CLASS 6: CHAPTER 2: WHOLE NUMBERS

Exercise 2.1

1.      Write the next three natural numbers after 10999.

Solution

The natural number after 10,999 can be obtained by adding 1.

10,999 + 1 = 11,000

11,000 + 1 = 11,001

11,001 + 1 = 11,002

The next three natural numbers after 10,999 are 11,000, 11,001 and 11,002

2.      Write the three whole numbers occurring just before 10001.

Solution

The whole number before 10,001 is obtained by subtracting 1.

10,001 – 1 = 10,000

10,000 – 1 = 9,999

9,999 – 1 = 9,998

The three whole numbers occurring just before 10,001 are 10,000, 9,999 and 9,998

3.      Which is the smallest whole number?

Solution

The numbers 0, 1, 2, 3… form the collection of whole numbers.

The smallest whole number is 0.

4.      How many whole numbers are there between 32 and 53?

Solution

Let us write the whole numbers.

32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53

Now count the numbers between 32 and 53:

32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53

There are 20 whole numbers between 32 and 53.

5.      Write the successor of :

           

            Solution

            By adding 1 to a number, we get its successor.

(a)    2440701

Successor: 2440701 + 1 = 24,40,702

(b)   100199

Successor: 100199 + 1 = 1,00,200

(c)    1099999

Successor: 1099999 + 1 = 1,100,000

(d)   2345670

Successor: 2345670 + 1 = 2,345,671

                

6.      Write the predecessor of :

Solution:

By subtracting 1 from a number, we get its predecessor.

(a)    94

Predecessor of 94 is 94 -1 = 93

(b)   10000

Predecessor of 10000 – 1 = 9,999

(c)   208090

Predecessor of 208090 is 208089

(d)   7654321

Predecessor of 7654321 is 7654320

7.      In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.

Solution

Draw a line, mark a point on it and label it 0. Mark out points to the right of 0, at equal intervals. Label them as 1, 2, 3,... Thus, we have a number line with the whole numbers represented on it.

Observe that, out of any two whole numbers, the number on the right of the other number is the greater number. [3 is to the right of 2. So 3 is the greater number and 2 is smaller]

The number line can be used to represent larger numbers.

a)      530, 503

503 is to the left of 530.

So, 503 < 530

b)      370, 307

307 is to the left of 370.

370 > 307

c)      98765, 56789

56789 is to the left of 98765

98765 > 56789

           

d)     9830415, 10023001

98,30,415 is to the left of 1,00,23,001

9830415 < 10023001

8.      Which of the following statements are true (T) and which are false (F) ?

a)      Zero is the smallest natural number.

False, zero is not a natural number. It is a whole number.

b)      400 is the predecessor of 399.

False, 400 is the successor of 399,

c)      Zero is the smallest whole number.

True

d)     600 is the successor of 599.

True

e)      All natural numbers are whole numbers.

True

f)       All whole numbers are natural numbers.

False, every whole number except 0 is a natural number.

g)      The predecessor of a two digit number is never a single digit number.

False, the predecessor of 10 is 9.

h)      1 is the smallest whole number.

False, 0 is the smallest whole number.

i)        The natural number 1 has no predecessor.

True

j)        The whole number 1 has no predecessor.

False, 0 is a predecessor of the whole number 1.

k)      The whole number 13 lies between 11 and 12.

False, 13 is the successor of 12.

l)        The whole number 0 has no predecessor.

True

m)   The successor of a two digit number is always a two digit number.

                        False, it can be a three-digit number.

Exercise 2.2

1. Find the sum by suitable rearrangement:

a)      837 + 208 + 363

Solution

Rearrange the sum 837 + 208 + 363 as (837 + 363) + 208.

(837 + 363) + 208 = 1,200 + 208

= 1408

b)      1962 + 453 + 1538 + 647

Solution

Rearrange the sum 1962 + 453 + 1538 + 647 as (1962 + 1538) + (453 + 647)

(1962 + 1538) + (453 + 647) = 3500 + 1100 = 4,600

           

2. Find the product by suitable rearrangement:

(a)    2 × 1768 × 50

Solution

2 × 1768 × 50 = (2 × 50) × 1768 = 100 × 1768 = 1,76,800

(b)   4 × 166 × 25

Solution

4 × 166 × 25 = (4 × 25) × 166 = 100 × 166 = 16,600

(c)    8 × 291 × 125

Solution

8 × 291 × 125 = (8 × 125) × 291 = 1000 × 291 = 2,91,000

(d)   625 × 279 × 16

Solution

625 × 279 × 16 = (625 × 16) × 279 = 10,000 × 279 = 2,790,000

(e)    285 × 5 × 60

Solution

285 × 5 × 60 = 285 × (5 × 60) = 285 × 300 = 85,500

(f)    125 × 40 × 8 × 25

Solution

125 × 40 × 8 × 25 = (125 × 40) × (8 × 25) = 5,000 × 200 = 10,00,000

3. Find the value of the following:

(a)    297 × 17 + 297 × 3

Solution

297 × 17 + 297 × 3 = 297 × (17 + 3)

                                = 297 × (20)

                                = 5,940

(b)   54279 × 92 + 8 × 54279

Solution

54279 × 92 + 8 × 54279 = 54279 × [92 + 8]

                = 54279 × [100]

                = 54,27,900

(c)    81265 × 169 – 81265 × 69

Solution

81265 × 169 – 81265 × 69 = 81265 × [169 – 69]

                                            = 81265 × [100]

                                            = 81,26,500

(d)   3845 × 5 × 782 + 769 × 25 × 218

Solution

3845 × 5 × 782 + 769 × 25 × 218 = (3845 × 5) × 782 + (769 × 25) × 218

      = (19225) × 782 + (19225) × 218

      = 19225 × (782 + 218)

      = 19225 × (1000)

      = 1,92,25,000

4. Find the product using suitable properties.

(a)    738 × 103

Solution

738 × 103 = 738 × (100 + 3)

= 738 × 100 + 738 × 3 [Distributivity of multiplication over addition]

= 73800 + 2214

= 76,014

(b)   854 × 102

Solution

854 × 102 = 854 × (100 + 2)

= 854 × 100 + 854 × 2 [Distributivity of multiplication over addition]

= 85400 + 1708

= 87,108

(c)    258 × 1008

Solution

258 × 1008 = 258 × (1000 + 8)

= 258 × 1000 + 258 × 8

= 258000 + 2064

= 2,60,064

(d)   1005 × 168

Solution

1005 × 168 = (1000 + 5) × 168

= 1000 × 168 + 5 × 168

= 168000 + 840

= 1,68,840

5. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs Rs 44 per litre, how much did he spend in all on petrol?

     

Solution

Method 1: find the amount spent for each day and then add.

Amount spent for petrol on Monday = 40 × 44 = Rs. 1760

Amount spent for petrol on Tuesday = 50 × 44 = Rs. 2200

Total amount spent in all = Rs. 1760 + Rs.2200  =  Rs. 3960

Method 2: Find the total litres of petrol filled for both days and multiply by the cost per litre.

Total petrol filled for both days = 40 + 50 = 90 litres

Cost per litre of petrol = Rs. 44

Amount spent in all for petrol = 44 × 90 = Rs. 3960

6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If  the milk costs Rs 15 per litre, how much money is due to the vendor per day?

      Solution

      Cost of milk supplied in the morning = 32 × 15

      Cost of milk supplied in the evening = 68 × 15

      Total cost of milk supplied for the day = 32 × 15 + 68 × 15

                                                             = (32 + 68) × 15

                                                             = (100) × 15 = Rs. 1500

      Amount due per day is Rs. 1500

7. Match the following:

(i) 425 × 136 = 425 × (6 + 30 +100)	(a) Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49	(b) Commutativity under addition.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005	(c) Distributivity of multiplication over addition.

Solution:

        i.            425 × 136 = 425 × (6 + 30 +100)

We see that the number 136 is rewritten as 6 + 30 + 100

Distributivity of multiplication over addition (c)

      ii.            2 × 49 × 50 = 2 × 50 × 49

We see that the numbers are rearranged under the multiplication operation.

Commutativity under multiplication (a)

    iii.            80 + 2005 + 20 = 80 + 20 + 2005

The numbers are rearranged under the addition operation.

Commutativity under addition (b)

Exercise 2.3

1.      Which of the following will not represent zero:

a)      1 + 0

b)      0 × 0

c)     

d)    

Solution

Choice a does not represent 0, as 1 + 0 is 1.

Choice b is 0, as any number multiplied by 0 becomes zero.

Choice c is 0, as 0 cannot be divided.

Choice d is 0, as 10 – 10 is 0 and 0 cannot be divided.

2.      If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

Solution

Yes, if the product of two numbers is 0, then one or both of them will be zero.

Let us consider two numbers, for example 12 and 15.

12 × 15 = 180

The numbers are non-zero and so is the product.

Similarly, if we consider any two non-zero numbers, the product is a non-zero number.

Let us now consider numbers 12 and 0 OR 0 and 135.

We know that any number multiplied by 0 becomes 0, and hence the product of both 12 and 0 AND 0 and 135 are 0.

Also 0 × 0 = 0.

Hence, if the product of any two whole numbers is zero, then one or both of them will be zero.

3.      If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

Solution

If the product of two whole numbers is 1, then both the numbers should be 1.

Let us consider two numbers, for example 20 and 1.

20 × 1 = 20

We see that the product is not 1.

But, 1 × 1 = 1.

Hence, if the product of any two whole numbers is 1, then both of them should be 1.

4.      Find using distributive property :

a)      728 × 101

Solution

728 × 101 = 728 × (100 + 1)

                 = 728 × 100 + 728

                 = 72800 + 728  

                 = 73,528

b)      5437 × 1001

Solution

5437 × 1001 = 5437 × (1000 + 1)

                     = 5437 × 1000 + 5437 × 1

                     = 5437000 + 5437

                     = 54,42,437

c)      824 × 25

Solution

824 × 25 = 824 × (20 + 5)

               = 824 × 20 + 824 × 5

               = 16480 + 4120

               = 20,600

d)     4275 × 125

Solution

4275 × 125 = 4275 × (100 + 20 + 5)

                   = 4275 × 100 + 4275 × 20 + 4275 × 5

                   = 427500 + 85,500 + 21375

                   = 5,34,375

e)      504 × 35

Solution

504 × 35 = (500 + 4) × 35

   = 500 × 35 + 4 × 35

   = 17500 + 140

   = 17,640

5.      Study the pattern :

1 × 8 + 1 = 9

12 × 8 + 2 = 98

123 × 8 + 3 = 987

1234 × 8 + 4 = 9876

12345 × 8 + 5 = 98765

Write the next two steps. Can you say how the pattern works?

(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

Solution
Let us see how the pattern works.

1 × 8 + 1 = 9

12 × 8 + 2 = 98
i.e. (11 + 1) × 8 + 2 = 11 × 8 + 1 × 8 + 2 = 88 + 8 + 2 = 98

123 × 8 + 3 = 987 ; 

i.e. (111 + 11 + 1) × 8 + 3 = 111 × 8 + 11 × 8 + 1 × 8 + 2 = 888 + 88 + 8 + 3 = 987

1234 × 8 + 4 = 9876 ;
i.e. (1111 + 111 + 11 + 1) × 8 + 4 = 1111 × 8 + 111 × 8 + 11× 8 + 1 × 8 + 4
                                                 = 8888 + 888 + 88 + 8 + 4 = 9876

12345 × 8 + 5 = 98765
i.e. (11111 + 1111 + 111 + 11 + 1) × 8 + 5 = 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 5
                                                               = 88888 + 8888 + 888 + 88 + 8 + 5 = 98765

Extending the pattern, we have

123456 × 8 + 6 =
(111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 5
                                     = 111111 × 8 + 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 5
                                     = 888888 + 88888 + 8888 + 888 + 88 + 8 + 6 = 987654

           1234567 × 8 + 7 =
                (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 7
                                                          = 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 7
                                                          = 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8 + 7 = 9876543