Page 1 Exercise 4.4 page: 4.19 1. Does there exist a whole number a such that a ÷ a = a? Solution: Yes. There exists a whole number ‘a’ such that a ÷ a = a. We know that the whole number is 1 where 1 ÷ 1 = 1. 2. Find the value of: (i) 23457 ÷ 1 (ii) 0 ÷ 97 (iii) 476 + (840 ÷ 84) (iv) 964 – (425 ÷ 425) (v) (2758 ÷ 2758) – (2758 ÷ 2758) (vi) 72450 ÷ (583 – 58) Solution: (i) 23457 ÷ 1 By division 23457 ÷ 1 = 23457 (ii) 0 ÷ 97 By division 0 ÷ 97 = 0 (iii) 476 + (840 ÷ 84) On further calculation 476 + (840 ÷ 84) = 476 + 10 = 486 (iv) 964 – (425 ÷ 425) On further calculation 964 – (425 ÷ 425) = 964 – 1 = 963 (v) (2758 ÷ 2758) – (2758 ÷ 2758) On further calculation (2758 ÷ 2758) – (2758 ÷ 2758) = 1 – 1 = 0 (vi) 72450 ÷ (583 – 58) On further calculation 72450 ÷ (583 – 58) = 72450 ÷ 525 = 138 3. Which of the following statements are true: (i) 10 ÷ (5 × 2) = (10 ÷ 5) × (10 ÷ 2) (ii) (35 – 14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5 Page 2 Exercise 4.4 page: 4.19 1. Does there exist a whole number a such that a ÷ a = a? Solution: Yes. There exists a whole number ‘a’ such that a ÷ a = a. We know that the whole number is 1 where 1 ÷ 1 = 1. 2. Find the value of: (i) 23457 ÷ 1 (ii) 0 ÷ 97 (iii) 476 + (840 ÷ 84) (iv) 964 – (425 ÷ 425) (v) (2758 ÷ 2758) – (2758 ÷ 2758) (vi) 72450 ÷ (583 – 58) Solution: (i) 23457 ÷ 1 By division 23457 ÷ 1 = 23457 (ii) 0 ÷ 97 By division 0 ÷ 97 = 0 (iii) 476 + (840 ÷ 84) On further calculation 476 + (840 ÷ 84) = 476 + 10 = 486 (iv) 964 – (425 ÷ 425) On further calculation 964 – (425 ÷ 425) = 964 – 1 = 963 (v) (2758 ÷ 2758) – (2758 ÷ 2758) On further calculation (2758 ÷ 2758) – (2758 ÷ 2758) = 1 – 1 = 0 (vi) 72450 ÷ (583 – 58) On further calculation 72450 ÷ (583 – 58) = 72450 ÷ 525 = 138 3. Which of the following statements are true: (i) 10 ÷ (5 × 2) = (10 ÷ 5) × (10 ÷ 2) (ii) (35 – 14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5 (v) 12 × (14 ÷ 7) = (12 × 14) ÷ (12 × 7) (vi) (20 ÷ 5) ÷ 2 = (20 ÷ 2) ÷ 5 Solution: (i) False. We know that LHS = 10 ÷ (5 × 2) So we get = 10 ÷ 10 = 1 RHS = (10 ÷ 5) × (10 ÷ 2) So we get = 2 × 5 = 10 (ii) True. We know that LHS = (35 – 14) ÷ 7 So we get = 21 ÷ 7 = 3 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iii) False. We know that LHS = 35 – 14 ÷ 7 So we get = 35 – 2 = 33 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iv) False. We know that LHS = (20 – 5) ÷ 5 So we get = 15 ÷ 5 = 3 RHS = 20 ÷ 5 – 5 So we get = 4 – 5 Page 3 Exercise 4.4 page: 4.19 1. Does there exist a whole number a such that a ÷ a = a? Solution: Yes. There exists a whole number ‘a’ such that a ÷ a = a. We know that the whole number is 1 where 1 ÷ 1 = 1. 2. Find the value of: (i) 23457 ÷ 1 (ii) 0 ÷ 97 (iii) 476 + (840 ÷ 84) (iv) 964 – (425 ÷ 425) (v) (2758 ÷ 2758) – (2758 ÷ 2758) (vi) 72450 ÷ (583 – 58) Solution: (i) 23457 ÷ 1 By division 23457 ÷ 1 = 23457 (ii) 0 ÷ 97 By division 0 ÷ 97 = 0 (iii) 476 + (840 ÷ 84) On further calculation 476 + (840 ÷ 84) = 476 + 10 = 486 (iv) 964 – (425 ÷ 425) On further calculation 964 – (425 ÷ 425) = 964 – 1 = 963 (v) (2758 ÷ 2758) – (2758 ÷ 2758) On further calculation (2758 ÷ 2758) – (2758 ÷ 2758) = 1 – 1 = 0 (vi) 72450 ÷ (583 – 58) On further calculation 72450 ÷ (583 – 58) = 72450 ÷ 525 = 138 3. Which of the following statements are true: (i) 10 ÷ (5 × 2) = (10 ÷ 5) × (10 ÷ 2) (ii) (35 – 14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5 (v) 12 × (14 ÷ 7) = (12 × 14) ÷ (12 × 7) (vi) (20 ÷ 5) ÷ 2 = (20 ÷ 2) ÷ 5 Solution: (i) False. We know that LHS = 10 ÷ (5 × 2) So we get = 10 ÷ 10 = 1 RHS = (10 ÷ 5) × (10 ÷ 2) So we get = 2 × 5 = 10 (ii) True. We know that LHS = (35 – 14) ÷ 7 So we get = 21 ÷ 7 = 3 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iii) False. We know that LHS = 35 – 14 ÷ 7 So we get = 35 – 2 = 33 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iv) False. We know that LHS = (20 – 5) ÷ 5 So we get = 15 ÷ 5 = 3 RHS = 20 ÷ 5 – 5 So we get = 4 – 5 = - 1 (v) False. We know that LHS = 12 × (14 ÷ 7) So we get = 12 × 2 = 24 RHS = (12 × 14) ÷ (12 × 7) So we get = 168 ÷ 84 = 2 (vi) True. We know that LHS = (20 ÷ 5) ÷ 2 So we get = 4 ÷ 2 = 2 RHS = (20 ÷ 2) ÷ 5 So we get = 10 ÷ 5 = 2 4. Divide and check the quotient and remainder: (i) 7772 ÷ 58 (ii) 6906 ÷ 35 (iii) 16135 ÷ 875 (iv) 16025 ÷ 1000 Solution: (i) 7772 ÷ 58 So we get 7772 ÷ 58 = 134 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 7772 = 58 × 134 + 0 Page 4 Exercise 4.4 page: 4.19 1. Does there exist a whole number a such that a ÷ a = a? Solution: Yes. There exists a whole number ‘a’ such that a ÷ a = a. We know that the whole number is 1 where 1 ÷ 1 = 1. 2. Find the value of: (i) 23457 ÷ 1 (ii) 0 ÷ 97 (iii) 476 + (840 ÷ 84) (iv) 964 – (425 ÷ 425) (v) (2758 ÷ 2758) – (2758 ÷ 2758) (vi) 72450 ÷ (583 – 58) Solution: (i) 23457 ÷ 1 By division 23457 ÷ 1 = 23457 (ii) 0 ÷ 97 By division 0 ÷ 97 = 0 (iii) 476 + (840 ÷ 84) On further calculation 476 + (840 ÷ 84) = 476 + 10 = 486 (iv) 964 – (425 ÷ 425) On further calculation 964 – (425 ÷ 425) = 964 – 1 = 963 (v) (2758 ÷ 2758) – (2758 ÷ 2758) On further calculation (2758 ÷ 2758) – (2758 ÷ 2758) = 1 – 1 = 0 (vi) 72450 ÷ (583 – 58) On further calculation 72450 ÷ (583 – 58) = 72450 ÷ 525 = 138 3. Which of the following statements are true: (i) 10 ÷ (5 × 2) = (10 ÷ 5) × (10 ÷ 2) (ii) (35 – 14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5 (v) 12 × (14 ÷ 7) = (12 × 14) ÷ (12 × 7) (vi) (20 ÷ 5) ÷ 2 = (20 ÷ 2) ÷ 5 Solution: (i) False. We know that LHS = 10 ÷ (5 × 2) So we get = 10 ÷ 10 = 1 RHS = (10 ÷ 5) × (10 ÷ 2) So we get = 2 × 5 = 10 (ii) True. We know that LHS = (35 – 14) ÷ 7 So we get = 21 ÷ 7 = 3 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iii) False. We know that LHS = 35 – 14 ÷ 7 So we get = 35 – 2 = 33 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iv) False. We know that LHS = (20 – 5) ÷ 5 So we get = 15 ÷ 5 = 3 RHS = 20 ÷ 5 – 5 So we get = 4 – 5 = - 1 (v) False. We know that LHS = 12 × (14 ÷ 7) So we get = 12 × 2 = 24 RHS = (12 × 14) ÷ (12 × 7) So we get = 168 ÷ 84 = 2 (vi) True. We know that LHS = (20 ÷ 5) ÷ 2 So we get = 4 ÷ 2 = 2 RHS = (20 ÷ 2) ÷ 5 So we get = 10 ÷ 5 = 2 4. Divide and check the quotient and remainder: (i) 7772 ÷ 58 (ii) 6906 ÷ 35 (iii) 16135 ÷ 875 (iv) 16025 ÷ 1000 Solution: (i) 7772 ÷ 58 So we get 7772 ÷ 58 = 134 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 7772 = 58 × 134 + 0 So we get 7772 = 7772 LHS = RHS (ii) 6906 ÷ 35 So we get quotient = 197 and remainder = 11 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 6906 = 35 × 197 + 11 On further calculation 6906 = 6895 + 11 We get 6906 = 6906 LHS = RHS (iii) 16135 ÷ 875 So we get quotient = 18 and remainder = 385 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 16135 = 875 × 18 + 385 On further calculation 16135 = 15750 + 385 We get 16135 = 16135 LHS = RHS (iv) 16025 ÷ 1000 Page 5 Exercise 4.4 page: 4.19 1. Does there exist a whole number a such that a ÷ a = a? Solution: Yes. There exists a whole number ‘a’ such that a ÷ a = a. We know that the whole number is 1 where 1 ÷ 1 = 1. 2. Find the value of: (i) 23457 ÷ 1 (ii) 0 ÷ 97 (iii) 476 + (840 ÷ 84) (iv) 964 – (425 ÷ 425) (v) (2758 ÷ 2758) – (2758 ÷ 2758) (vi) 72450 ÷ (583 – 58) Solution: (i) 23457 ÷ 1 By division 23457 ÷ 1 = 23457 (ii) 0 ÷ 97 By division 0 ÷ 97 = 0 (iii) 476 + (840 ÷ 84) On further calculation 476 + (840 ÷ 84) = 476 + 10 = 486 (iv) 964 – (425 ÷ 425) On further calculation 964 – (425 ÷ 425) = 964 – 1 = 963 (v) (2758 ÷ 2758) – (2758 ÷ 2758) On further calculation (2758 ÷ 2758) – (2758 ÷ 2758) = 1 – 1 = 0 (vi) 72450 ÷ (583 – 58) On further calculation 72450 ÷ (583 – 58) = 72450 ÷ 525 = 138 3. Which of the following statements are true: (i) 10 ÷ (5 × 2) = (10 ÷ 5) × (10 ÷ 2) (ii) (35 – 14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7 (iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5 (v) 12 × (14 ÷ 7) = (12 × 14) ÷ (12 × 7) (vi) (20 ÷ 5) ÷ 2 = (20 ÷ 2) ÷ 5 Solution: (i) False. We know that LHS = 10 ÷ (5 × 2) So we get = 10 ÷ 10 = 1 RHS = (10 ÷ 5) × (10 ÷ 2) So we get = 2 × 5 = 10 (ii) True. We know that LHS = (35 – 14) ÷ 7 So we get = 21 ÷ 7 = 3 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iii) False. We know that LHS = 35 – 14 ÷ 7 So we get = 35 – 2 = 33 RHS = 35 ÷ 7 – 14 ÷ 7 So we get = 5 – 2 = 3 (iv) False. We know that LHS = (20 – 5) ÷ 5 So we get = 15 ÷ 5 = 3 RHS = 20 ÷ 5 – 5 So we get = 4 – 5 = - 1 (v) False. We know that LHS = 12 × (14 ÷ 7) So we get = 12 × 2 = 24 RHS = (12 × 14) ÷ (12 × 7) So we get = 168 ÷ 84 = 2 (vi) True. We know that LHS = (20 ÷ 5) ÷ 2 So we get = 4 ÷ 2 = 2 RHS = (20 ÷ 2) ÷ 5 So we get = 10 ÷ 5 = 2 4. Divide and check the quotient and remainder: (i) 7772 ÷ 58 (ii) 6906 ÷ 35 (iii) 16135 ÷ 875 (iv) 16025 ÷ 1000 Solution: (i) 7772 ÷ 58 So we get 7772 ÷ 58 = 134 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 7772 = 58 × 134 + 0 So we get 7772 = 7772 LHS = RHS (ii) 6906 ÷ 35 So we get quotient = 197 and remainder = 11 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 6906 = 35 × 197 + 11 On further calculation 6906 = 6895 + 11 We get 6906 = 6906 LHS = RHS (iii) 16135 ÷ 875 So we get quotient = 18 and remainder = 385 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 16135 = 875 × 18 + 385 On further calculation 16135 = 15750 + 385 We get 16135 = 16135 LHS = RHS (iv) 16025 ÷ 1000 So we get quotient = 16 and remainder = 25 By verifying We know that Dividend = Divisor × Quotient + Remainder By substituting values 16025 = 1000 × 16 + 25 On further calculation 16025 = 16000 + 25 We get 16025 = 16025 LHS = RHS 5. Find a number which when divided by 35 gives the quotient 20 and remainder 18. Solution: We know that Dividend = Divisor × Quotient + Remainder By substituting values Dividend = 35 × 20 + 18 On further calculation Dividend = 700 + 18 So we get Dividend = 718 6. Find the number which when divided by 58 gives a quotient 40 and remainder 31. Solution: We know that Dividend = Divisor × Quotient + Remainder By substituting values Dividend = 58 × 40 + 31 On further calculation Dividend = 2320 + 31 So we get Dividend = 2351 7. The product of two numbers is 504347. If one of the numbers is 1591, find the other. Solution: The product of two numbers = 504347 One of the numbers = 1591 Consider A as the number A × 1591 = 504347Read More

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