Class 6 - Maths - Whole Numbers

Exercise 2.1

Question 1.

Write the next three natural numbers after 10999.

Answer:

Given number is 10999

Now, 10,999 + 1 = 11,000

11,000 + 1 = 11,001

11,001 + 1 = 11,002

So, the next three natural numbers after 10999 are 11,000, 11,001 and 11,002

Question 2.

Write the three whole numbers occurring just before 10001.

Answer:

Given number is 10001

Now, 10,001 – 1 = 10,000

10,000 – 1 = 9,999

9,999 – 1 = 9,998

So, the three whole numbers occurring just before 10001 are 10,000, 9,999 and 9,998

Question 3.

Which is the smallest whole number?

Answer:

All counting numbers (1, 2, 3… etc.) and 0 form the set of whole numbers.

Thus, 0, 1, 2, 3… etc. are whole numbers.

So, 0 (zero) is the smallest whole number.

Question 4.

How many whole numbers are there between 32 and 53?

Answer:

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

So, there are 20 whole numbers between 32 and 53.

Question 5.

Write the successor of:

(a) 2440701

(b) 100199

(c) 1099999

(d) 2345670

Answer:

The successor of a number is the number obtained by adding 1 to it.

(a) Successor of 2440701 is 2440701 + 1 = 2440702

(b) Successor of 100199 is 100199 + 1 = 100200

(c) Successor of 1099999 is 1099999 + 1 = 1100000

(d) Successor of 2345670 is 2345670 + 1 = 2345671

Question 6.

Write the predecessor of:

(a) 94

(b) 10000

(c) 208090

(d) 7654321

Answer:

The predecessor of a whole number is 1 less than the given number.

(a) The predecessor of 94 is 94 – 1 = 93

(b) The predecessor of 10000 is 10000 – 1 = 9999

(c) The predecessor of 208090 is 208090 – 1 = 208089

(d)  The predecessor of 7654321 is 7654321 – 1 = 7654320

Question 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.

(a) 530, 503

(b)370, 307

(c)98765, 56789

(d)9830415, 10023001

Answer:

(a) Since 530 > 503;

So, 503 appear on left side of 530 on number line.

(b) 370 > 307;

So, 307 appear on left side of 370 on number line.

(c) 98765 > 56789;

So, 56789 appear on left side of 98765 on number line.

(d) 9830415 < 10023001;

So 9830415 appear on left side of 10023001 on number line.

Question 8.

Which of the following statements are true (T) and which are false (F):

(a) Zero is the smallest natural number.

(b) 400 is the predecessor of 399.

(c) Zero is the smallest whole number.

(d) 600 is the successor of 599.

(e) All natural numbers are whole numbers.

(f) All whole numbers are natural numbers.

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

(h) 1 is the smallest whole number.

(i) The natural number 1 has no predecessor.

(j) The whole number 1 has no predecessor.

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

(l) The whole number 0 has no predecessor.

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

Answer:

(a) False, 0 is not a natural number.

(b) False, as predecessor of 399 is 398 (399 – 1 = 398)

(c)True, 0 is the smallest whole number.

(d) True, as 599 + 1 = 600

(e) True, all natural numbers are whole numbers.

(f) False, as 0 is a whole number but not a natural number.

(g) False, as predecessor of 10 is 9 (10 – 1 = 9)

(h) False, 0 is the smallest whole number.

(i)True, as 0 is the predecessor of 1 but it is not a natural number.

(j) False, as 0 is the predecessor of 1 but it is a whole number.]

(k) False, 13 do not lie between 11 and 12

(l) True, the predecessor of 0 is -1, which is not a whole number.

(m) False, as successor of 99 is 100 (99 + 1 = 100 which is a 3 digit number)

Exercise 2.2

Question 1.

Find the sum by suitable rearrangement:

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

Answer:

(a) 837 + 208 + 363 = (837 + 363) + 208 = 1200 + 208 = 1408

(b) 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600

Question 2.

Find the product by suitable arrangement:

(a) 2 * 1768 * 50

(b) 4 * 166 * 25

(c) 8 * 291 * 125

(d) 625 * 279 * 16

(e) 285 * 5 * 60

(f) 125 * 4 * 8 * 25

Answer:

(a) 2 * 1768 * 50 = (2 * 50) * 1768 = 100 * 1768 = 176800

(b) 4 * 166 * 25 =  = (4 * 25) * 166 = 100 * 166 = 16600

(c) 8 * 291 * 125 = (8 * 125) * 291 = 1000 * 291 = 291000

(d) 625 * 279 * 16 = (625 * 16) * 279 = 10000 * 279 = 2790000

(e) 285 * 5 * 60 = 285 * (5 * 60) = 285 * 300 = 85500

(f) 125 * 40 * 8 * 25 = (125 * 8) * (40 * 25) = 1000 * 1000 = 1000000

Question 3.

Find the value of the following:

(a) 297 * 17 + 297 * 3

(b) 54279 * 92 + 8 * 54279

(c) 81265 * 169 – 81265 * 69

(d) 3845 * 5 * 782 + 769 * 25 * 218

Answer:

(a) 297 * 17 + 297 * 3 = 297 * (17 + 3) = 297 * 20 = 5940

(b) 54279 * 92 + 8 * 54279 = 54279 * (92 + 8) = 54279 * 100 = 5427900

(c) 81265 * 169 – 81265 * 69 = 81265 * (169 – 69) = 81265 * 100 = 8126500

(d) 3845 * 5 * 782 + 769 * 25 * 218 = 3845 * 5 * 782 + 769 * 5 * 5 * 218

= 3845 * 5 * 782 + 3845 * 5 * 218

= 3845 * 5 * (782 + 218)

= 3845 * 5 * 1000

= 19225 * 1000

= 19225000

Question 4.

Find the product using suitable properties:

(a) 738 * 103

(b) 854 * 102

(c) 258 * 1008

(d) 1005 * 168

Answer:

(a) 738 * 103 = 738 * (100 + 3) = 738 * 100 + 738 * 3 = 73800 + 2214 = 76014

(b) 854 * 102 = 854 x (100 + 2) = 854 * 100 + 854 * 2 = 85400 + 1708 = 87108

(c) 258 * 1008 = 258 x (1000 + 8) = 258 * 1000 + 258 * 8 = 258000 + 2064 = 260064

(d) 1005 * 168 = (1000 + 8) * 168 = 1000 * 168 + 8 * 168 = 168000 + 840 = 168840

Question 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?

Answer:

Petrol filled on Monday = 40 litres

Petrol filled on next day = 50 litres

So, total petrol filled = 40 + 50 = 90 litres

Now, cost of 1 litre petrol = Rs 44

So, cost of 90 litres petrol = 44 * 90

= 44 * (100 – 10)

= 44 * 100 – 44 * 10

= 4400 – 440

= Rs 3960

Therefore, the taxi driver spent Rs 3960 on petrol.

Question 6.

A vendor supplies 32 litres of milk to a hotel in a 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?

Answer :

Supply of milk in morning = 32 litres

Supply of milk in evening = 68 litres

Total supply of milk = 32 + 68 = 100 litres

Now, the cost of 1 litre milk = Rs 15

So, the cost of 100 litres milk = 15 * 100 = Rs1500

Therefore, Rs 1500 is due to the vendor per day.

Question 7:

Match the following:

(i) 425 x 136 = 425 x (6 + 30 + 100)      (a)Commutativity under multiplication

(ii) 2 x 48 x 50 = 2 x 50 x 49                    (b) Commutativity under addition

(iii) 80 + 2005 + 20 = 80 + 20 + 2005    (c) Distributivity multiplication under addition

Answer:

We can add two or more whole numbers in any order and this property is known as

Commutativity under addition.

Example:

2 + 3 + 5 = 2 + 5 + 3 = 3 + 5 + 2

We can multiply two or more whole numbers in any order and this property is known as

Commutativity under multiplication.

Example:

2 * 3 * 5 = 2 * 5 * 3 = 3 * 5 * 2

Multiplication is distributive over addition for whole numbers.

Example: 4 * 36 = 4  * (6 + 30)

(i) 425 x 136 = 425 x (6 + 30 + 100)      (c) Distributivity multiplication under addition

(ii) 2 x 48 x 50 = 2 x 50 x 49                    (a) Commutativity under multiplication

(iii) 80 + 2005 + 20 = 80 + 20 + 2005    (b) Commutativity under addition

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