Operations on Whole Numbers - Exercise 4.4 Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : Operations on Whole Numbers - Exercise 4.4 Class 6 Notes | EduRev

 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 = 504347 
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Operations on Whole Numbers - Exercise 4.4 Class 6 Notes | EduRev

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