Whole numbers are numbers that do not contain a decimal. In addition to this, it does not contain any negative numbers. In this chapter, the students will learn about the properties of whole numbers like whole number multiplication, addition, subtraction, and division. To practice questions apart from NCERT textbook students can refer to RS Aggarwal Class 6 solutions Chapter 3 whole numbers.
RS Aggarwal is one of the best reference material for Class 6 students if they want to master in mathematics. In our BYJUâ€™S website, we have provided Chapter 3 RS Aggarwal solutions to all the students so that they can use it as a ready reference and prepare for their exams. The solutions are solved in a proper step by step manner for all the questions mentioned in the exercises. Solving RS Aggarwal Maths solution will increase the confidence level of the students.
Download PDF of RS Aggarwal Class 6 Chapter 3- Whole Numbers
All the exercise questions with solutions in Chapter 3 – Whole Numbers is given below:
Q1: Write the next three whole numbers after 30999.
Sol:
The next three whole numbers after 30999 are 31000, 31001 and 31002
Q2: Write the three whole numbers occurring just before 10001.
Sol:
The whole number occurring just before 10001 are as follows:
10001 – 1 = 10000
10000 – 1 = 9999
9999 – 1 = 9998
Therefore, The three whole numbers just before 10001 are 10000, 9999, 9998.
Q3: How many whole numbers are there between 1032 and 1209?
Sol:
Number of whole numbers between 1032 and 1209 are
(1209 – 1032) – 1
= 177 – 1
= 176
Q4: Which is the smallest whole number?
Sol:
Zero (0) is the smallest whole number.
All the natural numbers along with 0 are called whole numbers.
Q5: Write the successor of:
(i) 2540801
(ii) 9999
(iii) 50904
(iv) 61639
(v) 687890
(vi) 5386700
(vii) 6475999
(viii) 9999999
Sol:
(i) Successor of 2540801 = 250801 + 1 = 2540802
(ii) Successor of 9999 = 9999 + 1 = 10000
(iii) Successor of 50904 = 50904 + 1 = 50905
(iv) Successor of 61639 = 61639 + 1 = 61640
(v) Successor of 687890 = 5386700 + 1 = 687891
(vi) Successor of 5386700 = 5386700 + 1 = 5386701
(vii) Successor of 6475999 = 6475999 + 1 = 6476000
(viii) Successor of 9999999 = 9999999 + 1 = 10000000
Q6: Write the predecessor of :
(i) 97
(ii) 10000
(iii) 36900
(iv) 7684320
(v) 1566391
(vi) 2456800
(vii) 100000
(viii) 1000000
Sol:
(i) Predecessor of 97 = 97 – 1 = 96
(ii) Predecessor of 10000 = 10000 – 1 = 9999
(iii) Predecessor of 36900 = 36900 – 1 = 36899
(iv) Predecessor of 7684320 = 7684320 – 1 = 7684319
(v) Predecessor of 1566391 = 1566391 – 1 = 1566390
(vi) Predecessor of 2456800 = 2456800 – 1 = 2456799
(vii) Predecessor of 100000 = 100000 – 1 = 99999
(viii) Predecessor of 1000000 = 1000000 – 1 = 999999
Q7: Write down three consecutive whole numbers just preceeding 7510001.
Sol:
The three consecutive numbers just preceding 751001 are as follows:
751001 – 1 = 751000
751000 – 1 = 750999
750999 – 1 = 750998
Therefore, The three consecutive numbers just preceding 7510001 are 7510000, 7509999, 7509998.
Q8: Write (T) for true and (F) for false against each of the following statements:
(i) Zero is the smallest natural number.
(ii) Zero is the smallest whole number.
(iii) Every whole number is a natural number.
(iv) Every natural number is a whole number.
(v) 1 is the smallest whole number.
(vi) The natural number 1 has no predecessor.
(vii) The whole number 1 has no predecessor.
(viii) The whole number 0 has no predecessor.
(ix) The predecessor of a two digit number is never a single digit number.
(x) The successor of a two digit number is always a two digit number.
(xi) 500 is the predecessor of 499.
(xii) 7000 is the successor of 6999.
Sol:
(i) False. 0 is not a natural number. 1 is the smallest natural number.
(ii) True
(iii) False. 0 is a whole number but not a natural number.
(iv) True. Natural numbers include 1,2,3,…. Which are whole numbers.
(v) False. 0 is the smallest whole number.
(vi) True, The predecessor of 1 is 1 – 1 = 0, Â which is not a natural number.
(vii) False. The predecessor of 1 is 1 – 1 = 0, Â which is a whole number.
(viii) True. The predecessor of 0 is 0 – 1 = -1, which is not a whole number.
(ix) False. The predecessor of a two digit number can be a single digit number. For example, the predecessor of 10 is 10 – 1 = 9
(x) False. The successor of a two digit number is not always a two digit number. For example, the successor of 99 is 99 + 1 = 100
(xi) False. The predecessor of 499 is 499 – 1 = 498
(xii) True. The successor of 6999 is 6999 + 1 = 7000.
Exercise 3B
Q1: Fill in the blanks to make each of the following a true statement:
(i) 458 + 639 = 639 + ______
(ii) 864 + 2006 = 2006 + ______
(iii) 1946 + _____ = 984 + 1946
(iv) 8063 + 0 = 0 + _____
(v) 53501 + (574 + 799) = 574 + (54501 + _____)
Sol:
(i) 458 + 639 = 639 + 458
(ii) 864 + 2006 = 2006 + 864
(iii) 1946 + 984 = 984 + 1946
(iv) 8063 + 0 = 0 + 8063
(v) 53501 + (574 + 799) = 574 + (54501 + 799)
Q2: Add the following number and check by reversing the order of the addends:
(i) 16509 + 114
(ii) 2359 + 548
(iii) 19753 + 2867
Sol:
(i) 16509 + 114 = 16623
By reversing the order of the addends,we get:
114 + 16509 = 16623
Therefore, 16509 + 114 = 114 + 16509
(ii) 2359 + 548 = 2907
By reversing the order of the addends,we get:
548 + 2359 = 2907
Therefore, 2359 + 548 = 548 + 2359
(iii) 19753 + 2867 = 22620
By reversing the order of the addends,we get:
2867 + 19753 = 22620
Therefore, 19753 + 2867 = 2867 + 19753
Q3: Find the sum: (1546 + 498) + 3589
Also, find the sum:1546 + (498 + 3589)
Are they equal ?
State the property satisfied.
Sol:
We have :
(1546 + 498) + 3589 = 2044 + 3589 = 5633
Also, 1546 + (498 + 3589) = 1546 + 4087 = 5633
Yes, the two sums are equal.
The associative property of addition is satisfied.
Q4: Determine each of the sums given below using suitable rearrangement.
(i) 953 + 707 + 647
(ii) 1983 + 647 + 217 + 353
(iii) 15409 + 278 + 691 + 422
(iv) 3259 + 1001 + 2641 + 9999
(v) 1 + 2 + 3 + 4 + 96 + 97 + 98 + 99
(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48
Sol:
(i) 953 + 707 + 647
953 + (707 + 647) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( Using associative property of addition )
= 953 + 1354
= 2307
(ii) 1983 + 647 + 217 + 353
(1983 + 647) + (217 + 353) Â Â Â Â Â Â Â Â Â ( Using associative property of addition )
= 2630 + 570
= 3200
(iii) 15409 + 278 + 691 + 422
(15409 + 278) + (691 + 422) Â Â Â Â Â Â Â ( Using associative property of addition )
= 15689 + 1113
= 16800
(iv) 3259 + 1001 + 2641 + 9999
(3259 + 1001) + (2641 + 9999) Â Â Â Â Â ( Using associative property of addition )
= 13260 + 12640
= 25900
(v) 1 + 2 + 3 + 4 + 96 + 97 + 98 + 99
(1 + 2 + 3 + 4) + (96 + 97 + 98 + 99) Â Â Â Â Â ( Using associative property of addition )
= 10 + 390
= 400
(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48
(2 + 3 + 4 + 5) + (45 + 46 + 47 + 48) Â Â Â Â Â ( Using associative property of addition )
= 14 + 186
= 200
Q5 : Find the sum by short method:
(i) 6784 + 9999
(ii) 10578 + 99999
Sol:
(i) 6784 + 9999
= 6784 + (10000 – 1)
= (6784 + 10000) – 1 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( Using associative property of addition )
= 16784 – 1
= 16783
(ii) 10578 + 99999
= 10578 + (100000 – 1)
= (10587 + 100000) – 1 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( Using associative property of addition )
= 110578 – 1
= 110577
Q6: For any whole number a, b, c is it true that (a + b) + c = a + (b + c)? Give reasons.
Sol:
For any whole numbers a,b and c we have:
(a + b) + c = a + (b + c)
Let a = 5, b = 1 and c = 9 Â Â [ We can take any values for a, b and c ]
LHS = (a + b) + c
= (5 + 1) + 9
= (6) + 9
= 15
RHS = a + (b + c)
= 5 + (1 + 9)
= 5 + 10
= 15
Therefore, This shows that associativity in addition is one of the properties of whole numbers.
Q7: Complete each of the following magic squares by supplying the missing numbers:
(i)
9 | ||
5 | 7 | |
8 |
(ii)
16 | 9 | |
5 | ||
6 |
(iii)
2 | 15 | 16 | |
9 | 12 | ||
8 | 7 | 10 | |
14 | 17 |
(iv)
18 | 17 | 4 | |
14 | 11 | ||
9 | 10 | ||
19 | 16 |
Sol:
In a magic square, the sum of each row is equal to the sum of each column and the sum of each main diagonals. By using this concept we have:
(i)
4 | 9 | 2 |
3 | 5 | 7 |
8 | 1 | 6 |
(ii)
16 | 9 | 2 |
3 | 5 | 7 |
8 | 1 | 6 |
(iii)
2 | 15 | 16 | 5 |
9 | 12 | 11 | 6 |
13 | 8 | 7 | 10 |
14 | 3 | 4 | 17 |
(iv)
7 | 18 | 17 | 4 |
8 | 13 | 14 | 11 |
12 | 9 | 10 | 15 |
19 | 6 | 5 | 16 |
Q8: Write (T) for true and (F) for false against each of the following statements:
(i) The sum of two odd numbers is an odd number.
(ii) The sum of two even number is an even number.
(iii) The sum of an even number and an odd number is an odd number.
Sol:
(i) F (false). The sum of two odd numbers may not be an odd number. For example: 5 + 11 = 16, which is an even number.
(ii) T (true). The sum of two even numbers is an even number. Example: 4+ 8 = 12, which is an even number.
(iii) T (true). The sum of an even and odd number is an odd number. Example: 7 + 10 = 17, which is an odd number.
RS Aggarwal Class 6 Solutions Chapter 3 – Whole Numbers
Solving questions from RS Aggarwal Class 6 solutions Chapter 3 – Whole Numbers will strengthen your knowledge of the concepts and also increase your speed. Practicing these solutions will make you feel confident while attempting the final question paper. Maths is a subject that needs lots of practice and these RS Aggarwal solutions will help you to do so as it provides different types of questions from each topic.