Euclids Algorithm And Real Numbers Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 1.1
  1. Prove that 16 divides n^4 + 4n^2 + 11, if n is an odd integer.
  2. Prove that if n is a positive even integer, then 24 divides n(n + 1)(n + 2).…
  3. Prove that if either of 2a + 3b and 9a + 5b is divisible by 17, so is the other.…
  4. Prove that every natural number can be written in the form 5k or 5k ± 1 or 5k ±…
  5. Prove that if 6 has no common factor with n, n^2 — 1 is divisible by 6.…
  6. Prove that product of four consecutive positive integers is divisible by 24.…
Exercise 1.2
  1. 144, 233 Find g. c. d
  2. 765, 65 Find g. c. d
  3. 10211, 2517 Find g. c. d
  4. Find g. c. d. of 736 and 85 by using Euclid's algorithm.
  5. Prove g. c. d (a - b, a + b) = 1 or 2, if g. c. d. (a, b) = 1
  6. Using the fact that g. c. d (a, b) l.c.m.. (a, b) = ab, find l.c.m.. (115, 25)…
Exercise 1.3
  1. 7007 Express as a product of primes:
  2. 7500 Express as a product of primes:
  3. 10101 Express as a product of primes:
  4. 15422 Express as a product of primes:
  5. 250 and 336 Find g. c. d. and l.c.m.. using the fundamental theorem of…
  6. 4000 and 25 Find g. c. d. and l.c.m.. using the fundamental theorem of…
  7. 225 and 145 Find g. c. d. and l.c.m.. using the fundamental theorem of…
  8. 175 and 1001 Find g. c. d. and l.c.m.. using the fundamental theorem of…
  9. 15, 21, 35 Find g. c. d. and l.c.m.. :
  10. 40, 60, 80 Find g. c. d. and l.c.m.. :
  11. 49, 42, 91 Find g. c. d. and l.c.m.. :
  12. root 5 Prove that following numbers are irrational:
  13. root 15 Prove that following numbers are irrational:
  14. root 3+1 . Prove that following numbers are irrational:
  15. root 5 + root 7 Prove that following numbers are irrational:
  16. 5 root 2 Prove that following numbers are irrational:
  17. Find l.c.m.. (105, 91) using g. c. d. (a, b) (a, b) = ab
  18. Using (√7 + √3)(√7 — √3) = 4 and the fact that (√7 + √3) is irrational prove…
  19. Two buses start from the same spot for the same circular root. One is a BRTS bus…
Exercise 1.4
  1. 12/625 State whether following rational number have terminating decimal…
  2. 17/3125 State whether following rational number have terminating decimal…
  3. 13/6250 State whether following rational number have terminating decimal…
  4. 14/15625 State whether following rational number have terminating decimal…
  5. 47/500 State whether following rational number have terminating decimal…
  6. 9/1600 State whether following rational number have terminating decimal…
  7. 42/52 State whether following rational number have terminating decimal…
  8. 26/65 State whether following rational number have terminating decimal…
  9. 8/43 State whether following rational number have terminating decimal expansion…
  10. 5/128 State whether following rational number have terminating decimal…
  11. 0. 01001000100001…. Following real numbers are expressed in decimal form. Find…
  12. 3.456789123 Following real numbers are expressed in decimal form. Find whether…
  13. 5. 123456789 Following real numbers are expressed in decimal form. Find whether…
  14. 2.3 bar 2 Following real numbers are expressed in decimal form. Find whether…
  15. 0. bar 142857 Following real numbers are expressed in decimal form. Find…
  16. 0. bar 9 Following real numbers are expressed in decimal form. Find whether…
  17. 5. 781 Following real numbers are expressed in decimal form. Find whether they…
  18. 2. 312 Following real numbers are expressed in decimal form. Find whether they…
  19. 0. 12345 Following real numbers are expressed in decimal form. Find whether…
Exercise 1.5
  1. 5+2 root 6 Find the square roots of following surds:
  2. 9+2 root 14 Find the square roots of following surds:
  3. 2 - root 3 Find the square roots of following surds:
  4. a + root a^2 - b^2 Find the square roots of following surds:
  5. 7 + root 48 Find the square roots of following surds:
  6. 6+4 root 2 Find the square roots of following surds:
  7. 5 + root 21 Find the square roots of following surds:
  8. Simplify: 1/root 12-2 root 35 + 1/root 8-2 root 15 - 2/root 10-2 root 21…
Exercise 1
  1. 25, 35 Find g. c. d and l.c.m.
  2. 105, 125 Find g. c. d and l.c.m.
  3. 220, 132 Find g. c. d and l.c.m.
  4. 3125, 625 Find g. c. d and l.c.m.
  5. 15625, 35 Find g. c. d and l.c.m.
  6. 15, 25, 35 Find g. c. d and l.c.m.
  7. 18, 12, 16 Find g. c. d and l.c.m.
  8. 16, 24, 36 Find g. c. d and l.c.m.
  9. 35, 28, 63 Find g. c. d and l.c.m.
  10. 112, 128, 144 Find g. c. d and l.c.m.
  11. root 3 + root 5 Prove following numbers are irrational.
  12. 5 root 3 Prove following numbers are irrational.
  13. 1/root 5 - root 3 Prove following numbers are irrational.
  14. root 7 + root 3 Prove following numbers are irrational.
  15. root 2+1 Prove following numbers are irrational.
  16. 10 root 2+7 root 3 Prove following numbers are irrational.
  17. root 5 - root 2 Prove following numbers are irrational.
  18. root 12 Prove following numbers are irrational.
  19. root 18 Prove following numbers are irrational.
  20. root 37 Prove following numbers are irrational.
  21. 211/125 Which of the following numbers have terminating decimal expansion and…
  22. 156/625 Which of the following numbers have terminating decimal expansion and…
  23. 337/35 Which of the following numbers have terminating decimal expansion and…
  24. 132/49 Which of the following numbers have terminating decimal expansion and…
  25. 235/16 Which of the following numbers have terminating decimal expansion and…
  26. 12+2 root 35 Find the square root of the following in the form of a binomial…
  27. 8+2 root 7 Find the square root of the following in the form of a binomial surd…
  28. 2 + 2/3 root 5 Find the square root of the following in the form of a binomial…
  29. 14+6 root 5 Find the square root of the following in the form of a binomial…
  30. n + root n^2 - 1 Find the square root of the following in the form of a…
  31. 1/root 3 + root 2 + 1/root 4 + root 3 + root 2 Simplify :
  32. 6/root 24-2 root 135 - root 15 Simplify :
  33. Find the largest number dividing 230 and 142 and leaving remainders 5 and 7…
  34. Find the largest number dividing 110, 62, 92 and leaving remainders 5, 6 and 1…
  35. The length and the breadth and the height of a room are 735 cm, 625 cm and 415…
  36. A milk man has 150 litres of milk of higher fat and 240 litres of milk of lower…
  37. Find the smallest number which decreased by 15 is a multiple of 125 and 225.…
  38. Find the smallest number of six digits divisible by 18, 24 and 30…
  39. Prove if 3 | (a^2 + b^2) then 3 |a and 3| b, a e N, b e N.
  40. Prove n^4 + 4 is a composite number for n 1
  41. In a morning walk a man, a woman and a child step off together. Their steps…
  42. Find the number nearest to 24001 and between 24001 and 25000 divisible by 16,…
  43. Product of any four consecutive positive integers is divisible by ……A. 16 B.…
  44. √4 + 3 is……..A. irrational B. rational but not integer. C. non-recurring…
  45. If g. c. d of two numbers is 8 and their product is 384, then their l.c.m..…
  46. If l.c.m.. of two numbers (greater than 1) is the product of them, then their…
  47. If p1 and p2 are distinct primes, their g. c. d isA. P1 B. P2 C. P1P2 D. 1…
  48. If p, q, r are distinct primes, their l.c.m.. is ……A. pqr B. pq C. 1 D. pq +…
  49. g. c. d (15, 24, 40) = …….A. 40 B. 1 C. 1 D. 15 × 24 × 40
  50. l.c.m. (15, 24, 40) = …….A. 1 B. 15 × 24 × 40 C. 120 D. 60
  51. 0. 02222... is a……A. rational number B. integer C. irrational number D. zero…
  52. root 3 + root 5 = ……A. root 3 + root 2 B. root 5+1 C. root 5+1/root 2 D. does…
  53. root 9 + root 141 = …….A. does not exist as a real number B. does not exist as…
  54. g. c. d (136, 221, 391) = …….A. 136 B. 17 C. 221 D. 391
  55. l.c.m.. (136, 221, 391) = ………A. 40664 B. 136 × 221 × 391 C. g. c. d(136, 221,…
  56. If g. c. d (a, b) = 8, l.c.m.. (a, b) = 64 and a b then a = ……..A. 64 B. 8 C.…
  57. If g. c. d (a, b) = 1, then g. c. d (a — b, a b) = ………A. 1 or 2 B. a or b C. a…
  58. 1f n 1, n^4 + 4 is. n E NA. a prime B. a composite integer C. 1 D. infinite…
  59. If g. c. d (a, b) = 18, 1. c. m. (a, b) not equal …….A. 36 B. 72 C. 48 D. 108…
  60. 18/5^3 has …. digits after decimal point.A. 5 B. 4 C. 3 D. 2
  61. The decimal expansion of 2517/6250 will terminate after …….. digits.A. 4 B. 5…
  62. 5n (n e N) ends with ………..A. 0 B. 5 C. 25 D. 10
  63. 2m5n (m, n e N) ends with ………..A. 0 B. 5 C. 25 D. 125
  64. 317/3125 representsA. a terminating decimal B. a non-recurring decimal C. a…
  65. (5k + 1)^2 leaves remainder ……….. on dividing by 5.A. 2 B. 0 C. -1 or 1 D. 1…
  66. On division by 6, a^2 cannot leave remainder (a e N)A. 1 B. 4 C. 5 D. 3…
  67. Product of three consecutive integers is divisible by ………..A. 24 B. 8 but not…

Exercise 1.1
Question 1.

Prove that 16 divides n4 + 4n2 + 11, if n is an odd integer.


Answer:

Given here, n is an odd integer for some k Z

where Z is the set of all integers.


Since, we know that every odd integer is of the form 4k + 1 and 4k – 1.


Consider two cases:


Case 1: For n = 4k + 1






)


Therefore, it is divisible by 16.


Case 2: For n = 4k – 1






)


Therefore, it is divisible by 16.


Thus, n4 + 4n2 + 11 is divisible by 16.



Question 2.

Prove that if n is a positive even integer, then 24 divides n(n + 1)(n + 2).


Answer:

Given here, n is a positive even integer.

We know that any positive even integer can be expressed as n = 2k


Therefore, n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)……eq (1)


Now, considering the following cases,


Case 1: k = 1


n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)


= 2 × 1(2 × 1 + 1)(2 × 1 + 2)


= 2 × (3) × (4)


= 24


Hence, n(n + 1)(n + 2) is divisible by 24. (From (1))


Case 2: k = 2


n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)


= 2 × 2(2 × 2 + 1)(2 × 2 + 2)


= 4 × (5) × (6)


= 24 × 5


Hence, n(n + 1)(n + 2) is divisible by 24. (from (1))


Case 3: k3


Here, k and k + 1 being even integers, one of them will always be odd and the other will be even.


So, 2k(2k + 1)(2k + 2) will always be divisible by 2.


Also as k≥ 3 is a positive integer, so for some l∈N,


k = 3a or k = 3a + 1 or k = 3a + 2


For k = 3a


k(k + 1)(2k + 1) = 3a(3a + 1)(2(3a) + 1)


= 3(a(3a + 1)(6a + 1)


Therefore, it is divisible by 3.


For k = 3a + 1


k(k + 1)(2k + 1) = (3a + 1)(3a + 1 + 1)(2(3a + 1) + 1)


= (3a + 1)(3a + 2)(6a + 3)


= 3(3a + 1)(3a + 2)(2a + 1)


Therefore, it is divisible by 3.


For k = 3a + 2


k(k + 1)(2k + 1) = (3a + 2)(3a + 2 + 1)(2(3a + 2) + 1)


= (3a + 2)(3a + 3)(6a + 5)


= 3(3a + 2)(a + 1)(6a + 5)


Therefore, it is divisible by 3.


We can see that in any case k(k + 1)(2k + 1) is divisible by 2 and 3.


Also, we know that 2 and 3 are mutually prime numbers.


Therefore the expression must be divisible by 2 × 3 = 6


So, k(k + 1)(2k + 1) is divisible by 24. (From (1))



Question 3.

Prove that if either of 2a + 3b and 9a + 5b is divisible by 17, so is the other. a, b E N (Hint: 4(2a + 3b) + 9a + 5b = 17a + 17b)


Answer:

Let 2a + 3b be divisible by 17.

Therefore, for some integer k, 2a + 3b = 17k….. eq (1)


Now,


9a + 5b = 17a + 17b – 4(2a + 3b)


= 17(a + b) – 4(17k) (from eq (1))


= 17(a + b – 4k)


Therefore, we can say that 9a + 5b is divisible by 17.



Question 4.

Prove that every natural number can be written in the form 5k or 5k ± 1 or 5k ± 2, k E N V {0}.


Answer:

We know that from Euclid’s division lemma, for every natural number a if we take b = 5 then unique non-negative integers k and m can be obtained so that a = 5k + m

Where 0 ≤ m ≤ 5


Therefore, m = 0, 1, 2, 3, 4 and 5


So, we can say that a = 5k or a = 5k + 1 or a = 5k + 2 or a = 5k + 3 or a = 5k + 4 where k E N V {0}


Now, a = 5k + 3 = 5k + 5 – 2 = 5(k + 1) – 2


= 5k1 – 2 where (k1 = k + 1)


Therefore, a = 5k + 3 can be written as a = 5k – 2


Now, a = 5k + 4 = 5k + 5 – 1 = 5(k + 1) – 1


= 5k1 – 1 where (k1 = k + 1)


Therefore, a = 5k + 4 can be written as a = 5k – 1


So, every natural number can be written in the form 5k or 5k ± 1 or 5k ± 2, k E N V {0}.



Question 5.

Prove that if 6 has no common factor with n, n2 — 1 is divisible by 6.


Answer:

Given here, there is no common factor between 6 and n.

Therefore, 6 and n are two distinct natural numbers.


We know that 6 has 2 and 3 as prime factors.


n can be written as n = 2k + 1 for all 








Therefore, we can see that  is divisible by 2.


Similarly, n can be written as n = 3k + 1 for all 








Therefore, we can see that  is divisible by 3.


Similarly, n can also be written as n = 3k–1 for all 








Therefore, we can see that  is divisible by 3.


So, n2 — 1 is divisible by 2 and 3 both.


Since, 2 and 3 are prime numbers.


Therefore, n2 — 1 is divisible by 2 × 3 = 6.



Question 6.

Prove that product of four consecutive positive integers is divisible by 24.


Answer:

Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).

Product = n(n + 1)(n + 2)(n + 3)


We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.


For, n = 3k


Product = n(n + 1)(n + 2)(n + 3)


= 3k(3k + 1)(3k + 2)(3k + 3)


= 33k(3k + 1)(3k + 2)(k + 1)


Therefore it is divisible by 3.


For, n = 3k + 1


Product = n(n + 1)(n + 2)(n + 3)


= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)


= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)


Therefore it is divisible by 3.


For, n = 3k + 2


Product = n(n + 1)(n + 2)(n + 3)


= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)


= 3(3k + 2)(k + 1)(3k + 4)(3k + 5)


Therefore it is divisible by 3.


n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2


For n = 4p


Product = n(n + 1)(n + 2)(n + 3)


= 4p(4p + 1)(4p + 2)(4p + 3)


= 24p(4p + 1)(2p + 1)(4p + 3)


= 8p(4p + 1)(2p + 1)(4p + 3)


Therefore it is divisible by 8.


For n = 4p + 1


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)


= 24(4p + 1)(2p + 1)(4p + 3)(p + 1)


= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 3


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)


= 24(4p + 3)(p + 1)(4p + 5)(2p + 3)


= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)


Therefore, it is divisible by 8.


Since it is divisible by both 3 and 8 and both are mutually prime numbers .


Therefore, it will be divisible by 38 = 24


So, product of four consecutive positive integers is divisible by 24.




Exercise 1.2
Question 1.

Find g. c. d

144, 233


Answer:

144, 233

Here, 233 > 144


233 = 144 × 1 + 89


144 = 89 × 1 + 55


89 = 55 × 1 + 34


55 = 34 × 1 + 21


34 = 21 × 1 + 13


21 = 13 × 1 + 8


13 = 8 × 1 + 5


8 = 5 × 1 + 3


5 = 3 × 1 + 2


3 = 2 × 1 + 1


2 = 1 × 2 + 0


The last non- zero remainder is 1.


Therefore, g. c. d (144, 233) = 1



Question 2.

Find g. c. d

765, 65


Answer:

765, 65


Here 765 > 65


765 = 65 × 11 + 50


65 = 50 × 1 + 15


50 = 15 × 3 + 5


15 = 5 × 3 + 0


The last non- zero remainder is 5.


Therefore, g. c. d (765, 65) = 5



Question 3.

Find g. c. d

10211, 2517


Answer:

10211, 2517


Here, 10211 > 2517


10211 = 3517 × 4 + 143


2517 = 143 × 17 + 86


143 = 86 × 1 + 57


86 = 57 × 1 + 29


57 = 29 × 1 + 28


29 = 28 × 1 + 1


28 = 1 × 28 + 0


The last non- zero remainder is 1.


Therefore, g. c. d (10211, 2517) = 1



Question 4.

Find g. c. d. of 736 and 85 by using Euclid's algorithm.


Answer:

Here, 736 > 85

736 = 85 × 8 + 56


85 = 56 × 1 + 29


56 = 29 × 1 + 27


29 = 27 × 1 + 2


27 = 2 × 13 + 1


2 = 1 × 2 + 0


The last non- zero remainder is 1.


Therefore, g. c. d (736, 85) = 1



Question 5.

Prove g. c. d (a - b, a + b) = 1 or 2, if g. c. d. (a, b) = 1


Answer:

Given g.c.d. (a,b) = 1

From Euclid’s algorithm we can say that if b|a, then g.c.d(a,b) = b


Let g. c. d (a - b, a + b) = k


Therefore we can say that k is a factor of both (a–b) and (a + b).


We can write a–b = rk for some r N


And also a + b = sk for some sN


Now, (a + b) + (a–b) = rk + sk


 a + a + b–b = k(r + s)


2a = k(r + s)………..eq(1)


(a + b) – (a – b) = rk – sk


 a + b–a + b = k(r–s)


2b = k(r–s)………..eq(2)


Also, g. c. d (a, b) = 1


Therefore, 2 × g. c. d (a, b) = 2 × 1


g. c. d (2a, 2b) = 2


g. c. d [(r + s)k, (r – s)k] = 2 (from eq (1) and eq (2))


k × g. c. d(r + s, r – s) = 2


= 2 × 1


So, k × g. c. d(r + s, r – s) = 2


k × g. c. d(r + s, r – s) = 2 × 1 = 2 × g.c.d(a,b)


By comparing we get k = 2


We know that 1 and 2 are co–prime numbers.


Similarly, we get k = 1


So, g. c. d (a—b, a + b) = k = 2


or g. c. d (a—b, a + b) = k = 1


g. c. d (a—b, a + b) = 1 or 2


Question 6.

Using the fact that g. c. d (a, b) l.c.m.. (a, b) = ab, find l.c.m.. (115, 25)


Answer:

Here, 115 > 25

115 = 25 × 4 + 15


25 = 15 × 1 + 10


15 = 10 × 1 + 5


10 = 5 × 2 + 0


The last non- zero remainder is 5.


Therefore, g. c. d(115, 25) = 5


Now, by Euclid’s Algorithm


g. c. d(a, b) × l.c.m.(a, b) = ab


Here, a = 115, b = 25


g. c. d(115, 25) × l.c.m.(115, 25) = 115 × 25


⇒ 5 × l.c.m.(115, 25) = 115 × 25


⇒ l.c.m.(115, 25) = 


⇒ l.c.m.(115, 25) = 115 × 5


Therefore, l.c.m.(115, 25) = 575




Exercise 1.3
Question 1.

Express as a product of primes:

7007


Answer:

7007 = 7 × 1001

= 7 × 7 × 143


= 7 × 7 × 11 × 13


= 72 × 11 × 13


Therefore, 7007 can be expressed as a product of prime numbers 7, 11 and 13.



Question 2.

Express as a product of primes:

7500


Answer:

7500 = 2 × 3750

= 2 × 2 × 1875


= 2 × 2 × 3 × 625


= 2 × 2 × 3 × 5 × 125


= 2 × 2 × 3 × 5 × 5 × 25


= 2 × 2 × 3 × 5 × 5 × 5 × 5


= 22 × 3 × 54


Therefore, 7500 can be expressed as a product of prime numbers 2, 3 and 5.



Question 3.

Express as a product of primes:

10101


Answer:

10101 = 3 × 3367

= 3 × 7 × 481


= 3 × 7 × 13 × 37


Therefore, 10101 can be expressed as a product of prime numbers 3, 7, 13 and 37



Question 4.

Express as a product of primes:

15422


Answer:

15422 = 2 × 7711

= 2 × 11 × 701


Therefore, 15422 can be expressed as a product of prime numbers 2, 11 and 701



Question 5.

Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:

250 and 336


Answer:

250 = 2 × 125

= 2 × 5 × 25


= 2 × 5 × 5 × 5 = 2 × 55


336 = 2 × 168


= 2 × 2 × 84


= 2 × 2 × 2 × 42


= 2 × 2 × 2 × 2 × 21


= 2 × 2 × 2 × 2 × 3 × 7


= 24 × 3 × 7


Therefore, g. c. d(250, 336) = 2


l.c.m.(250, 336) = 24 × 3 × 7 × 53


= 16 × 21 × 125


= 336 × 125 = 42000



Question 6.

Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:

4000 and 25


Answer:

4000 = 2 × 2000

= 2 × 2 × 1000


= 2 × 2 × 2 × 500


= 2 × 2 × 2 × 2 × 250


= 2 × 2 × 2 × 2 × 2 × 125


= 2 × 2 × 2 × 2 × 2 × 5 × 25


= 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5


= 25 × 53


25 = 5 × 5


g. c. d(4000, 25) = 5 × 5 = 25


l.c.m.(4000, 25) = 25 × 53 = 4000



Question 7.

Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:

225 and 145


Answer:

225 = 5 × 45 = 5 × 5 × 9 = 5 × 5 × 3 × 3

145 = 5 × 29


g. c. d(225, 145) = 5


l.c.m.(225, 145) = 52 × 32 × 29


= 25 × 9 × 29= 6525



Question 8.

Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:

175 and 1001


Answer:

175 = 5 × 35 = 5 × 5 × 7

1001 = 7 × 143 = 7 × 11 × 13


g. c. d(175, 1001) = 7


l.c.m.(175, 1001) = 52 × 7 × 11 × 13


= 25025



Question 9.

Find g. c. d. and l.c.m.. :

15, 21, 35


Answer:

15 = 1 × 3 × 5


21 = 1 × 3 × 7


35 = 1 × 5 × 7


Therefore, g. c. d(15, 21, 35) = 1


l.c.m.(15, 21, 35) = 3 × 5 × 7 = 15 × 7 = 105



Question 10.

Find g. c. d. and l.c.m.. :

40, 60, 80


Answer:

40 = 2 × 20 = 2 × 2 × 10

= 2 × 2 × 2 × 5


60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 5 × 3


80 = 2 × 40 = 2 × 2 × 20 = 2 × 2 × 2 × 10


= 2 × 2 × 2 × 2 × 5


Therefore, g. c. d(40, 60, 80) = 2 × 2 × 5 = 20


l.c.m.(40, 60, 80) = 24 × 3 × 5 = 16 × 15 = 240



Question 11.

Find g. c. d. and l.c.m.. :

49, 42, 91


Answer:

49 = 7 × 7

42 = 2 × 21 = 2 × 3 × 7


91 = 7 × 13


Therefore, g. c. d(49, 42, 91) = 7


l.c.m.(49, 42, 91) = 72 × 2 × 3 × 13


= 49 × 78 = 3822



Question 12.

Prove that following numbers are irrational:



Answer:

Let  be a rational number.

Therefore,, where g. c. d(a, b) = 1 and a, b  N.




Now, squaring both sides


……. eq (1)


 5 | a2


 5 | a


Now, let a = 5k, k  N



So, b = 5k


 5 | b



It is a contradiction as g. c. d (a, b) = 1


So, our assumption is wrong.


Therefore,  is irrational.



Question 13.

Prove that following numbers are irrational:



Answer:

Let  be a rational number.

Therefore , where g. c. d(a, b) = 1 and a, b  N.




Now, squaring both sides


……. eq (1)


 15 | a2


 15 | a


Now, let a = 15k, k  N



So, b = 15k


 15 | b



It is a contradiction as g. c. d(a, b) = 1


So, our assumption is wrong.


Therefore,  is irrational.



Question 14.

Prove that following numbers are irrational:

.


Answer:

Let k =  is a rational number

Therefore,  = 1 which is also a rational number


But,  is irrational and we know that subtraction of a rational and irrational is always irrational.


So, our assumption is wrong.


 is an irrational number.



Question 15.

Prove that following numbers are irrational:



Answer:

Let k =  is a rational number

Therefore, 



 is rational



 is also rational


But, we know that  is irrational.


So, our assumption is wrong.


 is an irrational number.



Question 16.

Prove that following numbers are irrational:



Answer:

Let k =  is a rational number

Therefore,  =  is also a rational number


But,  is irrational.


So, our assumption is wrong.


 is an irrational number



Question 17.

Find l.c.m.. (105, 91) using g. c. d. (a, b) (a, b) = ab


Answer:

Here, 105 > 91

105 = 91 × 1 + 14


91 = 14 × 6 + 7


14 = 7 × 2 + 0


Last non-negative remainder is 7.


Therefore, g. c. d(105, 91) = 7


Now, we know the relation


g. c. d(a, b) × l.c.m.(a, b) = ab


putting a = 105 and b = 91


g. c. d(105, 91) × l. c. d(105, 91) = 105 × 91


7 × l. c. d(105, 91) = 105 × 91


l. c. d(105, 91) = 


l. c. d(105, 91) = 105 × 13 = 1365



Question 18.

Using (√7 + √3)(√7 — √3) = 4 and the fact that (√7 + √3) is irrational prove that √7 — √3 is irrational.


Answer:

Given ( + )( — ) = 4 and ( + ) is irrational.


(multiplying numerator and denominator by )



(we know that (a – b) × (a + b) = (a2 – b2))


 = 


Here, it is given that  is irrational and we know that 4 is rational.


Therefore, 


We know that rational divided by irrational is always irrational.


So,  is irrational.



Question 19.

Two buses start from the same spot for the same circular root. One is a BRTS bus returning in 35 minutes. The other is a regular express bus taking 42 minutes to return. After how many minutes will they meet again at the same initial spot?


Answer:

Given that the BRTS bus returns in 35 minutes and the express bus returns in 42 minutes.

So, we can say that the BRTS returns in the time intervals which are multiples of 35 and the express bus returns in the time intervals which are multiples of 42.


Therefore, both the buses meet at the time interval of l.c.m. of 35 and 42.


35 = 5 × 7


42 = 2 × 3 × 7


l.c.m.(35, 42) = 5 × 2 × 3 × 7 = 210 minutes


So, both the buses will meet after 210 minutes at the initial spot.




Exercise 1.4
Question 1.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.


If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


Prime factorization of 625 = 5 × 5 × 5 × 5 = 54 × 1 = 54 × 20


i.e. it is in the power of 5


So, the given rational number is terminating in nature.




Question 2.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


Prime factorization of 3125 = 5 × 5 × 5 × 5 × 5 = 55


i.e. it is in the power of 5


So, the given rational number is terminating in nature.




Question 3.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


So, the given rational number is terminating in nature.


Prime factorization of 6250 = 5 × 5 × 5 × 5 × 5 × 2 = 55 × 21


i.e. it is in the power of 5 & 2




Question 4.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


Prime factorization of 15625 = 5 × 5 × 5 × 5 × 5 × 5 × 5 = 57


i.e. it is in the power of 5


So, the given rational number is terminating in nature




Question 5.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


Prime factorization of 500 = 5 × 5 × 5 × 2 × 2 = 


i.e. it is in the power of 5 & 2


So, the given rational number is terminating in nature




Question 6.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


Prime factorization of 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 


i.e. it is in the power of 5 & 2


So, the given rational number is terminating in nature




Question 7.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


 can be written as 


Prime factorization of 26 = 2 × 13


Here, denominator has got a factor other than 2 &5


So, it is non-terminating in nature.



Question 8.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


 can be written as 


As, its denominator is itself 5. So, it is terminating in nature .




Question 9.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


Prime factorization of 343 = 7 × 7 × 7


i.e. it is not in the power of 5 & 2


So, the given rational number is non- terminating in nature.



Question 10.

State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:



Answer:

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.


Prime factorization of 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27


i.e. it is in the power of 2.


So, the given rational number is terminating in nature .




Question 11.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:

0. 01001000100001….


Answer:

The given decimal number is non-recurring and non-terminating Therefore, it’s an irrational number.



Question 12.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:



Answer:

Let × = 3. 456789123456…………eq(1)


As there are 9 repeating digits. So we multiply by 109


109 × = 3456789123. 45678912345678…. …..eq(2)


Subtracting (1)from(2):


99999999 × = 3456789120




Question 13.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:

5. 123456789


Answer:

This is a terminating number so its rational for 



Question 14.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:



Answer:

Let 10 × = 23. 12121212…. ….eq(1)


As there are 2 repeating digits. So we multiply by 102


1000 × = 2312. 121212……….eq(2)


Subtracting (1) from (2):


990 × = 2289





Question 15.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:



Answer:

Let × = 0. 142857142857142…. …eq(1)


As there are 6 repeating digits. So we multiply by 106


106 × = 142857. 14285714…. …..eq(2)


Subtracting (1) from (2):


99999 × = 142857





Question 16.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:



Answer:

Let × = 0. 9999…. ….eq(1)


As there are repeating digits.


So we multiply by 10


10 × = 9. 9999…. ……..eq(2)


Subtracting (1) from (2):


9 × = 9


× = 1



Question 17.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:

5. 781


Answer:

As the given number is non-terminating and non-recurring so its rational number can be shown as, 



Question 18.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:

2. 312


Answer:

As the given number is non-terminating and non-recurring so its rational number can be shown as, 



Question 19.

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:

0. 12345


Answer:

As the given number is non-terminating and non-recurring so its rational number can be shown as, 




Exercise 1.5
Question 1.

Find the square roots of following surds:



Answer:

Let  = 

Now, analysing × and y such that × + y = 5 and × y = 6


Therefore, it can be found that × = 3 and y = 2


So,  = 



Question 2.

Find the square roots of following surds:



Answer:

Let  = 

Now, analysing × and y such that × + y = 9 and × y = 14


Therefore, it can be found that × = 7 and y = 2


So,  = 



Question 3.

Find the square roots of following surds:



Answer:

Let 

Now, analysing × and y such that × + y = 2 and × y = 


Therefore, it can be found that × = 3 and y = 2


So,  = 



Question 4.

Find the square roots of following surds:



Answer:






Question 5.

Find the square roots of following surds:



Answer:








Question 6.

Find the square roots of following surds:



Answer:








Question 7.

Find the square roots of following surds:



Answer:






Question 8.

Simplify: 


Answer:

 = 



………….. eq (1)


 = 




………….. eq (2)


 = 




 ………………eq (3)


Now, subtracting eq (3) from the addition of eq (1) and eq (2).


 +  =


 + 


Now, rationalising the terms


 + 


 + 


 + 


 + 


 + 


 + 



 = 




Exercise 1
Question 1.

Find g. c. d and l.c.m.

25, 35


Answer:

25 = 5 × 5

35 = 5 7


Therefore, g. c. d(25, 35) = 5


And l.c.m. (25, 35) = 5 × 7 × 5 = 175



Question 2.

Find g. c. d and l.c.m.

105, 125


Answer:

105 = 3 × 35 = 3 × 5 × 7

125 = 5 × 25 = 5 × 5 × 5


Therefore, g. c. d(105, 125) = 5


And l.c.m.(105, 125) = 3 × 5 × 5 × 5 × 7 = 2675



Question 3.

Find g. c. d and l.c.m.

220, 132


Answer:

220 = 2 × 110 = 2 × 2 × 55 = 2 × 2 × 5 × 11

132 = 2 × 66 = 2 × 2 × 33 = 2 × 2 × 3 × 11


Therefore, g. c. d(220, 132) = 2 × 2 × 11 = 44


And l.c.m.(220, 132) = 2 × 2 × 3 × 11 × 5 = 660



Question 4.

Find g. c. d and l.c.m.

3125, 625


Answer:

3125 = 5 × 625 = 5 × 5 × 125 = 5 × 5 × 5 × 5 × 5

625 = 5 × 125 = 5 × 5 × 5 × 5


Therefore, g. c. d(3125, 625) = 5 × 5 × 5 × 5 = 625


And l.c.m.(3125, 625) = 5 × 5 × 5 × 5 × 5 = 3125



Question 5.

Find g. c. d and l.c.m.

15625, 35


Answer:

15625 = 5 × 5 × 5 × 5 × 5 × 5

35 = 5 × 7


Therefore, g. c. d(15625, 625) = 5


And l.c.m.(15625, 625) = 5 × 5 × 5 × 5 × 5 × 5 × 7 = 109375



Question 6.

Find g. c. d and l.c.m.

15, 25, 35


Answer:

15 = 5 × 3

25 = 5 × 5


35 = 5 × 7


Therefore, g. c. d(15, 25, 35) = 5


And l.c.m.(15, 25, 35) = 5 × 5 × 3 × 7 = 525



Question 7.

Find g. c. d and l.c.m.

18, 12, 16


Answer:

18 = 3 × 3 × 2

12 = 3 × 2 × 2


16 = 2 × 2 × 2 × 2


Therefore, g. c. d(18, 12, 16) = 2


And l.c.m.(18, 12, 16) = 2 × 2 × 2 × 2 × 3 × 3 = 144



Question 8.

Find g. c. d and l.c.m.

16, 24, 36


Answer:

16 = 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3


36 = 2 × 3 × 2 × 3


Therefore, g. c. d(16, 24, 36) = 2 × 2 = 4


And l.c.m.(16, 24, 36) = 2 × 2 × 2 × 2 × 3 × 3 = 144



Question 9.

Find g. c. d and l.c.m.

35, 28, 63


Answer:

35 = 5 × 7

28 = 2 × 2 × 7


63 = 3 × 3 × 7


Therefore, g. c. d(35, 28, 63) = 7


And l.c.m.(35, 28, 63) = 3 × 3 × 2 × 2 × 7 × 5 = 1260



Question 10.

Find g. c. d and l.c.m.

112, 128, 144


Answer:

For L.C.M or G.C.D we first need to find the prime factors of the numbers

112 = 2 × 2 × 2 × 2 × 7


128 = 2 × 2 × 2 × 2 × 2 × 2 × 2


144 = 2 × 2 × 2 × 2 × 3 × 3

g.c.d of a group of numbers is the product of numbers that are common to the prime factors of numbers 

Therefore, g. c. d(112, 128, 144) = 2 × 2 × 2 × 2 = 16

L.C.M of a group of numbers is the product of all the factors of three numbers without replicating the number more than one which is common to all numbers

Therefore, l.c.m.(112, 128, 144) = 27 × 3 × 3 × 7 = 8064
 
Question 11.

Prove following numbers are irrational.



Answer:

Let k =  is a rational number

Therefore, 


Squaring both sides we have,



 is rational



 is also rational


But, we know that  is irrational.


So, our assumption is wrong.


 is an irrational number.



Question 12.

Prove following numbers are irrational.



Answer:

Let k =  is a rational number

Therefore,  =  is also a rational number


But,  is irrational.


So, our assumption is wrong.


 is an irrational number



Question 13.

Prove following numbers are irrational.



Answer:

Let  is a rational number

Therefore, k = 



Squaring both sides we have,



 is rational



 is also rational


But, we know that  is irrational.


So, our assumption is wrong.


 is an irrational number.


So,  is also an irrational number.



Question 14.

Prove following numbers are irrational.



Answer:

Let k =  is a rational number

Therefore, 


Squaring both sides we have,



 is rational



 is also rational


But, we know that  is irrational.


So, our assumption is wrong.


 is an irrational number.



Question 15.

Prove following numbers are irrational.



Answer:

Let k =  is a rational number

Therefore,  is a rational number


But, we know that  is irrational.


So, our assumption is wrong.


 is an irrational number.



Question 16.

Prove following numbers are irrational.



Answer:

Let k =  is a rational number

Therefore, 


Squaring both sides we have,



 is rational



 is also rational


But, we know that  is irrational.


So, our assumption is wrong.


 is an irrational number.



Question 17.

Prove following numbers are irrational.



Answer:

Let k = 


Squaring both sides we have,



 is rational



 is also rational


But, we know that  is irrational.


So, our assumption is wrong.


Therefore,  is an irrational number.



Question 18.

Prove following numbers are irrational.



Answer:


Let k =  is a rational number


Therefore,  =  is also a rational number


But,  is irrational.


So, our assumption is wrong.


 is an irrational number.


So,  is an irrational number.



Question 19.

Prove following numbers are irrational.



Answer:


Let k =  is a rational number


Therefore,  =  is also a rational number


But,  is irrational.


So, our assumption is wrong.


 is an irrational number.


So,  is also an irrational number.



Question 20.

Prove following numbers are irrational.



Answer:

We know that 37 is a prime number.

Therefore  is an irrational number as it has only two factors 37 and 1.



Question 21.

Which of the following numbers have terminating decimal expansion and why?



Answer:

Here, the denominator is 125.

125 = 5 × 5 × 5


We can see that that 125 contains 5 as its prime factors.


Also, 211 is a prime number.


So, g. c. d(211, 125) = 1


Therefore,
 has a terminating decimal expansion.



Question 22.

Which of the following numbers have terminating decimal expansion and why?



Answer:

Here, the denominator is 625.

625 = 5 × 5 × 5 × 5 × 20


We can see that that 625 contains 5 as its prime factors.


Also, 156 = 2 × 2 × 3 × 13


So, g. c. d(156, 625) = 1


Therefore,
 has a terminating decimal expansion.



Question 23.

Which of the following numbers have terminating decimal expansion and why?



Answer:

Here, the denominator is 35.

35 = 5 × 7 × 20


We can see that that 35 contains 7 as its prime factors other than 5 and 2.


Also, 337 is a prime number.


So, g. c. d(337, 35) = 1


Therefore,
 does not have a terminating decimal expansion.



Question 24.

Which of the following numbers have terminating decimal expansion and why?



Answer:

Here, the denominator is 49.

49 = 7 × 7 × 20 × 50


We can see that that 49 contains 7 as its prime factors other than 2 and 5.


Also, 132 = 2 × 2 × 3 × 11


So, g. c. d(132, 49) = 1


Therefore,
 does not have a terminating decimal expansion



Question 25.

Which of the following numbers have terminating decimal expansion and why?



Answer:

Here, the denominator is 16.

16 = 2 × 2 × 2 × 2 × 50


We can see that that 16 contains 2 as its prime factors.


Also, 235 = 5 × 47


So, g. c. d(235, 16) = 1


Therefore,
 has a terminating decimal expansion.



Question 26.

Find the square root of the following in the form of a binomial surd :



Answer:

Let  = 

Now, analysing x and y such that x + y = 12 and xy = 35


Therefore, it can be found that x = 7 and y = 5


So,  = 



Question 27.

Find the square root of the following in the form of a binomial surd :



Answer:

Let  = 

Now, analysing x and y such that x + y = 8 and xy = 7


Therefore, it can be found that x = 7 and y = 1


So,  = 



Question 28.

Find the square root of the following in the form of a binomial surd :



Answer:

Let  = 

Now, analysing x and y such that x + y = 6 and xy = 5


Therefore, it can be found that x = 5 and y = 1


So,  = 



Question 29.

Find the square root of the following in the form of a binomial surd :



Answer:

Let  = 

 = 


Now, analysing x and y such that x + y = 14 and xy = 45


Therefore, it can be found that x = 9 and y = 5


So,  = 



Question 30.

Find the square root of the following in the form of a binomial surd :



Answer:






Question 31.

Simplify :



Answer:

 + 


Now, rationalising the terms


 + 


 + 


 + 





Question 32.

Simplify :



Answer:

 = 











Question 33.

Find the largest number dividing 230 and 142 and leaving remainders 5 and 7 respectively.


Answer:

Let k be the largest number dividing 230 and 142 leaving remainders 5 and 7 respectively.

Therefore, we can write 230 = ka + 5 and 142 = kb + 7


Where, a, b 


230 = ka + 5 142 = kb + 7


⇒ ka = 230 – 5 ⇒ kb = 142 – 7


ka = 225 kb = 135


But, k is the largest number.


So, k becomes the largest divisor, g. c. d of 225 and 135


225 = 5 × 5 × 3 × 3


135 = 5 × 3 × 3 × 3


Therefore, g. c. d(225, 135) = k


g. c. d(225, 135) = 5 × 3 × 3 = 45


k = 45


Therefore, the largest number dividing 230 and 142 leaving remainders 5 and 7 respectively is 45.



Question 34.

Find the largest number dividing 110, 62, 92 and leaving remainders 5, 6 and 1 respectively.


Answer:

Let k be the largest number dividing 110, 62 and 92 leaving remainders 5, 6 and 1 respectively.

Therefore, we can write 110 = ka + 5, 62 = kb + 6 and 92 = kc + 1


Where, a, b and c 


110 = ka + 5 62 = kb + 6 92 = kc + 1


⇒ ka = 110 – 5 ⇒ kb = 62 – 6 ⇒ kc = 92 – 1


ka = 105 kb = 56 kc = 91


But, k is the largest number.


So, k becomes the largest divisor, g. c. d of 105, 56 and 91


105 = 5 × 3 × 7


56 = 7 × 2 × 2 × 2


91 = 7 × 13


Therefore, g. c. d(105, 56, 91) = k


g. c. d(105, 56, 91) = 7


k = 7


Therefore, the largest number dividing 110, 62 and 92 leaving remainders 5, 6 and 1 respectively is 7.



Question 35.

The length and the breadth and the height of a room are 735 cm, 625 cm and 415 cm. Find the length of the largest scale measuring instrument which can measure all the three dimensions.


Answer:

The length of the largest scale measuring instrument which can measure all the three dimensions is the g. c. d of their measures.

735 = 5 × 3 × 7 × 7


635 = 5 × 5 × 5 × 5


415 = 5 × 83


Therefore, g. c. d(735, 625, 415) = 5


So, the length of the largest scale is 5 cm.



Question 36.

A milk man has 150 litres of milk of higher fat and 240 litres of milk of lower fat. He wants to pack the milk in tins of equal capacity. What should be the capacity of each tin?


Answer:

For packing 150 litres of milk of higher fat and 240 litres of milk of lower fat in tins of equal capacity, we need to find the g. c. d of 150 and 240.

150 = 5 × 3 × 5 × 2


240 = 3 × 2 × 2 × 2 × 2 × 5


Therefore, g. c. d(150, 240) = 5 × 3 × 2 = 30


So, the capacity of the tin should be 30 litres.



Question 37.

Find the smallest number which decreased by 15 is a multiple of 125 and 225.


Answer:

The smallest number which is a multiple of 125 and 225 is the l.c.m. of them.

125 = 5 × 5 × 5


225 = 5 × 5 × 3 × 3


Therefore, l.c.m.(125, 225) = 5 × 5 × 5 × 3 × 3 = 1125


Now, we have to find the smallest number which is decreased by 15.


Therefore, the required number = 1125 + 15 = 1140



Question 38.

Find the smallest number of six digits divisible by 18, 24 and 30


Answer:

To find the smallest number of six digits divisible by 18, 24 and 30 we find the l.c.m. of them.

18 = 3 × 3 × 2


24 = 2 × 2 × 2 × 3


30 = 3 × 2 × 5


Therefore, l.c.m.(18, 24, 30) = 2 × 2 × 2 × 3 × 3 × 5 = 360


We know that the smallest six digit number is 100000.


Dividing 100000 by 18 =  = 277. 77


So, the integer bigger and nearest to 277. 77 is 278.


Therefore, required integer = 278 × 360 = 100080


So, the smallest number of six digits divisible by 18, 24 and 30


is 100080.



Question 39.

Prove if 3 | (a2 + b2) then 3 |a and 3| b, a  N, bN.


Answer:

Let us suppose 3 is a factor of a and b.


Therefore, we can write a = 3k and b = 3m


Now, a2 + b2 = (3k)2 + (3m)2


= 9k2 + 9m2


= 3(3k2 + 3m2)


Therefore it is divisible by 3.


So, 3 is a factor of a2 + b2


Similarly, the reverse condition is also true.



Question 40.

Prove n4 + 4 is a composite number for n > 1


Answer:



 (as a2–b2 = (a + b)(a–b))


….. eq (1)


Now, n > 1


n – 1 > 0


Also, 


Therefore,  and  are distinct positive integers.


So, we can say that


n–1 > 0




Similarly, n + 1 > 0




Therefore,  are also distinct.


Thus, from eq(1) n4 + 4 has two distinct factors  and 1 as a factor.


So, n4 + 4 is a composite number for n > 1.



Question 41.

In a morning walk a man, a woman and a child step off together. Their steps measure 90 cm, 80 cm and 60 cm. What is the minimum distance each should walk to cover the distance in complete steps?


Answer:

Steps of the man measure = 90 cm

Steps of the woman measures = 80 cm


Steps of the child measures = 60 cm


To find the minimum distance each man, woman and child should walk we have to find the l.c.m. of their steps.


90 = 2 × 5 × 3 × 3


80 = 2 × 2 × 2 × 2 × 5


60 = 2 × 5 × 2 × 3


Therefore, l.c.m.(90, 80, 60) = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720


So, the minimum distance each should walk to cover the distance in complete steps is 720 cm.



Question 42.

Find the number nearest to 24001 and between 24001 and 25000 divisible by 16, 24, 40


Answer:

To find the required number we must find the l.c.m. of 16, 24 and 40.

16 = 2 × 2 × 2 × 2


24 = 2 × 2 × 2 × 3


40 = 2 × 2 × 2 × 5


l.c.m.(16, 24, 40) = 2 × 2 × 2 × 2 × 3 × 5 = 240


To find the number between 24001 and 25000,


Divide 24001 by 240


 = 100. 004


So, the integer nearer to and greater than it is 101.


Hence, a number divisible by 240 and between 24001 and 25000 = 101 × 240 = 24240



Question 43.

Product of any four consecutive positive integers is divisible by ……
A. 16

B. 48

C. 24

D. 32


Answer:

Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).

Product = n(n + 1)(n + 2)(n + 3)


We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.


For, n = 3k


Product = n(n + 1)(n + 2)(n + 3)


= 3k(3k + 1)(3k + 2)(3k + 3)


= 33k(3k + 1)(3k + 2)(k + 1)


Therefore it is divisible by 3.


For, n = 3k + 1


Product = n(n + 1)(n + 2)(n + 3)


= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)


= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)


Therefore it is divisible by 3.


For, n = 3k + 2


Product = n(n + 1)(n + 2)(n + 3)


= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)


= 3(3k + 2)(k + 1)(3k + 4)(3k + 5)


Therefore it is divisible by 3.


n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2


For n = 4p


Product = n(n + 1)(n + 2)(n + 3)


= 4p(4p + 1)(4p + 2)(4p + 3)


= 24p(4p + 1)(2p + 1)(4p + 3)


= 8p(4p + 1)(2p + 1)(4p + 3)


Therefore it is divisible by 8.


For n = 4p + 1


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)


= 24(4p + 1)(2p + 1)(4p + 3)(p + 1)


= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 2


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)


= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)


= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)


Therefore it is divisible by 8.


For n = 4p + 3


Product = n(n + 1)(n + 2)(n + 3)


= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)


= 24(4p + 3)(p + 1)(4p + 5)(2p + 3)


= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)


Therefore it is divisible by 8.


Since it is divisible by both 3 and 8 and both are mutually prime numbers .


Therefore, it will be divisible by 38 = 24


So, product of four consecutive positive integers is divisible by 24.


Question 44.

√4 + 3 is……..
A. irrational

B. rational but not integer.

C. non-recurring decimal

D. integer


Answer:

 which is an integer.


Question 45.

If g. c. d of two numbers is 8 and their product is 384, then their l.c.m.. is
A. 24

B. 16

C. 48

D. 32


Answer:

l.c.m. of two numbers = 


Question 46.

If l.c.m.. of two numbers (greater than 1) is the product of them, then their g. c. d is ……
A. 1

B. 2

C. one of the numbers

D. a prime


Answer:

g. c. d of two numbers = 


Question 47.

If p1 and p2 are distinct primes, their g. c. d is
A. P1

B. P2

C. P1P2

D. 1


Answer:

p1 and p2 are distinct prime numbers.

So, 1 is the only common factor among them.


Therefore, g. c. d of them is 1.


Question 48.

If p, q, r are distinct primes, their l.c.m.. is ……
A. pqr

B. pq

C. 1

D. pq + qr + pr


Answer:

p,q and r are distinct primes.

So, they have no common factor other than 1.


Then their g. c. d is 1.


Therefore, l.c.m. of two numbers = 


 = pqr


Question 49.

g. c. d (15, 24, 40) = …….
A. 40

B. 1

C. 1

D. 15 × 24 × 40


Answer:

15 = 5 × 3 × 1

24 = 2 × 2 × 2 × 3 × 1


40 = 2 × 2 × 2 × 5 × 1


Therefore, g. c. d(15, 24, 40) = 1


Question 50.

l.c.m. (15, 24, 40) = …….
A. 1

B. 15 × 24 × 40

C. 120

D. 60


Answer:

15 = 3 × 5

24 = 2 × 2 × 2 × 3


40 = 2 × 2 × 2 × 5


Therefore, l.c.m.(15, 24, 40) = 2 × 2 × 2 × 3 × 5 = 120


Question 51.

0. 02222... is a……
A. rational number

B. integer

C. irrational number

D. zero


Answer:

It is repeating and terminating.


Question 52.

 = ……
A. 

B. 

C. 

D. does not exist


Answer:





Question 53.

 = …….
A. does not exist as a real number

B. does not exist as a binomial surd

C. 

D. 


Answer:

 = 


Therefore, x = 9, y = 


x2 – 4y =  = 81 – 141 = – 60


x2 – 4y can not be the square of a rational number as we know that square of a rational number is always positive.


Therefore, the given expression does not exist as a binomial surd.


Question 54.

g. c. d (136, 221, 391) = …….
A. 136

B. 17

C. 221

D. 391


Answer:

136 = 2 × 2 × 2 × 17


221 = 13 × 17


391 = 17 × 23


So, g. c. d. (136, 221, 391) = 17


Question 55.

l.c.m.. (136, 221, 391) = ………
A. 40664

B. 136 × 221 × 391

C. g. c. d(136, 221, 391)

D. 136 × 221


Answer:

136 = 2 × 2 × 2 × 17


221 = 13 × 17


391 = 17 × 23


So, l.c.m.. (136, 221, 391) = 2 × 2 × 2 × 23 × 13 × 17 = 40664


Question 56.

If g. c. d (a, b) = 8, l.c.m.. (a, b) = 64 and a > b then a = ……..
A. 64

B. 8

C. 16

D. 32


Answer:

We know that g. c. d. (a, b) × l.c.m.. (a, b) = ab


8 × 64 = ab


ab = 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2


Given, g. c. d (a, b) = 8 = 2 × 2 × 2


So, greatest divisor is 8.


Therefore, one of the number is 8.


Therefore the other number is 


Question 57.

If g. c. d (a, b) = 1, then g. c. d (a — b, a b) = ………
A. 1 or 2

B. a or b

C. a + b or a— b

D. 4


Answer:

Given g.c.d. (a,b) = 1


From Euclid’s algorithm we can say that if b|a, then g.c.d(a,b) = b


Let g. c. d (a—b, a + b) = k


Therefore we can say that k is a factor of both (a–b) and (a + b).


We can write a–b = rk for some r N


And also a + b = sk for some sN


Now, (a + b) + (a–b) = rk + sk


 a + a + b–b = k(r + s)


2a = k(r + s)………..eq(1)


(a + b) – (a – b) = rk – sk


 a + b–a + b = k(r–s)


2b = k(r–s)………..eq(2)


Also, g. c. d (a, b) = 1


Therefore, 2 × g. c. d (a, b) = 2 × 1


g. c. d (2a, 2b) = 2


g. c. d [(r + s)k, (r – s)k] = 2 (from eq (1) and eq (2))


k × g. c. d(r + s, r – s) = 2


= 2 × 1


So, k × g. c. d(r + s, r – s) = 2


k × g. c. d(r + s, r – s) = 2 × 1 = 2 × g.c.d(a,b)


By comparing we get k = 2


We know that 1 and 2 are co–prime numbers.


Similarly, we get k = 1


So, g. c. d (a—b, a + b) = k = 2


or g. c. d (a—b, a + b) = k = 1


g. c. d (a—b, a + b) = 1 or 2


Question 58.

1f n > 1, n4 + 4 is. n E N
A. a prime

B. a composite integer

C. 1

D. infinite


Answer:



 (as a2–b2 = (a + b)(a–b))


….. eq (1)


Now, n > 1


n – 1 > 0


Also, 


Therefore,  and  are distinct positive integers.


So, we can say that


n–1 > 0




Similarly, n + 1 > 0




Therefore,  are also distinct.


Thus, from eq(1) n4 + 4 has two distinct factors  and 1 as a factor.


So, n4 + 4 is a composite number for n > 1.


Question 59.

If g. c. d (a, b) = 18, 1. c. m. (a, b)  …….
A. 36

B. 72

C. 48

D. 108


Answer:

Except 48 all the numbers are divisible by 18.


Question 60.

 has …. digits after decimal point.
A. 5

B. 4

C. 3

D. 2


Answer:


Question 61.

The decimal expansion of  will terminate after …….. digits.
A. 4

B. 5

C. 3

D. 6


Answer:


55 means the decimal expansion will terminate after 5 digits.


Question 62.

5n (n  N) ends with ………..
A. 0

B. 5

C. 25

D. 10


Answer:

Any power of 5 ends with 5.


Question 63.

2m5n (m, n  N) ends with ………..
A. 0

B. 5

C. 25

D. 125


Answer:

2m5n = 2m – 1 × 5n – 1 × 2 × 5 = 2m – 1 × 5n – 1 × 10


(m, n  N so (m – 1), (n – 1)  N – {0} )


2m × 5n ends with 0


Question 64.

 represents
A. a terminating decimal

B. a non-recurring decimal

C. a recurring decimal

D. an integer


Answer:


Question 65.

(5k + 1)2 leaves remainder ……….. on dividing by 5.
A. 2

B. 0

C. –1 or 1

D. 1


Answer:

(5k + 1)2 = 25k2 + 10k + 1 = 5(5k2 + 2k) + 1


(5k + 1)2 leaves remainder 1 when divided by 5


Question 66.

On division by 6, a2 cannot leave remainder (a  N)
A. 1

B. 4

C. 5

D. 3


Answer:

From Euclid’s Division lemma, for any positive integer aN.


a = 6m + r where, 0<r<6 and m N


a = 6m, a = 6m + 1, a = 6m + 2, a = 6m + 3, a = 6m + 4 or a = 6m + 5


Now, a = 6m


Then a2 = 62m2


= 36m2 = 6 × (6m)2 + 0


Here, remainder is 0.


If a = 6m + 1


Then a2 = (6m + 1)2


= 36m2 + 12m + 1


= 6(6m2 + 2m) + 1


Here, remainder is 1.


If a = 6m + 2


Then a2 = (6m + 2)2


= 36m2 + 24m + 4


= 6(6m2 + 4m) + 4


Here, remainder is 4.


If a = 6m + 3


Then a2 = (6m + 3)2


= 36m2 + 32m + 9


= 6(6m2 + 6m + 1) + 3


Here, remainder is 3.


If a = 6m + 4


Then a2 = (6m + 4)2


= 36m2 + 48m + 16


= 6(6m2 + 8m + 2) + 4


Here, remainder is 4.


If a = 6m + 5


Then a2 = (6m + 5)2


= 36m2 + 60m + 25


= 6(6m2 + 10m + 4) + 1


Here, remainder is 1 = .


Thus in any case, if a2 is divided by 6, then the remainder is 0, 1, 3 or 4 but not 5.


Question 67.

Product of three consecutive integers is divisible by ………..
A. 24

B. 8 but not by 24

C. 6

D. 20


Answer:

Product of three consecutive integers = odd even odd

Or Product of three consecutive integers = even odd even


In both the cases atleast one is even.


So, It is divisible by 2


Also, in three consecutive integers one of them is a multiple of 3.


Also, 2&3 are mutually prime numbers.


So, product of three consecutive integers is divisible by 6.


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