RS Aggarwal Solutions for Class 8 Maths Chapter 5 Exercise 5.3: The Physics Wallah academic team has produced a comprehensive answer for Chapter 5: Playing with Numbers in the RS Aggarwal class 8 textbook. One should read Chapter 5 Playing with Numbers theory before attempting to solve all of the numerical problems in exercise 5C.
This will ensure that you have a firm understanding of Chapter 5 Playing With Numbers. For class 8 maths students, the NCERT textbook is a highly recommended resource for solving numerical problems and referencing NCERT solutions.RS Aggarwal Solutions for Class 8 Maths Chapter 5 Exercise 5.3 PDF
Replace A, B, C by suitable numerals.
Question (1) Solution: A + 7 = 13
⇒ A = 13 – 7 = 6 So, 1 is carried over. (1 + 5 + 8) = 14. So, B = 4 and 1 is carried over. And C = 1. ∴ A = 6, B = 4 and C = 1.Question (2) Solution: A + 6 = 13
⇒ A = 16 – 6 = 7 So, 1 is carried over. 1 + B + 9 = 17 ⇒ B = 17 – 10 = 7, B = 7 and 1 is carried over. 1 + C + 6 = 11 ⇒ C = 11 – 7 = 4. ∴ A = 7, B = 7 and C = 4.Question (3) Solution: A + A + A = A, 1 is carried over.
When, A = 5 A + A + A = 15, (1 is carried over) ⇒ B = 1. ∴ A = 5, B = 1.Question (4) Solution: 6 – A = 3
This implies that the maximum value of A can be 3. A ≤ 3 …..(i) The next column has the following: A – B = 7 To reconcile this with equation (1), borrowing is involved. We know: 12 – 5 = 7 ∴ A = 2 and b = 5Question (5) 5 – A = 9
This implies that 1 is borrowed. We know, 15 – 6 = 9 ⇒ A = 6 B – 5 = 8 This implies that 1 is borrowed. 13 – 5 = 8 But 1 has also been lent. ∴ B = 4 C – 2 = 2 This implies that 1 has been lent. ∴ C = 5 ∴ A = 6, B = 4 and C = 5.Question (6) (B × 3) = B
Then, B can either be 0 or 5. If B is 5, then 1 will be carried. Then, A × 3 + 1 = A will not be possible for any number. ∴ B = 0 A × 3 = A is possible for either 0 or 5. If we take A = 0, then all number will become 0. However, this not a possible. ∴ A = 5 Then, 1 will be carried. ∴ C = 1 ∴ A = 5, B = 0 and C = 1Question (7) A × B = B
⇒ A = 1 In the question: First digit = B + 1 Thus, 1 will be carried from 1 + B 2 and becomes (B + 1) (B 2 – 9)B. ∴ C = B 2 – 1 Now, all b, B + 1 and B 2 – 9 are one digit number. This condition is satisfied for B = 3 or B = 4. For, B<3, B – 9 will be negative. For, B > 3, B 2 – 9 will become digit number. For B = 3, C = 3 2 – 9 = 9 – 9 = 0 For, B = 4, C = 4 2 – 9 = 16 – 9 = 7 ∴ A = 1, B = 4, C = 7.(8) Solution: (A – 4) = 3
⇒ A = 7 Also, 6 × 6 = 36 ⇒ 36 – 36 = 0 ⇒ B = 6 ∴ A = 7 B = C = 6.Question (9) Find two numbers whose product is a 1- digit number and the sum is a 2-digit number.
Solution: 1 and 9 are two numbers whose product is a single digit number. ∴ 1 × 9 = 9 Sum of the numbers is a two digit number. ∴ 1 + 9 = 10Question (10) Find three whole numbers whose product and sum are equal.
Solution: The three whole numbers are 1, 2 and 3. 1 + 2 + 3 = 6 = 1 × 2 × 3Question (11) complete the magic square given below, so that the sum of the numbers in each roe or in each column or along each diagonal is 15.
Solution: 6 + 5 + x = 15 ⇒ x = 4 Now taking the first row: 6 + 1 + x = 15 ⇒ x = 15 – 7 = 8 Taking last column: 8 + x + 4 = 15 ⇒ x = 15 – 12 = 3 Taking second column: 1 + 5 + x = 15 ⇒ x = 15 – 6 = 9 Taking second row: X + 5 + 3 = 15 ⇒ x = 15 – 8 = 7 Taking diagonal that begins with 8: 8 + 5 + x = 15 ⇒ x = 2Question (12) Fill in the numbers from 1 to 6 without repetition, so that each side of the triangle adds up to 12.
Solution: 6 + 2 + 4 = 12 4 + 3 + 5 = 12 6 + 1 + 5 = 12Question (13) Fibonacci numbers Take 10 numbers shown below:
A, b, (a + b), (a + 2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21a + 34b). Sum of all these numbers = 11(5a + 8b) = 11 × 7 th number. Solution: a = 8 and b = 13 The numbers are 8, 13, 21, 34, 55, 89, 144, 233, 377 and 610. Sum of the numbers = 8 + 13 + 21 + 34 + 55+ 89 + 144 + 233 + 377 + 610 = 1584 11 × 7th number = 11 × 144 = 1584Question (14) Complete the magic square:
14 | 0 | ||
8 | 6 | 11 | |
4 | 7 | ||
2 | 1 | 12 |
3 | 14 | 13 | 0 |
8 | 5 | 6 | 11 |
4 | 9 | 10 | 7 |
15 | 2 | 1 | 12 |
Conceptual Clarity: The solutions provide detailed explanations and step-by-step approaches, helping students grasp the underlying concepts of divisibility, prime numbers, and factorization. This clarity is crucial for building a strong mathematical foundation.
Enhanced Problem-Solving Skills: By practicing various problems, students develop critical problem-solving skills. The exercise encourages logical reasoning and analytical thinking, enabling students to tackle complex mathematical challenges.
Time Management: Practicing these exercises helps students improve their speed and accuracy in solving mathematical problems, which is beneficial for time management during exams.
Application of Theoretical Knowledge: The solutions bridge the gap between theoretical knowledge and practical application, allowing students to see how mathematical concepts are used in real-world scenarios.
Confidence Building: Regular practice with well-structured solutions boosts students' confidence in handling different types of numerical problems, preparing them for higher-level math courses.