Class 10: Maths Chapter 1 solutions. Complete Class 10 Maths Chapter 1 Notes.
RS Aggarwal Solutions for Class 10 Maths Chapter 1–Real Numbers
RS Aggarwal 10th Maths Chapter 1, Class 10 Maths Chapter 1 solutions
Exercise 1A Solutions
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 Solutions
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 Solutions
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
RS Aggarwal Solutions for Class 10 Maths Chapter 1: Download PDF
RS Aggarwal Solutions for Class 10 Maths Chapter 1–Real Numbers
Download PDF: RS Aggarwal Solutions for Class 10 Maths Chapter 1–Real Numbers PDF
Chapterwise RS Aggarwal Solutions for Class 10 Maths :
- Chapter 1–Real Numbers
- Chapter 2–Polynomials
- Chapter 3–Linear Equations In Two Variables
- Chapter 4–Quadratic Equations
- Chapter 5–Arithmetic Progression
- Chapter 6–Coordinate Geometry
- Chapter 7–Triangles
- Chapter 8–Circles
- Chapter 9–Constructions
- Chapter 10–Trigonometric Ratios
- Chapter 11–T Ratios Of Some Particular Angles
- Chapter 12–Trigonometric Ratios Of Some Complementary Angles
- Chapter 13–Trigonometric Identities
- Chapter 14–Height and Distance
- Chapter 15–Perimeter and Areas of Plane Figures
- Chapter 16–Areas of Circle, Sector and Segment
- Chapter 17–Volume and Surface Areas of Solids
- Chapter 18–Mean, Median, Mode of Grouped Data
- Chapter 19–Probability
About RS Aggarwal Class 10 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal's book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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