Key Concept: Rational Numbers, Equations

b) $\frac{2}{5}$

[Solution Description]

First, add $\frac{1}{4}$ to both sides of the equation:

$\frac{5}{2}x = \frac{3}{4} + \frac{1}{4}$

Simplify the right side:

$\frac{5}{2}x = \frac{4}{4} = 1$

Now, multiply both sides by $\frac{2}{5}$ to solve for $x$:

$x = 1 \times \frac{2}{5} = \frac{2}{5}$

The solution is $x = \frac{2}{5}$.

Key Concept: Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the equation $2x = 3$ has no solution in the set of integers. To verify this, we solve the equation:

$2x = 3 \implies x = \frac{3}{2}$

The solution $\frac{3}{2}$ is not an integer, so the assertion is true.

The reason states that rational numbers are required to solve equations of the form $ax = b$ where $a$ and $b$ are integers and $a \neq 0$. This is also true because, as seen in the example, the solution $\frac{3}{2}$ is a rational number. Furthermore, the reason correctly explains why the assertion is true since the solution requires the use of rational numbers.

Key Concept: Definition of Rational Numbers

d) $\frac{\sqrt{4}}{2}$

[Solution Description]

A number which can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, is called a rational number.

$\sqrt{3}$ cannot be expressed as a fraction of integers, so it is not a rational number.

0 can be written as $\frac{0}{1}$, making it a rational number.

$\pi$ is an irrational number that cannot be expressed as a simple fraction.

$\frac{\sqrt{4}}{2} = \frac{2}{2} = 1$, which is a rational number because it can be expressed as $\frac{1}{1}$.

Therefore, the correct answer is $\frac{\sqrt{4}}{2}$.

Key Concept: Definition of Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that $-\frac{7}{5}$ is a rational number. This is true because it can be expressed in the form $\frac{p}{q}$ where $p = -7$ and $q = 5$, both integers with $q \neq 0$. The reason defines a rational number as any number that can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. This is also true and correctly explains why $-\frac{7}{5}$ is a rational number.

Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Need for Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that the equation $2x = 3$ cannot be solved using integers. This is true because when we solve the equation, we get $x = \frac{3}{2}$, which is not an integer but a rational number. The reason states that the solution to the equation $2x = 3$ requires a rational number $\frac{3}{2}$. This is also true and correctly explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Rational Numbers, Equations

c) $\frac{1}{20}$

[Solution Description]

To solve the equation $\frac{5}{2}x + \frac{3}{4} = \frac{7}{8}$, follow these steps:

Subtract $\frac{3}{4}$ from both sides: $\frac{5}{2}x = \frac{7}{8} - \frac{3}{4}$

Convert $\frac{3}{4}$ to eighths: $\frac{3}{4} = \frac{6}{8}$

Subtract: $\frac{7}{8} - \frac{6}{8} = \frac{1}{8}$

Now, the equation is $\frac{5}{2}x = \frac{1}{8}$

Solve for $x$ by multiplying both sides by $\frac{2}{5}$: $x = \frac{1}{8} \times \frac{2}{5} = \frac{2}{40} = \frac{1}{20}$

The solution is $x = \frac{1}{20}$.

Key Concept: Rational Numbers

b) $\frac{3}{2}$

[Solution Description]

To solve for $x$, divide both sides of the equation by $2$: $x = \frac{3}{2}$. The solution $\frac{3}{2}$ is a rational number.

Key Concept: Rational Numbers

c) $\frac{7}{8}$

[Solution Description]

A rational number is any number that can be expressed as the quotient $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

Key Concept: Representation of Rational Numbers on a Number Line

c) $\frac{7}{8}$

[Solution Description]

To determine which rational number is closest to $1$, we calculate the absolute difference between each number and $1$:

- For $\frac{3}{4}$, the difference is $|1 - \frac{3}{4}| = \frac{1}{4}$.

- For $\frac{5}{4}$, the difference is $|\frac{5}{4} - 1| = \frac{1}{4}$.

- For $\frac{7}{8}$, the difference is $|1 - \frac{7}{8}| = \frac{1}{8}$.

- For $\frac{9}{8}$, the difference is $|\frac{9}{8} - 1| = \frac{1}{8}$.

Since both $\frac{7}{8}$ and $\frac{9}{8}$ have the smallest difference of $\frac{1}{8}$, they are equally close to $1$. However, considering only one option, the closest is $\frac{7}{8}$.

Key Concept: Representation of Rational Numbers on a Number Line

c) Assertion is true, but Reason is false.

[Solution Description]

To determine whether the assertion and reason are correct, let's analyze the given rational numbers. First, compare $\frac{3}{4}$ and $\frac{5}{6}$. Convert these fractions to have a common denominator for easier comparison. The least common denominator (LCD) of 4 and 6 is 12.

Now, convert $\frac{3}{4}$ to twelfths: $\frac{3}{4} = \frac{9}{12}$. Similarly, convert $\frac{5}{6}$ to twelfths: $\frac{5}{6} = \frac{10}{12}$. Since $\frac{9}{12} < \frac{10}{12}$, it follows that $\frac{3}{4}$ lies to the left of $\frac{5}{6}$ on the number line. Thus, the Assertion is true.

However, the Reason states that when two rational numbers are plotted on the number line, the one with the smaller denominator is always to the left. This is not universally true. For example, consider $\frac{1}{2}$ and $\frac{1}{3}$. Here, $\frac{1}{2}$ has a smaller denominator than $\frac{1}{3}$, but $\frac{1}{2}$ lies to the right of $\frac{1}{3}$ on the number line. Therefore, the Reason is false.

In conclusion, the Assertion is true, but the Reason is false.

Key Concept: Additive Identity, Multiplicative Identity

a) $0$

[Solution Description]

Here, we use the multiplicative identity ($q = 1$) and the additive identity ($0$). First, compute $p \times q$:

$p \times q = \frac{9}{11} \times 1 = \frac{9}{11}$

Now, subtract $p$:

$p \times q - p = \frac{9}{11} - \frac{9}{11} = 0$

Finally, add $0$:

$p \times q - p + 0 = 0 + 0 = 0$

So, the result is $0$.

Key Concept: Closure Property of Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the sum of any two rational numbers is always a rational number. This is true because, by definition, rational numbers are closed under addition, meaning that adding any two rational numbers will result in another rational number. The reason provided is also true as it correctly states that rational numbers are closed under addition. Furthermore, the reason directly explains why the assertion is true. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Closure under Addition

b) $\frac{-1}{12}$

[Solution Description]

To find the sum of two rational numbers, we add them by finding a common denominator. Here, the denominators are 4 and 6. The least common denominator (LCD) is 12. Convert each fraction to have the LCD:

$a = \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

$b = \frac{-5}{6} = \frac{-5 \times 2}{6 \times 2} = \frac{-10}{12}$

Now, add the two fractions:

$a + b = \frac{9}{12} + \frac{-10}{12} = \frac{9 - 10}{12} = \frac{-1}{12}$

Key Concept: Closure Property, Multiplication of Rational Numbers

b) $\frac{-5}{12}$

[Solution Description]

To find the product of $\frac{2}{3}$ and $\frac{-5}{8}$, multiply the numerators together and the denominators together: $\frac{2}{3} \times \frac{-5}{8} = \frac{2 \times (-5)}{3 \times 8} = \frac{-10}{24}$. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor (GCD), which is 2: $\frac{-10}{24} = \frac{-5}{12}$. Thus, the product of $\frac{2}{3}$ and $\frac{-5}{8}$ is $\frac{-5}{12}$.

Key Concept: Closure under Addition, Rational Numbers

a) $\frac{1}{10}$

[Solution Description]

To find the sum of $\frac{7}{10}$ and $\frac{-3}{5}$, first find a common denominator. The least common multiple of 10 and 5 is 10. Convert both fractions:

$\frac{7}{10} = \frac{7 \times 1}{10 \times 1} = \frac{7}{10}$

$\frac{-3}{5} = \frac{-3 \times 2}{5 \times 2} = \frac{-6}{10}$

Now, add the two fractions:

$\frac{7}{10} + \frac{-6}{10} = \frac{1}{10}$

Thus, the value of $p + q$ is $\frac{1}{10}$.

Key Concept: Closure Property of Rational Numbers

a) $\frac{16}{15}$

[Solution Description] To find the sum of $\frac{3}{5}$ and $\frac{7}{15}$, we first convert them to have a common denominator. The least common denominator (LCD) for 5 and 15 is 15. Convert $\frac{3}{5}$ to $\frac{9}{15}$. Now, add $\frac{9}{15}$ and $\frac{7}{15}$: $\frac{9 + 7}{15} = \frac{16}{15}$. Thus, the sum is $\frac{16}{15}$, which is a rational number.

Key Concept: Closure under Subtraction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

According to the property of rational numbers, the difference between any two rational numbers $a$ and $b$ is also a rational number. This means that if you subtract one rational number from another, the result will still be a rational number. Therefore, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Closure under Subtraction

c) $\frac{13}{12}$

[Solution Description] To calculate the difference between $\frac{11}{6}$ and $\frac{3}{4}$, we first find a common denominator. The LCD of 6 and 4 is 12. Converting $\frac{11}{6}$ to a denominator of 12, we get $\frac{11 \times 2}{6 \times 2} = \frac{22}{12}$. Similarly, converting $\frac{3}{4}$ to a denominator of 12, we get $\frac{3 \times 3}{4 \times 3} = \frac{9}{12}$. Now, subtract $\frac{9}{12}$ from $\frac{22}{12}$: $\frac{22}{12} - \frac{9}{12} = \frac{13}{12}$. Therefore, the difference is $\frac{13}{12}$.

Key Concept: Closure under Multiplication

a) $\frac{10}{33}$

[Solution Description] To find the product of two rational numbers, multiply the numerators together and the denominators together. So, $\frac{6}{11} \times \frac{5}{9} = \frac{6 \times 5}{11 \times 9} = \frac{30}{99}$. Simplifying $\frac{30}{99}$ by dividing numerator and denominator by 3 gives $\frac{10}{33}$. Thus, the product is $\frac{10}{33}$, which is a rational number.

Key Concept: Closure under Multiplication

b) $\frac{10}{21}$

[Solution Description] To find the product of two rational numbers, multiply the numerators together and the denominators together. So, $\frac{2}{3} \times \frac{5}{7} = \frac{2 \times 5}{3 \times 7} = \frac{10}{21}$. Thus, the product is $\frac{10}{21}$, which is a rational number.

Key Concept: Closure under Division

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the set of rational numbers is closed under division when zero is excluded. This is true because division by zero is undefined, and excluding zero ensures that every non-zero rational number has a rational reciprocal, making the set closed under division. The reason correctly explains why exclusion of zero is necessary for closure under division. Both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Associative Property of Division, Rational Numbers

c) $p > q$

[Solution Description]

First, calculate $p$:

$p = \left( \frac{3}{4} \div \frac{2}{5} \right) \div \frac{1}{6} = \left( \frac{3}{4} \times \frac{5}{2} \right) \div \frac{1}{6} = \frac{15}{8} \div \frac{1}{6} = \frac{15}{8} \times 6 = \frac{90}{8} = \frac{45}{4}$

Next, calculate $q$:

$q = \frac{3}{4} \div \left( \frac{2}{5} \div \frac{1}{6} \right) = \frac{3}{4} \div \left( \frac{2}{5} \times 6 \right) = \frac{3}{4} \div \frac{12}{5} = \frac{3}{4} \times \frac{5}{12} = \frac{15}{48} = \frac{5}{16}$

Comparing $p$ and $q$, we find that $p \neq q$. This demonstrates that division is not associative for rational numbers.

Key Concept: Commutativity of Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The syllabus states that addition is commutative for rational numbers, meaning that the order of addition does not affect the result. Therefore, the assertion is true. The reason provided correctly explains this property by stating that for any two rational numbers $a$ and $b$, $a + b = b + a$. This is the definition of commutativity in addition. Hence, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Subtraction, Commutativity

d) $\frac{7}{8} - \frac{2}{3} \neq \frac{2}{3} - \frac{7}{8}$

[Solution Description]

To show that subtraction is not commutative for rational numbers, we need to demonstrate that $x - y \neq y - x$. Let's compute both sides:

$x - y = \frac{7}{8} - \frac{2}{3} = \frac{21}{24} - \frac{16}{24} = \frac{5}{24}$

$y - x = \frac{2}{3} - \frac{7}{8} = \frac{16}{24} - \frac{21}{24} = \frac{-5}{24}$

Since $\frac{5}{24} \neq \frac{-5}{24}$, subtraction is not commutative for rational numbers. The correct expression is $\frac{7}{8} - \frac{2}{3} \neq \frac{2}{3} - \frac{7}{8}$.

Key Concept: Commutative Property of Addition

a) $\frac{5}{12} + \frac{3}{4} = \frac{3}{4} + \frac{5}{12}$

[Solution Description]

According to the commutative property of addition, for any two rational numbers $a$ and $b$, $a + b = b + a$. Thus, for $\frac{5}{12}$ and $\frac{3}{4}$, the equation $\frac{5}{12} + \frac{3}{4} = \frac{3}{4} + \frac{5}{12}$ must be correct.

Key Concept: Commutative Property of Addition, Rational Numbers

c) $0$

[Solution Description]

By the commutative property of addition, $p + q = q + p$. Therefore, $(p + q) - (q + p) = 0$.

Key Concept: Commutative Property of Multiplication for Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that for any two rational numbers $a$ and $b$, $a \times b = b \times a$. This is the definition of the commutative property of multiplication, which holds true for all rational numbers. The reason explains the commutative property by stating that the order of multiplication does not affect the product, which is a correct explanation of the assertion. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Commutative Property of Multiplication

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The commutative property of multiplication for rational numbers states that for any two rational numbers $a$ and $b$, the product $a \times b$ is equal to $b \times a$. This property is fundamental in arithmetic and algebra, ensuring that the order in which numbers are multiplied does not affect the result. The assertion correctly describes this property, and the reason accurately explains why the assertion holds true.

Key Concept: Non-commutativity of Subtraction

b) $(x - y) - (y - x) = 1$

[Solution Description]

To determine which statement is true, we evaluate each option step by step.

First, compute $x - y$:

$x - y = \frac{5}{6} - \frac{1}{3} = \frac{5}{6} - \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$

Next, compute $y - x$:

$y - x = \frac{1}{3} - \frac{5}{6} = \frac{2}{6} - \frac{5}{6} = -\frac{3}{6} = -\frac{1}{2}$

Now, compute $(x - y) + (y - x)$:

$(x - y) + (y - x) = \frac{1}{2} + \left(-\frac{1}{2}\right) = 0$

Finally, compute $(x - y) - (y - x)$:

$(x - y) - (y - x) = \frac{1}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1$

Therefore, the statement $(x - y) - (y - x) = 1$ is true.

Key Concept: Non-commutativity of Subtraction

d) Assertion is false, but Reason is true.

[Solution Description]

To check if subtraction is commutative for rational numbers, let's evaluate both expressions:

$\frac{2}{5} - \frac{5}{2} = \frac{4}{10} - \frac{25}{10} = \frac{4 - 25}{10} = \frac{-21}{10}$

$\frac{5}{2} - \frac{2}{5} = \frac{25}{10} - \frac{4}{10} = \frac{25 - 4}{10} = \frac{21}{10}$

Since $\frac{-21}{10} \neq \frac{21}{10}$, subtraction is not commutative for rational numbers. Therefore, the Assertion is false, but the Reason is true.

Key Concept: Non-commutativity of Division

b) No

[Solution Description]

Division is not commutative for rational numbers. For example, $\frac{5}{4} \div \frac{3}{7} = \frac{35}{12}$ and $\frac{3}{7} \div \frac{5}{4} = \frac{12}{35}$. Since $\frac{35}{12} \neq \frac{12}{35}$, division is not commutative for rational numbers.

Key Concept: Non-commutativity of Division for Rational Numbers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

Let us verify the assertion using the example $\frac{5}{4} \div \frac{3}{7}$ and $\frac{3}{7} \div \frac{5}{4}$.

Step 1: Calculate $\frac{5}{4} \div \frac{3}{7}$. To divide by a fraction, multiply by its reciprocal:

$\frac{5}{4} \div \frac{3}{7} = \frac{5}{4} \times \frac{7}{3} = \frac{35}{12}$

Step 2: Calculate $\frac{3}{7} \div \frac{5}{4}$. Again, multiply by the reciprocal:

$\frac{3}{7} \div \frac{5}{4} = \frac{3}{7} \times \frac{4}{5} = \frac{12}{35}$

Step 3: Compare the results:

$\frac{35}{12} \neq \frac{12}{35}$

Thus, division is not commutative for rational numbers. The reason is correct because taking the reciprocal does change the order of division, but it is not the complete explanation of why division is non-commutative. The assertion is true, and the reason is also true but does not fully explain the assertion.

Key Concept: Associativity of Multiplication and Subtraction

a) $(x \times y) \times z = x \times (y \times z)$

[Solution Description]
We need to determine which expression is correct based on the associativity properties of multiplication and subtraction.

From the syllabus:
- Multiplication is associative: $(x \times y) \times z = x \times (y \times z)$.
- Subtraction is not associative: $(x - y) - z \neq x - (y - z)$.

Let's evaluate each expression:

Expression A: $(x \times y) \times z = x \times (y \times z)$
$(x \times y) \times z = \left(\frac{2}{3} \times \frac{4}{5}\right) \times \frac{6}{7} = \frac{8}{15} \times \frac{6}{7} = \frac{48}{105}$
$x \times (y \times z) = \frac{2}{3} \times \left(\frac{4}{5} \times \frac{6}{7}\right) = \frac{2}{3} \times \frac{24}{35} = \frac{48}{105}$
This expression is correct.

Expression B: $(x - y) - z = x - (y - z)$
$(x - y) - z = \left(\frac{2}{3} - \frac{4}{5}\right) - \frac{6}{7} = \left(\frac{10 - 12}{15}\right) - \frac{6}{7} = \frac{-2}{15} - \frac{6}{7} = \frac{-14 - 90}{105} = \frac{-104}{105}$
$x - (y - z) = \frac{2}{3} - \left(\frac{4}{5} - \frac{6}{7}\right) = \frac{2}{3} - \left(\frac{28 - 30}{35}\right) = \frac{2}{3} - \frac{-2}{35} = \frac{70 + 6}{105} = \frac{76}{105}$
This expression is incorrect.

Thus, only Expression A is correct.

Key Concept: Associativity of Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that addition is associative for rational numbers, which means that for any three rational numbers $a$, $b$, and $c$, the equation $a + (b + c) = (a + b) + c$ holds true. This is a fundamental property of rational numbers under addition. The reason explains that this is because the grouping of numbers does not affect the sum. This explanation aligns with the mathematical definition of associativity, where the way in which numbers are grouped during addition does not change the result. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Associative Property of Addition

a) $(3 + (-4)) + 5 = 3 + (-4 + 5)$

[Solution Description]

The associative property of addition states that for any three numbers $a$, $b$, and $c$, $(a + b) + c = a + (b + c)$. Using the given values, we can verify this property:

Step 1: Calculate $(a + b) + c$.

$(a + b) + c = (3 + (-4)) + 5 = (-1) + 5 = 4$

Step 2: Calculate $a + (b + c)$.

$a + (b + c) = 3 + (-4 + 5) = 3 + 1 = 4$

Both expressions yield the same result, confirming the associative property.

Key Concept: Associative Property of Addition

c) Associative Property

[Solution Description]

The equation $\frac{2}{3} + \left( \frac{5}{2} + \frac{7}{27} \right) = \left( \frac{2}{3} + \frac{5}{2} \right) + \frac{7}{27}$ demonstrates that the way in which rational numbers are grouped during addition does not affect the sum. This is known as the Associative Property of Addition.

Key Concept: Associative Property of Multiplication

a) $\frac{5}{16}$

[Solution Description]

First, calculate $y \times z$:

$\frac{3}{4} \times \frac{-5}{6} = \frac{(3 \times -5)}{(4 \times 6)} = \frac{-15}{24}$

Simplify $\frac{-15}{24}$ by dividing numerator and denominator by 3:

$\frac{-15 \div 3}{24 \div 3} = \frac{-5}{8}$

Next, multiply the result by $x$:

$\frac{-1}{2} \times \frac{-5}{8} = \frac{(-1 \times -5)}{(2 \times 8)} = \frac{5}{16}$

Key Concept: Associative Property of Multiplication

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

[Solution Description]

The associative property of multiplication states that for any three rational numbers $a$, $b$, and $c$, $a \times (b \times c) = (a \times b) \times c$. This means the grouping of numbers does not change the product.

Key Concept: Non-associativity of Subtraction, Rational Numbers

b) $-\frac{5}{144}$

[Solution Description]

First, calculate the value of $E$. Start by solving $\frac{5}{6} - \frac{2}{3}$. Convert $\frac{2}{3}$ to $\frac{4}{6}$ to have a common denominator. So, $\frac{5}{6} - \frac{4}{6} = \frac{1}{6}$. Now subtract $\frac{1}{4}$ from $\frac{1}{6}$. Convert both fractions to twelfths: $\frac{1}{6} = \frac{2}{12}$ and $\frac{1}{4} = \frac{3}{12}$. So, $\frac{2}{12} - \frac{3}{12} = -\frac{1}{12}$. Therefore, $E = -\frac{1}{12}$.

Next, calculate the value of $F$. Start by solving $\frac{2}{3} - \frac{1}{4}$. Convert both fractions to twelfths: $\frac{2}{3} = \frac{8}{12}$ and $\frac{1}{4} = \frac{3}{12}$. So, $\frac{8}{12} - \frac{3}{12} = \frac{5}{12}$. Now subtract $\frac{5}{12}$ from $\frac{5}{6}$. Convert $\frac{5}{6}$ to $\frac{10}{12}$. So, $\frac{10}{12} - \frac{5}{12} = \frac{5}{12}$. Therefore, $F = \frac{5}{12}$.

Finally, calculate $E \times F$:

$E \times F = -\frac{1}{12} \times \frac{5}{12} = -\frac{5}{144}$

Key Concept: Non-associativity of Division

b) No, division is not associative for rational numbers.

[Solution Description]

Division is not associative for rational numbers. This means that $(a \div b) \div c$ is not equal to $a \div (b \div c)$. For example, consider $a = \frac{1}{2}$, $b = \frac{1}{3}$, and $c = \frac{2}{5}$.

Step 1: Calculate $\frac{1}{2} \div \left(\frac{1}{3} \div \frac{2}{5}\right)$.

$\frac{1}{3} \div \frac{2}{5} = \frac{1}{3} \times \frac{5}{2} = \frac{5}{6}$

$\frac{1}{2} \div \frac{5}{6} = \frac{1}{2} \times \frac{6}{5} = \frac{3}{5}$

Step 2: Calculate $\left(\frac{1}{2} \div \frac{1}{3}\right) \div \frac{2}{5}$.

$\frac{1}{2} \div \frac{1}{3} = \frac{1}{2} \times 3 = \frac{3}{2}$

$\frac{3}{2} \div \frac{2}{5} = \frac{3}{2} \times \frac{5}{2} = \frac{15}{4}$

Step 3: Compare the results.

$\frac{3}{5} \neq \frac{15}{4}$

Therefore, division is not associative for rational numbers.

Key Concept: Non-associativity of Division

b) $\frac{15}{16}$

[Solution Description]

First, compute $\left(a \div b\right) \div c$:

$a \div b = \frac{3}{4} \div \frac{2}{5} = \frac{3}{4} \times \frac{5}{2} = \frac{15}{8}$

Then,

$\left(\frac{15}{8}\right) \div c = \frac{15}{8} \div \frac{1}{2} = \frac{15}{8} \times 2 = \frac{15}{4}$

Next, compute $a \div \left(b \div c\right)$:

$b \div c = \frac{2}{5} \div \frac{1}{2} = \frac{2}{5} \times 2 = \frac{4}{5}$

Then,

$a \div \left(\frac{4}{5}\right) = \frac{3}{4} \div \frac{4}{5} = \frac{3}{4} \times \frac{5}{4} = \frac{15}{16}$

Since $\frac{15}{4}$ is not equal to $\frac{15}{16}$, the expressions are not equal.

Key Concept: Non-associativity of Division

b) $\frac{5}{16}$

[Solution Description]

First, compute $\left(p \div q\right) \div r$:

$p \div q = \frac{5}{6} \div \frac{2}{3} = \frac{5}{6} \times \frac{3}{2} = \frac{15}{12} = \frac{5}{4}$

Then,

$\left(\frac{5}{4}\right) \div r = \frac{5}{4} \div \frac{1}{4} = \frac{5}{4} \times 4 = 5$

Next, compute $p \div \left(q \div r\right)$:

$q \div r = \frac{2}{3} \div \frac{1}{4} = \frac{2}{3} \times 4 = \frac{8}{3}$

Then,

$p \div \left(\frac{8}{3}\right) = \frac{5}{6} \div \frac{8}{3} = \frac{5}{6} \times \frac{3}{8} = \frac{15}{48} = \frac{5}{16}$

Since $5$ is not equal to $\frac{5}{16}$, the expressions are not equal.

Key Concept: Additive Identity

c) $x$

[Solution Description]

The additive identity for rational numbers is 0. Adding 0 to any rational number does not change its value. Therefore, $x + 0 = x$.

Key Concept: Multiplicative Identity

b) $1$

[Solution Description]

The multiplicative identity for rational numbers is 1. Multiplying any rational number by 1 does not change its value. Hence, the multiplicative identity is 1.

Key Concept: Additive Identity of Integers

a) $-8$

[Solution Description]

The additive identity for integers states that $b + 0 = b$, where $b$ is an integer. Here, $b = -8$. Therefore, $-8 + 0 = -8$.

Key Concept: Additive Identity of Whole Numbers

b) $15$

[Solution Description]

For whole numbers, the additive identity property is $a + 0 = a$, where $a$ is a whole number. Here, $a = 15$. So, $15 + 0 = 15$.

Key Concept: Additive Identity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that adding zero to any rational number leaves the number unchanged. This is true because, according to the concept of additive identity, $a + 0 = a$ for any rational number $a$. The reason provided is also correct as zero is indeed called the additive identity for rational numbers. Moreover, the reason correctly explains why the assertion is true, as the property of zero being the additive identity directly leads to the conclusion that adding zero to a rational number leaves it unchanged.

Key Concept: Additive Identity in Integers

b) $-8$

[Solution Description]

Adding zero to any integer does not change its value. This is because zero is the additive identity for integers. Therefore, $-8 + 0 = -8$.

Key Concept: Additive Identity, Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that zero is called the additive identity for rational numbers. This is true because, by definition, the additive identity of a set is an element that does not change other elements when added to them. For rational numbers, this property holds as $c + 0 = c$ for any rational number $c$. The reason provided is that when 0 is added to any rational number, the result remains unchanged. This is also true and directly explains why zero is called the additive identity for rational numbers. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Rational Numbers, Zero as Additive Identity

c) $a$ and $b$ are equal

[Solution Description]

The property of zero as the additive identity states that adding zero to any rational number does not change its value. Therefore, $a + 0 = a$. This implies that $b$ must be equal to $a$. Thus, the correct statement is that $a$ and $b$ are equal.

Key Concept: Zero in Multiplication

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The assertion states that any number multiplied by zero results in zero, which is true based on the properties of multiplication. The reason given is that zero is the additive identity, and multiplication by zero always yields zero because it represents the absence of quantity. While it is true that zero is the additive identity, this property does not directly explain why multiplication by zero results in zero. The correct explanation for this is rooted in the definition of multiplication as repeated addition. For example, $5 \times 0$ means adding 5 zero times, which results in 0. Therefore, the assertion is true, but the reason provided is not the correct explanation for it.

Key Concept: Zero in Multiplication, Complex Equations

a) $a = 0$ and $b = 0$

[Solution Description]

Expand the equation $(a - b)^2 + (a + b)^2 = 0$:

$(a - b)^2 = a^2 - 2ab + b^2$

$(a + b)^2 = a^2 + 2ab + b^2$

Adding them: $a^2 - 2ab + b^2 + a^2 + 2ab + b^2 = 2a^2 + 2b^2 = 0$. Since $a^2$ and $b^2$ are non-negative, the only solution is $a = 0$ and $b = 0$.

Key Concept: Division of Rational Numbers

a) Yes

[Solution Description]

To determine if $\frac{7}{8} \div \frac{2}{5}$ is a rational number, perform the division: $\frac{7}{8} \times \frac{5}{2} = \frac{35}{16}$. Since $\frac{35}{16}$ is a ratio of two integers with a non-zero denominator, it is a rational number.

Key Concept: Zero in Division

a) $5 \div 0$

[Solution Description]

In division, any number divided by zero is not defined. Hence, options involving division by zero are undefined.

Key Concept: Multiplicative Identity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that for any rational number $a$, multiplying it by 1 gives back $a$. This is correct because, by definition, 1 is the multiplicative identity for rational numbers. The reason provided is also correct as it defines 1 as the multiplicative identity for rational numbers. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Multiplicative Identity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that 1 is the multiplicative identity for all real numbers. The reason explains that multiplying any real number by 1 yields the same real number. Let's verify this with examples:

1. For a whole number: $5 \times 1 = 5$.

2. For a rational number: $-\frac{2}{7} \times 1 = -\frac{2}{7}$.

3. For an integer: $-3 \times 1 = -3$.

These examples confirm that multiplying any real number by 1 results in the same number. Therefore, both the assertion and reason are true, and the reason correctly explains why 1 is the multiplicative identity for real numbers.

Key Concept: Role of Special Numbers: One in Multiplication

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that multiplying any number by one leaves the number unchanged, which is a true statement because one acts as the multiplicative identity. The reason explains that one is the multiplicative identity element, which directly supports the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Role of Special Numbers: One in Multiplication

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that multiplying any number by one leaves the number unchanged. This is a fundamental property of multiplication, where one acts as the multiplicative identity. The reason given is that one is the multiplicative identity element, which is a mathematical fact. The reason correctly explains why multiplying any number by one leaves it unchanged. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Multiplicative Identity for Rational Numbers

b) $\frac{4}{5}$

[Solution Description]

The multiplicative identity property states that any rational number multiplied by 1 gives the same rational number as the product. Therefore, $a \times 1 = \frac{4}{5} \times 1 = \frac{4}{5}$.

Key Concept: Multiplicative Identity

a) $a \times 1 = a$

[Solution Description]

The multiplicative identity property states that multiplying any rational number by 1 does not change its value. Therefore, $a \times 1 = a$ is correct for any rational number $a$.

Key Concept: Distributivity of Multiplication over Subtraction

b) $-\frac{1}{2}$

[Solution Description]

To simplify the expression, we apply the distributive property:

$-4 \times \left( \frac{3}{8} - \frac{1}{4} \right) = -4 \times \frac{3}{8} - (-4) \times \frac{1}{4}$

Now, calculate each term separately:

$-4 \times \frac{3}{8} = -\frac{12}{8} = -\frac{3}{2}$

$-(-4) \times \frac{1}{4} = 4 \times \frac{1}{4} = 1$

Add the two results:

$-\frac{3}{2} + 1 = -\frac{1}{2}$

Therefore, the simplified form is $-\frac{1}{2}$.

Key Concept: Distributive Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that for any rational numbers $a, b,$ and $c$, the expression $a(b - c) = ab - ac$ is always true. This is a direct application of the distributive property of multiplication over subtraction, which is a fundamental property in arithmetic. The reason correctly explains that the distributive property allows multiplication to distribute over subtraction, making the assertion true. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Distributive Property of Multiplication Over Addition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that for rational numbers $a$, $b$, and $c$, the expression $a(b + c)$ can be rewritten as $ab + ac$. This is a direct application of the distributive property of multiplication over addition, which is a fundamental property in algebra. The reason provided is also true as it correctly explains the distributive property. Since the reason directly explains why the assertion is true, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Distributive Property of Multiplication Over Addition and Subtraction

a) $\frac{21}{80}$

[Solution Description]

First, apply the distributive property to the expression:

$\frac{3}{4} \times \left( \frac{2}{5} + \frac{1}{10} - \frac{3}{20} \right) = \frac{3}{4} \times \frac{2}{5} + \frac{3}{4} \times \frac{1}{10} - \frac{3}{4} \times \frac{3}{20}$

Now, perform each multiplication separately:

$\frac{3}{4} \times \frac{2}{5} = \frac{6}{20}$

$\frac{3}{4} \times \frac{1}{10} = \frac{3}{40}$

$\frac{3}{4} \times \frac{3}{20} = \frac{9}{80}$

Combine these results:

$\frac{6}{20} + \frac{3}{40} - \frac{9}{80}$

Convert all fractions to a common denominator (80):

$\frac{24}{80} + \frac{6}{80} - \frac{9}{80} = \frac{21}{80}$

The simplified form of the expression is $\frac{21}{80}$.

Key Concept: Distributive Property of Multiplication Over Subtraction

b) 2

[Solution Description]

Step 1: Apply the distributive property to simplify the expression:

$5 \times \left( \frac{7}{10} - \frac{3}{10} \right) = 5 \times \frac{7}{10} - 5 \times \frac{3}{10}$.

Step 2: Multiply each term:

$5 \times \frac{7}{10} = \frac{35}{10}$, and $5 \times \frac{3}{10} = \frac{15}{10}$.

Step 3: Subtract the results:

$\frac{35}{10} - \frac{15}{10} = \frac{20}{10} = 2$.

Key Concept: Distributive Property of Multiplication Over Subtraction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion claims that the expression $5 \times (8 - 3)$ can be simplified using the distributive property as $5 \times 8 - 5 \times 3$. This is correct because the distributive property allows us to multiply $5$ with each term inside the parentheses separately. Let's verify this:

Step 1: Apply the distributive property: $5 \times (8 - 3) = 5 \times 8 - 5 \times 3$.

Step 2: Calculate the multiplication: $5 \times 8 = 40$ and $5 \times 3 = 15$.

Step 3: Subtract the results: $40 - 15 = 25$.

Therefore, $5 \times (8 - 3) = 25$, which matches the result obtained by applying the distributive property. The reason correctly explains the assertion by stating the general rule of the distributive property.

Key Concept: Rational numbers between two rational numbers

b) $\frac{5}{12}$

[Solution Description]

To find a rational number between $\frac{1}{3}$ and $\frac{1}{2}$, we can take the average of the two numbers. First, convert them to have a common denominator, which is 6:

$\frac{1}{3} = \frac{2}{6} \quad \text{and} \quad \frac{1}{2} = \frac{3}{6}$

Now, calculate the average:

$\frac{\frac{2}{6} + \frac{3}{6}}{2} = \frac{\frac{5}{6}}{2} = \frac{5}{12}$

The rational number between $\frac{1}{3}$ and $\frac{1}{2}$ is $\frac{5}{12}$.

Key Concept: Rational Numbers

c) $1$

[Solution Description]

The reciprocal of a rational number $\frac{p}{q}$ is $\frac{q}{p}$. Therefore, the reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$.

Step 1: Multiply $\frac{3}{4}$ by its reciprocal $\frac{4}{3}$.

Step 2: $\frac{3}{4} \times \frac{4}{3} = \frac{12}{12} = 1$.

Thus, the result of multiplying $\frac{3}{4}$ by its reciprocal is $1$.

Key Concept: Rational Numbers on a Number Line

b) $\frac{11}{16}$

[Solution Description] First, find the difference between $\frac{3}{4}$ and $\frac{1}{2}$. Convert $\frac{1}{2}$ to $\frac{2}{4}$ to have a common denominator. The difference is $\frac{3}{4} - \frac{2}{4} = \frac{1}{4}$. Three-fourths of this distance is $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$. Add this to $\frac{1}{2}$ (or $\frac{8}{16}$) to get the desired rational number: $\frac{8}{16} + \frac{3}{16} = \frac{11}{16}$.

Key Concept: Rational Numbers on a Number Line

c) Assertion is true, but Reason is false.

[Solution Description] The assertion states that $\frac{1}{2}$ lies exactly halfway between 0 and 1 on the number line, which is correct because $\frac{1}{2}$ is the midpoint of the interval [0, 1]. However, the reason claims that a rational number is always equidistant from two consecutive integers on the number line, which is false. For example, $\frac{1}{3}$ is not equidistant from 0 and 1. Therefore, the assertion is true, but the reason is false.

Key Concept: Finding Rational Numbers Between Two Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find a rational number between $\frac{1}{2}$ and $\frac{3}{4}$, we can use the concept of mean. The mean of two rational numbers $a$ and $b$ is given by $\frac{a + b}{2}$. Applying this to $\frac{1}{2}$ and $\frac{3}{4}$, we get:

$\text{Mean} = \frac{\frac{1}{2} + \frac{3}{4}}{2} = \frac{\frac{2}{4} + \frac{3}{4}}{2} = \frac{\frac{5}{4}}{2} = \frac{5}{8}.$

Since $\frac{5}{8}$ is a rational number, the assertion is true. The reason states that the mean of two rational numbers is always a rational number, which is also true. Moreover, the reason correctly explains why the method used in the assertion works. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Finding Rational Numbers Between Two Rational Numbers

b) $\frac{7}{24}$

[Solution Description]

To find a rational number between $\frac{1}{4}$ and $\frac{1}{3}$, we can use the mean of the two numbers. The mean of two numbers $a$ and $b$ is given by $\frac{a + b}{2}$. Here, $a = \frac{1}{4}$ and $b = \frac{1}{3}$. So, the mean is $\frac{\frac{1}{4} + \frac{1}{3}}{2}$. To add $\frac{1}{4}$ and $\frac{1}{3}$, we need a common denominator, which is 12. So, $\frac{1}{4} = \frac{3}{12}$ and $\frac{1}{3} = \frac{4}{12}$. Adding these gives $\frac{3}{12} + \frac{4}{12} = \frac{7}{12}$. Now, divide this sum by 2: $\frac{\frac{7}{12}}{2} = \frac{7}{24}$. So, $\frac{7}{24}$ is a rational number between $\frac{1}{4}$ and $\frac{1}{3}$.

Key Concept: Rational Number Identification

c) $\sqrt{5}$

[Solution Description]

A rational number can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. The number $\sqrt{5}$ cannot be expressed as such a fraction, making it an irrational number. Therefore, $\sqrt{5}$ is not a rational number.

Key Concept: Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the sum of two rational numbers is always a rational number. This is true because rational numbers are defined as numbers of the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. When you add two rational numbers, the result is also a rational number. The reason given is that rational numbers are closed under addition, which is also true. The closure property of rational numbers under addition means that adding any two rational numbers will result in another rational number. Therefore, the reason correctly explains the assertion.

Key Concept: Standard Form of a Rational Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the standard form of $\frac{6}{8}$ is $\frac{3}{4}$, which is true because both the numerator and denominator are divided by their greatest common divisor, which is 2. The reason states that a rational number is in its standard form when the numerator and denominator have no common factors other than 1, which is also true. Moreover, the reason correctly explains why $\frac{6}{8}$ simplifies to $\frac{3}{4}$. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Standard Form of a Rational Number

b) $\frac{-2}{3}$

[Solution Description]

To find the standard form of $\frac{-12}{18}$, we need to simplify the fraction by dividing both numerator and denominator by their greatest common divisor (GCD), which is 6.

$\frac{-12}{18} = \frac{-12 \div 6}{18 \div 6} = \frac{-2}{3}$

The fraction $\frac{-2}{3}$ is in its standard form because the numerator and denominator are coprime, and the denominator is positive.

Therefore, the correct answer is $\frac{-2}{3}$.

Key Concept: Standard Form of a Rational Number, Converting Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the standard form of $\frac{-18}{24}$ is $\frac{-3}{4}$. This is correct because the GCD of 18 and 24 is 6. Dividing both the numerator and denominator by 6 gives $\frac{-18 \div 6}{24 \div 6} = \frac{-3}{4}$. The reason explains that to convert a rational number into its standard form, we divide both the numerator and denominator by their GCD. This is also correct and directly explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Standard Form of a Rational Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To verify the assertion, we need to convert $\frac{24}{36}$ into its standard form. The GCD of 24 and 36 is 12. Dividing both the numerator and denominator by 12 gives $\frac{24 \div 12}{36 \div 12} = \frac{2}{3}$. Thus, the assertion is true because $\frac{2}{3}$ is indeed the standard form of $\frac{24}{36}$. The reason provided is also true, as the standard form of a rational number is obtained by dividing both the numerator and denominator by their GCD. Moreover, the reason correctly explains the assertion.

Key Concept: Closure Property, Multiplication

b) $\frac{15}{77}$

[Solution Description]

To find the product of $a \times b$, we multiply the numerators and denominators:

$a \times b = \frac{3}{7} \times \frac{5}{11} = \frac{3 \times 5}{7 \times 11} = \frac{15}{77}.$

The product of the two rational numbers is $\frac{15}{77}$.

Key Concept: Distributivity, Simplification

c) $-\frac{9}{8}$

[Solution Description]

To simplify the expression $-\frac{3}{4} \times \left( \frac{2}{3} + \frac{5}{6} \right)$, we first solve the addition inside the parenthesis:

$\frac{2}{3} + \frac{5}{6} = \frac{4}{6} + \frac{5}{6} = \frac{9}{6} = \frac{3}{2}.$

Now multiply by $-\frac{3}{4}$:

$-\frac{3}{4} \times \frac{3}{2} = -\frac{9}{8}.$

The simplified form of the expression is $-\frac{9}{8}$.

Key Concept: Closure under Subtraction

a) $\frac{5}{4} - \frac{8}{5}$

[Solution Description]

Rational numbers are closed under subtraction, meaning that the difference of any two rational numbers is also a rational number. Let's verify this with an example: $\frac{5}{4} - \frac{8}{5}$.

Calculate $\frac{5}{4} - \frac{8}{5}$:

$$\frac{5}{4} - \frac{8}{5} = \frac{25 - 32}{20} = -\frac{7}{20}$$

The result $-\frac{7}{20}$ is a rational number. Therefore, the difference of two rational numbers is always a rational number.

Key Concept: Addition of Rational Numbers

a) $\frac{24}{7}$

[Solution Description]

To add $\frac{3}{7} + \frac{21}{7}$, since the denominators are the same, we simply add the numerators: $3 + 21 = 24$. The denominator remains 7. Thus, the sum is $\frac{24}{7}$.

Key Concept: Comparison of Rational Numbers, Multi-step Solutions, Conceptual Complexity

b) $\frac{4}{5}$ is greater than $\frac{2}{3}$ but less than $\frac{6}{7}$

[Solution Description] To compare the rational numbers $\frac{2}{3}$, $\frac{4}{5}$, and $\frac{6}{7}$, we first find a common denominator. The denominators are 3, 5, and 7. The least common multiple (LCM) of 3, 5, and 7 is 105. Now, convert each fraction to have this common denominator:

$\frac{2}{3} = \frac{2 \times 35}{3 \times 35} = \frac{70}{105}$

$\frac{4}{5} = \frac{4 \times 21}{5 \times 21} = \frac{84}{105}$

$\frac{6}{7} = \frac{6 \times 15}{7 \times 15} = \frac{90}{105}$

Now, we compare the numerators of the fractions with the same denominator. Since $70 < 84 < 90$, the correct order is $\frac{2}{3} < \frac{4}{5} < \frac{6}{7}$. Therefore, the statement "$\frac{4}{5}$ is greater than $\frac{2}{3}$ but less than $\frac{6}{7}$" is true.

Key Concept: Comparison of Rational Numbers, Multi-step Solutions, Conceptual Complexity

a) $\frac{3}{4}, \frac{5}{6}, \frac{7}{8}$

[Solution Description] To compare the rational numbers $\frac{3}{4}$, $\frac{5}{6}$, and $\frac{7}{8}$, we first find a common denominator. The denominators are 4, 6, and 8. The least common multiple (LCM) of 4, 6, and 8 is 24. Now, convert each fraction to have this common denominator:

$\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}$

$\frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}$

$\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}$

Now, we compare the numerators of the fractions with the same denominator. Since $18 < 20 < 21$, the ascending order is $\frac{3}{4} < \frac{5}{6} < \frac{7}{8}$.

Key Concept: Converting to Same Denominator

a) $\frac{3}{4} \rightarrow \frac{15}{20}$

[Solution Description]

To convert a fraction to a denominator of 20, we must ensure that the conversion is done by multiplying both the numerator and the denominator by the same number. For example, $\frac{3}{4}$ can be converted to $\frac{15}{20}$ by multiplying both the numerator and the denominator by 5. Similarly, $\frac{7}{10}$ can be converted to $\frac{14}{20}$ by multiplying both the numerator and the denominator by 2. Among the options, $\frac{3}{4}$ converted to $\frac{15}{20}$ is correct.

Key Concept: Division of Rational Numbers, Non-Associativity

a) $\frac{8}{15}$

[Solution Description] First, solve the left-hand side (LHS): $\frac{2}{3} \div \left( \frac{1}{4} \div \frac{1}{5} \right)$.

Step 1: Simplify the inner division $\frac{1}{4} \div \frac{1}{5} = \frac{1}{4} \times \frac{5}{1} = \frac{5}{4}$.

Step 2: Now solve $\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$.

Next, solve the right-hand side (RHS): $\left( \frac{2}{3} \div \frac{1}{4} \right) \div \frac{1}{5}$.

Step 1: Solve $\frac{2}{3} \div \frac{1}{4} = \frac{2}{3} \times \frac{4}{1} = \frac{8}{3}$.

Step 2: Now solve $\frac{8}{3} \div \frac{1}{5} = \frac{8}{3} \times \frac{5}{1} = \frac{40}{3}$.

LHS $= \frac{8}{15}$, RHS $= \frac{40}{3}$. Since $\frac{8}{15} \neq \frac{40}{3}$, the expression is not associative.

Key Concept: Comparison of Rational Numbers

a) $\frac{1}{2}$, $\frac{2}{3}$, $\frac{3}{4}$

[Solution Description]

To arrange the rational numbers $\frac{2}{3}$, $\frac{1}{2}$, and $\frac{3}{4}$ in ascending order, we first convert them to have a common denominator. The least common denominator (LCD) of 3, 2, and 4 is 12.

Convert $\frac{2}{3}$ to twelfths: $\frac{2}{3} = \frac{8}{12}$.

Convert $\frac{1}{2}$ to twelfths: $\frac{1}{2} = \frac{6}{12}$.

Convert $\frac{3}{4}$ to twelfths: $\frac{3}{4} = \frac{9}{12}$.

Now compare $\frac{6}{12}$, $\frac{8}{12}$, and $\frac{9}{12}$. The ascending order is $\frac{6}{12} < \frac{8}{12} < \frac{9}{12}$, which corresponds to $\frac{1}{2} < \frac{2}{3} < \frac{3}{4}$.

Key Concept: Comparing Rational Numbers

c) $a < b$

[Solution Description]

To compare $a = -\frac{2}{3}$ and $b = -\frac{1}{2}$, we first convert them to decimal form: $a \approx -0.666$ and $b = -0.5$. On the number line, $-0.666$ is to the left of $-0.5$, meaning $a < b$.

Key Concept: Distributivity, Rational Numbers

a) $\frac{1}{5}$

[Solution Description]

Apply the distributive property:

$x(y - z) = x \cdot y - x \cdot z$

Calculate $x \cdot y$ and $x \cdot z$ separately:

$x \cdot y = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$

$x \cdot z = \frac{2}{3} \times \frac{1}{2} = \frac{2}{6} = \frac{1}{3}$

Now subtract the two results:

$x(y - z) = \frac{8}{15} - \frac{1}{3} = \frac{8}{15} - \frac{5}{15} = \frac{3}{15} = \frac{1}{5}$

The final answer is $\frac{1}{5}$.

Key Concept: Additive Identity

b) 0

[Solution Description]

The additive identity for rational numbers is the number that, when added to any rational number, leaves the original number unchanged. For rational numbers, this number is $0$.

Key Concept: Addition of Rational Numbers, Commutativity

c) 7

[Solution Description]

To solve this, first find the sum of $\frac{3}{4}$ and $\frac{5}{6}$.

The least common denominator (LCD) of 4 and 6 is 12.

Convert $\frac{3}{4}$ to $\frac{9}{12}$ and $\frac{5}{6}$ to $\frac{10}{12}$.

Now, add them: $\frac{9}{12} + \frac{10}{12} = \frac{19}{12}$.

Here, $a = 19$ and $b = 12$, so $a - b = 19 - 12 = 7$.

Key Concept: Addition of Rational Numbers, Associativity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that for any three rational numbers $a$, $b$, and $c$, the equation $a + (b + c) = (a + b) + c$ holds true. This is a fundamental property of rational numbers known as associativity of addition. The reason given is that addition of rational numbers is associative, which is a correct statement. Since the assertion is true and the reason correctly explains why the assertion is true, the correct answer is option a).

Key Concept: Subtraction of Rational Numbers, Closure Property

c) $\frac{23}{12}$

[Solution Description]

To find $a - b$, we subtract $\frac{7}{4}$ from $\frac{11}{3}$. The denominators are different, so we find the Least Common Multiple (LCM) of 3 and 4, which is 12. We convert both fractions to have a denominator of 12:

$\frac{11}{3} = \frac{11 \times 4}{3 \times 4} = \frac{44}{12}, \quad \frac{7}{4} = \frac{7 \times 3}{4 \times 3} = \frac{21}{12}$

Now, subtracting the two fractions:

$\frac{44}{12} - \frac{21}{12} = \frac{23}{12}$

So, $a - b = \frac{23}{12}$.

Key Concept: Subtraction of Rational Numbers, Associativity

d) Assertion is false, but Reason is true.

[Solution Description]

To determine whether the assertion and reason are true, we evaluate both sides of the equation separately.

First, calculate $\left(\frac{2}{3} - \frac{4}{5}\right) - \frac{1}{2}$:

$\left(\frac{2}{3} - \frac{4}{5}\right) = \frac{10 - 12}{15} = \frac{-2}{15}$

Then subtract $\frac{1}{2}$ from the result:

$\frac{-2}{15} - \frac{1}{2} = \frac{-4 - 15}{30} = \frac{-19}{30}$

Next, calculate $\frac{2}{3} - \left(\frac{4}{5} - \frac{1}{2}\right)$:

$\frac{4}{5} - \frac{1}{2} = \frac{8 - 5}{10} = \frac{3}{10}$

Then subtract this result from $\frac{2}{3}$:

$\frac{2}{3} - \frac{3}{10} = \frac{20 - 9}{30} = \frac{11}{30}$

Comparing the two results, $-19/30$ is not equal to $11/30$, so the Assertion is false.

The Reason states that subtraction of rational numbers is associative, which is incorrect. Subtraction is not associative for rational numbers, as demonstrated by the calculation above.

Therefore, Assertion is false, but Reason is also false.

Key Concept: Multiplication of Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the product of any two rational numbers is always a rational number. This is true because, by definition, a rational number can be expressed as a fraction of two integers. When you multiply two such fractions, the result is another fraction of integers, which is also a rational number. The reason given is that rational numbers are closed under multiplication. Closure under multiplication means that the product of any two rational numbers is also a rational number. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Commutativity, Multiplication of Rational Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that for any two rational numbers $a$ and $b$, the product $a \times b$ is always a rational number. This is true because, by definition, a rational number is any number that can be expressed as the quotient of two integers, and the product of two quotients of integers will also be a quotient of integers. The reason states that rational numbers are closed under multiplication, which means that the product of any two rational numbers is also a rational number. This is exactly what the assertion claims. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Division of Rational Numbers

b) $\frac{28}{27}$

[Solution Description]

To divide two rational numbers, we multiply the first fraction by the reciprocal of the second fraction. So, $\frac{7}{9} \div \frac{3}{4} = \frac{7}{9} \times \frac{4}{3}$. Multiplying the numerators and denominators gives $\frac{7 \times 4}{9 \times 3} = \frac{28}{27}$. Therefore, the answer is $\frac{28}{27}$.

Key Concept: Finding Rational Numbers Between Given Numbers

b) $\frac{1}{12}$

[Solution Description] To find a rational number between $-\frac{1}{3}$ and $\frac{1}{2}$, we use the mean method. First, add $-\frac{1}{3}$ and $\frac{1}{2}$. To add these fractions, convert them to a common denominator. The least common denominator for 3 and 2 is 6. Thus, $-\frac{1}{3} = -\frac{2}{6}$ and $\frac{1}{2} = \frac{3}{6}$. Now, $-\frac{2}{6} + \frac{3}{6} = \frac{1}{6}$. Next, divide $\frac{1}{6}$ by 2 to get the mean: $\frac{1}{6} \div 2 = \frac{1}{12}$. Therefore, $\frac{1}{12}$ is a rational number between $-\frac{1}{3}$ and $\frac{1}{2}$.

Key Concept: Mean Method, Rational Numbers

b) $\frac{5}{14}$

[Solution Description]

To find a rational number between $\frac{2}{7}$ and $\frac{3}{7}$, we use the mean method. The mean of two numbers $a$ and $b$ is given by $\text{Mean} = \frac{a + b}{2}$. Here, $a = \frac{2}{7}$ and $b = \frac{3}{7}$. So,

$\text{Mean} = \frac{\frac{2}{7} + \frac{3}{7}}{2} = \frac{\frac{5}{7}}{2} = \frac{5}{14}$

Thus, the rational number between $\frac{2}{7}$ and $\frac{3}{7}$ is $\frac{5}{14}$.

Key Concept: Finding Rational Numbers Between Two Numbers

b) $-\frac{1}{6}$

[Solution Description]

To find a rational number between $-\frac{1}{3}$ and $0$, we can calculate the mean of the two numbers. The mean is given by:

$\text{Mean} = \frac{-\frac{1}{3} + 0}{2}$

The sum of $-\frac{1}{3}$ and $0$ is $-\frac{1}{3}$. Now, divide the sum by 2:

$\frac{-\frac{1}{3}}{2} = -\frac{1}{6}$

Therefore, $-\frac{1}{6}$ is a rational number between $-\frac{1}{3}$ and $0$.