Rd Sharma Solutions Class 6 Chapter 4 Operation on Whole Numbers
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Rd Sharma Class 6 Mathematics, Chapter 4, Operation on Whole Numbers. Here students can easily find Exercise wise solution for chapter 4, Operation on Whole Numbers. Students will find proper solutions for Exercise 4.1, 4.2, 4.3, 4.4 and 4.5. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Operation on Whole Numbers Exercise 4.1 Solution
Question no – (1)
Solution :
(i) 359 + 476 = 476 + 359 …(Commutative law)
(ii) 2008 + 1952 = 1952 + 2008 …(Commutative)
(iii) 90758 + 0 = 9078 …(Additive identity)
(iv) 54321 + (489 + 699) 489 + (54321 + 699) …(Associative)
Question no – (2)
Solution :
(i) 5628 + 39784
= 45412
Now, by reversing,
= 39784+ 5628
= 45412
(ii) 923584 + 178
= 923762
Now, by reversing,
178 + 923584
= 923762
(iii) 15409 + 112
= 15521
Now, by reversing,
= 112 + 15409
= 15521
(iv) 2359 + 641
= 3000
Now, by reversing,
641 + 2359
= 3000
Question no – (4)
Solution :
(i) Given statement is False.
(ii) Given statement is True.
(ii) Given statement is True.
(iv) Given statement is False.
(v) Given statement is True.
(vi) Given statement is False.
(vii) Given statement is False.
(viii) Given statement is True.
(ix) Given statement is True.
(x) Given statement is False.
(xi) Given statement is False.
(xii) Given statement is True.
Operation on Whole Numbers Exercise 4.2 Solution
Question no – (1)
Solution :
(i)
| 15 | 8 | 13 |
| 10 | 12 | 14 |
| 11 | 16 | 9 |
∴ Diagonal values,
= 13 + 12 + 11
= 36
(ii)
| 22 | 29 | 6 | 13 | 20 |
| 28 | 10 | 12 | 19 | 21 |
| 9 | 11 | 18 | 25 | 27 |
| 15 | 17 | 24 | 26 | 8 |
| 16 | 23 | 30 | 7 | 14 |
∴ Consider the diagonal values,
= 20 + 15 + 18 + 17 + 16
= 90
Question no – (2)
Solution :
(i) 57839 – 2983 = 54856
Now, by addition,
= 54856 + 2983
= 57839
(ii) 92507- 10879 = 81628
Now, by addition,
= 81628 + 10879
= 92507
(iii) 400000 – 98798 = 301202
Now, by addition,
301202 + 98798
= 400000
(iv) 5050501 – 969696 = 4080805
Now, by addition,
= 4080805 + 969696
= 5050501
(v) 200000 – 97531 = 1024469
Now, by addition,
= 102469 + 97351
= 200000
Question no – (4)
Solution :
We know that,
Largest no of 5 digits = 99999
Largest no of 6 digits = 100000
∴ Difference is,
= (100000 – 99999)
= 1
Therefore, the difference will be 1.
Question no – (5)
Solution :
As we know that,
Largest number of 4 digits = 9999
Smallest number of 7 digits = 1000000
∴ Difference is,
= (1000000 – 9999)
= 990001
Therefore, the difference will be 990001.
Question no – (6)
Solution :
In the given question,
Rohit deposited = Rs 125000
He withdrew = Rs 35425
Money left = ?
∴ Money left in his account,
= (125000 – 25425)
= 89575 Rs
Therefore, Rs 89575 was left in Rohit’s account.
Question no – (7)
Solution :
According to the question,
Population of a town is 96209
Men = 29642
Women = 29167
Children = ?
∴ Total no of number + women,
= (29642 + 29167)
= 58809
∴ No of children,
= (96209 – 58809)
= 37400
Thus, the the number of children will be 37400.
Question no – (8)
Solution :
Original no = 39460
After interchanging 6 and 9
= 36490 (new number)
∴ Difference is,
= (39460 – 36490)
= 2790
Therefore, the difference will be 2790.
Question no – (9)
Solution :
In the given question,
population of a town = 59000.
Increased due to new births = 4536
persons died or left the town = 9218
∴ Population at the end of the year,
= 59000 + 4536 – 9218
= 54318
Therefore, the population at the end of the year will be 54318.
Operation on Whole Numbers Exercise 4.3 Solution
Question no – (1)
Solution :
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567
(iii) 475 × 129 = 129 × 475
(iv) 1243 × 8975 × 1243
(v) 10 × 100 × = 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
Question no – (2)
Solution :
(i) 2 × 1497 × 50
= (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
= (4 × 25) × 358
= 100 × 358
= 35800
(iii) 495 × 625 × 16
= (625 × 16) × 495
= 10000 × 495 ‘
= 4950000
(iv) 625 × 20 × 8 × 50
= 625 × 20 × 8 × 50
= (625 × 8) (20 × 500
= 5000000
Question no – (3)
Solution :
(i) 736 × 103
= 736 (100 + 3)
= (736 × 100) + (736 × 3)
= 73600 + (736 × 3)
= 788808
(ii) 258 × 1008
= 258 (1000 + 8)
= (258 × 1000) × (258 × 8)
= 258000 + 2064
= 260064
Question no – (4)
Solution :
(i) 736 × 93
= 736 × 93 = 736 (100 – 7)
= (736 × 100) – (736 × 7)
= 73600 – 5752
= 68448
So, the product will be 68448
(ii) 816 × 745
= 816 × 745 = 816 × (750 – 6)
= (816 × 750) – (816 × 5)
= 612000 – 4080
= 607920
So, the product will be 607920
(iii) 2032 × 613
= 2032 × 613 = 2032 (600 + 13)
= (2032 × 600) + (2032 × 13)
= 1219200 + 26416
= 1245616
So, the product will be 1245616
Question no – (8)
Solution :
In the question,
Dealer purchased = 125 colour television
Cost of each set = Rs 19820
Determine cost of all sets together = ?
∴ Cost of 125 television sets,
= (19820 × 125) Rs
= 2477500 Rs
Therefore, the cost of all sets together will be 2477500 Rs.
Question no – (9)
Solution :
In the given question,
The annual fee of class VI = Rs 8880.
Students in class VI = 235
∴ Annual fee changed for 235 students,
= (8880 × 235)
= 208600 Rs
Therefore, the total collection is 208600 Rs.
Question no – (10)
Solution :
In the question we get,
Constructed flats= 350
Cost of construction for each flat = Rs 993570
∴ Cost of Constriction,
= (993570 × 350) Rs
= 347749500 Rs
Hence, total cost of construction of all the flats will be 347749500 Rs.
Question no – (11)
Solution :
I conclude, The product of two whole numbers is Zero when at least one number or both now are Zero,
Question no – (12)
Solution :
Two numbers when multiplied with gives the same number 0 × 0 = 0, 1 × 1 = 1
Question no – (13)
Solution :
Total apartment in 1 large building,
= 10 × 2
= 20
1 small building, no of apartment,
= 12 × 3
= 36
∴ Total apartment,
= (22 × 20) + (15 × 36)
= 440 + 540
= 980
Hence, there will be total 980 apartments.
Operation on Whole Numbers Exercise 4.4 Solution
Question no – (1)
Solution :
Yes, there exist a whole number we know, the whole number is 1,
Where 1 ÷ 1 = 1.
Question no – (2)
Solution :
(i) 23457 ÷ 1
= 23457
(ii) 0 ÷ 97
= 0
(iii) 476 + (840 ÷ 84)
= 476 + 10
= 486
(iv) 964 – (425 ÷ 425)
= 964 – 1
= 963
(v) (2758 ÷ 2758) – (2758 ÷ 2758)
= 1 – 1
= 0
Question no – (3)
Solution :
(i) 10 ÷ (5×2) = (10÷5) × (10÷2)
= L.H.S 10 ÷ (5×2)
= 10 ÷ 10
= 1
R.H.S (10÷5) × (10÷2)
= 2 × 5
= 10
Therefore, L.H.S ≠ R.H.S
(ii) (53–14) ÷ 7 = 35 ÷ 7 – 14 ÷ 7
= L.H.S (35–14) ÷ 7
= 21 ÷ 7
= 3
R.H.S, 35 ÷ 7 – 14 ÷ 7
= 5 – 2
= 3
Therefore, L.H.S = R.H.S (Proved)
(iii) 35 – 14 ÷ 7 = 35 ÷ 7 – 14 ÷ 7
= L.H.S, 25 – 14 ÷ 7
= 35 – 2
= 33
R.H.S, 35 ÷ 7 – 14 ÷ 7
= 5 ÷ 2
= 3
Therefore, L.H.S ≠ R.H.S
(iv) (20 – 5) ÷ 5 = 20 ÷ 5 – 5
= L.H.S (20–5) ÷ 5
= 15 ÷ 5
= 3
R.H.S, 20 ÷ 5 – 5
= 4 – 5
= – 1
Therefore, L.H.S ≠ R.H.S
(v) 12×(14 ÷ 7) = (12×14) ÷ (12×7)
= L.H.S, 12 (14÷7)
= 12×2
= 24
R.H.S, (12×14) ÷ (12÷7)
= 168 ÷ 84
= 2
Therefore, L.H.S ≠ R.H.S
Question no – (4)
Solution :
(i) Given, 7772 ÷ 58
∴ Quotient = 134,
Remainder = 0
For check, We, know,
Divided = Divisor × Quotient + Remainder.
∴ 7772 = 58 × 134 + 0
= 7772
∴ L.H.S = R.H.S (proved)
(ii) Given, 6906 ÷ 35
Hence,
Quotient = 197
Remainder = 11
(iii) Given, 16135 ÷ 875
Here,
Quotient = 18,
Remainder = 385
Question no – (5)
Solution :
As we know that,
Divided = Divisor × Quotient + Remainder
∴ 35 × 20 + 18
= 700 + 18
= 718
Therefore, the required number is 718.
Question no – (6)
Solution :
As we know,
Divided = Divisor × Quotient + Remainder
Divided = 58 × 40 + 31
= 2320 + 31
= 2351
Therefore, the required number is 2351.
Question no – (7)
Solution :
In the question,
Product of two numbers = 504347
One number = 1591
Other number = ?
∴ Other number is,
= 504347/1591
= 317
Hence, the other number will be 317.
Question no – (8)
Solution :
As we know,
Divided = Divisor × Quotient + Remainder
∴ 59761 = Divisor × 189 + 37
= 59761 – 37 = 189 Divisor
∴ Now the required divisor,
= 59724/189
= 316
Therefore, the Divisor will be 316.
Question no – (9)
Solution :
As we know that,
Divided = Divisor × Quotient + Remainder
∴ 55390 = 299 × Quotient + 75
= 299× Quotient = 55390 – 75
∴ Now the required quotient,
= 55315/299
= 185
Hence, the quotient will be 185.
Operation on Whole Numbers Exercise 4.5 Solution
Question no – (1)
Solution :
(i) 10th square number
= 10th square number (10×10)
= (10)2
= 100
Therefore, 10th square number is 100.
(ii) 6th triangular number,
= 6 × (6 + 1)/2
= 6 × 7/2
= 42/2
= 21
Hence, the 6th triangular number is 21.
Question no – (2)
Solution :
(i) Yes, a rectangular number also be a square number.
(ii) Yes, a triangular number also be a square number.
Question no – (3)
Solution :
Here, we know that,
1 × 5 = 5
2 × 6 = 12
3 × 7 = 21
4 × 8 = 32
First four products of two numbers with difference 4.
= 5 – 1 = 4; 6 – 2 = 4
= 7 – 3 = 4; 8 – 4 = 4
Question no – (4)
Solution :
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 5 = 88888
98765 × 9 + 3 = 8888888
987654 × 9 + 2 = 8888888
9876543 × 9 + 1 = 88888888
Question no – (5)
Solution :
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
10 × 6 – 45 = 15
11 × 7 – 60 = 17
12 × 8 – 77 = 19
Question no – (6)
Solution :
By observing the above pattern,
(i) 1 + 3 + 5 + 7 + 9 + 11
= 6 × 6
= 36
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
= 8 × 8
= 64
(iii) 21 + 23 + 25 + __ + 51
= 262 – 102
= 676 – 100
= 576
Question no – (7)
Solution :
By observing the above pattern,
(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
= (10×11)/2
= 110/2
= 55
(ii) 50 + 51 + 52 + __ + 100
= 100 × 100/2 – 49 × 50/2
= 5050 – 1225
= 3825
(iii) 2 + 4 + 6 + 8+ 10 + __ + 100
= 2 (1 + 2 + 3 + 4 + 5 + ……)
= 2 × (50 ×57/2)
= 2550
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