Chapter 2 Whole Numbers (Additional Questions)

Solution:

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The set of whole numbers consists of all non-negative integers.

The whole numbers are $0, 1, 2, 3, \dots$

Comparing the given options:

(A) 1 is a whole number.

(B) 0 is a whole number.

(C) -1 is an integer but not a whole number as it is negative.

(D) The smallest whole number exists and can be determined.

Among the whole numbers $0, 1, 2, 3, \dots$, the smallest number is 0.

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Thus, the smallest whole number is 0.

The correct option is (B).

Solution:

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

To find the successor of 10,99,999, we need to calculate $10,99,999 + 1$.

Let's perform the addition:

$\begin{array}{ccccccc} & 1 & 0 &, & 9 & 9 & 9 & 9 & 9 \\ + & & & , & & & & & 1 \\ \hline & 1 & 1 &, & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}$

So, the successor of 10,99,999 is 11,00,000.

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Comparing this result with the given options:

(A) 10,99,998 - This is the predecessor.

(B) 11,00,000 - This matches our calculated successor.

(C) 10,00,000 - This is incorrect.

(D) 11,00,001 - This is incorrect.

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Therefore, the successor of 10,99,999 is 11,00,000.

The correct option is (B).

Solution:

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The predecessor of a whole number (other than 0) is the number obtained by subtracting 1 from it.

To find the predecessor of 500, we need to calculate $500 - 1$.

$500 - 1 = 499$

So, the predecessor of 500 is 499.

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Comparing this result with the given options:

(A) 501 - This is the successor.

(B) 499 - This matches our calculated predecessor.

(C) 500 - This is the number itself.

(D) 498 - This is incorrect.

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Therefore, the predecessor of 500 is 499.

The correct option is (B).

Solution:

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The set of natural numbers, denoted by $\mathbb{N}$, consists of counting numbers starting from 1:

$\mathbb{N} = \{1, 2, 3, \dots\}$

The set of whole numbers, denoted by $\mathbb{W}$, consists of all natural numbers along with zero:

$\mathbb{W} = \{0, 1, 2, 3, \dots\}$

The set of integers, denoted by $\mathbb{Z}$, includes all whole numbers and their negatives:

$\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$

The set of rational numbers, denoted by $\mathbb{Q}$, includes all numbers that can be expressed as a fraction $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$.

The set of real numbers, denoted by $\mathbb{R}$, includes all rational and irrational numbers.

The question asks for the collection of natural numbers along with zero. This collection is $\{0, 1, 2, 3, \dots\}$, which is the definition of whole numbers.

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Therefore, the collection of natural numbers along with zero is called Whole numbers.

The correct option is (B).

Solution:

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Let's analyze each statement based on the definitions of natural numbers and whole numbers.

The set of natural numbers is $\mathbb{N} = \{1, 2, 3, 4, \dots\}$.

The set of whole numbers is $\mathbb{W} = \{0, 1, 2, 3, 4, \dots\}$.

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(A) Every natural number is a whole number.

The set of natural numbers $\{1, 2, 3, \dots\}$ is a subset of the set of whole numbers $\{0, 1, 2, 3, \dots\}$. Every element in the set of natural numbers is also present in the set of whole numbers. Thus, this statement is true.

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(B) Every whole number is a natural number.

The number 0 is a whole number, but it is not a natural number. Thus, this statement is false.

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(C) Zero is a natural number.

The set of natural numbers starts from 1. Zero (0) is not included in the set of natural numbers. Thus, this statement is false.

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(D) The predecessor of a whole number is always a whole number.

The predecessor of a whole number $n$ is $n-1$. For $n=0$, the predecessor is $0-1 = -1$. The number -1 is an integer but not a whole number. Thus, this statement is false.

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Based on the analysis, only statement (A) is true.

The correct option is (A).

Solution:

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We need to find the number of whole numbers that are greater than 5 and less than 15.

The whole numbers greater than 5 are $6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, \dots$.

The whole numbers less than 15 are $\dots, 11, 12, 13, 14$.

The whole numbers that are both greater than 5 and less than 15 are the numbers from 6 up to 14, inclusive.

These numbers are: $6, 7, 8, 9, 10, 11, 12, 13, 14$.

Let's count these numbers. There are 9 numbers in this list.

Alternatively, the number of whole numbers between two distinct whole numbers $a$ and $b$ (where $a < b$), exclusive of $a$ and $b$, is given by $(b - a - 1)$.

Here, $a = 5$ and $b = 15$.

Number of whole numbers $= (15 - 5 - 1) = 10 - 1 = 9$.

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Therefore, there are 9 whole numbers between 5 and 15 (exclusive).

The correct option is (B).

Solution:

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The successor of a number is the number that comes immediately after it. It is obtained by adding 1 to the original number.

On a standard horizontal number line, numbers increase in value as you move from left to right.

For example, the successor of 3 is $3+1=4$. On the number line, 4 is to the right of 3.

The successor of any number will always be greater than the number itself.

Therefore, to find the successor of a number on a number line, you move towards the direction where the values increase, which is to the right.

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The correct option is (B).

Solution:

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The set of whole numbers is $\mathbb{W} = \{0, 1, 2, 3, \dots\}$.

The predecessor of a whole number $n$ (if it exists within the set of whole numbers) is the whole number $n-1$.

Let's examine the predecessor for each given option:

(A) For the number 1: The predecessor is $1 - 1 = 0$. The number 0 is a whole number. So, 1 has a predecessor (0) in the set of whole numbers.

(B) For the number 10: The predecessor is $10 - 1 = 9$. The number 9 is a whole number. So, 10 has a predecessor (9) in the set of whole numbers.

(C) For the number 0: The predecessor is $0 - 1 = -1$. The number -1 is an integer but not a whole number. So, 0 does not have a predecessor in the set of whole numbers.

(D) For the number 100: The predecessor is $100 - 1 = 99$. The number 99 is a whole number. So, 100 has a predecessor (99) in the set of whole numbers.

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From the analysis above, the only whole number among the options that does not have a predecessor in the set of whole numbers is 0.

The correct option is (C).

Solution:

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Let's examine the properties listed in the options for whole numbers:

(A) Associativity of addition: This property states that for any three whole numbers $a, b,$ and $c$, the way the numbers are grouped does not affect the sum. Mathematically, it is expressed as $(a + b) + c = a + (b + c)$.

(B) Commutativity of addition: This property states that the order in which two whole numbers are added does not affect the sum. Mathematically, it is expressed as $a + b = b + a$.

(C) Distributivity over addition: This property relates multiplication and addition. It states that for any three whole numbers $a, b,$ and $c$, $a \times (b + c) = (a \times b) + (a \times c)$.

(D) Closure under addition: This property states that the sum of any two whole numbers is always a whole number. If $a$ and $b$ are whole numbers, then $a + b$ is also a whole number.

The given property is $a + b = b + a$. This corresponds to the definition of the Commutativity property for addition.

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Therefore, the property $a + b = b + a$ for whole numbers is called Commutativity of addition.

The correct option is (B).

Solution:

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We need to evaluate the given expression: $15 \times (10 - 0)$.

According to the order of operations (PEMDAS/BODMAS), we first evaluate the expression inside the parentheses.

$10 - 0 = 10$

Now, substitute this result back into the original expression:

$15 \times 10$

Perform the multiplication:

$15 \times 10 = 150$

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The value of the expression $15 \times (10 - 0)$ is 150.

Comparing this result with the given options:

(A) 15

(B) 150

(C) 0

(D) 1510

The calculated value matches option (B).

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The correct option is (B).

Solution:

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The identity element for an operation in a set is an element which, when combined with any other element in the set using that operation, leaves the other element unchanged.

For addition of whole numbers, the identity element $e$ is a whole number such that for any whole number $a$,

$a + e = a$ and $e + a = a$.

Let's check the given options:

If $e = 1$, then $a + 1 = a$. This is not true for any whole number $a$, except in specific contexts not applicable here. For example, $5 + 1 = 6 \neq 5$.

If $e = 0$, then $a + 0 = a$ and $0 + a = a$. This is true for every whole number $a$. Adding zero to any whole number does not change the value of the whole number.

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Thus, 0 is the identity element for addition of whole numbers.

The correct option is (B).

Solution:

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Division is the inverse operation of multiplication. If we divide a number $a$ by a number $b$ (where $b \neq 0$) and get a result $c$, it means that $a = b \times c$.

In this question, we are asked to evaluate $12 \div 0$. Let's assume there is a result $c$ such that $12 \div 0 = c$.

According to the definition of division, this would mean that $12 = 0 \times c$.

However, we know that for any number $c$, $0 \times c = 0$.

So, the equation becomes $12 = 0$. This is a false statement.

There is no number $c$ that can satisfy the equation $12 = 0 \times c$.

Therefore, division by zero is not possible in the standard number system. It is considered undefined.

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Thus, the value of $12 \div 0$ is undefined.

The correct option is (D).

Solution:

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Given:

Quantity of milk supplied in the morning = 32 litres

Quantity of milk supplied in the evening = 68 litres

Cost of milk per litre = $\textsf{₹}$ 45

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To Find:

The total money due to the vendor per day.

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Solution:

First, calculate the total quantity of milk supplied by the vendor per day.

Total quantity of milk = Quantity in morning + Quantity in evening

$Total$ quantity $ = 32 + 68$ litres

$Total$ quantity $ = 100$ litres

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Next, calculate the total money due to the vendor per day using the total quantity and the cost per litre.

Total money due = Total quantity $\times$ Cost per litre

$Total$ money $ due = 100 \times 45$

$Total$ money $ due = 4500$

The total money due to the vendor per day is $\textsf{₹}$ 4500.

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Comparing the result with the given options:

(A) $\textsf{₹}$ 4,000

(B) $\textsf{₹}$ 4,500

(C) $\textsf{₹}$ 3,600

(D) $\textsf{₹}$ 4,200

The calculated amount is $\textsf{₹}$ 4500, which corresponds to option (B).

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The correct option is (B).

Solution:

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We need to find the product of $25 \times 1234 \times 4$ using a suitable property.

The commutative and associative properties of multiplication allow us to rearrange and group the numbers in a way that simplifies the calculation.

The given expression is $25 \times 1234 \times 4$.

We can group 25 and 4 together because their product is a round number (100), which makes further multiplication easier.

Using the associative property of multiplication, we have:

$(25 \times 4) \times 1234$

First, calculate the product inside the parentheses:

$25 \times 4 = 100$

Now, substitute this result back into the expression:

$100 \times 1234$

Multiplying by 100 is straightforward:

$100 \times 1234 = 123400$

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The value of the product $25 \times 1234 \times 4$ is 123400.

Comparing this result with the given options:

(A) 123400

(B) 12340

(C) 1234000

(D) 123450

The calculated value matches option (A).

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The correct option is (A).

Solution:

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An operation $*$ is commutative for a set of numbers if for any two numbers $a$ and $b$ in the set, $a * b = b * a$. We need to check this property for each given operation using whole numbers. The set of whole numbers is $\mathbb{W} = \{0, 1, 2, 3, \dots\}$.

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(A) Addition: Check if $a + b = b + a$ for all whole numbers $a, b$.

Let's take an example: $a = 5$, $b = 8$.

$5 + 8 = 13$

$8 + 5 = 13$

Since $13 = 13$, $5 + 8 = 8 + 5$. This holds true for all pairs of whole numbers.

So, addition is commutative for whole numbers.

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(B) Subtraction: Check if $a - b = b - a$ for all whole numbers $a, b$.

Let's take an example: $a = 7$, $b = 3$.

$7 - 3 = 4$

$3 - 7 = -4$

Since $4 \neq -4$, $7 - 3 \neq 3 - 7$. The commutative property does not hold for subtraction of whole numbers.

So, subtraction is NOT commutative for whole numbers.

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(C) Multiplication: Check if $a \times b = b \times a$ for all whole numbers $a, b$.

Let's take an example: $a = 4$, $b = 9$.

$4 \times 9 = 36$

$9 \times 4 = 36$

Since $36 = 36$, $4 \times 9 = 9 \times 4$. This holds true for all pairs of whole numbers.

So, multiplication is commutative for whole numbers.

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(D) Division: Check if $a \div b = b \div a$ for all whole numbers $a, b$ (where $b \neq 0$ and $a \neq 0$ if $b$ is also 0).

Let's take an example: $a = 10$, $b = 2$.

$10 \div 2 = 5$

$2 \div 10 = \frac{2}{10} = \frac{1}{5}$

Since $5 \neq \frac{1}{5}$, $10 \div 2 \neq 2 \div 10$. The commutative property does not hold for division of whole numbers.

So, division is NOT commutative for whole numbers.

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Based on the analysis, subtraction and division are not commutative for whole numbers.

The correct options are (B) and (D).

Solution:

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We are given the calculation steps using the distributive property:

$18 \times 101 = 18 \times (100 + 1)$

Using the distributive property, $a \times (b + c) = (a \times b) + (a \times c)$:

$18 \times (100 + 1) = (18 \times 100) + (18 \times 1)$

We are given the next step as:

$18 \times 100 + 18 \times 1 = 1800 + \text{_____}$

The term $18 \times 100$ is evaluated to $1800$.

The next term is $18 \times 1$.

Let's calculate $18 \times 1$:

$18 \times 1 = 18$

So, the expression becomes:

$1800 + 18$

The blank space in the equation is the value of $18 \times 1$.

Therefore, the number that fills the blank is 18.

The final step in the given equation is $1800 + 18 = 1818$, which is correct.

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The number that fills the blank is 18.

Comparing this with the given options:

(A) 1

(B) 100

(C) 18

(D) 1800

The value 18 matches option (C).

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The correct option is (C).

Solution:

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We need to evaluate the truthfulness of the assertion (A) and the reason (R) and determine if R correctly explains A.

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Assertion (A): Division is associative for whole numbers.

The associative property for division states that for any three whole numbers $a, b, c$, $(a \div b) \div c = a \div (b \div c)$. Note that for division, we must consider cases where the denominators are not zero.

Let's test this with an example. Let $a=12$, $b=6$, $c=2$.

Left side: $(a \div b) \div c = (12 \div 6) \div 2 = 2 \div 2 = 1$.

Right side: $a \div (b \div c) = 12 \div (6 \div 2) = 12 \div 3 = 4$.

Since $1 \neq 4$, $(12 \div 6) \div 2 \neq 12 \div (6 \div 2)$.

This single counterexample shows that the associative property does not hold for division of whole numbers in general.

Therefore, Assertion (A) is false.

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Reason (R): For any whole numbers $a, b, c$ (where $b, c \neq 0$), $(a \div b) \div c = a \div (b \div c)$ is always true.

This statement is the mathematical formulation of the associative property for division. As shown in the example above, the equality $(a \div b) \div c = a \div (b \div c)$ is not always true for whole numbers. The example with $a=12, b=6, c=2$ proves this. Even with the condition $b, c \neq 0$, the property does not hold universally for all $a, b, c$.

Therefore, Reason (R) is false.

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Since both the assertion and the reason are false, we look for the option that reflects this.

(A) Both A and R are true... (False)

(B) Both A and R are true... (False)

(C) A is true but R is false. (False, A is false)

(D) A is false but R is false. (True)

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The correct option is (D).

Solution:

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Given:

Number of dozen bananas bought = 12 dozen

Number of bananas in one dozen = 12 bananas

Number of bananas sold = 100 bananas

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To Find:

The number of bananas left with the shopkeeper.

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Solution:

First, calculate the total number of bananas the shopkeeper bought.

Total bananas bought = Number of dozen $\times$ Number of bananas per dozen

Total bananas bought = $12 \times 12$

Calculate the product:

$\begin{array}{cc}& & 1 & 2 \\ \times & & 1 & 2 \\ \hline && 2 & 4 \\ & 1 & 2 & \times \\ \hline & 1 & 4 & 4 \\ \hline \end{array}$

Total bananas bought = 144 bananas.

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Now, calculate the number of bananas left after selling 100 bananas.

Bananas left = Total bananas bought - Bananas sold

Bananas left = $144 - 100$

Bananas left = 44 bananas.

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The number of bananas left with the shopkeeper is 44.

Comparing this result with the given options:

(A) 44

(B) 40

(C) 36

(D) 24

The calculated number of bananas left matches option (A).

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The correct option is (A).

Solution:

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The distributive property of multiplication over addition states that for any three numbers $a, b,$ and $c$, $a \times (b + c) = (a \times b) + (a \times c)$ or $(a + b) \times c = (a \times c) + (b \times c)$.

The distributive property of multiplication over subtraction states that for any three numbers $a, b,$ and $c$, $a \times (b - c) = (a \times b) - (a \times c)$ or $(a - b) \times c = (a \times c) - (b \times c)$.

Let's examine each option:

(A) $5 \times (7 + 2) = 5 \times 7 + 5 \times 2$

This is of the form $a \times (b + c) = a \times b + a \times c$, which is an example of the distributive property of multiplication over addition.

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(B) $10 \times (12 - 3) = 10 \times 12 - 10 \times 3$

This is of the form $a \times (b - c) = a \times b - a \times c$, which is an example of the distributive property of multiplication over subtraction.

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(C) $8 + (4 + 6) = (8 + 4) + 6$

This is of the form $a + (b + c) = (a + b) + c$. This property states that the way in which numbers are grouped in an addition does not affect the sum. This is the associative property of addition.

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(D) $(15 + 5) \times 4 = 15 \times 4 + 5 \times 4$

This is of the form $(a + b) \times c = a \times c + b \times c$, which is an example of the distributive property of multiplication over addition (right distributive property).

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The question asks which option is NOT an example of the distributive property. Option (C) illustrates the associative property of addition.

Therefore, the correct option is (C).

Solution:

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We need to match each operation with the properties it satisfies for all whole numbers. Let's analyze the properties (Closure, Commutativity, and Associativity) for each operation on the set of whole numbers $\mathbb{W} = \{0, 1, 2, 3, \dots\}$.

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(i) Addition:

• Closure: For any whole numbers $a, b$, $a+b$ is always a whole number. (True)
• Commutativity: For any whole numbers $a, b$, $a+b = b+a$. (True)
• Associativity: For any whole numbers $a, b, c$, $(a+b)+c = a+(b+c)$. (True)

Addition of whole numbers is always closed, commutative, and associative.

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(ii) Subtraction:

• Closure: For any whole numbers $a, b$, $a-b$ is not always a whole number (e.g., $3-5 = -2$, which is not a whole number). (False)
• Commutativity: For any whole numbers $a, b$, $a-b = b-a$ is not always true (e.g., $7-3 = 4$, but $3-7 = -4$). (False)
• Associativity: For any whole numbers $a, b, c$, $(a-b)-c = a-(b-c)$ is not always true (e.g., $(10-5)-2 = 5-2=3$, but $10-(5-2) = 10-3=7$). (False)

Subtraction of whole numbers is not always closed, not commutative, and not associative.

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(iii) Multiplication:

• Closure: For any whole numbers $a, b$, $a \times b$ is always a whole number. (True)
• Commutativity: For any whole numbers $a, b$, $a \times b = b \times a$. (True)
• Associativity: For any whole numbers $a, b, c$, $(a \times b) \times c = a \times (b \times c)$. (True)

Multiplication of whole numbers is always closed, commutative, and associative.

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(iv) Division by a non-zero number:

• Closure: For any whole numbers $a, b$ with $b \neq 0$, $a \div b$ is not always a whole number (e.g., $5 \div 2 = 2.5$, which is not a whole number). (False)
• Commutativity: For any whole numbers $a, b$ with $b \neq 0$ and $a \neq 0$, $a \div b = b \div a$ is not always true (e.g., $10 \div 2 = 5$, but $2 \div 10 = 0.2$). (False)
• Associativity: For any whole numbers $a, b, c$ with $b \neq 0, c \neq 0$, $(a \div b) \div c = a \div (b \div c)$ is not always true (e.g., $(12 \div 6) \div 2 = 1$, but $12 \div (6 \div 2) = 4$). (False)

Division by a non-zero number for whole numbers is not always closed, not commutative, and not associative.

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Now let's match these findings with the property descriptions (a), (b), (c), (d):

Properties described in (a) and (c): Always closed, commutative, and associative.

Properties described in (b) and (d): Not always closed, not commutative, not associative.

Based on our analysis:

• (i) Addition satisfies the properties in (a) and (c).
• (ii) Subtraction satisfies the properties in (b) and (d).
• (iii) Multiplication satisfies the properties in (a) and (c).
• (iv) Division by a non-zero number satisfies the properties in (b) and (d).

Let's examine the options to find the correct pairing:

(A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d)

• (i) maps to (a): Addition (Set 1 properties) maps to (a) (Set 1 properties). Correct.
• (ii) maps to (b): Subtraction (Set 2 properties) maps to (b) (Set 2 properties). Correct.
• (iii) maps to (c): Multiplication (Set 1 properties) maps to (c) (Set 1 properties). Correct.
• (iv) maps to (d): Division (Set 2 properties) maps to (d) (Set 2 properties). Correct.

Option (A) provides a consistent mapping where operations satisfying Set 1 properties are mapped to letters describing Set 1 properties, and operations satisfying Set 2 properties are mapped to letters describing Set 2 properties. Given that options (a) and (c) have identical text, and (b) and (d) have identical text, multiple options might seem valid if only considering the text descriptions. However, assuming the letters themselves represent specific categories to be matched, Option (A) represents the straightforward pairing (i)->a, (ii)->b, (iii)->c, (iv)->d, which aligns with the properties each operation satisfies as described by the text in the respective letters.

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The correct option is (A).

Solution:

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Given:

Distance covered by car in 1 hour = 60 km

Total distance to be travelled = 180 km

The car travels at the same speed.

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To Find:

The time taken to travel 180 km.

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Solution:

We know the speed of the car is 60 km per hour.

Speed = $\frac{\text{Distance}}{\text{Time}}$

Given speed is 60 km/hour.

We need to find the time taken to travel 180 km.

Let $T$ be the time taken in hours.

Using the formula:

$Speed = \frac{Total \text{ distance}}{\text{Time}}$

$60 = \frac{180}{T}$

To find $T$, we can rearrange the equation:

$T = \frac{180}{60}$

Now, perform the division:

$T = \frac{\cancel{180}^{3}}{\cancel{60}_{1}}$

$T = 3$ hours.

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Alternatively, we can use proportionality.

If the car travels 60 km in 1 hour, and it travels at a constant speed, then the time taken is directly proportional to the distance travelled.

Let the time taken to travel 180 km be $T$ hours.

$\frac{\text{Distance}_1}{\text{Time}_1} = \frac{\text{Distance}_2}{\text{Time}_2}$

$\frac{60 \text{ km}}{1 \text{ hour}} = \frac{180 \text{ km}}{T \text{ hours}}$

Cross-multiply:

$60 \times T = 180 \times 1$

$60T = 180$

$T = \frac{180}{60}$

$T = 3$ hours.

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The time taken to travel 180 km at the same speed is 3 hours.

Comparing this result with the given options:

(A) 2 hours

(B) 3 hours

(C) 4 hours

(D) 5 hours

The calculated time matches option (B).

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The correct option is (B).

Solution:

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The identity element for an operation in a set is an element which, when combined with any other element in the set using that operation, leaves the other element unchanged.

For multiplication of whole numbers, the identity element $e$ is a whole number such that for any whole number $a$,

$a \times e = a$ and $e \times a = a$.

Let's check the given options:

If $e = 0$, then $a \times 0 = 0$. This is not equal to $a$ unless $a$ is specifically 0. For example, $5 \times 0 = 0 \neq 5$. So, 0 is not the identity element for multiplication.

If $e = 1$, then $a \times 1 = a$ and $1 \times a = a$. This is true for every whole number $a$. Multiplying any whole number by 1 does not change the value of the whole number.

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Thus, 1 is the identity element for multiplication of whole numbers.

The correct option is (B).

Solution:

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We need to examine each statement to determine which one is correct.

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(A) Subtraction is commutative for whole numbers.

For an operation to be commutative, $a - b = b - a$ must hold for all whole numbers $a, b$. We know this is not true. For example, $5 - 3 = 2$ but $3 - 5 = -2$. $2 \neq -2$. So, this statement is false.

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(B) Division by zero is allowed for any non-zero whole number.

Division by zero is undefined in the standard number system. Dividing any number (whether non-zero or zero) by zero is not allowed. So, this statement is false.

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(C) $15 \times 0 = 15$

The property of multiplication by zero states that the product of any number and zero is zero. So, $15 \times 0 = 0$. The statement says $15 \times 0 = 15$, which is incorrect. So, this statement is false.

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(D) $0 \div 5 = 0$

Division is the inverse of multiplication. If $0 \div 5 = q$, it means $0 = 5 \times q$. The only value of $q$ that satisfies this equation is $q=0$. Therefore, $0 \div 5 = 0$ is true. So, this statement is correct.

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Based on the analysis, only statement (D) is correct.

The correct option is (D).

Solution:

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We need to find the value of the expression $9876 \times 567 - 9876 \times 467$ using the distributive property.

The distributive property of multiplication over subtraction states that $a \times b - a \times c = a \times (b - c)$.

In the given expression, we can see that 9876 is a common factor in both terms.

Let $a = 9876$, $b = 567$, and $c = 467$.

The expression is in the form $a \times b - a \times c$.

Applying the distributive property, we can rewrite the expression as:

$9876 \times (567 - 467)$

First, evaluate the expression inside the parentheses:

$567 - 467$

$\begin{array}{cc} & 5 & 6 & 7 \\ - & 4 & 6 & 7 \\ \hline & 1 & 0 & 0 \\ \hline \end{array}$

So, $567 - 467 = 100$.

Now, substitute this result back into the expression:

$9876 \times 100$

Multiplying by 100 is straightforward:

$9876 \times 100 = 987600$

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The value of the expression is 987600.

Comparing this result with the given options:

(A) 987600

(B) 98760

(C) 9876

(D) 9876000

The calculated value matches option (A).

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The correct option is (A).

Solution:

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Multiplication is a fundamental arithmetic operation. It is often introduced as a shortcut for repeated addition.

For example, to calculate $3 \times 4$, we can interpret this as adding the number 4 to itself 3 times:

$3 \times 4 = 4 + 4 + 4 = 12$

Similarly, $5 \times 2$ means adding 2 to itself 5 times:

$5 \times 2 = 2 + 2 + 2 + 2 + 2 = 10$

In general, the product $a \times b$ (where $a$ is a positive integer) can be thought of as the sum of $b$ added to itself $a$ times.

Subtraction is related to repeated subtraction (which leads to division), division is related to repeated subtraction, and identity refers to the identity element for an operation.

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Therefore, the operation of multiplication can be represented as repeated addition.

Comparing this with the given options:

(A) Subtraction

(B) Division

(C) Addition

(D) Identity

The correct representation is repeated addition, which matches option (C).

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The correct option is (C).

Solution:

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Triangular numbers are figurate numbers that represent the number of dots needed to form an equilateral triangle.

The $n$-th triangular number, denoted by $T_n$, is the sum of the first $n$ natural numbers.

$T_n = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$

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We are asked to find the third triangular number, which corresponds to $n=3$.

Using the formula:

$T_3 = \frac{3(3+1)}{2} = \frac{3 \times 4}{2} = \frac{12}{2} = 6$

Alternatively, by definition, the third triangular number is the sum of the first 3 natural numbers:

$T_3 = 1 + 2 + 3 = 6$

The arrangement of dots for the first few triangular numbers is:

• First triangular number ($T_1$): 1 dot ($\bullet$)
• Second triangular number ($T_2$): 3 dots ($\begin{smallmatrix} & \bullet & \\ \bullet & & \bullet \end{smallmatrix}$)
• Third triangular number ($T_3$): 6 dots ($\begin{smallmatrix} & & \bullet & & \\ & \bullet & & \bullet & \\ \bullet & & \bullet & & \bullet \end{smallmatrix}$)

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Therefore, 6 dots are needed to form the third triangular number.

Comparing this result with the given options:

(A) 3

(B) 4

(C) 5

(D) 6

The number of dots needed is 6, which matches option (D).

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The correct option is (D).

Solution:

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A number can be arranged as a rectangle if it can be expressed as the product of two whole numbers greater than 1 (since we are excluding 1). Let the number be $N$. If $N = l \times w$, where $l > 1$ and $w > 1$, then $N$ can be arranged as a rectangle with length $l$ and width $w$.

A number can be arranged as a square if it can be expressed as the product of two equal whole numbers greater than 1, i.e., if it is a perfect square greater than 1. If $N = s \times s = s^2$, where $s > 1$, then $N$ can be arranged as a square with side length $s$.

We are looking for a number that can be arranged as a rectangle but NOT as a square (excluding 1). This means we are looking for a non-prime composite number that is not a perfect square (and is not 1).

Let's examine each option:

(A) 4:

• Can it be arranged as a rectangle? Yes, $4 = 2 \times 2$. Since $2 > 1$ and $2 > 1$, it can be a rectangle.
• Can it be arranged as a square? Yes, $4 = 2 \times 2 = 2^2$. Since $2 > 1$, it is a perfect square greater than 1, so it can be a square.

So, 4 can be arranged as both a rectangle and a square. This is not the number we are looking for.

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(B) 9:

• Can it be arranged as a rectangle? Yes, $9 = 3 \times 3$. Since $3 > 1$ and $3 > 1$, it can be a rectangle.
• Can it be arranged as a square? Yes, $9 = 3 \times 3 = 3^2$. Since $3 > 1$, it is a perfect square greater than 1, so it can be a square.

So, 9 can be arranged as both a rectangle and a square. This is not the number we are looking for.

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(C) 6:

• Can it be arranged as a rectangle? Yes, $6 = 2 \times 3$ or $6 = 3 \times 2$. Since $2 > 1$ and $3 > 1$, it can be a rectangle.
• Can it be arranged as a square? No, 6 is not a perfect square. There is no whole number $s$ such that $s^2 = 6$.

So, 6 can be arranged as a rectangle but not a square. This matches the condition.

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(D) 25:

• Can it be arranged as a rectangle? Yes, $25 = 5 \times 5$. Since $5 > 1$ and $5 > 1$, it can be a rectangle.
• Can it be arranged as a square? Yes, $25 = 5 \times 5 = 5^2$. Since $5 > 1$, it is a perfect square greater than 1, so it can be a square.

So, 25 can be arranged as both a rectangle and a square. This is not the number we are looking for.

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The only number among the options that can be arranged as a rectangle but not a square (excluding 1) is 6.

The correct option is (C).

Solution:

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We are given a sequence of numbers: 5, 10, 15, 20, _____. We need to find the next number in the pattern.

Let's look at the difference between consecutive terms:

• Difference between the second and first term: $10 - 5 = 5$
• Difference between the third and second term: $15 - 10 = 5$
• Difference between the fourth and third term: $20 - 15 = 5$

The difference between consecutive terms is constant and equal to 5. This indicates that the pattern is an arithmetic progression with a common difference of 5. Each term is obtained by adding 5 to the previous term.

The pattern can be described as follows:

• First term: 5
• Second term: $5 + 5 = 10$
• Third term: $10 + 5 = 15$
• Fourth term: $15 + 5 = 20$

To find the next term (the fifth term), we add 5 to the fourth term:

Fifth term = $20 + 5 = 25$.

Alternatively, the pattern represents the multiples of 5:

• $5 \times 1 = 5$
• $5 \times 2 = 10$
• $5 \times 3 = 15$
• $5 \times 4 = 20$

The next term would be the fifth multiple of 5:

$5 \times 5 = 25$.

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The next number in the pattern is 25.

Comparing this result with the given options:

(A) 24

(B) 25

(C) 30

(D) 22

The calculated number matches option (B).

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The correct option is (B).

Solution:

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The given equation is $91 \times 11 \times 7 = 7007$.

We know that $91 = 7 \times 13$.

So, the expression can be written as $(7 \times 13) \times 11 \times 7$.

Using the commutative and associative properties of multiplication, we can rearrange and group the terms:

$(7 \times 11 \times 13) \times 7$

Calculate the product inside the parentheses:

$7 \times 11 \times 13 = 77 \times 13 = 1001$.

So the expression becomes $1001 \times 7$.

$1001 \times 7 = 7007$.

The product $91 \times 11 \times 7$ is equal to $1001 \times 7$. The pattern involves multiplying by 1001, where $1001 = 7 \times 11 \times 13$.

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The pattern involved is multiplying by 1001.

The correct option is (B).

Solution:

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When a number is represented by dots:

• A number $N$ represented as a line requires $N$ dots.
• A number $N$ represented as a rectangle requires $l \times w$ dots, where $N = l \times w$ for integers $l > 1$ and $w > 1$.
• A number $N$ represented as a square requires $s \times s$ dots, where $N = s \times s$ for an integer $s > 1$.

The question asks which number needs 5 dots to represent it in one of these forms. We need to find a number from the options for which the number of dots required is exactly 5.

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Let's examine each option:

(A) 4:

As a line: Needs 4 dots.

As a rectangle: $4 = 2 \times 2$. Needs $2 \times 2 = 4$ dots.

As a square: $4 = 2^2$. Needs $2 \times 2 = 4$ dots.

None of these representations need 5 dots for the number 4.

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(B) 6:

As a line: Needs 6 dots.

As a rectangle: $6 = 2 \times 3$. Needs $2 \times 3 = 6$ dots.

As a square: 6 is not a perfect square. Cannot be represented as a square.

None of these representations need 5 dots for the number 6.

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(C) 5:

As a line: Needs 5 dots.

As a rectangle: 5 is a prime number ($5 = 5 \times 1$ or $1 \times 5$). It cannot be formed into a rectangle with length and width both greater than 1.

As a square: 5 is not a perfect square. Cannot be represented as a square.

Representing the number 5 as a line requires exactly 5 dots.

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(D) 8:

As a line: Needs 8 dots.

As a rectangle: $8 = 2 \times 4$. Needs $2 \times 4 = 8$ dots.

As a square: 8 is not a perfect square. Cannot be represented as a square.

None of these representations need 5 dots for the number 8.

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The only number among the options that requires exactly 5 dots for one of the given representations (specifically, as a line) is 5.

The correct option is (C).

Solution:

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The pattern for multiplying by numbers consisting of nines (like 9, 99, 999, etc.) involves rewriting the number with nines in terms of a power of 10 minus 1.

• $9 = 10 - 1$
• $99 = 100 - 1$
• $999 = 1000 - 1$
• $9999 = 10000 - 1$, and so on.

We need to calculate $6 \times 999$.

Using the pattern, we replace 999 with $1000 - 1$.

$6 \times 999 = 6 \times (1000 - 1)$

Now, we can use the distributive property of multiplication over subtraction, which states that $a \times (b - c) = a \times b - a \times c$.

Here, $a=6$, $b=1000$, and $c=1$.

$6 \times (1000 - 1) = (6 \times 1000) - (6 \times 1)$

$ = 6000 - 6$.

So, $6 \times 999 = 6000 - 6$.

Let's check the options:

(A) $6 \times (1000 + 1)$: This is $6 \times 1001 = 6006$. Incorrect.

(B) $6 \times 1000 - 6$: This is $6000 - 6$. This matches our result.

(C) $6 \times 1000 + 6$: This is $6000 + 6 = 6006$. Incorrect.

(D) $6000 - 1$: This is 5999. Incorrect.

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The expression $6 \times 999$ can be written as $6 \times 1000 - 6$ using the described pattern and the distributive property.

The correct option is (B).

Solution:

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A number can be represented as a square using dots if it is a perfect square. A perfect square is a number that can be expressed as the product of an integer with itself, i.e., $n \times n = n^2$ for some integer $n$.

Let's examine each option to see if it is a perfect square:

(A) 1: Can 1 be written as $n^2$ for some integer $n$? Yes, $1 = 1 \times 1 = 1^2$. Thus, 1 is a perfect square. It can be represented as a single dot, which is a 1x1 square.

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(B) 2: Can 2 be written as $n^2$ for some integer $n$? No, the integers squared are $1^2=1$, $2^2=4$, etc. 2 is between 1 and 4. Thus, 2 is not a perfect square. It cannot be represented as a square arrangement of dots.

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(C) 4: Can 4 be written as $n^2$ for some integer $n$? Yes, $4 = 2 \times 2 = 2^2$. Thus, 4 is a perfect square. It can be represented as a 2x2 square arrangement of dots:

$\begin{smallmatrix} \bullet & \bullet \\ \bullet & \bullet \end{smallmatrix}$

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(D) 9: Can 9 be written as $n^2$ for some integer $n$? Yes, $9 = 3 \times 3 = 3^2$. Thus, 9 is a perfect square. It can be represented as a 3x3 square arrangement of dots:

$\begin{smallmatrix} \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet \end{smallmatrix}$

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The numbers among the options that can be represented as a square using dots are the perfect squares. These are 1, 4, and 9.

The question asks to select all that apply.

The correct options are (A), (C), and (D).

Solution:

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Case Study Pattern:

The case study provides examples of multiplying whole numbers by 9 and observing the sum of the digits of the product:

• $1 \times 9 = 9$. Sum of digits = 9.
• $2 \times 9 = 18$. Sum of digits = $1 + 8 = 9$.
• $3 \times 9 = 27$. Sum of digits = $2 + 7 = 9$.
• $4 \times 9 = 36$. Sum of digits = $3 + 6 = 9$.

In each of these examples, the sum of the digits of the product is 9.

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General Property:

There is a general property related to the divisibility of numbers by 9. A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

The product of any whole number and 9 is always a multiple of 9.

Since the product is a multiple of 9, the sum of its digits must be divisible by 9.

The question asks for the sum of the digits of the product of any whole number and 9, specifically when this sum is a single digit.

The possible single-digit sums of digits that are divisible by 9 are 0 and 9.

Let's consider if the sum of digits can be 0. The only way the sum of digits of a non-negative whole number can be 0 is if the number itself is 0 (e.g., the sum of digits of 0 is 0). The product of a whole number and 9 is 0 only if the whole number is 0 ($0 \times 9 = 0$). In this case, the sum of digits is 0.

For any non-zero whole number multiplied by 9, the product will be a positive multiple of 9. The sum of the digits of any positive multiple of 9 is always a multiple of 9. If this sum is a single digit, the only positive single digit divisible by 9 is 9.

The pattern shown in the case study (for $1 \times 9$ to $4 \times 9$) consistently results in a sum of digits equal to 9.

Let's try another example: $5 \times 9 = 45$. Sum of digits $4+5 = 9$.

$10 \times 9 = 90$. Sum of digits $9+0 = 9$.

$11 \times 9 = 99$. Sum of digits $9+9 = 18$. (Note: the sum of digits here is not a single digit).

If we continue summing the digits of 18 ($1+8=9$), we get 9. This is related to the digital root.

However, the question specifically asks for the sum of the digits if that sum is a single digit.

For products where the sum of digits is a single digit (like 9, 18, 27, 36, 45, 54, 63, 72, 81, 90), the sum is always 9, unless the product is 0 (from $0 \times 9$). If the product is 0, the sum of digits is 0.

The pattern provided does not include $0 \times 9$. It starts from $1 \times 9$. The results 9, 18, 27, 36 have single-digit sums of digits equal to 9.

Based on the provided pattern and the constraint that the sum of digits is a single digit, the consistent result shown is 9. While 0 is a theoretical possibility if the original number is 0, the examples point towards the non-zero products.

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Based on the pattern and the typical context of such questions, the expected single-digit sum of the digits of the product of a whole number and 9 (for non-zero products) is 9.

Comparing with the options:

(A) 0 (Only for product = 0)

(B) 1 (Incorrect)

(C) 9 (Consistent with the pattern for non-zero products with single-digit sums of digits)

(D) Cannot be determined (The pattern strongly suggests 9)

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The correct option is (C).

Definition of Whole Numbers:

Whole numbers are the set of non-negative integers. They include all natural numbers along with zero.

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Definition of Natural Numbers:

Natural numbers (also known as counting numbers) are the set of positive integers. They start from 1 and go up infinitely.

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Difference between Whole Numbers and Natural Numbers:

The key difference between whole numbers and natural numbers is the inclusion of the number zero (0).

Natural numbers start from 1 ($1, 2, 3, 4, ...$) and do not include zero.

Whole numbers start from 0 ($0, 1, 2, 3, 4, ...$) and include all natural numbers plus zero.

In set notation:

Set of Natural Numbers (N) = $\{1, 2, 3, 4, ...\}$

Set of Whole Numbers (W) = $\{0, 1, 2, 3, 4, ...\}$

Therefore, whole numbers are the natural numbers combined with zero.

Smallest Whole Number:

The smallest whole number is 0 (zero).

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Largest Whole Number:

There is no largest whole number.

The set of whole numbers ($0, 1, 2, 3, ...$) extends infinitely. For any whole number you choose, you can always find a larger one by adding 1.

Answer:

Yes, all natural numbers are also whole numbers.

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Explanation:

The set of natural numbers consists of positive integers starting from 1:

Natural Numbers ($N$) = $\{1, 2, 3, 4, 5, ...\}$

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The set of whole numbers consists of all natural numbers along with zero:

Whole Numbers ($W$) = $\{0, 1, 2, 3, 4, 5, ...\}$

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By comparing the two sets, we can see that every number that is in the set of natural numbers (1, 2, 3, ...) is also present in the set of whole numbers (0, 1, 2, 3, ...).

The set of whole numbers essentially contains the set of natural numbers as a subset, with the addition of the number zero.

Therefore, every natural number is included in the collection of whole numbers, making the statement true.

Answer:

No, not all whole numbers are also natural numbers.

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Explanation:

The set of whole numbers includes zero ($0$) and all positive integers:

Whole Numbers ($W$) = $\{0, 1, 2, 3, 4, ...\}$

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The set of natural numbers includes only the positive integers, starting from one:

Natural Numbers ($N$) = $\{1, 2, 3, 4, 5, ...\}$

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By comparing these two sets, we can see that the number zero (0) is present in the set of whole numbers but is not present in the set of natural numbers.

Since zero is a whole number but not a natural number, it is not true that all whole numbers are also natural numbers.

The set of natural numbers is a subset of the set of whole numbers ($N \subset W$).

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

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To find the successor of $999$, we add 1 to $999$.

Successor of $999 = 999 + 1$

Let's perform the addition:

$\begin{array}{cc} & 9 & 9 & 9 \\ + & & & 1 \\ \hline 1 & 0 & 0 & 0 \\ \hline \end{array}$

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So, $999 + 1 = 1000$.

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The successor of $999$ is $1000$.

The predecessor of a whole number is the number obtained by subtracting 1 from it.

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To find the predecessor of $500$, we subtract 1 from $500$.

Predecessor of $500 = 500 - 1$

Let's perform the subtraction:

$\begin{array}{cc} & 5 & 0 & 0 \\ - & & & 1 \\ \hline & 4 & 9 & 9 \\ \hline \end{array}$

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So, $500 - 1 = 499$.

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The predecessor of $500$ is $499$.

Solution:

We need to find the whole numbers that are greater than $25$ and less than $35$.

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The whole numbers are $0, 1, 2, 3, ...$.

The whole numbers strictly between $25$ and $35$ are:

$26, 27, 28, 29, 30, 31, 32, 33, 34$.

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Now, we count how many numbers are in this list:

$26$ (1st)

$27$ (2nd)

$28$ (3rd)

$29$ (4th)

$30$ (5th)

$31$ (6th)

$32$ (7th)

$33$ (8th)

$34$ (9th)

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There are a total of 9 whole numbers between $25$ and $35$.

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Alternatively, the number of whole numbers between two distinct whole numbers $a$ and $b$ (where $a < b$) is given by the formula $b - a - 1$.

In this case, $a = 25$ and $b = 35$.

Number of whole numbers $= 35 - 25 - 1$

$= 10 - 1$

$= 9$

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Thus, there are 9 whole numbers between $25$ and $35$.

Solution:

To represent whole numbers on a number line, we draw a straight line and mark points at equal intervals.

We label the points starting from 0, 1, 2, and so on.

We will mark the points corresponding to the numbers $0, 3,$ and $7$.

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Here is a representation of the number line with the points $0, 3,$ and $7$ marked:

•——•————•———————•——>

$0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ ...

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In the representation above, the solid circles (•) indicate the positions of the numbers $0, 3,$ and $7$ on the number line.

Solution:

On a standard horizontal number line, numbers are arranged in increasing order from left to right.

This means that a number located to the left of another number is smaller than that number.

Conversely, a number located to the right of another number is larger than that number.

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We need to determine whether $12$ or $18$ is smaller than $15$.

Let's compare $12$ and $15$:

$12 < 15$

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Let's compare $18$ and $15$:

$18 > 15$

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Since $12$ is smaller than $15$ ($12 < 15$), $12$ is located to the left of $15$ on the number line.

Since $18$ is larger than $15$ ($18 > 15$), $18$ is located to the right of $15$ on the number line.

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Therefore, the number to the left of $15$ is $12$.

Solution:

On a number line, numbers increase in value as you move from left to right.

Therefore, if a number '$a$' is to the left of another number '$b$' on the number line, it means that $a$ is less than $b$ ($a < b$).

Conversely, if a number '$a$' is to the right of another number '$b$', it means that $a$ is greater than $b$ ($a > b$).

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We need to compare $89,456$ and $89,654$. Let's determine which number is smaller.

We compare the digits from left to right:

• The ten thousands digit is $8$ for both numbers ($8=8$).
• The thousands digit is $9$ for both numbers ($9=9$).
• The hundreds digit is $4$ for $89,456$ and $6$ for $89,654$.

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Since $4 < 6$, the number $89,456$ (which has $4$ in the hundreds place) is smaller than $89,654$ (which has $6$ in the hundreds place).

$89,456 < 89,654$

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Using the number line concept:

Since $89,456$ is smaller than $89,654$, $89,456$ would be located to the left of $89,654$ on the number line.

And $89,654$ would be located to the right of $89,456$ on the number line.

Solution:

We need to find the sum of $587$ and $398$. We can perform addition by aligning the numbers vertically according to their place values.

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Addition of $587$ and $398$:

$\begin{array}{cc} & 5 & 8 & 7 \\ + & 3 & 9 & 8 \\ \hline & 9 & 8 & 5 \\ \hline \end{array}$

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Steps:

1. Add the units digits: $7 + 8 = 15$. Write down 5 in the units place and carry over 1 to the tens place.

2. Add the tens digits: $8 + 9 + 1$ (carry-over) $= 18$. Write down 8 in the tens place and carry over 1 to the hundreds place.

3. Add the hundreds digits: $5 + 3 + 1$ (carry-over) $= 9$. Write down 9 in the hundreds place.

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The sum is $985$.

Thus, $587 + 398 = 985$.

Solution:

We need to find the difference between $905$ and $478$. We can perform subtraction by aligning the numbers vertically according to their place values.

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Subtraction of $478$ from $905$:

$\begin{array}{cc} & 9 & 0 & 5 \\ - & 4 & 7 & 8 \\ \hline & 4 & 2 & 7 \\ \hline \end{array}$

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Steps:

1. Subtract the units digits: $5 - 8$. We cannot subtract 8 from 5. We need to borrow from the tens place. The tens digit is 0, so we must borrow from the hundreds place.

2. Borrow from the hundreds: The hundreds digit 9 becomes 8. The tens digit 0 becomes 10.

3. Borrow from the tens: The tens digit 10 becomes 9. The units digit 5 becomes 15.

4. Now, subtract the units digits: $15 - 8 = 7$. Write down 7 in the units place.

5. Subtract the tens digits: $9 - 7 = 2$. Write down 2 in the tens place.

6. Subtract the hundreds digits: $8 - 4 = 4$. Write down 4 in the hundreds place.

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The difference is $427$.

Thus, $905 - 478 = 427$.

Solution:

We need to find the product of $15$ and $60$. We can perform multiplication by aligning the numbers vertically according to their place values.

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Multiplication of $15$ by $60$:

$\begin{array}{cc}& & 1 & 5 \\ \times & & 6 & 0 \\ \hline && 0 & 0 & \textsf{(15 } \times \textsf{ 0)} \\ & 9 & 0 & \times & \textsf{(15 } \times \textsf{ 6, shifted one place left)} \\ \hline & 9 & 0 & 0 \\ \hline \end{array}$

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Steps:

1. Multiply $15$ by the units digit of $60$, which is $0$. $15 \times 0 = 0$. Write down $00$ (or just $0$) in the first partial product row, starting from the units place.

2. Multiply $15$ by the tens digit of $60$, which is $6$. Since $6$ is in the tens place, we are essentially multiplying by $60$. First, write a $0$ in the units place of the second partial product row as a placeholder.

3. Then, multiply $15 \times 6$. $15 \times 6 = 90$. Write $90$ starting from the tens place in the second partial product row, to the left of the placeholder $0$. This gives $900$.

4. Add the partial products: $00 + 900 = 900$.

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Alternatively, we can observe that $60 = 6 \times 10$. So, $15 \times 60 = 15 \times (6 \times 10) = (15 \times 6) \times 10$.

First, calculate $15 \times 6$:

$\begin{array}{cc}& & 1 & 5 \\ \times & & & 6 \\ \hline & & 9 & 0 \\ \hline \end{array}$

$15 \times 6 = 90$.

Now, multiply the result by $10$: $90 \times 10 = 900$.

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The product is $900$.

Thus, $15 \times 60 = 900$.

Solution:

We need to divide $108$ by $9$. We can perform division using the long division method.

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Long division of $108$ by $9$:

$\begin{array}{r} 12 \\ 9{\overline{\smash{\big)}\,108}} \\ \underline{-~\phantom{(}(9}\phantom{0)} \\ 18 \\ \underline{-~\phantom{()}(18)} \\ 0\phantom{)} \end{array}$

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Steps:

1. Divide the first digit of the dividend (1) by the divisor (9). Since 1 is less than 9, we consider the first two digits, which is 10.

2. Divide 10 by 9. The largest multiple of 9 less than or equal to 10 is $9 \times 1 = 9$. Write 1 in the quotient above the digit 0.

3. Subtract 9 from 10: $10 - 9 = 1$. Write 1 below 10.

4. Bring down the next digit from the dividend, which is 8, next to 1. This forms the number 18.

5. Divide 18 by 9. The largest multiple of 9 less than or equal to 18 is $9 \times 2 = 18$. Write 2 in the quotient above the digit 8.

6. Subtract 18 from 18: $18 - 18 = 0$. Write 0 below 18.

The remainder is 0.

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The quotient is $12$ and the remainder is $0$.

Thus, $108 \div 9 = 12$.

Answer:

Yes, addition of whole numbers is commutative.

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Explanation:

An operation is said to be commutative if changing the order of the operands does not change the result.

For addition of whole numbers $a$ and $b$, commutativity means that $a + b = b + a$.

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Example:

Let's take two whole numbers, say $5$ and $8$.

Perform the addition in one order: $5 + 8$

$5 + 8 = 13$

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Now, perform the addition in the reverse order: $8 + 5$

$8 + 5 = 13$

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Since the result is the same in both cases ($13$), we can see that $5 + 8 = 8 + 5$.

This holds true for any pair of whole numbers.

Therefore, addition is commutative for whole numbers.

Answer:

No, subtraction of whole numbers is not commutative.

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Explanation:

For subtraction of whole numbers $a$ and $b$, commutativity would mean that $a - b = b - a$.

However, this is not generally true for whole numbers (unless $a = b$).

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Example:

Let's take two distinct whole numbers, say $10$ and $4$.

Perform the subtraction in one order: $10 - 4$

$10 - 4 = 6$

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Now, perform the subtraction in the reverse order: $4 - 10$

For whole numbers, we cannot subtract a larger number from a smaller number. $4 - 10$ would result in a negative integer ($-6$), which is not a whole number.

Even if we consider integers where subtraction is always possible, the result is different:

$4 - 10 = -6$

---

Since $10 - 4 = 6$ and $4 - 10 = -6$, and $6 \neq -6$, we have $10 - 4 \neq 4 - 10$.

This single example shows that subtraction is not commutative for whole numbers.

Role of 0 in Addition:

The number $0$ is called the additive identity for whole numbers.

---

Explanation:

When $0$ is added to any whole number, the sum is the whole number itself.

Similarly, when any whole number is added to $0$, the sum is the whole number itself.

This means that adding $0$ does not change the identity or value of the whole number.

For any whole number $a$, we have:

$a + 0 = a$

and

$0 + a = a$

---

Example:

Let's take the whole number $15$.

$15 + 0 = 15$

---

Let's take the whole number $0$.

$0 + 0 = 0$

---

Let's take the whole number $100$.

$100 + 0 = 100$

---

In all cases, adding $0$ to a whole number leaves the number unchanged.

Thus, the role of $0$ in the addition of whole numbers is that it serves as the additive identity element.

Role of 1 in Multiplication:

The number $1$ is called the multiplicative identity for whole numbers.

---

Explanation:

When any whole number is multiplied by $1$, the product is the whole number itself.

Similarly, when $1$ is multiplied by any whole number, the product is the whole number itself.

This means that multiplying by $1$ does not change the identity or value of the whole number.

For any whole number $a$, we have:

$a \times 1 = a$

and

$1 \times a = a$

---

Example:

Let's take the whole number $25$.

$25 \times 1 = 25$

---

Let's take the whole number $1$.

$1 \times 1 = 1$

---

Let's take the whole number $0$.

$0 \times 1 = 0$

---

In all cases, multiplying a whole number by $1$ leaves the number unchanged.

Thus, the role of $1$ in the multiplication of whole numbers is that it serves as the multiplicative identity element.

Property of Multiplication by Zero:

The property of multiplication by zero states that the product of any whole number and zero is always zero.

For any whole number $a$, $a \times 0 = 0$.

---

Solution:

We need to find the product of $5432 \times 0$.

Using the property of multiplication by zero, where $a = 5432$, we have:

$5432 \times 0 = 0$

---

Therefore, the product of $5432 \times 0$ is $0$.

Identifying the Pattern:

Let's look at the difference between consecutive numbers in the given sequence:

$20 - 10 = 10$

$30 - 20 = 10$

$40 - 30 = 10$

---

The pattern is that each number is obtained by adding $10$ to the previous number.

---

Writing the next three numbers:

Following this pattern, we add $10$ to the last given number ($40$) to find the next number.

Next number 1: $40 + 10 = 50$

Next number 2: $50 + 10 = 60$

Next number 3: $60 + 10 = 70$

---

The next three numbers in the sequence are $50, 60,$ and $70$.

Identifying the Pattern:

Let's look at the difference between consecutive numbers in the given sequence:

$4 - 1 = 3$

$7 - 4 = 3$

$10 - 7 = 3$

---

The pattern is that each number is obtained by adding $3$ to the previous number.

---

Writing the next three numbers:

Following this pattern, we add $3$ to the last given number ($10$) to find the next number, and repeat this process.

Next number 1: $10 + 3 = 13$

Next number 2: $13 + 3 = 16$

Next number 3: $16 + 3 = 19$

---

The next three numbers in the sequence are $13, 16,$ and $19$.

Identifying the Pattern:

Let's look at the relationship between consecutive numbers in the given sequence.

Check if there is a common difference (arithmetic progression):

$6 - 2 = 4$

$18 - 6 = 12$

$54 - 18 = 36$

The difference is not constant, so it's not an arithmetic progression.

---

Check if there is a common ratio (geometric progression):

$\frac{6}{2} = 3$

$\frac{18}{6} = 3$

$\frac{54}{18} = 3$

---

The pattern is that each number is obtained by multiplying the previous number by $3$.

---

Writing the next three numbers:

Following this pattern, we multiply the last given number ($54$) by $3$ to find the next number, and repeat this process.

Next number 1: $54 \times 3$

$\begin{array}{cc} & 5 & 4 \\ \times & & 3 \\ \hline 1 & 6 & 2 \\ \hline \end{array}$

$54 \times 3 = 162$

---

Next number 2: $162 \times 3$

$\begin{array}{cc} & 1 & 6 & 2 \\ \times & & & 3 \\ \hline & 4 & 8 & 6 \\ \hline \end{array}$

$162 \times 3 = 486$

---

Next number 3: $486 \times 3$

$\begin{array}{cc} & 4 & 8 & 6 \\ \times & & & 3 \\ \hline 1 & 4 & 5 & 8 \\ \hline \end{array}$

$486 \times 3 = 1458$

---

The next three numbers in the sequence are $162, 486,$ and $1458$.

Definition of a Triangular Number:

A triangular number is a number which can be represented as the sum of consecutive natural numbers starting from 1.

These numbers can be visualized as dots arranged in an equilateral triangle.

The $n$-th triangular number ($T_n$) is given by the sum of the first $n$ natural numbers, which can be calculated using the formula:

$T_n = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

---

First Triangular Number:

The first triangular number is obtained by taking the sum of the first 1 natural number, which is simply 1.

$T_1 = 1$

Using the formula: $T_1 = \frac{1(1+1)}{2} = \frac{1 \times 2}{2} = \frac{2}{2} = 1$

---

Diagram for the First Triangular Number:

The first triangular number, 1, can be represented by a single dot.

•

(Represents the number 1)

Definition of a Square Number:

A square number is a number which is the product of an integer multiplied by itself. In other words, it is the square of an integer.

These numbers can be visualized as dots arranged in the shape of a square.

The $n$-th square number is given by the square of the $n$-th natural number, which is $n \times n$ or $n^2$.

---

Second Square Number:

The second square number is obtained by squaring the second natural number, which is 2.

Second natural number = 2

Second square number = $2 \times 2 = 2^2 = 4$

---

Diagram for the Second Square Number:

The second square number, 4, can be represented by arranging 4 dots in a square shape (2 rows and 2 columns).

• •
• •

(Represents the number 4)

Distributive Property of Multiplication over Addition:

The distributive property states that for any three whole numbers $a$, $b$, and $c$, the following equality holds:

$a \times (b + c) = (a \times b) + (a \times c)$

---

Solution:

We need to find the product $8 \times (10 + 2)$ using the distributive property.

Here, we can consider $a = 8$, $b = 10$, and $c = 2$.

Applying the distributive property:

$8 \times (10 + 2) = (8 \times 10) + (8 \times 2)$

---

Now, we calculate the products within the parentheses:

$8 \times 10 = 80$

$8 \times 2 = 16$

---

Substitute these values back into the equation:

$8 \times (10 + 2) = 80 + 16$

---

Finally, perform the addition:

$80 + 16 = 96$

---

So, $8 \times (10 + 2) = 96$.

---

Alternatively, we can first calculate the sum inside the parentheses and then multiply:

$8 \times (10 + 2) = 8 \times 12$

$\begin{array}{cc}& & 1 & 2 \\ \times & & & 8 \\ \hline & 9 & 6 \\ \hline \end{array}$

$8 \times 12 = 96$

This confirms the result obtained using the distributive property.

---

The product of $8 \times (10 + 2)$ using the distributive property is $96$.

Concept of Whole Numbers:

Whole numbers are the set of non-negative integers. They are the natural numbers ($1, 2, 3, ...$) combined with the number zero ($0$).

The set of whole numbers is $\{0, 1, 2, 3, 4, ...\}$.

The smallest whole number is $0$. There is no largest whole number as they extend infinitely.

---

Representation on a Number Line:

We can represent whole numbers on a number line by drawing a straight line and marking points at equal distances. We start with $0$ and label the points $1, 2, 3,$ and so on, moving towards the right.

Here is the representation of whole numbers up to $10$ on a number line:

•——•——•——•——•——•——•——•——•——•———>

$0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$

---

Comparing $4, 9,$ and $2$ using the Number Line:

On a number line, a number is smaller than another number if it is located to its left. Conversely, a number is larger than another number if it is located to its right.

Let's locate the numbers $4, 9,$ and $2$ on the number line:

•——•——•——•——•——•——•——•——•——•———>

$0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$

Observing their positions:

• $2$ is to the left of $4$. This means $2 < 4$.
• $2$ is to the left of $9$. This means $2 < 9$.
• $4$ is to the left of $9$. This means $4 < 9$.

The order from left to right on the number line is $2, 4, 9$.

---

Writing in Ascending Order:

Ascending order means arranging numbers from smallest to largest.

Based on the comparison using the number line, the smallest number is $2$, followed by $4$, and the largest is $9$.

Therefore, the numbers $4, 9,$ and $2$ in ascending order are:

$2, 4, 9$

Successor:

The successor of a whole number is the number that comes immediately after it.

To find the successor of any whole number, we add $1$ to the number.

If $a$ is a whole number, its successor is $a + 1$.

---

Predecessor:

The predecessor of a whole number (other than $0$) is the number that comes immediately before it.

To find the predecessor of any whole number (greater than $0$), we subtract $1$ from the number.

If $a$ is a whole number greater than $0$, its predecessor is $a - 1$.

---

Finding the Successor of $5999$:

Successor of $5999 = 5999 + 1$

$\begin{array}{cc} & 5 & 9 & 9 & 9 \\ + & & & & 1 \\ \hline & 6 & 0 & 0 & 0 \\ \hline \end{array}$

The successor of $5999$ is $6000$.

---

Finding the Predecessor of $7000$:

Predecessor of $7000 = 7000 - 1$

$\begin{array}{cc} & 7 & 0 & 0 & 0 \\ - & & & & 1 \\ \hline & 6 & 9 & 9 & 9 \\ \hline \end{array}$

The predecessor of $7000$ is $6999$.

---

Whole numbers without a successor or predecessor:

Every whole number has a successor. If you take any whole number $n$, $n+1$ is always a whole number. Since whole numbers go on infinitely ($0, 1, 2, ...$), you can always find a number greater than any given whole number by adding 1.

Every whole number except $0$ has a predecessor that is also a whole number. For any whole number $n > 0$, $n-1$ is a whole number.

However, the whole number $0$ does not have a predecessor that is a whole number. If we subtract $1$ from $0$, we get $0 - 1 = -1$, which is an integer but not a whole number.

---

So, there are no whole numbers that do not have a successor.

There is one whole number that does not have a predecessor within the set of whole numbers, which is $0$.

Commutative Property:

The commutative property states that the order of operands does not affect the result of the operation.

For whole numbers $a$ and $b$:

Addition: $a + b = b + a$

Multiplication: $a \times b = b \times a$

---

Verification for $15$ and $23$:

Addition:

LHS: $15 + 23$

$\begin{array}{cc} & 1 & 5 \\ + & 2 & 3 \\ \hline & 3 & 8 \\ \hline \end{array}$

$15 + 23 = 38$

---

RHS: $23 + 15$

$\begin{array}{cc} & 2 & 3 \\ + & 1 & 5 \\ \hline & 3 & 8 \\ \hline \end{array}$

$23 + 15 = 38$

Since $15 + 23 = 23 + 15 = 38$, the commutative property of addition is verified for $15$ and $23$.

---

Multiplication:

LHS: $15 \times 23$

$\begin{array}{cc}& & 1 & 5 \\ \times & & 2 & 3 \\ \hline && 4 & 5 \textsf{ (15} \times \textsf{ 3)} \\ & 3 & 0 & \times \textsf{ (15} \times \textsf{ 2, shifted)} \\ \hline & 3 & 4 & 5 \\ \hline \end{array}$

$15 \times 23 = 345$

---

RHS: $23 \times 15$

$\begin{array}{cc}& & 2 & 3 \\ \times & & 1 & 5 \\ \hline & 1 & 1 & 5 \textsf{ (23} \times \textsf{ 5)} \\ & 2 & 3 & \times \textsf{ (23} \times \textsf{ 1, shifted)} \\ \hline & 3 & 4 & 5 \\ \hline \end{array}$

$23 \times 15 = 345$

Since $15 \times 23 = 23 \times 15 = 345$, the commutative property of multiplication is verified for $15$ and $23$.

---

Associative Property:

The associative property states that the way operands are grouped does not affect the result of the operation.

For whole numbers $a$, $b$, and $c$:

Addition: $(a + b) + c = a + (b + c)$

Multiplication: $(a \times b) \times c = a \times (b \times c)$

---

Verification for $5, 8,$ and $12$:

Addition:

LHS: $(5 + 8) + 12$

First, calculate $5 + 8 = 13$.

Then, calculate $13 + 12 = 25$.

$(5 + 8) + 12 = 25$

---

RHS: $5 + (8 + 12)$

First, calculate $8 + 12 = 20$.

Then, calculate $5 + 20 = 25$.

$5 + (8 + 12) = 25$

Since $(5 + 8) + 12 = 5 + (8 + 12) = 25$, the associative property of addition holds for $5, 8,$ and $12$.

---

Multiplication:

LHS: $(5 \times 8) \times 12$

First, calculate $5 \times 8 = 40$.

Then, calculate $40 \times 12$.

$\begin{array}{cc}& & 4 & 0 \\ \times & & 1 & 2 \\ \hline && 8 & 0 \textsf{ (40} \times \textsf{ 2)} \\ & 4 & 0 & \times \textsf{ (40} \times \textsf{ 1, shifted)} \\ \hline & 4 & 8 & 0 \\ \hline \end{array}$

$(5 \times 8) \times 12 = 480$

---

RHS: $5 \times (8 \times 12)$

First, calculate $8 \times 12 = 96$.

Then, calculate $5 \times 96$.

$\begin{array}{cc}& & 9 & 6 \\ \times & & & 5 \\ \hline & 4 & 8 & 0 \\ \hline \end{array}$

$5 \times (8 \times 12) = 480$

Since $(5 \times 8) \times 12 = 5 \times (8 \times 12) = 480$, the associative property of multiplication holds for $5, 8,$ and $12$.

Distributive Property of Multiplication over Addition:

The distributive property states that for any three whole numbers $a$, $b$, and $c$, the multiplication of a number by the sum of two other numbers is equal to the sum of the products of the number with each of the other two numbers separately.

Mathematically, it is expressed as:

$a \times (b + c) = (a \times b) + (a \times c)$

The property also extends to subtraction: $a \times (b - c) = (a \times b) - (a \times c)$.

---

Calculation of $25 \times 102$ using the Distributive Property:

Way 1: Split $102$ as $100 + 2$

$25 \times 102 = 25 \times (100 + 2)$

Using the distributive property $a \times (b + c) = (a \times b) + (a \times c)$, with $a=25$, $b=100$, and $c=2$:

$25 \times (100 + 2) = (25 \times 100) + (25 \times 2)$

$= 2500 + 50$

$= 2550$

---

Way 2: Split $25$ as $20 + 5$

$25 \times 102 = (20 + 5) \times 102$

Using the distributive property $(a + b) \times c = (a \times c) + (b \times c)$, with $a=20$, $b=5$, and $c=102$:

$(20 + 5) \times 102 = (20 \times 102) + (5 \times 102)$

$= (20 \times (100 + 2)) + (5 \times (100 + 2))$

$= (20 \times 100 + 20 \times 2) + (5 \times 100 + 5 \times 2)$

$= (2000 + 40) + (500 + 10)$

$= 2040 + 510$

$= 2550$

---

The product $25 \times 102$ is $2550$.

---

Calculation of $18 \times 99$ using the Distributive Property:

Way 1: Split $99$ as $100 - 1$

$18 \times 99 = 18 \times (100 - 1)$

Using the distributive property $a \times (b - c) = (a \times b) - (a \times c)$, with $a=18$, $b=100$, and $c=1$:

$18 \times (100 - 1) = (18 \times 100) - (18 \times 1)$

$= 1800 - 18$

$\begin{array}{cc} & 1 & 8 & 0 & 0 \\ - & & & 1 & 8 \\ \hline & 1 & 7 & 8 & 2 \\ \hline \end{array}$

$= 1782$

---

Way 2: Split $18$ as $10 + 8$

$18 \times 99 = (10 + 8) \times 99$

Using the distributive property $(a + b) \times c = (a \times c) + (b \times c)$, with $a=10$, $b=8$, and $c=99$:

$(10 + 8) \times 99 = (10 \times 99) + (8 \times 99)$

$= 990 + (8 \times (100 - 1))$

$= 990 + (8 \times 100 - 8 \times 1)$

$= 990 + (800 - 8)$

$= 990 + 792$

$\begin{array}{cc} & 9 & 9 & 0 \\ + & 7 & 9 & 2 \\ \hline 1 & 7 & 8 & 2 \\ \hline \end{array}$

$= 1782$

---

The product $18 \times 99$ is $1782$.

Division Algorithm for Whole Numbers:

The division algorithm states that for any two whole numbers $a$ (dividend) and $b$ (divisor), where $b \neq 0$, there exist unique whole numbers $q$ (quotient) and $r$ (remainder) such that:

$a = bq + r$

where $0 \leq r < b$.

---

In this equation:

• $a$ is the dividend (the number being divided).
• $b$ is the divisor (the number by which we are dividing).
• $q$ is the quotient (the whole number of times the divisor goes into the dividend).
• $r$ is the remainder (the amount left over after the division).

The condition $0 \leq r < b$ means that the remainder must be a whole number and must be strictly less than the divisor.

---

Division of $157$ by $12$:

We need to divide $157$ by $12$. Here, $a = 157$ and $b = 12$. We will use long division to find $q$ and $r$.

$\begin{array}{r} 13\phantom{)} \\ 12{\overline{\smash{\big)}\,157\phantom{)}} }\\ \underline{-~\phantom{(}(12}\phantom{7)} \\ 37\phantom{)} \\ \underline{-~\phantom{()}(36)} \\ 1\phantom{)} \end{array}$

---

From the long division:

• The divisor $12$ goes into $15$ one time ($12 \times 1 = 12$). $15 - 12 = 3$.
• Bring down the next digit, $7$, to get $37$.
• The divisor $12$ goes into $37$ three times ($12 \times 3 = 36$). $37 - 36 = 1$.
• The remainder is $1$.

So, the quotient $q = 13$ and the remainder $r = 1$. The dividend $a = 157$ and the divisor $b = 12$.

---

We can verify this with the division algorithm formula:

$a = bq + r$

$157 = 12 \times 13 + 1$

$157 = 156 + 1$

$157 = 157$

This confirms our division is correct, and $0 \leq 1 < 12$, satisfying the remainder condition.

---

Identifying the parts:

• Dividend ($a$): $157$
• Divisor ($b$): $12$
• Quotient ($q$): $13$
• Remainder ($r$): $1$

---

Division of a whole number by $0$:

Division by zero is undefined.

---

Reasoning:

Let's consider the definition of division. Division is the inverse operation of multiplication.

When we say $a \div b = q$, it means that $b \times q = a$.

Now, let's consider dividing a whole number $a$ by $0$. Suppose $a \div 0 = q$. This would imply that $0 \times q = a$.

We know that the product of any number and zero is always zero ($0 \times q = 0$).

So, the equation $0 \times q = a$ becomes $0 = a$.

---

Case 1: If $a$ is a non-zero whole number (e.g., $5 \div 0$).

If $5 \div 0 = q$, then $0 \times q = 5$. But $0 \times q = 0$. So, $0 = 5$, which is false. There is no number $q$ that can satisfy this condition. Thus, division of a non-zero number by zero is impossible.

---

Case 2: If $a$ is zero (i.e., $0 \div 0$).

If $0 \div 0 = q$, then $0 \times q = 0$. This equation is true for any value of $q$ (any number multiplied by 0 is 0). This means there is no unique answer for $q$. Thus, division of zero by zero is indeterminate (meaning it can be any number, but a single value cannot be determined).

---

Because division by zero either leads to a contradiction (for non-zero dividend) or infinitely many possibilities (for zero dividend), it is not allowed and is considered undefined in mathematics.

Given:

Starting quantity of rice = $1000$ kg

Quantity of rice sold on Monday = $350$ kg

Quantity of rice sold on Tuesday = $425$ kg

---

To Find:

Quantity of rice left at the end of Tuesday.

---

Solution:

First, we need to calculate the total quantity of rice sold over the two days.

Total rice sold = Rice sold on Monday + Rice sold on Tuesday

Then, subtract the total rice sold from the initial quantity to find the remaining rice.

Rice left = Starting quantity - Total rice sold

---

Calculations:

Total quantity of rice sold:

Total sold = $350 + 425$ kg

$\begin{array}{cc} & 3 & 5 & 0 \\ + & 4 & 2 & 5 \\ \hline & 7 & 7 & 5 \\ \hline \end{array}$

Total quantity of rice sold = $775$ kg

---

Quantity of rice left:

Rice left = $1000 - 775$ kg

$\begin{array}{cc} & 1 & 0 & 0 & 0 \\ - & & 7 & 7 & 5 \\ \hline & & 2 & 2 & 5 \\ \hline \end{array}$

Quantity of rice left = $225$ kg

---

Answer:

The shopkeeper is left with $225$ kg of rice at the end of Tuesday.

Given:

Number of television sets produced per day = $150$

Number of working days in a month = $25$ days

Number of large containers = $5$

---

To Find:

1. Total television sets produced in the month.

2. Number of television sets in each container.

---

Solution:

Part 1: Total production in the month

Total production = Production per day $\times$ Number of working days

Total production = $150 \times 25$

$\begin{array}{cc}& & 1 & 5 & 0 \\ \times & & & 2 & 5 \\ \hline && 7 & 5 & 0 \textsf{ (150} \times \textsf{ 5)} \\ & 3 & 0 & 0 & \times \textsf{ (150} \times \textsf{ 2, shifted)} \\ \hline 3 & 7 & 5 & 0 \\ \hline \end{array}$

Total television sets produced in the month = $3750$

---

Part 2: Television sets per container

Since the total production is packed equally into $5$ containers, we divide the total production by the number of containers.

Television sets per container = Total production $\div$ Number of containers

Television sets per container = $3750 \div 5$

$\begin{array}{r} 750\phantom{)} \\ 5{\overline{\smash{\big)}\,3750\phantom{)}} }\\ \underline{-~\phantom{(}(35}\phantom{50)} \\ 25\phantom{0)} \\ \underline{-~\phantom{()}(25}\phantom{0)} \\ 00\phantom{)} \\ \underline{-~\phantom{()}(00)} \\ 0\phantom{)} \end{array}$

Television sets per container = $750$

---

Answer:

The factory produces $3750$ television sets in that month.

There are $750$ television sets in each container.

Expression to simplify:

$125 \times 45 \times 8$

---

Property Used:

We will use the Associative Property of Multiplication. This property states that the grouping of factors in a multiplication does not change the product. For whole numbers $a$, $b$, and $c$, $(a \times b) \times c = a \times (b \times c)$.

We can also implicitly use the Commutative Property of Multiplication ($a \times b = b \times a$) to rearrange the factors before grouping, although the associative property is the primary tool for changing the grouping.

---

Simplification:

We can rearrange the factors to group $125$ and $8$ together, as their product ($125 \times 8$) results in a round number ($1000$), which makes the subsequent multiplication easier.

Original expression: $125 \times 45 \times 8$

Rearrange using Commutative Property: $125 \times 8 \times 45$

Group using Associative Property: $(125 \times 8) \times 45$

---

Now, calculate the product inside the parentheses:

$125 \times 8 = 1000$

---

Substitute this value back into the expression:

$(125 \times 8) \times 45 = 1000 \times 45$}

---

Now, perform the final multiplication:

$1000 \times 45 = 45000$

---

Thus, the simplified value of the expression is $45000$.

Identifying the Rule:

Let's examine the differences between consecutive terms in the pattern $1, 3, 6, 10, 15, ...$:

$3 - 1 = 2$

$6 - 3 = 3$

$10 - 6 = 4$

$15 - 10 = 5$

---

The difference between consecutive terms increases by $1$ each time ($2, 3, 4, 5, ...$).

This means that each term is obtained by adding a consecutive natural number to the previous term, starting with adding $2$ to the first term.

Alternatively, the pattern can be described as the sum of consecutive natural numbers starting from 1:

1st term: $1 = 1$

2nd term: $3 = 1 + 2$

3rd term: $6 = 1 + 2 + 3$}

4th term: $10 = 1 + 2 + 3 + 4$}

5th term: $15 = 1 + 2 + 3 + 4 + 5$}

The rule is that the $n$-th term is the sum of the first $n$ natural numbers, which can be calculated by the formula $T_n = \frac{n(n+1)}{2}$.

---

Diagrams for the first four terms:

These numbers can be represented visually as arrangements of dots forming equilateral triangles.

1st term ($1$):

•

---

2nd term ($3$):

•
• •

---

3rd term ($6$):

•
• •
• • •

---

4th term ($10$):

•
• •
• • •
• • • •

---

Type of Numbers:

Numbers that can be represented as the sum of consecutive natural numbers starting from 1 and can be arranged in the shape of an equilateral triangle are called triangular numbers.

Identifying the Rule:

Let's examine the numbers in the pattern: $1, 4, 9, 16, 25, ...$

We can see that these are the results of squaring consecutive natural numbers:

• $1 = 1 \times 1 = 1^2$ (1st term)
• $4 = 2 \times 2 = 2^2$ (2nd term)
• $9 = 3 \times 3 = 3^2$ (3rd term)
• $16 = 4 \times 4 = 4^2$ (4th term)
• $25 = 5 \times 5 = 5^2$ (5th term)

---

The rule used to form this pattern is that the $n$-th term is the square of the natural number $n$.

The rule can be written as: $n$-th term $= n^2$.}

---

Finding the next three terms:

The last given term is the 5th term ($5^2 = 25$). The next three terms will be the 6th, 7th, and 8th terms.

6th term: $6^2 = 6 \times 6 = 36$

7th term: $7^2 = 7 \times 7 = 49$

8th term: $8^2 = 8 \times 8 = 64$

---

The next three terms in this pattern are $36, 49,$ and $64$.

---

Geometric Shape formed by dots:

Numbers in this pattern are called square numbers because they can be represented by arranging dots in the shape of a perfect square.

• 1 dot forms a $1 \times 1$ square.
• 4 dots form a $2 \times 2$ square.
• 9 dots form a $3 \times 3$ square.
• 16 dots form a $4 \times 4$ square, and so on.

The geometric shape that can be formed with dots corresponding to these numbers is a square.