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

c) $5x + 7 = 0$

[Solution Description]

We need to identify an equation that cannot be solved with integers alone and requires rational numbers. Consider the equation $3x + 4 = 1$. To solve this, we subtract 4 from both sides: $3x = 1 - 4$ which gives $3x = -3$. Then, divide both sides by 3: $x = \frac{-3}{3} = -1$. Since the solution is an integer, it does not require rational numbers. Now consider the equation $5x + 7 = 0$. Subtracting 7 from both sides: $5x = -7$. Dividing both sides by 5: $x = \frac{-7}{5}$. This solution is a rational number, so this equation requires rational numbers for its solution.

Key Concept: Definition of Rational Numbers

c) $\sqrt{3}$

[Solution Description]

A rational number can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

$-4$ can be written as $\frac{-4}{1}$, so it is a rational number.

$-\frac{7}{9}$ is already in the form of $\frac{p}{q}$, so it is a rational number.

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

$\frac{0}{8} = 0$ can be written as $\frac{0}{1}$, making it a rational number.

Therefore, $\sqrt{3}$ is not a rational number.

Key Concept: Definition of Rational Numbers

c) $3x = 7$

[Solution Description]

We need to solve each equation and check if the solution is a rational number.

For equation $x + 5 = 5$, subtracting 5 from both sides gives $x = 0$. 0 is a rational number because it can be written as $\frac{0}{1}$.

For equation $x^2 = -1$, there is no real number solution, let alone a rational one.

For equation $3x = 7$, dividing both sides by 3 gives $x = \frac{7}{3}$. $\frac{7}{3}$ is a rational number.

For equation $x^2 = 2$, taking the square root of both sides gives $x = \pm\sqrt{2}$. $\sqrt{2}$ is an irrational number.

Therefore, the equation $3x = 7$ has a rational number solution.

Key Concept: Rational Numbers, Equations

a) $1$

[Solution Description]

To solve the equation $-\frac{3}{4}x + \frac{5}{6} = \frac{1}{12}$, follow these steps:

\begin{itemize}

Key Concept: Rational Numbers, Additive Identity

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

[Solution Description]

The assertion states that adding $0$ to any rational number $x$ results in $x$. This is a fundamental property of the number $0$, which is known as the additive identity. The reason correctly identifies that $0$ is the additive identity for rational numbers. Since both statements are true and the reason correctly explains why the assertion is valid, the correct option is (a).

Key Concept: Rational Numbers

b) $2x = 5$

[Solution Description]

Let's analyze each equation to determine if it has a rational number solution:

\begin{itemize}

Key Concept: Rational Numbers, Multiplicative Identity

a) $\frac{3}{5} \times 1 = \frac{3}{5}$

[Solution Description]

The multiplicative identity property states that any rational number multiplied by 1 remains unchanged. Therefore, the expression $\frac{3}{5} \times 1 = \frac{3}{5}$ demonstrates this property.

Key Concept: Rational Numbers, Number Line

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

[Solution Description]

To find the rational number that lies exactly midway between $\frac{1}{4}$ and $\frac{3}{4}$, we calculate the average of these two numbers. The average is given by:

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

Therefore, the rational number midway between $\frac{1}{4}$ and $\frac{3}{4}$ is $\frac{1}{2}$.

Key Concept: Representation of Rational Numbers on a Number Line

a) $-\frac{3}{4}$

[Solution Description]

To determine which rational number lies between $-1$ and $0$, we need to find a number that is greater than $-1$ but less than $0$. Let's evaluate each option:

- Option 1: $-\frac{3}{4}$ is greater than $-1$ and less than $0$.

- Option 2: $\frac{1}{2}$ is greater than $0$, so it does not lie between $-1$ and $0$.

- Option 3: $-\frac{5}{4}$ is less than $-1$, so it does not lie between $-1$ and $0$.

- Option 4: $0$ is equal to $0$, so it does not lie between $-1$ and $0$.

Therefore, the correct answer is $-\frac{3}{4}$.

Key Concept: Distributivity, Rational Numbers

a) $\frac{116}{105}$

[Solution Description]

We apply the distributive property of multiplication over addition for rational numbers. First, compute $y + z$:

$y + z = \frac{4}{5} + \frac{6}{7} = \frac{28}{35} + \frac{30}{35} = \frac{58}{35}$

Now, multiply by $x$:

$x(y + z) = \frac{2}{3} \times \frac{58}{35} = \frac{116}{105}$

So, the value of $x(y + z)$ is $\frac{116}{105}$.

Key Concept: Commutative Property of Rational Numbers

d) Commutative Property

[Solution Description]

The commutative property states that changing the order of the operands does not change the result. For rational numbers, addition is commutative. This means that for any two rational numbers $a$ and $b$, $a + b = b + a$. Therefore, the correct answer is Commutative Property.

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 two rational numbers is always a rational number. This is true because, according to the closure property of rational numbers, when you add any two rational numbers, the result is always another rational number. For example, adding $\frac{3}{8}$ and $\frac{-5}{7}$ gives $\frac{-19}{56}$, which is still a rational number. The reason given is that rational numbers are closed under addition, which is also true and directly explains why the assertion is correct. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

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 closure property of rational numbers states that the sum of any two rational numbers is also a rational number. This means if $a$ and $b$ are rational numbers, then $a + b$ is also a rational number. The assertion states that the sum of two rational numbers is always a rational number, which is true based on the closure property. The reason given is that rational numbers are closed under addition, which is also true. Furthermore, the reason correctly explains the assertion. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Closure under Addition

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

[Solution Description]

Let us consider two rational numbers, $\frac{p}{q}$ and $\frac{r}{s}$, where $p$, $q$, $r$, and $s$ are integers, and $q \neq 0$, $s \neq 0$. The sum of these two rational numbers is given by:

$\frac{p}{q} + \frac{r}{s} = \frac{ps + rq}{qs}$

Here, the numerator ($ps + rq$) and the denominator ($qs$) are both integers, and since $q \neq 0$ and $s \neq 0$, the denominator $qs \neq 0$. Therefore, the result is also a rational number. This proves that the sum of any two rational numbers is indeed a rational number, confirming the assertion. The reason states that rational numbers are closed under addition, which is a direct consequence of the assertion being true. Hence, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Closure under Addition

a) $\frac{2}{45}$

[Solution Description]

To add the rational numbers $\frac{-2}{9}$ and $\frac{4}{15}$, we first find a common denominator. The least common multiple (LCM) of 9 and 15 is 45.

Convert $\frac{-2}{9}$ to a fraction with denominator 45:

$\frac{-2}{9} = \frac{-2 \times 5}{9 \times 5} = \frac{-10}{45}$

Convert $\frac{4}{15}$ to a fraction with denominator 45:

$\frac{4}{15} = \frac{4 \times 3}{15 \times 3} = \frac{12}{45}$

Now, add the two fractions:

$\frac{-10}{45} + \frac{12}{45} = \frac{-10 + 12}{45} = \frac{2}{45}$

Thus, the sum of $\frac{-2}{9}$ and $\frac{4}{15}$ is $\frac{2}{45}$.

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 Subtraction

b) $\frac{1}{4}$

[Solution Description]

To solve $\frac{3}{4} - \frac{1}{2}$, first convert $\frac{1}{2}$ to a fraction with denominator 4: $\frac{1}{2} = \frac{2}{4}$. Now subtract the numerators: $\frac{3}{4} - \frac{2}{4} = \frac{1}{4}$. Therefore, the result is $\frac{1}{4}$.

Key Concept: Closure under Multiplication, Rational Numbers

b) $\frac{2}{9}$

[Solution Description]

To find the product of $x$ and $y$, we multiply the numerators together and the denominators together:

$x \times y = \frac{5}{12} \times \frac{8}{15} = \frac{5 \times 8}{12 \times 15} = \frac{40}{180}$

Now, simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 20:

$\frac{40 \div 20}{180 \div 20} = \frac{2}{9}$

Therefore, the product of $x$ and $y$ is $\frac{2}{9}$.

Key Concept: Closure under Multiplication

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

[Solution Description]

The assertion states that the product of two rational numbers is always a rational number. This is true because, by definition, rational numbers can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. When multiplying two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, the result is $\frac{a \times c}{b \times d}$, which is also a rational number since $a \times c$ and $b \times d$ are integers and $b \times d \neq 0$. 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 the correct explanation for the assertion. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Closure under Division, Rational Numbers

b) The result is always a rational number.

[Solution Description]

To determine whether the result of dividing two rational numbers is a rational number, we use the property of closure under division for non-zero rational numbers. Given two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, their division is calculated as:

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$

Since $b, c, d \neq 0$, $bc \neq 0$. Therefore, $\frac{ad}{bc}$ is a rational number. Thus, the collection of rational numbers is closed under division when zero is excluded.

Key Concept: Associativity of Division

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

[Solution Description] First, compute $(a \div b) \div c$.

$\left( \frac{1}{2} \div \frac{2}{3} \right) = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.

Now, $\frac{3}{4} \div \frac{5}{2} = \frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \frac{3}{10}$.

Next, compute $a \div (b \div c)$.

$\left( \frac{2}{3} \div \frac{5}{2} \right) = \frac{2}{3} \times \frac{2}{5} = \frac{4}{15}$.

Now, $\frac{1}{2} \div \frac{4}{15} = \frac{1}{2} \times \frac{15}{4} = \frac{15}{8}$.

Since $\frac{3}{10} \neq \frac{15}{8}$, division is not associative for rational numbers.

Key Concept: Commutativity of Rational Numbers

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that division is commutative for rational numbers. The reason provided is that for any two rational numbers $a$ and $b$, $a \div b = b \div a$. To check this, let's take an example: consider $a = \frac{5}{3}$ and $b = \frac{4}{7}$. Now, calculate $a \div b$ and $b \div a$:

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

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

Clearly, $\frac{35}{12} \neq \frac{12}{35}$, which means $a \div b \neq b \div a$. Therefore, the assertion is false. However, the reason states that $a \div b = b \div a$ for any two rational numbers, which is also false based on our calculation. But since the assertion is false, we need to see if the reason is true or not. In this case, the reason is also false because it claims something that is not true for all rational numbers.

Therefore, both the assertion and the reason are false, but in the given options, the closest match is that the assertion is false, but the reason is true. However, in reality, the reason is also false. Hence, the correct answer is that the assertion is false, but the reason is true.

Key Concept: Multiplication, Commutativity

a) $\frac{-3}{5} \times \frac{4}{7} = \frac{4}{7} \times \frac{-3}{5}$

[Solution Description]

To illustrate the commutativity of multiplication for rational numbers, we need to show that $p \times q = q \times p$. Let's compute both sides:

$p \times q = \frac{-3}{5} \times \frac{4}{7} = \frac{-12}{35}$

$q \times p = \frac{4}{7} \times \frac{-3}{5} = \frac{-12}{35}$

Since $p \times q = q \times p$, the expression $\frac{-3}{5} \times \frac{4}{7} = \frac{4}{7} \times \frac{-3}{5}$ illustrates commutativity.

Key Concept: Commutative Property of Addition, 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 $b$, $a + b = b + a$. This is true because the commutative property of addition applies to all rational numbers, as explained in the syllabus. The reason correctly explains the assertion by stating that the order in which two numbers are added does not change the sum, which is the definition of the commutative property of addition.

Both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Commutative Property of Addition

a) $\frac{7}{8} + \frac{2}{3} = \frac{2}{3} + \frac{7}{8}$

[Solution Description]

The commutative property of addition for rational numbers states that $a + b = b + a$. Therefore, for $\frac{7}{8}$ and $\frac{2}{3}$, the equation $\frac{7}{8} + \frac{2}{3} = \frac{2}{3} + \frac{7}{8}$ must hold true.

Key Concept: Commutative Property of Multiplication

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

[Solution Description]

To solve this, we use the commutative property of multiplication for rational numbers. According to the property, $a \times b = b \times a$. Here, $a = \frac{3}{4}$ and $b = \frac{5}{6}$. Multiply $a$ and $b$ as follows:

$a \times b = \frac{3}{4} \times \frac{5}{6} = \frac{15}{24} = \frac{5}{8}$

Therefore, the result of $a \times b$ is $\frac{5}{8}$.

Key Concept: Commutative Property of Multiplication

c) $\frac{3}{8}$

[Solution Description]

The commutative property allows us to multiply rational numbers in any order. Let $a = \frac{1}{2}$ and $b = \frac{3}{4}$. Then:

$a \times b = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$

Since multiplication is commutative, $a \times b = b \times a$, which also equals $\frac{3}{8}$. Hence, the answer is $\frac{3}{8}$.

Key Concept: Non-commutativity of Subtraction

b) $(p - q) - (q - p)$

[Solution Description]

To determine which expression results in a positive value, we evaluate each option step by step.

First, compute $p - q$:

$p - q = \frac{7}{8} - \frac{3}{4} = \frac{7}{8} - \frac{6}{8} = \frac{1}{8}$

Next, compute $q - p$:

$q - p = \frac{3}{4} - \frac{7}{8} = \frac{6}{8} - \frac{7}{8} = -\frac{1}{8}$

Now, compute $(p - q) + (q - p)$:

$(p - q) + (q - p) = \frac{1}{8} + \left(-\frac{1}{8}\right) = 0$

Finally, compute $(p - q) - (q - p)$:

$(p - q) - (q - p) = \frac{1}{8} - \left(-\frac{1}{8}\right) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$

Therefore, the expression $(p - q) - (q - p)$ results in a positive value.

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) $c \div d \neq d \div c$

[Solution Description] To check if division is commutative, we compare $c \div d$ with $d \div c$. First, calculate $c \div d = \frac{7}{9} \div \frac{1}{3} = \frac{7}{9} \times \frac{3}{1} = \frac{21}{9} = \frac{7}{3}$. Next, calculate $d \div c = \frac{1}{3} \div \frac{7}{9} = \frac{1}{3} \times \frac{9}{7} = \frac{9}{21} = \frac{3}{7}$. Since $\frac{7}{3} \neq \frac{3}{7}$, division is not commutative for rational numbers.

Key Concept: Non-commutativity of Division, Rational Numbers

b) $\frac{70}{81}$

[Solution Description]

Let's evaluate both expressions:

1. First expression: $(a \div b) \div c$

- Start with $a \div b$.

$\frac{7}{9} \div \frac{3}{5} = \frac{7}{9} \times \frac{5}{3} = \frac{35}{27}$

- Now, divide the result by $c$.

$\frac{35}{27} \div \frac{2}{3} = \frac{35}{27} \times \frac{3}{2} = \frac{105}{54} = \frac{35}{18}$

2. Second expression: $a \div (b \div c)$

- Start with $b \div c$.

$\frac{3}{5} \div \frac{2}{3} = \frac{3}{5} \times \frac{3}{2} = \frac{9}{10}$

- Now, divide $a$ by the result.

$\frac{7}{9} \div \frac{9}{10} = \frac{7}{9} \times \frac{10}{9} = \frac{70}{81}$

Comparing the two results, $\frac{35}{18}$ and $\frac{70}{81}$, we see that they are not equal. Therefore, division is not associative for rational numbers.

Key Concept: Associativity of Multiplication and Subtraction

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

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

Key Concept: Non-associativity of Subtraction

c) $\left( \frac{1}{2} - \frac{1}{3} \right) - \frac{1}{4}$

[Solution Description]

The associative property does not hold for subtraction of rational numbers. For instance, $\left( \frac{1}{2} - \frac{1}{3} \right) - \frac{1}{4}$ is not equal to $\frac{1}{2} - \left( \frac{1}{3} - \frac{1}{4} \right)$. Hence, the correct example is the one involving subtraction.

Key Concept: Associative Property of Addition

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

[Solution Description]

The assertion is true because, as per the syllabus, the associative property of addition holds for rational numbers. This means that the way in which numbers are grouped during addition does not change the result. The reason is also true because it correctly defines the associative property, which applies to rational numbers. Moreover, the reason provides the correct explanation for the assertion as it directly relates to the property being discussed.

Key Concept: Associative Property of Addition

a) $(-5 + 8) + (-3) = -5 + (8 + (-3))$

[Solution Description]

The associative property of addition ensures that $(x + y) + z = x + (y + z)$. Using the given values, we can verify this property:

Step 1: Calculate $(x + y) + z$.

$(x + y) + z = (-5 + 8) + (-3) = 3 + (-3) = 0$

Step 2: Calculate $x + (y + z)$.

$x + (y + z) = -5 + (8 + (-3)) = -5 + 5 = 0$

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

Key Concept: Associative Property of Multiplication

b) Associative Property of Multiplication

[Solution Description]

The given equation shows that the product remains the same regardless of how the numbers are grouped. This is the associative property of multiplication for rational numbers, which states $a \times (b \times c) = (a \times b) \times c$.

Key Concept: Associative Property of Multiplication

c) $(x \times y) \times z$

[Solution Description]

According to the associative property of multiplication, $x \times (y \times z) = (x \times y) \times z$. Thus, both expressions will yield the same result regardless of the grouping.

Key Concept: Non-associativity of Subtraction

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

[Solution Description]

Subtraction is not associative for rational numbers. This means that $(a - b) - c$ is not equal to $a - (b - c)$. For example, consider $a = \frac{2}{3}$, $b = \frac{4}{5}$, and $c = \frac{1}{2}$.

Step 1: Calculate $\left(\frac{2}{3} - \frac{4}{5}\right) - \frac{1}{2}$.

$\frac{2}{3} - \frac{4}{5} = \frac{10}{15} - \frac{12}{15} = -\frac{2}{15}$

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

Step 2: Calculate $\frac{2}{3} - \left(\frac{4}{5} - \frac{1}{2}\right)$.

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

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

Step 3: Compare the results.

$-\frac{19}{30} \neq \frac{11}{30}$

Therefore, subtraction is not associative for rational numbers.

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, the results are unequal

[Solution Description]

Let’s start by evaluating $(a \div b) \div c$. Substituting the values of $a$, $b$, and $c$, we get:

$\left(\frac{2}{3} \div -\frac{1}{4}\right) \div \frac{1}{2} = \left(\frac{2}{3} \times -4\right) \div \frac{1}{2} = -\frac{8}{3} \div \frac{1}{2} = -\frac{8}{3} \times 2 = -\frac{16}{3}$

Next, evaluate $a \div (b \div c)$, which gives:

$\frac{2}{3} \div \left(-\frac{1}{4} \div \frac{1}{2}\right) = \frac{2}{3} \div \left(-\frac{1}{4} \times 2\right) = \frac{2}{3} \div -\frac{1}{2} = \frac{2}{3} \times -2 = -\frac{4}{3}$

The two results are $-\frac{16}{3}$ and $-\frac{4}{3}$, respectively. Since they are different, division is not associative.

Key Concept: Non-associativity of Division

b) $\frac{7}{12}$

[Solution Description]

First, compute $\left(x \div y\right) \div z$:

$x \div y = \frac{7}{8} \div \frac{3}{4} = \frac{7}{8} \times \frac{4}{3} = \frac{28}{24} = \frac{7}{6}$

Then,

$\left(\frac{7}{6}\right) \div z = \frac{7}{6} \div \frac{1}{2} = \frac{7}{6} \times 2 = \frac{14}{6} = \frac{7}{3}$

Next, compute $x \div \left(y \div z\right)$:

$y \div z = \frac{3}{4} \div \frac{1}{2} = \frac{3}{4} \times 2 = \frac{6}{4} = \frac{3}{2}$

Then,

$x \div \left(\frac{3}{2}\right) = \frac{7}{8} \div \frac{3}{2} = \frac{7}{8} \times \frac{2}{3} = \frac{14}{24} = \frac{7}{12}$

Since $\frac{7}{3}$ is not equal to $\frac{7}{12}$, the expressions are not equal.

Key Concept: Role of special numbers: Addition and Multiplication identities

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

[Solution Description]

The assertion states that 0 is the additive identity for rational numbers, which is correct. Adding 0 to any rational number $a$ results in $a + 0 = a$. The reason states that 1 is the multiplicative identity for rational numbers, which is also correct. Multiplying any rational number $b$ by 1 results in $b \times 1 = b$. However, the reason does not explain why 0 is the additive identity; it introduces a different concept (multiplicative identity). Therefore, both the assertion and the reason are true, but the reason is NOT the correct explanation of the assertion.

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: Rational Numbers, Additive Identity

a) $\frac{3}{4}$

[Solution Description]

The additive identity property states that for any rational number $x$, $x + 0 = x$. Given $x + 0 = \frac{3}{4}$, it follows that $x = \frac{3}{4}$.

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 adding zero to any rational number results in the same rational number. This is a direct application of the property of additive identity, which is defined as an element that, when added to any other element, leaves the latter unchanged. The reason correctly identifies zero as the additive identity for rational numbers. Since the assertion is a specific instance of the general property stated in the reason, the reason 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: Additive Identity, Integers, Rational Numbers

a) $-7$ and $\frac{1}{4}$

[Solution Description]

Given $a + 0 = -7$. Since zero is the additive identity for integers, the value of $a$ remains unchanged when zero is added to it. Thus, $a = -7$. Similarly, for $b + 0 = \frac{1}{4}$, since zero is the additive identity for rational numbers, the value of $b$ remains unchanged. Thus, $b = \frac{1}{4}$.

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: Rational Numbers, Zero as Additive Identity

c) $y$ is equal to $x$

[Solution Description]

Since zero is the additive identity for rational numbers, adding zero to any rational number does not change its value. Therefore, if $x$ is a rational number, then $x + 0 = x$. Hence, $y = x$.

Key Concept: Additive Identity

b) $\frac{-7}{9}$

[Solution Description]

According to the additive identity property, adding zero to a rational number does not change its value. So, $\frac{-7}{9} + 0 = \frac{-7}{9}$.

Key Concept: Zero in Multiplication

c) 0

[Solution Description]

Multiplying any real number $x$ by zero results in zero. This is because zero acts as an absorbing element in multiplication.

Key Concept: Zero in Multiplication, Complex Equations

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

[Solution Description]

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

\begin{itemize}

Key Concept: Rational numbers, Division of fractions

b) $\frac{4}{3}$

[Solution Description]

To divide two rational numbers, multiply the first fraction by the reciprocal of the second fraction. So, $b \div c = \frac{3}{4} \times \frac{16}{9} = \frac{48}{36} = \frac{4}{3}$.

Key Concept: Rational Numbers and Division by Zero

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

[Solution Description]

The assertion states that the division of any rational number by zero is undefined. This is correct because dividing any number by zero does not yield a meaningful result in mathematics. The reason given is that zero has no multiplicative inverse. This is also correct because there is no number that can be multiplied by zero to give 1, which is the definition of a multiplicative inverse. The reason directly explains why division by zero is undefined, as the operation of division relies on the concept of multiplication by the reciprocal (multiplicative inverse). Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Multiplicative Identity

b) 7

[Solution Description]

The multiplicative identity states that multiplying any number by 1 gives the same number. So, $7 \times 1 = 7$.

Key Concept: Multiplicative Identity, Rational Numbers, Integers

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 both rational numbers and integers. This is true because multiplying any rational number or integer by 1 yields the same number. For example, $5 \times 1 = 5$ and $-\frac{2}{7} \times 1 = -\frac{2}{7}$ for rational numbers, and similarly, for integers like $-3 \times 1 = -3$. The reason provided explains why 1 is the multiplicative identity: it leaves the number unchanged when multiplied. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Basic Concept

a) $a$

[Solution Description]

Multiplying any real number $a$ by one will always result in the same number $a$. This property is fundamental to the concept of multiplication.

Key Concept: One in Multiplication

a) $x$

[Solution Description]

When any number $x$ is multiplied by 1, the result remains the same as the original number $x$. This property of multiplication by one is known as the multiplicative identity.

Key Concept: Multiplicative Identity for Whole Numbers

c) $12$

[Solution Description]

The multiplicative identity property applies to whole numbers as well. When a whole number is multiplied by 1, the result is the same whole number. Thus, $c \times 1 = 12 \times 1 = 12$.

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: Distributive Property, Rational Numbers

a) $\frac{35}{48}$

[Solution Description]

Step 1: Apply the distributive property to the expression.

$\frac{7}{8} \times \left( \frac{5}{6} + \frac{1}{3} \right) - \frac{7}{8} \times \left( \frac{1}{3} \right) = \frac{7}{8} \times \frac{5}{6} + \frac{7}{8} \times \frac{1}{3} - \frac{7}{8} \times \frac{1}{3}$

Step 2: Simplify the terms.

$\frac{7}{8} \times \frac{5}{6} = \frac{35}{48}$

$\frac{7}{8} \times \frac{1}{3} = \frac{7}{24}$

$-\frac{7}{8} \times \frac{1}{3} = -\frac{7}{24}$

Step 3: Combine the terms.

$\frac{35}{48} + \frac{7}{24} - \frac{7}{24} = \frac{35}{48}$

The final answer is $\frac{35}{48}$.

Key Concept: Distributive Property of Multiplication over Subtraction

b) 2

[Solution Description]

To simplify $5 \times \left( \frac{7}{10} - \frac{3}{10} \right)$, first subtract the fractions inside the parentheses:

$\frac{7}{10} - \frac{3}{10} = \frac{4}{10}$

Now multiply by 5:

$5 \times \frac{4}{10} = \frac{20}{10} = 2$

Therefore, the simplified form is $2$.

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

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

[Solution Description]

The assertion states that $\frac{3}{4} \times \left( \frac{2}{3} + \frac{5}{6} \right)$ can be simplified using the distributive property, which is mathematically correct. The reason correctly explains the distributive property as $a (b + c) = ab + ac$. Since both the assertion and the reason are true, and the reason correctly explains the assertion, the correct answer is:

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

Key Concept: Distributive Property of Multiplication Over Subtraction

a) -2

[Solution Description]

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

$-6 \times \left( \frac{8}{15} - \frac{3}{15} \right) = -6 \times \frac{8}{15} - (-6) \times \frac{3}{15}$.

Step 2: Multiply each term:

$-6 \times \frac{8}{15} = -\frac{48}{15}$, and $-6 \times \frac{3}{15} = -\frac{18}{15}$.

Step 3: Simplify the subtraction:

$-\frac{48}{15} - \left(-\frac{18}{15}\right) = -\frac{48}{15} + \frac{18}{15} = -\frac{30}{15} = -2$.

Key Concept: Distributive Property of Multiplication Over Subtraction

b) $\frac{15}{7}$

[Solution Description]

First, apply the distributive property:

$5 \times \left( \frac{6}{7} - \frac{3}{7} \right) = 5 \times \frac{6}{7} - 5 \times \frac{3}{7}$.

Next, perform the multiplication:

$5 \times \frac{6}{7} = \frac{30}{7}$ and $5 \times \frac{3}{7} = \frac{15}{7}$.

Now, subtract the two results:

$\frac{30}{7} - \frac{15}{7} = \frac{15}{7}$.

So, the simplified form is $\frac{15}{7}$.

Key Concept: Rational Numbers

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

[Solution Description]

The additive inverse of a rational number $\frac{p}{q}$ is the number that when added to $\frac{p}{q}$ gives zero. Therefore, the additive inverse of $\frac{2}{5}$ is $-\frac{2}{5}$.

Step 1: Let the additive inverse be $x$.

Step 2: Then, $\frac{2}{5} + x = 0$.

Step 3: Solving for $x$, we get $x = -\frac{2}{5}$.

Thus, the additive inverse of $\frac{2}{5}$ is $-\frac{2}{5}$.

Key Concept: 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 these two numbers. The mean of two numbers $a$ and $b$ is given by $\frac{a + b}{2}$. Here, $a = \frac{1}{4}$ and $b = \frac{1}{3}$.

Step 1: Convert $\frac{1}{4}$ and $\frac{1}{3}$ to have a common denominator. The least common denominator (LCD) of 4 and 3 is 12. So, $\frac{1}{4} = \frac{3}{12}$ and $\frac{1}{3} = \frac{4}{12}$.

Step 2: Add the two fractions: $\frac{3}{12} + \frac{4}{12} = \frac{7}{12}$.

Step 3: Divide the sum by 2 to find the mean: $\frac{7}{12} \div 2 = \frac{7}{24}$.

Thus, a rational number between $\frac{1}{4}$ and $\frac{1}{3}$ is $\frac{7}{24}$.

Key Concept: Rational Numbers on a Number Line

b) \$\frac{16}{45}\$

[Solution Description] First, find the distance between $\frac{1}{3}$ and $\frac{2}{5}$. Convert both fractions to have a common denominator of 15: $\frac{1}{3} = \frac{5}{15}$ and $\frac{2}{5} = \frac{6}{15}$. The distance between them is $\frac{6}{15} - \frac{5}{15} = \frac{1}{15}$. Dividing this distance into three equal parts gives $\frac{1}{45}$. The point closest to $\frac{1}{3}$ is $\frac{5}{15} + \frac{1}{45} = \frac{15}{45} + \frac{1}{45} = \frac{16}{45}$.

Key Concept: Rational Numbers on a Number Line

c) 2 units

[Solution Description]

To find the distance between two points on a number line, subtract the smaller number from the larger one. Here, $\frac{7}{6}$ is greater than $-\frac{5}{6}$.

Step 1: Subtract $-\frac{5}{6}$ from $\frac{7}{6}$: $\frac{7}{6} - (-\frac{5}{6}) = \frac{7}{6} + \frac{5}{6}$.

Step 2: Add the fractions: $\frac{7}{6} + \frac{5}{6} = \frac{12}{6} = 2$.

So, the distance is 2 units.

Key Concept: Rational Numbers, Additive Identity, Multiplicative Identity

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

[Solution Description]

Let's analyze the assertion and reason step by step.

**Assertion:** The sum of any rational number and its additive inverse is always 0.

- Let $x$ be a rational number. Its additive inverse is $-x$.

- Adding them: $x + (-x) = 0$. This is true because 0 is the additive identity for rational numbers.

**Reason:** The product of any rational number and its multiplicative inverse is always 1.

- Let $y$ be a non-zero rational number. Its multiplicative inverse is $\frac{1}{y}$.

- Multiplying them: $y \times \frac{1}{y} = 1$. This is true because 1 is the multiplicative identity for rational numbers.

Now, let's see the relationship between the assertion and reason:

- Both are true, but they are independent concepts. The assertion is about additive inverses and the reason is about multiplicative inverses. They do not explain each other.

Key Concept: Rational Numbers, Distributive Property

c) $\frac{17}{24}$

[Solution Description]

First, compute the sum of $b$ and $c$: $b + c = \frac{2}{3} + \frac{3}{4}$. To add these fractions, find a common denominator, which is 12. Convert each fraction: $\frac{2}{3} = \frac{8}{12}$ and $\frac{3}{4} = \frac{9}{12}$. Now, add them: $b + c = \frac{8}{12} + \frac{9}{12} = \frac{17}{12}$. Next, multiply $a$ with the sum $(b + c)$: $a(b + c) = \frac{1}{2} \times \frac{17}{12} = \frac{17}{24}$. Therefore, the value of $a(b + c)$ is $\frac{17}{24}$.

Key Concept: Closure Property of Rational Numbers

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

[Solution Description]

Let $a$ and $b$ be two rational numbers. By definition, they can be expressed as $\frac{p}{q}$ and $\frac{r}{s}$, where $p$, $q$, $r$, and $s$ are integers with $q \neq 0$ and $s \neq 0$.

The sum of $a$ and $b$ is given by:

$a + b = \frac{p}{q} + \frac{r}{s} = \frac{ps + rq}{qs}$

Since $ps + rq$ and $qs$ are both integers (as integers are closed under addition and multiplication), and $qs \neq 0$ (since $q$ and $s$ are non-zero), the result $\frac{ps + rq}{qs}$ is also a rational number.

Therefore, the assertion that the sum of two rational numbers is always a rational number is true. The reason provided is that rational numbers are closed under addition, which is also true and correctly explains the assertion.

Key Concept: Rational Number Definition

c) $0$

[Solution Description]

A rational number is defined as a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. The number $0$ can be written as $\frac{0}{1}$, which satisfies the definition of a rational number.

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: Basic Concept

d) Lowest terms

[Solution Description]

The standard form of a rational number is when the numerator and denominator are coprime, meaning they have no common factors other than 1. This ensures the fraction is in its simplest form.

Key Concept: Standard Form of a Rational Number

b) $\frac{3}{4}$

[Solution Description]

To convert $\frac{36}{48}$ into its standard form, we need to simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor (GCD). The GCD of 36 and 48 is 12. Dividing both numerator and denominator by 12 gives us $\frac{36 \div 12}{48 \div 12} = \frac{3}{4}$. Therefore, the standard form of $\frac{36}{48}$ is $\frac{3}{4}$.

Key Concept: Standard Form of a Rational Number

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

[Solution Description]

To convert $\frac{20}{25}$ into its standard form, we need to simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor (GCD). The GCD of 20 and 25 is 5. Dividing both numerator and denominator by 5 gives us $\frac{20 \div 5}{25 \div 5} = \frac{4}{5}$. Therefore, the standard form of $\frac{20}{25}$ is $\frac{4}{5}$.

Key Concept: Rational Numbers: Closure Property

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, which is true because rational numbers are closed under addition. The reason correctly explains this property of rational numbers. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Commutativity, Addition

c) $\frac{2}{5} + \frac{1}{5} = \frac{1}{5} + \frac{2}{5}$

[Solution Description]

Commutative property of addition states that changing the order of addends does not change the sum. For any two rational numbers $a$ and $b$,

$a + b = b + a.$

The expression $\frac{2}{5} + \frac{1}{5} = \frac{1}{5} + \frac{2}{5}$ correctly demonstrates this property.

Key Concept: Commutativity of Addition

a) $\frac{2}{5} + \frac{3}{7} = \frac{3}{7} + \frac{2}{5}$

[Solution Description]

The commutative property of addition states that changing the order of the addends does not change the sum. For any two rational numbers $a$ and $b$, $a + b = b + a$.

Let's verify this with an example: $\frac{2}{5} + \frac{3}{7} = \frac{3}{7} + \frac{2}{5}$. First, find the sum of $\frac{2}{5} + \frac{3}{7}$:

$$ \\ \frac{2}{5} + \frac{3}{7} = \frac{14 + 15}{35} = \frac{29}{35}$

$

Now, calculate $\frac{3}{7} + \frac{2}{5}$:

$$ \\ \frac{3}{7} + \frac{2}{5} = \frac{15 + 14}{35} = \frac{29}{35}$

$

Both sums are equal, so the commutative property holds.

Key Concept: Rational Number Multiplication

a) $\frac{2}{4} \times \frac{3}{5}$

[Solution Description]

Rational numbers are closed under multiplication, meaning that the product of any two rational numbers is also a rational number. Let's verify this with an example: $\frac{2}{4} \times \frac{3}{5}$.

Calculate $\frac{2}{4} \times \frac{3}{5}$:

$$ \\ \frac{2}{4} \times \frac{3}{5} = \frac{6}{20} = \frac{3}{10}$

$

The result $\frac{3}{10}$ is a rational number. Therefore, the product of two rational numbers is always a rational number.

Key Concept: Comparison of Rational Numbers

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

[Solution Description]

To compare $\frac{7}{8}$ and $\frac{9}{10}$, we need to find a common denominator. The least common denominator (LCD) of 8 and 10 is 40. Convert both fractions to have the denominator 40.

$\frac{7}{8} = \frac{7 \times 5}{8 \times 5} = \frac{35}{40}$

$\frac{9}{10} = \frac{9 \times 4}{10 \times 4} = \frac{36}{40}$

Now, compare $\frac{35}{40}$ and $\frac{36}{40}$. Since $35 < 36$, $\frac{7}{8}$ is smaller than $\frac{9}{10}$.

Key Concept: Comparison of Rational Numbers

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

[Solution Description] To compare the rational numbers $-\frac{7}{5}$ and $-\frac{3}{2}$, we first find their absolute values. The absolute value of $-\frac{7}{5}$ is $\frac{7}{5} = 1.4$, and the absolute value of $-\frac{3}{2}$ is $\frac{3}{2} = 1.5$. Since $1.5 > 1.4$, the rational number $-\frac{3}{2}$ has a larger absolute value. According to the rule for comparing negative rational numbers, the one with the larger absolute value is smaller. Therefore, $-\frac{7}{5}$ is greater than $-\frac{3}{2}$. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Division of Rational Numbers, Associative Property

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

[Solution Description]

To verify whether division is associative for rational numbers, let us consider three rational numbers, say $a = \frac{1}{2}$, $b = \frac{1}{3}$, and $c = \frac{2}{5}$. Now, calculate $a \div (b \div c)$ and $(a \div b) \div c$.

First, compute $b \div c$:

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

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

$a \div (b \div c) = \frac{1}{2} \div \frac{5}{6} = \frac{1}{2} \times \frac{6}{5} = \frac{6}{10} = \frac{3}{5}.$

Now, compute $a \div b$:

$a \div b = \frac{1}{2} \div \frac{1}{3} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2}.$

Finally, compute $(a \div b) \div c$:

$(a \div b) \div c = \frac{3}{2} \div \frac{2}{5} = \frac{3}{2} \times \frac{5}{2} = \frac{15}{4}.$

Since $\frac{3}{5} \neq \frac{15}{4}$, division is not associative for rational numbers. Thus, the Assertion is true, and the Reason is also true, providing the correct explanation.

Key Concept: Division of Rational Numbers, Associativity

b) $6$

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

Step 1: Simplify the inner division $\frac{1}{3} \div \frac{1}{4} = \frac{1}{3} \times \frac{4}{1} = \frac{4}{3}$.

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

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

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

Step 2: Now solve $\frac{3}{2} \div \frac{1}{4} = \frac{3}{2} \times \frac{4}{1} = 6$.

LHS $= \frac{3}{8}$, RHS $= 6$. Since $\frac{3}{8} \neq 6$, the expression is not associative.

Key Concept: Comparing Rational Numbers on a Number Line

b) $\frac{7}{8}$ is to the left of $\frac{9}{10}$

[Solution Description]

Convert $\frac{7}{8}$ and $\frac{9}{10}$ to decimals: $\frac{7}{8} = 0.875$ and $\frac{9}{10} = 0.9$. On a number line, $\frac{7}{8}$ is to the left of $\frac{9}{10}$ because $0.875 < 0.9$.

Key Concept: Rational Numbers, Number Line

d) Assertion is false, but Reason is true.

[Solution Description]

To compare $-\frac{3}{4}$ and $\frac{1}{2}$, plot them on a number line. The number $-\frac{3}{4}$ lies to the left of 0, while $\frac{1}{2}$ lies to the right of 0. Since numbers increase as we move from left to right on a number line, $\frac{1}{2}$ is greater than $-\frac{3}{4}$. Thus, the Assertion is false. However, the Reason is true because, on a number line, numbers indeed increase from left to right.

Key Concept: Closure Property, Multiplicative Identity

a) $\frac{13}{8}$

[Solution Description]

First, calculate the product of $a$ and $b$. Given $a = \frac{3}{4}$ and $b = \frac{5}{6}$, multiply them:

$a \times b = \frac{3}{4} \times \frac{5}{6} = \frac{15}{24} = \frac{5}{8}$

Now, add $1$ to the result. Since $1$ is the multiplicative identity:

$(a \times b) + 1 = \frac{5}{8} + 1 = \frac{5}{8} + \frac{8}{8} = \frac{13}{8}$

The final answer is $\frac{13}{8}$.

Key Concept: Rational Numbers Between Two Rational Numbers

a) $\frac{5}{12}, \frac{1}{2}, \frac{7}{12}$

[Solution Description]

To find rational numbers between $\frac{1}{3}$ and $\frac{2}{3}$, we can use the concept of mean. First, find the mean of $\frac{1}{3}$ and $\frac{2}{3}$:

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

Next, find the mean of $\frac{1}{3}$ and $\frac{1}{2}$:

$\text{Mean} = \frac{\frac{1}{3} + \frac{1}{2}}{2} = \frac{\frac{2}{6} + \frac{3}{6}}{2} = \frac{\frac{5}{6}}{2} = \frac{5}{12}$

Then, find the mean of $\frac{1}{2}$ and $\frac{2}{3}$:

$\text{Mean} = \frac{\frac{1}{2} + \frac{2}{3}}{2} = \frac{\frac{3}{6} + \frac{4}{6}}{2} = \frac{\frac{7}{6}}{2} = \frac{7}{12}$

The three rational numbers are $\frac{5}{12}$, $\frac{1}{2}$, and $\frac{7}{12}$. The correct option lists these values in order.

Key Concept: Addition of Rational Numbers

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

[Solution Description]

Let us evaluate the Assertion and Reason separately.

\begin{itemize}

Key Concept: Addition of Rational Numbers

c) $1$

[Solution Description]

To add $\frac{1}{3}$ and $\frac{2}{3}$, since they have the same denominator, we simply add the numerators:

$\frac{1 + 2}{3} = \frac{3}{3}$. Simplifying $\frac{3}{3}$ gives $1$.

Key Concept: Subtraction of Rational Numbers

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

[Solution Description]

To determine if the Assertion and Reason are true, let's analyze them step by step.

Step 1: A rational number is defined as any number that can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Step 2: Let’s consider two rational numbers, $\frac{a}{b}$ and $\frac{c}{d}$, where $a$, $b$, $c$, and $d$ are integers, and $b \neq 0$, $d \neq 0$.

Step 3: The difference of these two rational numbers is calculated as:

$\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}$.

Step 4: Since $a$, $b$, $c$, and $d$ are integers, $ad - bc$ and $bd$ are also integers. Additionally, $bd \neq 0$ because neither $b$ nor $d$ is zero.

Step 5: Therefore, $\frac{ad - bc}{bd}$ is a rational number. This shows that the difference of two rational numbers is always a rational number.

Step 6: The Reason states that rational numbers are closed under subtraction. This means that the result of subtracting any two rational numbers will always be a rational number.

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

Key Concept: Subtraction of Rational Numbers, Closure Property

c) $\frac{17}{12}$

[Solution Description]

To find $x - y$, we first need to subtract $\frac{5}{6}$ from $\frac{9}{4}$. Since the denominators are different, we find the Least Common Multiple (LCM) of 4 and 6, which is 12. Now, we convert both fractions to have a denominator of 12:

$\frac{9}{4} = \frac{9 \times 3}{4 \times 3} = \frac{27}{12}, \quad \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$

Now, subtracting the two fractions:

$\frac{27}{12} - \frac{10}{12} = \frac{17}{12}$

So, $x - y = \frac{17}{12}$.

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: Multiplication of Rational Numbers, Commutativity

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

[Solution Description]

To find the product of $\frac{3}{4}$ and $\frac{5}{6}$, multiply the numerators together and the denominators together:

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

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor (3):

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

Therefore, the product is $\frac{5}{8}$.

Key Concept: Division of Rational Numbers, Associativity

d) Assertion is false, but Reason is true.

[Solution Description] To verify whether division is associative for rational numbers, let us consider three rational numbers $\frac{1}{2}$, $\frac{2}{3}$, and $\frac{5}{2}$. We will evaluate both sides of the equation:

First, calculate the left-hand side (LHS):

$\left( \frac{1}{2} \div \frac{2}{3} \right) \div \frac{5}{2} = \left( \frac{1}{2} \times \frac{3}{2} \right) \div \frac{5}{2} = \frac{3}{4} \div \frac{5}{2} = \frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \frac{3}{10}$

Next, calculate the right-hand side (RHS):

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

Since $\frac{3}{10} \neq \frac{15}{8}$, division is not associative for rational numbers. Therefore, the Assertion is false. However, the Reason states that the expression $(a \div b) \div c$ equals $a \div (b \div c)$ for any three rational numbers, which is incorrect as shown by the calculation. Thus, the Reason is also false.

Hence, Assertion is false, but Reason is true.

Key Concept: Rational Numbers, Mean

b) $\frac{5}{8}$

[Solution Description]

To find a rational number between $\frac{1}{2}$ and $\frac{3}{4}$, we can use the concept of mean. First, add the two fractions: $\frac{1}{2} + \frac{3}{4}$. To add them, convert $\frac{1}{2}$ to $\frac{2}{4}$ so that both fractions have the same denominator. Now, $\frac{2}{4} + \frac{3}{4} = \frac{5}{4}$. Next, divide the sum by 2 to find the mean: $\frac{5}{4} \div 2 = \frac{5}{8}$. Therefore, the rational number between $\frac{1}{2}$ and $\frac{3}{4}$ is $\frac{5}{8}$.

Key Concept: Rational Numbers, Mean Method

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

[Solution Description]

To verify the assertion, we first calculate the mean of $\frac{1}{3}$ and $\frac{5}{8}$.

$\text{Mean} = \frac{\frac{1}{3} + \frac{5}{8}}{2} = \frac{\frac{8}{24} + \frac{15}{24}}{2} = \frac{\frac{23}{24}}{2} = \frac{23}{48}$

Now, compare $\frac{7}{12}$ with $\frac{23}{48}$:

$\frac{7}{12} = \frac{28}{48} \quad \text{and} \quad \frac{23}{48} = \frac{23}{48}$

Since $\frac{28}{48} > \frac{23}{48}$, it is clear that the mean $\frac{23}{48}$ does not lie between $\frac{1}{3}$ and $\frac{5}{8}$. However, $\frac{7}{12}$ does lie between these two numbers because $\frac{1}{3} < \frac{7}{12} < \frac{5}{8}$. Thus, the Assertion is true.

The Reason states that the mean of two rational numbers always lies between them. This is a fundamental property of the mean method and is correct. Therefore, both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion since the mean itself does not lie between the given numbers in this case.

Key Concept: Finding Rational Numbers Between Two Numbers

b) $0.375$

[Solution Description]

To find a rational number between $0.25$ and $0.5$, we can calculate the mean of the two numbers. The mean is given by:

$\text{Mean} = \frac{0.25 + 0.5}{2}$

Add $0.25$ and $0.5$:

$0.25 + 0.5 = 0.75$

Now, divide the sum by 2:

$\frac{0.75}{2} = 0.375$

Therefore, $0.375$ is a rational number between $0.25$ and $0.5$.