## Table of Contents

## Exercise 2.1

### Q1. Write the next three natural numbers after 10999.

**Ans.**

The next three natural numbers after 10999 are:

- 11000
- 11001
- 11002

### Q2.Write the three whole numbers occurring just before 10001.

**Ans.**

The three whole numbers occurring just before 10001 are:

- 10000
- 9999
- 9998

### Q3. Which is the smallest whole number?

**Ans.** The smallest whole number is 0.

### Q4. How many whole numbers are there between 32 and 53?

**Ans.** There are 20 whole numbers between 32 and 53.

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

### Q5. Write the successor of :

(a) 2440701

(b) 100199

(c) 1099999

(d) 2345670

**Ans.**

The successors of the given numbers are:

(a) 2440701 + 1 = 2440702

(b) 100199 + 1 = 100200

(c) 1099999 + 1 = 1100000

(d) 2345670 + 1 = 2345671

### Q6. Write the predecessor of :

(a) 94

(b) 10000

(c) 208090

(d) 7654321

**Ans.**

The predecessors of the given numbers are:

(a) 94 – 1 = 93

(b) 10000 – 1 = 9999

(c) 208090 – 1 = 208089

(d) 7654321 – 1 = 7654320

### Q7. 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

**Ans.**

The pairs of numbers with the appropriate signs are:

(a) 503 < 530

(b) 307 < 370

(c) 56789 < 98765

(d) 9830415 < 10023001

### Q8. 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.

**Ans.**

The statements can be categorized as true (T) or false (F) as follows:

(a) F – Zero is not considered a natural number.

(b) F – 400 is not the predecessor of 399. The predecessor of 399 is 398.

(c) T – Zero is the smallest whole number.

(d) F – 600 is not the successor of 599. The successor of 599 is 600.

(e) T – All natural numbers are also whole numbers.

(f) T – All whole numbers are also natural numbers.

(g) F – The predecessor of a two-digit number can be a single-digit number, such as 9 as the predecessor of 10.

(h) T – 1 is the smallest whole number.

(i) T – The natural number 1 has no predecessor.

(j) F – The whole number 1 has no predecessor, but it does have a successor, which is 2.

(k) F – The whole number 13 does not lie between 11 and 12.

(l) T – The whole number 0 has no predecessor.

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

Therefore, the true statements are: (c), (e), (f), (h), (i), (l), and (m).

## Exercise 2.2

### Q1. Find the sum by suitable rearrangement:

(a) 837 + 208 + 363

(b) 1962 + 453 + 1538 + 647

**Ans.**

(a) 837 + 208 + 363:

Rearranging the numbers, we can add them in any order.

Let’s add 208 and 363 first:

208 + 363 = 571

Now, we add 571 and 837:

571 + 837 = 1408

Therefore, the sum is 1408.

(b) 1962 + 453 + 1538 + 647:

Again, we can rearrange the numbers and add them in any order.

Let’s add 647 and 1962 first:

647 + 1962 = 2609

Now, we add 2609 and 453:

2609 + 453 = 3062

Finally, we add 3062 and 1538:

3062 + 1538 = 4600

Therefore, the sum is 4600.

### Q2. Find the product by suitable rearrangement:

(a) 2 × 1768 × 50

(b) 4 × 166 × 25

(c) 8 × 291 × 125

(d) 625 × 279 × 16

(e) 285 × 5 × 60

(f) 125 × 40 × 8 × 25

**Ans.**

Find the product by suitable rearrangement:

(a) 2 × 1768 × 50:

Rearranging the numbers, we can multiply them in any order.

Let’s multiply 1768 and 50 first:

1768 × 50 = 88400

Now, we multiply 88400 by 2:

88400 × 2 = 176800

Therefore, the product is 176800.

(b) 4 × 166 × 25:

We can multiply these numbers in any order.

Let’s multiply 166 and 25 first:

166 × 25 = 4150

Now, we multiply 4150 by 4:

4150 × 4 = 16600

Therefore, the product is 16600.

(c) 8 × 291 × 125:

Rearranging the numbers, we can multiply them in any order.

Let’s multiply 291 and 125 first:

291 × 125 = 36375

Now, we multiply 36375 by 8:

36375 × 8 = 291000

Therefore, the product is 291000.

(d) 625 × 279 × 16:

We can multiply these numbers in any order.

Let’s multiply 279 and 16 first:

279 × 16 = 4464

Now, we multiply 4464 by 625:

4464 × 625 = 2790000

Therefore, the product is 2790000.

(e) 285 × 5 × 60:

Rearranging the numbers, we can multiply them in any order.

Let’s multiply 5 and 60 first:

5 × 60 = 300

Now, we multiply 300 by 285:

300 × 285 = 85500

Therefore, the product is 85500.

(f) 125 × 40 × 8 × 25:

We can multiply these numbers in any order.

Let’s multiply 40 and 8 first:

40 × 8 = 320

Now, we multiply 320 by 25:

320 × 25 = 8000

Finally, we multiply 8000 by 125:

8000 × 125 = 1000000

Therefore, the product is 1000000.

### Q3. 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

**Ans.**

Find the value of the following:

(a) 297 × 17 + 297 × 3:

We can factor out 297 from both terms:

297 × 17 + 297 × 3 = 297 × (17 + 3)

Now, we can simplify the expression inside the parentheses: 17 + 3 = 20

Therefore, the value is 297 × 20 = 5940.

(b) 54279 × 92 + 8 × 54279:

We can factor out 54279 from both terms:

54279 × 92 + 8 × 54279 = 54279 × (92 + 8)

Now, we can simplify the expression inside the parentheses:

92 + 8 = 100

Therefore, the value is 54279 × 100 = 5427900.

(c) 81265 × 169 – 81265 × 69:

We can factor out 81265 from both terms:

81265 × 169 – 81265 × 69 = 81265 × (169 – 69)

Now, we can simplify the expression inside the parentheses: 169 – 69 = 100

Therefore, the value is 81265 × 100 = 8126500.

(d) 3845 × 5 × 782 + 769 × 25 × 218:

We can multiply each term separately and then add them:

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

Now, we can simplify each multiplication:

3845 × 5 × 782 = 9576250

769 × 25 × 218 = 4196570

Therefore, the value is 9576250 + 4196570 = 13762820.

### Q4. Find the product using suitable properties.

(a) 738 × 103

(b) 854 × 102

(c) 258 × 1008

(d) 1005 × 168

**Ans.**

Find the product using suitable properties:

(a) 738 × 103:

We can multiply the hundreds separately and then combine the results:

738 × 103 = (700 × 100) + (30 × 100) + (8 × 100) + (700 × 3) + (30 × 3) + (8 × 3)

= 70000 + 3000 + 800 + 2100 + 90 + 24

= 76314

(b) 854 × 102:

We can multiply the tens separately and then combine the results:

854 × 102 = (800 × 100) + (50 × 100) + (4 × 100) + (800 × 2) + (50 × 2) + (4 × 2)

= 80000 + 5000 + 400 + 1600 + 100 + 8

= 87108

(c) 258 × 1008:

We can multiply the thousands separately and then combine the results:

258 × 1008 = (200 × 1000) + (50 × 1000) + (8 × 1000) + (200 × 8) + (50 × 8) + (8 × 8)

= 200000 + 50000 + 8000 + 1600 + 400 + 64

= 257064

(d) 1005 × 168:

We can multiply the hundreds separately and then combine the results:

1005 × 168 = (1000 × 100) + (5 × 100) + (1000 × 60) + (5 × 60) + (1000 × 8) + (5 × 8)

= 100000 + 500 + 60000 + 30 + 8000 + 40

= 166570

### Q5. 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 44 per litre, how much did he spend in all on petrol?

**Ans.** To find out how much the taxi driver spent on petrol in total, we need to calculate the cost of the petrol for each day and then add them together.

On Monday, the taxi driver filled the tank with 40 litres of petrol. Since petrol costs 44 per litre, the cost of petrol on Monday would be 40 × 44 = 1760.

The next day, the taxi driver filled the tank with 50 litres of petrol. Using the same cost per litre (44), the cost of petrol on the next day would be 50 × 44 = 2200.

To find the total cost, we add the costs of petrol on both days: Total cost = 1760 + 2200 = 3960.

Therefore, the taxi driver spent 3960 in total on petrol.

### Q6. 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 45 per litre, how much money is due to the vendor per day?

**Ans.** To calculate the amount of money due to the vendor per day, we need to determine the total amount of milk supplied and then multiply it by the cost per litre.

In the morning, the vendor supplies 32 litres of milk to the hotel. Since the cost of milk is 45 per litre, the amount due for the morning supply is 32 × 45 = 1440.

In the evening, the vendor supplies 68 litres of milk. Again, using the cost per litre of 45, the amount due for the evening supply is 68 × 45 = 3060.

To find the total amount due per day, we add the amounts due for the morning and evening supplies: Total amount due = 1440 + 3060 = 4500.

Therefore, the vendor is due 4500 per day for the milk supplies to the hotel.

### Q7. 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. |

**Ans.**

(i) 425 × 136 = 425 × (6 + 30 + 100) | (c) Distributivity of multiplication over addition. |

(ii) 2 × 49 × 50 = 2 × 50 × 49 | (a) Commutativity under multiplication. |

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

Explanation:

- The distributivity of multiplication over addition states that the product of a number multiplied by the sum of two or more numbers is equal to the sum of the products of the number multiplied by each individual number. In this case, 425 × 136 can be calculated by breaking down 136 as (6 + 30 + 100) and applying distributivity: 425 × (6 + 30 + 100) = (425 × 6) + (425 × 30) + (425 × 100).
- Commutativity under multiplication states that changing the order of the numbers being multiplied does not affect the result. In the example, 2 × 49 × 50 and 2 × 50 × 49 are equal because the order of multiplication does not matter.
- Commutativity under addition states that changing the order of the numbers being added does not affect the result. In the example, 80 + 2005 + 20 and 80 + 20 + 2005 are equal because the order of addition does not matter.

## Exercise 2.3

### Q1. Which of the following will not represent zero:

(a) 1 + 0

(b) 0 × 0

(c) 0 / 2

(d) 10 – 10 / 2

**Ans.**

(a) 1 + 0 represents the addition of 1 and 0, which equals 1.

(b) 0 × 0 represents the multiplication of 0 and 0, which equals 0.

(c) 0 / 2 represents the division of 0 by 2, which equals 0.

(d) (10 – 10) / 2 represents the subtraction of 10 and 10, which equals 0, and then dividing it by 2, resulting in 0 / 2 = 0. Therefore, this expression does represent zero.

### Q2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

**Ans.**

Yes, if the product of two whole numbers is zero, it implies that one or both of the numbers will be zero. This can be justified through examples.

Let’s consider two whole numbers, a and b, such that their product is zero.

**Example 1**: Suppose a = 0 and b = 5. In this case, a is zero, and when we multiply it by 5, the product is zero.

**Example 2**: Suppose a = 3 and b = 0. In this case, b is zero, and when we multiply it by 3, the product is zero.

**Example 3**: Suppose a = 0 and b = 0. In this case, both a and b are zero, and when we multiply them together, the product is zero.

From these examples, we can see that whenever the product of two whole numbers is zero, at least one of the numbers has to be zero. It is not necessary for both numbers to be zero, but if either of them is zero, the product will always be zero.

### Q3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

**Ans.**

No, if the product of two whole numbers is 1, we cannot say that one or both of them will be 1. In fact, for the product of two whole numbers to be 1, both numbers must be either 1 or -1.

Let’s consider examples to illustrate this:

**Example 1**: Suppose a = 1 and b = 1. When we multiply them together, the product is 1 × 1 = 1.

**Example 2**: Suppose a = -1 and b = -1. When we multiply them together, the product is (-1) × (-1) = 1.

In both examples, the product of the two whole numbers is 1, but only one of the numbers is 1. The other number can be -1.

Therefore, if the product of two whole numbers is 1, it does not necessarily mean that one or both of the numbers will be 1. It is possible for one or both of the numbers to be -1.

### Q4. Find using distributive property :

(a) 728 × 101

(b) 5437 × 1001

(c) 824 × 25

(d) 4275 × 125

(e) 504 × 35

**Ans.**

Using the distributive property, we can expand the given expressions to simplify the multiplication. Let’s calculate the following:

(a) 728 × 101:

We can write 728 as 700 + 20 + 8.

Using the distributive property, we have:

728 × 101 = (700 + 20 + 8) × 101

= 700 × 101 + 20 × 101 + 8 × 101

= 70700 + 2020 + 808

= 72928

(b) 5437 × 1001:

We can write 1001 as 1000 + 1.

Using the distributive property, we have:

5437 × 1001 = 5437 × (1000 + 1)

= 5437 × 1000 + 5437 × 1

= 5437000 + 5437

= 5442437

(c) 824 × 25:

We can write 824 as 800 + 20 + 4.

Using the distributive property, we have:

824 × 25 = (800 + 20 + 4) × 25

= 800 × 25 + 20 × 25 + 4 × 25

= 20000 + 500 + 100

= 20600

(d) 4275 × 125:

We can write 125 as 100 + 20 + 5.

Using the distributive property, we have:

4275 × 125 = 4275 × (100 + 20 + 5)

= 4275 × 100 + 4275 × 20 + 4275 × 5

= 427500 + 85500 + 21375

= 534375

(e) 504 × 35:

We can write 504 as 500 + 4.

Using the distributive property, we have:

504 × 35 = (500 + 4) × 35

= 500 × 35 + 4 × 35

= 17500 + 140

= 17640

By using the distributive property, we have simplified the given multiplications.

### Q5. 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)

**Ans.**

Continuing the pattern:

123456 × 8 + 6 = 987654

1234567 × 8 + 7 = 9876543

In this pattern, each step involves multiplying a consecutive sequence of numbers by 8 and adding a specific number. The multiplication factor is always 8, and the added number increases by 1 with each step. The key to understanding the pattern lies in recognizing that the number being multiplied follows a specific sequence.

The number being multiplied follows the pattern: 1, 12, 123, 1234, 12345, …

If we look closely, we can observe that each number in the sequence is formed by adding the next digit to the previous number. For example: 12 = 1 + 2 123 = 12 + 3 1234 = 123 + 4 12345 = 1234 + 5

By understanding this pattern, we can easily predict the next numbers in the sequence and continue the pattern.