RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers

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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

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:

Exercise 1B

Questions 1:

Questions 2:

Questions 3:

Questions 4:
(i) 53.123456789 is a rational number since it is a terminating decimal.
(ii)  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:

Questions 2:

Questions 3:

Questions 4:

Questions 5:

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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