{"id":29193,"date":"2018-07-18T06:05:50","date_gmt":"2018-07-18T06:05:50","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=29193"},"modified":"2020-12-02T12:32:06","modified_gmt":"2020-12-02T07:02:06","slug":"rs-aggarwal-solutions-class-10-chapter-1-real-numbers","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/rs-aggarwal-solutions-class-10-chapter-1-real-numbers\/","title":{"rendered":"RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers"},"content":{"rendered":"
These Solutions are part of RS Aggarwal Solutions Class 10<\/a>. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers<\/p>\n Questions 1:<\/strong><\/span><\/p>\n For any two given positive integers a and b there exist unique whole numbers q and r such that More Resources<\/strong><\/p>\n Questions 2:<\/strong><\/span><\/p>\n By Euclid’s Division algorithm we have: Questions 3:<\/strong><\/span><\/p>\n By Euclid’s Division Algorithm, we have: Questions 1:<\/strong><\/span> Questions 1:<\/strong><\/span> Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers<\/a> are helpful to complete your math homework.<\/p>\n If you have any doubts, please comment below. A Plus Topper<\/a> try to provide online math tutoring for you.<\/p>\n","protected":false},"excerpt":{"rendered":" RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Exercise 1A Questions 1: For any two given positive integers a and b there exist unique whole numbers q and r such that … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6805],"tags":[],"yoast_head":"\nExercise 1A<\/span><\/h3>\n
\nHere, we call ‘a’ as dividend, b as divisor, q as quotient and r as remainder.
\nDividend = (divisor quotient) + remainder<\/p>\n\n
\nDividend = (divisor * quotient) + remainder
\n= (61 * 27) + 32 = 1647 + 32 = 1679<\/p>\n
\nDividend = (divisor quotient) + remainder<\/p>\n
\nQuestions 4:<\/strong><\/span>
\n(i) On dividing 2520 by 405, we get
\nQuotient = 6, remainder = 90
\n2520 = (405 x 6) + 90
\nDividing 405 by 90, we get\u00a0Quotient = 4,\u00a0Remainder = 45
\n405 = 90 x 4 + 45
\nDividing 90 by 45 We get\u00a0Quotient = 2, remainder = 0
\n90 = 45 x 2
\nH.C.F. of 405 and 2520 is 45
\n
\n(ii) Dividing 1188 by 504, we get\u00a0Quotient = 2, remainder = 180
\n1188 = 504 x 2+ 180
\nDividing 504 by 180 \u00a0Quotient = 2, remainder = 144
\n504 = 180 x 2 + 144
\nDividing 180 by 144, we get\u00a0Quotient = 1, remainder = 36
\nDividing 144 by 36
\nQuotient = 4, remainder = 0
\nH.C.F. of 1188 and 504 is 36
\n
\n(iii) Dividing 1575 by 960, we get
\nQuotient = 1, remainder = 615
\n1575 = 960 x 1 + 615
\nDividing 960 by 615, we get\u00a0Quotient = 1, remainder = 345
\n960 = 615 x 1 + 345
\nDividing 615 by 345\u00a0Quotient = 1, remainder = 270
\n615 = 345 x 1 + 270
\nDividing 345 by 270, we get\u00a0Quotient = 1, remainder = 75
\n345 = 270 x 1 + 75
\nDividing 270 by 75, we get\u00a0Quotient = 3, remainder =45
\n270 = 75 x 3 + 45
\nDividing 75 by 45, we get\u00a0Quotient = 1, remainder = 30
\n75 = 45 x 1 + 30
\nDividing 45 by 30, we get\u00a0Remainder = 15, quotient = 1
\n45 = 30 x 1 + 15
\nDividing 30 by 15, we get\u00a0Quotient = 2, remainder = 0
\nH.C.F. of 1575 and 960 is 15
\n
\nQuestions 5:<\/strong><\/span>
\n(i) By prime factorization, we get
\n
\n
\n(ii) By prime factorization. We get
\n
\n
\n(iii) By prime factorization, we get
\n
\n
\nQuestions 6:<\/strong><\/span>
\n(i) By prime factorization, we get
\n
\n
\n(ii) By prime factorization, we get
\n
\n
\n(iii) By prime factorization, we get
\n
\n
\nQuestions 7:<\/strong><\/span>
\n
\nQuestions 8:<\/strong><\/span>
\nH.C.F. of two numbers = 11, their L.C.M = 7700
\nOne number = 275, let the other number be b
\nNow, 275 x b = 11 x 7700
\n
\nQuestions 9:<\/strong><\/span>
\nBy going upward
\n5 x 11= 55
\n55 x 3 = 165
\n165 x 2 = 330
\n330 x 2 = 660
\n
\nQuestions 10:<\/strong><\/span>
\nSubtracting 6 from each number:
\n378 – 6 = 372, 510 – 6 = 504
\nLet us now find the HCF of 372 and 504 through prime factorization:
\n372 = 2 x 2 x 3 x 31
\n
\n504 = 2 x 2 x 2 x 3 x 3 x 7
\n
\nThe required number is 12.
\n
\nQuestions 11:<\/strong><\/span>
\nSubtracting 5 and 7 from 320 and 457 respectively:
\n320 – 5 = 315,
\n457 – 7 = 450
\nLet us now find the HCF of 315 and 405 through prime factorization:
\n
\n
\nThe required number is 45.
\nQuestions 12:<\/strong><\/span>
\n
\nQuestions 13:<\/strong><\/span>
\nThe prime factorization of 42, 49 and 63 are:
\n42 = 2 x 3 x 7, 49 = 7 x 7, 63 = 3 x 3 x 7
\nTherefore, H.C.F. of 42, 49, 63 is 7
\nHence, greatest possible length of each plank = 7 m
\nQuestions 14:<\/strong><\/span>
\n7 m = 700cm, 3m 85cm = 385 cm
\n12 m 95 cm = 1295 cm
\nLet us find the prime factorization of 700, 385 and 1295:
\n
\n
\nGreatest possible length = 35cm.
\nQuestions 15:<\/strong><\/span>
\nLet us find the prime factorization of 1001 and 910:
\n1001 = 11 x 7 x 13
\n910 = 2 x 5 x 7 x 13
\n
\nH.C.F. of 1001 and 910 is 7 x 13 = 91
\nMaximum number of students = 91
\nQuestions 16:<\/strong><\/span>
\nLet us find the HCF of 336, 240 and 96 through prime factorization:
\n
\n
\nEach stack of book will contain 48 books
\nNumber of stacks of the same height
\n
\nQuestions 17:<\/strong><\/span>
\n
\nQuestions 18:<\/strong><\/span>
\nLet us find the LCM of 64, 80 and 96 through prime factorization:
\n
\n
\nL.C.M of 64, 80 and 96\u00a0=
\n
\nTherefore, the least length of the cloth that can be measured an exact number of times by the rods of 64cm, 80cm and 96cm = 9.6m
\nQuestions 19:<\/strong><\/span>
\n
\nQuestions 20:<\/strong><\/span>
\nInterval of beeping together = LCM (60 seconds, 62 seconds)
\nThe prime factorization of 60 and 62:
\n60 = 30 x 2, 62 = 31 x 2
\nL.C.M of 60 and 62 is 30 x 31 x 2 = 1860 sec = 31min
\nelectronic device will beep after every 31minutes
\nAfter 10 a.m., it will beep at 10 hrs 31 minutes
\nQuestions 21:<\/strong><\/span>
\n<\/p>\nExercise 1B<\/span><\/h3>\n
\n
\n
\nQuestions 2:<\/strong><\/span>
\n
\nQuestions 3:<\/strong><\/span>
\n
\n
\nQuestions 4:<\/strong><\/span>
\n(i) 53.123456789 is a rational number since it is a terminating decimal.
\n(ii) \u00a0is a rational number because it is a non – terminating repeating decimal.
\n(iii) 0.12012001200012…… is not a rational number as it is a non-terminating, non – repeating decimal.<\/p>\nExercise 1C<\/span><\/h3>\n
\n
\nQuestions 2:<\/strong><\/span>
\n
\nQuestions 3:<\/strong><\/span>
\n
\n
\n
\n
\n
\nQuestions 4:<\/strong><\/span>
\n
\nQuestions 5:<\/strong><\/span>
\n
\nQuestions 6:<\/strong><\/span>
\n(i) The sum of two rationals is always rational – True
\n(ii) The product of two rationals is always rational – True
\n(iii) The sum of two irrationals is an irrational – False
\n(iv) The product of two irrationals is an irrational – False
\n(v) The sum of a rational and an irrational is irrational – True
\n(vi) The product of a rational and an irrational is irrational – True<\/p>\n