RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers EX 1A

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 1 Real Numbers EX 1A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 1 Real Numbers EX 1A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers EX 1A Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 1
Chapter NameReal Numbers
Exercise1A
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 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 Solutions Class 10 Chapter 1 Real Numbers 1a 3.1
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 Solutions Class 10 Chapter 1 Real Numbers 1a 4.1
(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 Solutions Class 10 Chapter 1 Real Numbers 1a 4.2
(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 Solutions Class 10 Chapter 1 Real Numbers 1a 4.3
Questions 5:
(i) By prime factorization, we get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 5.1
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 5.2
(ii) By prime factorization. We get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 5.3
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 5.4
(iii) By prime factorization, we get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 5.5
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 5.6
Questions 6:
(i) By prime factorization, we get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 6.1
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 6.2
(ii) By prime factorization, we get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 6.3
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 6.4
(iii) By prime factorization, we get
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 6.5
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 6.6
Questions 7:
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 7.1
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 Chapter 1 Real Numbers 1a 8.1
Questions 9:
By going upward
5 x 11= 55
55 x 3 = 165
165 x 2 = 330
330 x 2 = 660
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 9.1
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 Solutions Class 10 Chapter 1 Real Numbers 1a 10.1
504 = 2 x 2 x 2 x 3 x 3 x 7
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1c 9.1
The required number is 12.
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1c 9.2
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:
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 11.1
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 11.2
The required number is 45.
Questions 12:
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 12.1
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 Solutions Class 10 Chapter 1 Real Numbers 1a 14.1
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 14.2
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 Solutions Class 10 Chapter 1 Real Numbers 1a 15.1
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 Solutions Class 10 Chapter 1 Real Numbers 1a 16.1
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 16.2
Each stack of book will contain 48 books
Number of stacks of the same height
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 16.3
Questions 17:
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 17.1
Questions 18:
Let us find the LCM of 64, 80 and 96 through prime factorization:
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 18.1
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 18.2
L.C.M of 64, 80 and 96 =
RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers 1a 18.3
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 Solutions Class 10 Chapter 1 Real Numbers 1a 19.1
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 Solutions Class 10 Chapter 1 Real Numbers 1a 21.1

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