RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers


Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers download pdf.





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
Here, we call ‘a’ as dividend, b as divisor, q as quotient and r as remainder.
Dividend = (divisor quotient) + remainder





Questions 2:





By Euclid’s Division algorithm we have:
Dividend = (divisor * quotient) + remainder
= (61 * 27) + 32 = 1647 + 32 = 1679





Questions 3:





By Euclid’s Division Algorithm, we have:
Dividend = (divisor quotient) + remainder





rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q3
Questions 4:
(i) On dividing 2520 by 405, we get
Quotient = 6, remainder = 90
2520 = (405 x 6) + 90
Dividing 405 by 90, we get Quotient = 4, Remainder = 45
405 = 90 x 4 + 45
Dividing 90 by 45 We get Quotient = 2, remainder = 0
90 = 45 x 2
H.C.F. of 405 and 2520 is 45
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q4-1
(ii) Dividing 1188 by 504, we get Quotient = 2, remainder = 180
1188 = 504 x 2+ 180
Dividing 504 by 180  Quotient = 2, remainder = 144
504 = 180 x 2 + 144
Dividing 180 by 144, we get Quotient = 1, remainder = 36
Dividing 144 by 36
Quotient = 4, remainder = 0
H.C.F. of 1188 and 504 is 36
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q4-2
(iii) Dividing 1575 by 960, we get
Quotient = 1, remainder = 615
1575 = 960 x 1 + 615
Dividing 960 by 615, we get Quotient = 1, remainder = 345
960 = 615 x 1 + 345
Dividing 615 by 345 Quotient = 1, remainder = 270
615 = 345 x 1 + 270
Dividing 345 by 270, we get Quotient = 1, remainder = 75
345 = 270 x 1 + 75
Dividing 270 by 75, we get Quotient = 3, remainder =45
270 = 75 x 3 + 45
Dividing 75 by 45, we get Quotient = 1, remainder = 30
75 = 45 x 1 + 30
Dividing 45 by 30, we get Remainder = 15, quotient = 1
45 = 30 x 1 + 15
Dividing 30 by 15, we get Quotient = 2, remainder = 0
H.C.F. of 1575 and 960 is 15
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q4-3
Questions 5:
(i) By prime factorization, we get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers ex 1a-q5-1
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers ex 1aq5-2
(ii) By prime factorization. We get
Solution of RS Aggarwal Class 10 Chapter 1 Real Numbers ex 1a-q5-4
RS Aggarwal Solutions Class 10 2017 Chapter 1 Real Numbers ex 1a-q5-3
(iii) By prime factorization, we get
Class 10 RS Aggarwal Solutions Chapter 1 Real Numbers ex 1a-q5-5
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 1 Real Numbers ex 1a-q5-6
Questions 6:
(i) By prime factorization, we get
RS Aggarwal Class 10 Book pdf download Chapter 1 Real Numbers ex 1a-q6-1
Class 10 Maths RS Aggarwal Solutions Chapter 1 Real Numbers ex 1a-q6-2
(ii) By prime factorization, we get
Maths RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers ex 1a
RS Aggarwal Solutions Class 10 2018 Chapter 1 Real Numbers ex 1a
(iii) By prime factorization, we get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers ex 1a
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers ex 1a
Questions 7:
Solution of RS Aggarwal Class 10 Chapter 1 Real Numbers ex 1a
Questions 8:
H.C.F. of two numbers = 11, their L.C.M = 7700
One number = 275, let the other number be b
Now, 275 x b = 11 x 7700
RS Aggarwal Solutions Class 10 2017 Chapter 1 Real Numbers ex 1a
Questions 9:
By going upward
5 x 11= 55
55 x 3 = 165
165 x 2 = 330
330 x 2 = 660
Class 10 RS Aggarwal Solutions Chapter 1 Real Numbers ex 1a
Questions 10:
Subtracting 6 from each number:
378 – 6 = 372, 510 – 6 = 504
Let us now find the HCF of 372 and 504 through prime factorization:
372 = 2 x 2 x 3 x 31
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 1 Real Numbers ex 1a
504 = 2 x 2 x 2 x 3 x 3 x 7
RS Aggarwal Class 10 Book pdf download Chapter 1 Real Numbers ex 1a
The required number is 12.
Maths RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers ex 1a
Questions 11:
Subtracting 5 and 7 from 320 and 457 respectively:
320 – 5 = 315,
457 – 7 = 450
Let us now find the HCF of 315 and 405 through prime factorization:
Class 10 Maths RS Aggarwal Solutions Chapter 1 Real Numbers ex 1a
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q11-2
The required number is 45.
Questions 12:
RS Aggarwal Solutions Class 10 2018 Chapter 1 Real Numbers ex 1a
Questions 13:
The prime factorization of 42, 49 and 63 are:
42 = 2 x 3 x 7, 49 = 7 x 7, 63 = 3 x 3 x 7
Therefore, H.C.F. of 42, 49, 63 is 7
Hence, greatest possible length of each plank = 7 m
Questions 14:
7 m = 700cm, 3m 85cm = 385 cm
12 m 95 cm = 1295 cm
Let us find the prime factorization of 700, 385 and 1295:
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q13-1
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q13-2
Greatest possible length = 35cm.
Questions 15:
Let us find the prime factorization of 1001 and 910:
1001 = 11 x 7 x 13
910 = 2 x 5 x 7 x 13
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q15
H.C.F. of 1001 and 910 is 7 x 13 = 91
Maximum number of students = 91
Questions 16:
Let us find the HCF of 336, 240 and 96 through prime factorization:
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q16-1
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q16-2
Each stack of book will contain 48 books
Number of stacks of the same height
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q16-3
Questions 17:
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q17
Questions 18:
Let us find the LCM of 64, 80 and 96 through prime factorization:
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q18-1
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q18-2
L.C.M of 64, 80 and 96 =
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q18-3
Therefore, 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
Questions 19:
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q19
Questions 20:
Interval of beeping together = LCM (60 seconds, 62 seconds)
The prime factorization of 60 and 62:
60 = 30 x 2, 62 = 31 x 2
L.C.M of 60 and 62 is 30 x 31 x 2 = 1860 sec = 31min
electronic device will beep after every 31minutes
After 10 a.m., it will beep at 10 hrs 31 minutes
Questions 21:
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1a-q21





Exercise 1B





Questions 1:
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers ex 1b
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers ex 1b
Questions 2:
RS Aggarwal Solutions Class 10 2017 Chapter 1 Real Numbers ex 1b
Questions 3:
Class 10 RS Aggarwal Solutions Chapter 1 Real Numbers ex 1b
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 1 Real Numbers ex 1b
Questions 4:
(i) 53.123456789 is a rational number since it is a terminating decimal.
(ii) RS Aggarwal Class 10 Book pdf download Chapter 1 Real Numbers ex 1b is a rational number because it is a non – terminating repeating decimal.
(iii) 0.12012001200012…… is not a rational number as it is a non-terminating, non – repeating decimal.





Exercise 1C





Questions 1:
Class 10 Maths RS Aggarwal Solutions Chapter 1 Real Numbers ex 1b
Questions 2:
rs-aggarwal-class-10-solutions-real-numbers-2-ex-1c-q2
Questions 3:
Maths RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers ex 1b
RS Aggarwal Solutions Class 10 2018 Chapter 1 Real Numbers ex 1b
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers ex 1c
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers ex 1c
Class 10 RS Aggarwal Solutions Chapter 1 Real Numbers ex 1c
Questions 4:
RS Aggarwal Solutions Class 10 2017 Chapter 1 Real Numbers ex 1c
Questions 5:
Solution of RS Aggarwal Class 10 Chapter 1 Real Numbers ex 1c
Questions 6:
(i) The sum of two rationals is always rational – True
(ii) The product of two rationals is always rational – True
(iii) The sum of two irrationals is an irrational – False
(iv) The product of two irrationals is an irrational – False
(v) The sum of a rational and an irrational is irrational – True
(vi) The product of a rational and an irrational is irrational – True





Complete RS Aggarwal Solutions Class 10





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