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RS Aggarwal Solutions: Real Numbers - 1 | Mathematics (Maths) Class 10 PDF Download

Q.1. What do you mean by Euclid's division lemma?

Solution: For any two given positive integers a and b there exist unique whole numbers q and r such that

a = bq + r, when 0 ≤ r < b. 
Here, we call 'a' as a dividend, b as a divisor, q as the quotient and r as the remainder.
Dividend = (Divisor × Quotient) + Remainder

Q.2. A number when divided by 61 gives 27 as quotient and 32 as remainder. Find the number.
Solution: Using Euclid’s division Lemma
Dividend = (Divisor × Quotient) + Remainder
= (61 × 27) + 32
= 1647 + 32
= 1679
Required number = 1679

Q.3. By what number should 1365 be divided to get 31 as the quotient and 32 as the remainder?
Solution: By Euclid's Division Algorithm, we have:
Dividend = (Divisor X Quotient) + Remainder
⇒ 1365 = (Divisor × 31) + 32
⇒ (1365 - 32) / 31 = Divisor
⇒ 1333 / 31 = Divisor
Therefore, Divisor = 43

Q.4. Using Euclid's algorithm, find the HCF of:
(i) 405 and 2520
(ii) 504 and 1188
(iii) 960 and 1575
Solution:
(i) On dividing 2520 by 405, we get

  •  Quotient = 6, remainder = 90
  • Therefore, 2520 = (405 × 6) + 90
  • Dividing 405 by 90, we get
  • Quotient = 4, Remainder = 45
  • Therefore, 405 = 90 × 4 + 45
  • Dividing 90 by 45
  • Quotient = 2, remainder = 0
  • Therefore, 90 = 45 × 2
  •  H.C.F. of 405 and 2520 is 45

RS Aggarwal Solutions: Real Numbers - 1 | Mathematics (Maths) Class 10

(ii) Dividing 1188 by 504, we get

  • Quotient = 2, remainder = 180
  • Therefore, 1188 = 504 × 2 + 180
  • Dividing 504 by 180
  • Quotient = 2, remainder = 144
  • Therefore, 504 = 180 × 2 + 144
  • Dividing 180 by 144, we get
  • Quotient = 1, remainder = 36
  • Dividing 144 by 36
  • Quotient = 4, remainder = 0
  • Therefore, H.C.F. of 1188 and 504 is 36
    RS Aggarwal Solutions: Real Numbers - 1 | Mathematics (Maths) Class 10

(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: Real Numbers - 1 | Mathematics (Maths) Class 10

Q.5. Show that every positive integer is either even or odd.
Solution: Let a be a given positive integer.
On dividing a by 2, let q be the quotient and r be the remainder.
Then, by Eudid's algorithm, we have
a = 2q+ r, where 0 < r < 2
⇒ a = 2q + r, where r = 0,1
⇒ a = 2q or a = 2q + 1
When a = 2q for some integer q, then clearly a is even.

When a = 2q + 1 for some integer q, then dearly a is odd.

Thus, every positive integer is either even or odd.

Q.6. Show that any positive odd integer is of the form (6m + 1), (6m + 3) or (6m + 5), where m is some integer.
Solution: Let a be a given positive integer.
On dividing a by 6, let q be the quotient and r be the remainder.
Then, by Eudid's algorithm, we have
a = 6q + r, where 0 < r < 6
⇒ a = 6q + r, where r = 0,1,2,3,4,5,6
⇒ a = 6q or a = 6q + 1 or a = 6q + 2 or a = 6q + 3 or a = 6q + 4 or a = 6q + 5
But, a = 6q, a = 6q + 2, a = 6q + 4 gives even values of a.
Thus, when a is odd, it is of the form
6q + 1, 6q + 3 or 6q + 5 for some integer q.

Q.7. Show that any positive odd integer is of the form (4m + 1) or (4m + 3), wherein is some integer.
Solution: 
Let a be a given positive odd integer.
On dividing by 4, let q be the quotient and r be the remainder.
Then, by Eudid's algorithm, we have
a = 4m + r, where 0 < r < 4
⇒ a = 4m + r, where r = 0,1,2,3
⇒ a = 4m or a = 4m + 1 or a = 4m + 2 or a = 4m + 3
But, a = 4m and a = 4m + 2 = 2(2m + 1) are clearly even.
Thus, when a is odd, it is of the form a = (4m + 1)or (4m + 3) for some integer m.

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FAQs on RS Aggarwal Solutions: Real Numbers - 1 - Mathematics (Maths) Class 10

1. What are real numbers, and how are they classified?
Ans. Real numbers are the set of numbers that include all the rational and irrational numbers. They can be classified into various categories: natural numbers (positive integers), whole numbers (natural numbers including zero), integers (whole numbers including negative numbers), rational numbers (numbers that can be expressed as fractions), and irrational numbers (numbers that cannot be expressed as fractions, such as √2 or π).
2. How do you perform operations with real numbers?
Ans. Operations with real numbers include addition, subtraction, multiplication, and division. To perform these operations, you must follow the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)). It is important to ensure that the numbers you are working with are in the same form when necessary, for example, converting fractions to decimals.
3. What is the significance of the number line in understanding real numbers?
Ans. The number line is a visual representation of real numbers, where each point on the line corresponds to a real number. It helps in understanding the relative position of numbers, comparing them, and performing operations. The number line extends infinitely in both directions, with negative numbers on the left and positive numbers on the right, providing a clear framework for visualizing the concept of real numbers and their relationships.
4. What are some common properties of real numbers?
Ans. Real numbers possess several important properties, including the commutative property (a + b = b + a and ab = ba), associative property ((a + b) + c = a + (b + c) and (ab)c = a(bc)), distributive property (a(b + c) = ab + ac), and the existence of identity elements (0 for addition and 1 for multiplication). These properties are fundamental in solving mathematical problems involving real numbers.
5. How can I practice problems related to real numbers effectively?
Ans. To practice problems related to real numbers effectively, you can use textbooks like RS Aggarwal, online resources, and worksheets that provide a variety of exercises. Start with simple problems to build your understanding, then gradually move to more complex ones. Additionally, practicing with peers or tutors can help clarify concepts and improve problem-solving skills. Regular practice and reviewing solutions will enhance your proficiency in handling real numbers.
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