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.

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

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

(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

(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

Questions 5:
(i) By prime factorization, we get


(ii) By prime factorization. We get


(iii) By prime factorization, we get


Questions 6:
(i) By prime factorization, we get


(ii) By prime factorization, we get


(iii) By prime factorization, we get


Questions 7:

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

Questions 9:
By going upward
5 x 11= 55
55 x 3 = 165
165 x 2 = 330
330 x 2 = 660

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

504 = 2 x 2 x 2 x 3 x 3 x 7

The required number is 12.

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:


The required number is 45.
Questions 12:

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:


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

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:


Each stack of book will contain 48 books
Number of stacks of the same height

Questions 17:

Questions 18:
Let us find the LCM of 64, 80 and 96 through prime factorization:


L.C.M of 64, 80 and 96 =

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:

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:

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