385 = 5 × 277 From above prime factorization we got that the highest common factor of 155 and 1385 is 5 (iv) Prime factorization of 100 and 190 are: 100 = 2 × 2 × 5 × 5 190 = 2 × 5 ×19 From above prime factorization we got that the highest common factor of 155 and 1385 is 2×5 ⇒ 10 (v) Prime factorization of 105 and 120 are: 105 = 3 × 5 × 7 120 = 2 × 2 × 2 × 3 × 5 From above prime factorization we got that the highest common factor of 105 and 120 is 3×5 ⇒ 15 |
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| 125. |
Find the HCF of the pairs of numbers 56 and 88. |
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Answer» Now, apply Euclid’s Division lemma on 56 and 88 88 = 56 x 1 + 32. Since remainder ≠ 0, apply division lemma on 56 and remainder 32 56 = 32 x 1 + 24. Since remainder ≠ 0, apply division lemma on 32 and remainder 24 32 = 24 x 1 + 8. Since remainder ≠ 0, apply division lemma on 24 and remainder 8 24 = 8 x 3 + 0. Therefore, H.C.F. of 56 and 88 is 8. |
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| 126. |
Find the HCF of the pairs of numbers 32 and 54. |
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Answer» Now, apply Euclid’s Division Lemma on 54 and 32 54 = 32 x 1 + 22 Since remainder ≠ 0, apply division lemma on 32 and remainder 22 32 = 22 x 1 + 10 Since remainder ≠ 0, apply division lemma on 22 and remainder 10 22 = 10 x 2 + 2 Since remainder ≠ 0, apply division lemma on 10 and 2 10 = 2 x 5 + 0 Therefore, the H.C.F. of 32 and 54 is 2 |
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| 127. |
Define HOE of two positive integers and find the HCF of the following pairs of numbers:(i) 32 and 54 (ii) 18 and 24 (iii) 70 and 30 (iv) 56 and 88(v) 475 and 495 (vi) 75 and 243 (vii) 240 and 6552 (viii) 155 and 1385(ix) 100 and 190 (x) 105 and 120 |
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Answer» By applying Euclid’s division lemma 20 = 15 × 1 + 5 |
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| 128. |
Write in scientific notation : (0.00000004)3 |
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Answer» (0.00000004)3 = (4.0 x 10-8)3 = (4.0)3 x (10-8)3 = 64 x 10-24 = 6.4 x 10 x 10-24 = 6.4 x 10-23 |
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| 129. |
Represent the following numbers in the scientific notation: (i) 569430000000(ii) 2000.57(iii) 0.0000006000(iv) 0.0009000002 |
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Answer» (i) 569430000000 = 5.6943 x 1011 (ii) 2000.57 = 2.00057 x 1013 (iii) 0.0000006000 = 6.0 x 10-7 (iv) 0.0009000002 = 9.000002 x 10-4 |
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| 130. |
When (2√5 – √2)2 is simplified, we get(1) 4√5 + 2√2(2) 22 – 4√10(3) 8 – 4√10(4) 2√10 – 2 |
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Answer» (2) 22 – 4√10 (2√5 – √2)2 = (2√5)2 – 2 x 2√5 x √2 + √22 = 4 x 5 – 4√10 + 2 = 22 – 4√10 |
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| 131. |
\(\cfrac{1+\sqrt2}{3-2\sqrt2}\) = a + b√2 then (a, b) = …………………A) (7, 5) B) (-7, 5) C (3, 5) D) (-7, -5) |
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Answer» Correct option is (A) (7, 5) Given that \(\frac{1+\sqrt2}{3-2\sqrt2}=a+b\sqrt2\) \(\Rightarrow\) \(\frac{1+\sqrt2}{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2}=a+b\sqrt2\) \(\Rightarrow\) \(\frac{(1+\sqrt2)\,(3+2\sqrt2)}{3^2-(2\sqrt2)^2}\) \(=a+b\sqrt2\) \(\Rightarrow\) \(\frac{3+3\sqrt2+2\sqrt2+4}{9-8}\) \(=a+b\sqrt2\) \(\Rightarrow\) \(a+b\sqrt2=7+5\sqrt2\) \(\therefore\) a = 7 & b = 5 (By comparing rational and irrational parts of both equal real number) \(\therefore\) (a, b) = (7, 5) Correct option is (A) (7, 5) |
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| 132. |
Write a rational number between √3 and 2 |
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Answer» Since, √3 = 1.732…. So, we may take 1.8 as the required rational number between √3 and 2. Thus, the required rational number is 1.8. |
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| 133. |
Write a rational number between √3 and 2 |
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Answer» Since, √3 = 1.732…. So, we may take 1.8 as the required rational number between √3 and 2. Thus, the required rational number is 1.8. |
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| 134. |
HCF of (23 × 32 × 5), (22 × 33 × 52 ) and (24 × 3 × 53 × 7) is (a) 30 (b) 48 (c) 60 (d) 105 |
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Answer» (c) 60 HCF = (23 × 32 × 5, 22 × 33 × 52 , 24 × 3 × 53 × 7) HCF = Product of smallest power of each common prime factor in the numbers = 22 × 3 × 5 = 60 |
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| 135. |
Explain why 3. 1416 bar is a rational number ? |
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Answer» Since, 3. 1416 bar is a non-terminating repeating decimal. Hence, is a rational number. |
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| 136. |
If a = (22 × 33 × 54 ) and b = (23 × 32 × 5), then HCF (a, b) = ?(a) 90 (b) 180 (c) 360 (d) 540 |
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Answer» (b) 180 It is given that: a = (22 × 33 × 54 ) and b = (23 × 32 × 5) ∴ HCF (a,b) = Product of smallest power of each common prime factor in the numbers. = 22 × 32 × 5 = 180 |
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| 137. |
Which of the following rational numbers is expressible as a non-terminating decimal?(a) \(\frac{1351}{1250}\) (b) \(\frac{2017}{250}\) (c) \(\frac{3219}{1800}\) |
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Answer» (i) \(\frac{1351}{1250}\) = \(\frac{1351}{5^4\times2}\) We know 2 and 5 are not the factors of 1351. So, the given rational is in its simplest form. And it is of the form (2m × 5n) for some integers m, n. So, the given number is a terminating decimal. ∴ \(\frac{1351}{5^4\times2}\) = \(\frac{1351\times2}{5^4\times2^4}\) = \(\frac{10808}{10000}\) = 1.0808 (ii) \(\frac{2017}{250}\) = \(\frac{2017}{5^3\times2}\) We know 2 and 5 are not the factors of 2017. So, the given rational is in its simplest form. And it is of the form (2m × 5n) for some integers m, n. So, the given rational number is a terminating decimal. ∴ \(\frac{2017}{5^3\times2}\) = \(\frac{2017\times2^2}{5^3\times2^3}\) = \(\frac{8068}{1000}\) = 8.068 (iii) \(\frac{3219}{1800}\) = \(\frac{3219}{2^3\times5^2\times3^2}\) We know 2, 3 and 5 are not the factors of 3219. So, the given rational is in its simplest form. ∴ (23 × 52 × 32 ) ≠ (2m × 5n) for some integers m, n. Hence, \(\frac{3219}{1800}\) is not a terminating decimal. \(\frac{3219}{1800}\) = 1.78833333…. Thus, it is a repeating decimal. \(\frac{1723}{625}\) = \(\frac{1723}{5^4}\) We know 5 is not a factor of 1723. So, the given rational number is in its simplest form. And it is not of the form (2m × 5n) Hence, \(\frac{1723}{625}\)is not a terminating decimal. |
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| 138. |
Which of the following rational numbers is expressible as a non-terminating decimal?1351/1250 |
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Answer» 1351/1250 = 1351/ 5 4 × 2 We know 2 and 5 are not the factors of 1351. So, the given rational is in its simplest form. And it is of the form (2m × 5n ) for some integers m, n. So, the given number is a terminating decimal. ∴ 1351/ 54 × 2 = 1351 ×2/ 54 × 24 = 10808/10000 = 1.0808 |
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| 139. |
Which of the following rational numbers have terminating decimal?(i) 16/225(ii) 5/18(iii) 2/21(iv) 7/250A. (i) and (ii)B. (ii) and (iii)C. (i) and (iii)D. (i) and (iv) |
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Answer» Let x = p/q be a rational number, such that the prime factorization of q is of the form 2n 5m, where n, m are non - negative integers. Then x has a decimal expansion which terminates. (i) Here q = 225 225 can be written as 32×52 Since it is in the form of 5m, it is a terminating decimal. (ii) Here q = 18 18 can be written as 2 × 32 Since 3 is also there and it is not in the form of 2n5m, it is not a terminating decimal. (iii) Here q = 21 21 can be written as 3 × 7 Since it is not in the form of 2n5m, it is not a terminating decimal. (iv) Here q = 250 250 can be written as 2 × 53 Since it is in the form of 2n5m, it is a terminating decimal. |
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| 140. |
Which of the following numbers have terminating decimal expansion?(a) 8/225(b) 5/18(c) 11/21(d) 21/150 |
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Answer» The correct option is: (d) 21/150 Explanation: If the prime factorisation of g in rational number P/q is of the form of 2n x 5m, where m and n are non-negative integers, then the number has a terminating decimal expansion. |
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| 141. |
Prove that the product of three consecutive positive integer is divisible by 6. |
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Answer» Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or, 6q + 3 or 6q + 4 or 6q + 5. = 6[(6q + 4)(3q + 5)(2q + 1)] |
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| 142. |
For any positive integer n, prove that n3 − n divisible by 6. |
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Answer» n3 – n = n (n2 – 1) = n (n – 1) (n + 1) Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or, 6q + |
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| 143. |
Classify the following numbers as rational or irrational:(i) 22/7(ii) 3.1416(iii) \(\pi\)(iv) \(3.\overline{142857}\)(v) 5.63636363.... |
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Answer» (i) 22/7 is a rational number. (ii) 3.1416 is a rational number. It is a terminating decimal and non-repeating decimal. (iii) \(\pi\) is an irrational number. It is a non-terminating and non-repeating decimal. (iv) \(3.\overline{142857}\) A rational number. Non-terminating repeating decimal. (v) 5.636363… A rational number. A non-terminating repeating decimal. |
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| 144. |
Classify the following numbers as rational or irrational:(i) 2.040040004....(ii) 1.535335333.....(iii) 3.121221222...(iv) \(\sqrt{21}\)(v) \(3\sqrt3\) |
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Answer» (i) 2.040040004… An irrational number. It is a non-terminating and non-repeating decimal. (ii) 1.535335333… An irrational number. A non-terminating and non-repeating decimal. (iii) 3.121221222… An irrational number. A non-terminating and non-repeating decimal. (iv) \(\sqrt{21}\) An irrational number. 21 = 3 × 7 is an irrational number. And 3 and 7 are prime and irrational numbers. (v) \(3\sqrt3\) An irrational number. 3 is a prime number. So, 3 is an irrational number. |
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| 145. |
Prove that √6 is irrational. |
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Answer» Let us suppose that √6 is a rational number. There exists two co-prime numbers, say p and q So, √6 = p/q Squaring both sides, we get 6 = p2/q2 or 6q2 = p2 …(1) Which shows that, p2 is divisible by 6 This implies, p is divisible by 6 Let p = 6a for some integer a Equation (1) implies: 6q2 = 36a2 q2 = 6a2 q2 is also divisible by 6 q is divisible by 6 6 is common factors of p and q But this contradicts the fact that p and q have no common factor Our assumption is wrong. Thus, √6 is irrational |
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| 146. |
Without actual division show that each of the following rational numbers is a non-terminating repeating decimal9/35 |
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Answer» 9/35 = 9/5 x 7 We know either 5 or 7 is not a factor of 9, so it is in its simplest form. Moreover, (5 × 7) ≠ (2m × 5n ) Hence, the given rational is non-terminating repeating decimal. |
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| 147. |
Without actual division show that each of the following rational numbers is a non-terminating repeating decimal129/22 x 57 x 75 |
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Answer» 129/22 x 57 x 75 We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form. Moreover, (2 2× 5 7× 7 5 ) ≠ (2m × 5n ) Hence, the given rational is non-terminating repeating decimal. |
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| 148. |
Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form.(i) \(\frac{23}{2^3\times5^2}\)(ii) \(\frac{24}{125}\)(iii) \(\frac{171}{800}\)(iv) \(\frac{15}{1600}\)(v) \(\frac{17}{320}\)(vi) \(\frac{19}{3125}\) |
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Answer» (i) \(\frac{23}{2^3\times5^2}\) = \(\frac{23\times5}{2^3\times5^3}\) = \(\frac{115}{1000}\) = 0.115 We know either 2 or 5 is not a factor of 23, so it is in its simplest form Moreover, it is in the form of (2m × 5n). Hence, the given rational is terminating (ii) \(\frac{24}{125}\) = \(\frac{24}{5^3}\) = \(\frac{24\times2^3}{5^3\times2^3}\) = \(\frac{192}{1000}\) = 0.192 We know 5 is not a factor of 23, so it is in its simplest form. Moreover, it is in the form of (2m × 5n ). Hence, the given rational is terminating. (iii) \(\frac{171}{800}\) = \(\frac{171}{2^5\times5^2}\) = \(\frac{171\times5^3}{2^5\times5^5}\) = \(\frac{21375}{100000}\) = 0.21375 We know either 2 or 5 is not a factor of 171, so it is in its simplest form. Moreover, it is in the form of (2m × 5n ). Hence, the given rational is terminating. (iv) \(\frac{15}{1600}\) = \(\frac{15}{2^6\times5^2}\) = \(\frac{15\times5^4}{2^6\times5^6}\) = \(\frac{9375}{1000000}\) = 0.009375 We know either 2 or 5 is not a factor of 15, so it is in its simplest form. Moreover, it is in the form of (2m × 5n). Hence, the given rational is terminating. (v) \(\frac{17}{320}\) = \(\frac{17}{2^6\times5}\) = \(\frac{17\times5^5}{2^6\times5^6}\) = \(\frac{53125}{1000000}\) = 0.053125 We know either 2 or 5 is not a factor of 17, so it is in its simplest form. Moreover, it is in the form of (2m × 5n). Hence, the given rational is terminating (vi) \(\frac{19}{3125}\) = \(\frac{19}{5^5}\) = \(\frac{19\times2^5}{5^5\times2^5}\) = \(\frac{608}{100000}\) = 0.00608 We know either 2 or 5 is not a factor of 19, so it is in its simplest form. Moreover, it is in the form of (2m × 5n). Hence, the given rational is terminating. |
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| 149. |
If a number has a non-terminating and non-recurring decimal expansion, then it is (1) a rational number (2) a natural number(3) an irrational number (4) an integer |
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Answer» (3) an irrational number Rational number gave terminating or recurring and non-terminating decimal expansion. |
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| 150. |
Write the decimal form of rational numbers without actual division 8/25 , 9/50 |
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Answer» \(\frac{8}{25}\) \(=\frac{8}{25}\times\frac{4}{4}\) \(=\frac{32}{100}\) = 0.32 \(\frac{9}{50}\) \(=\frac{9}{50}\times\frac{2}{2}\) \(=\frac{18}{100}\) = 0.18 |
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