Class 8 Mathematics Chapter 1 Rational Numbers

Key Concept: Rational Numbers, Equations

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

[Solution Description]

First, subtract $\frac{1}{2}$ from both sides of the equation:

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

Simplify the right side:

$\frac{3}{4}x = \frac{4}{2} = 2$

Now, multiply both sides by $\frac{4}{3}$ to solve for $x$:

$x = 2 \times \frac{4}{3} = \frac{8}{3}$

The solution is $x = \frac{8}{3}$.

Key Concept: Rational Numbers, Equations

b) 1

[Solution Description]

First, subtract $\frac{1}{2}$ from both sides of the equation:

$3x = \frac{7}{2} – \frac{1}{2}$

Simplify the right side:

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

Now, divide both sides by 3 to solve for $x$:

$x = \frac{3}{3} = 1$

The solution is $x = 1$.

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

b) $x = 1$

[Solution Description]

To solve the equation $\frac{2}{3}x – \frac{1}{4} = \frac{5}{12}$, follow these steps:

Step 1: Add $\frac{1}{4}$ to both sides of the equation to isolate the term with $x$:

$\frac{2}{3}x – \frac{1}{4} + \frac{1}{4} = \frac{5}{12} + \frac{1}{4}$

This simplifies to:

$\frac{2}{3}x = \frac{5}{12} + \frac{3}{12} = \frac{8}{12} = \frac{2}{3}$

Step 2: Multiply both sides by $\frac{3}{2}$ to solve for $x$:

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

The solution $x = 1$ is a rational number since it can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

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

c) $-7$

[Solution Description]

A rational number is any number that can be expressed as $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. Among the options, $-7$ can be written as $\frac{-7}{1}$, which fits the definition of a rational number.

Key Concept: Rational Numbers

b) $2x = 5$

[Solution Description]

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

For $x^2 = 2$, the solutions are $x = \sqrt{2}$ and $x = -\sqrt{2}$. Both are irrational numbers.

For $2x = 5$, solving for $x$ gives $x = \frac{5}{2}$, which is a rational number.

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

For $x^2 = \pi$, the solutions are $x = \sqrt{\pi}$ and $x = -\sqrt{\pi}$, both of which are irrational numbers.

Thus, the equation $2x = 5$ has a rational number solution.

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: Representation of Rational Numbers on a Number Line

c) Assertion is true, but Reason is false.

[Solution Description] The assertion states that the rational number $\frac{3}{4}$ lies to the left of 1 on the number line, which is true because $\frac{3}{4}$ is less than 1. However, the reason states that a rational number is always less than 1, which is false since rational numbers can be greater than, equal to, or less than 1. Therefore, the assertion is true, but the reason is false.

Key Concept: Representation of Rational Numbers on a Number Line

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

[Solution Description]

The assertion states that the rational number $\frac{1}{2}$ lies exactly midway between 0 and 1 on the number line, which is true because $\frac{1}{2}$ is exactly halfway between 0 and 1. The reason states that rational numbers can always be represented as points that divide the distance between two integers into equal parts, which is also true. However, the reason is not the correct explanation for the assertion, as the assertion specifically refers to $\frac{1}{2}$, while the reason is a general statement about all rational numbers.

Key Concept: Additive Identity of Rational Numbers

d) 0

[Solution Description]

The additive identity is an element that, when added to any rational number, leaves it unchanged. For rational numbers, the additive identity is 0. This means that for any rational number $a$, $a + 0 = a$. Therefore, the correct answer is 0.

Key Concept: Commutativity of Addition

c) $\frac{19}{12}$

[Solution Description]

To find the sum of $\frac{3}{4}$ and $\frac{5}{6}$, first find a common denominator. The least common denominator (LCD) of 4 and 6 is 12. Convert both fractions to have this denominator:

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

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

Now, add the two fractions:

$\frac{9}{12} + \frac{10}{12} = \frac{19}{12}$

Therefore, $a + b = \frac{19}{12}$.

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 under Multiplication

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

[Solution Description]

To find the product of two rational numbers, multiply the numerators together and the denominators together:

$a \times b = \frac{-2}{3} \times \frac{5}{7} = \frac{-2 \times 5}{3 \times 7} = \frac{-10}{21}$

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, 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 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, 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, Rational Numbers

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

[Solution Description]

Multiply the numerators and denominators to find the product:

$m \times n = \frac{3}{8} \times \frac{-4}{9} = \frac{3 \times -4}{8 \times 9} = \frac{-12}{72}$

Simplify the fraction by dividing both numerator and denominator by 12:

$\frac{-12 \div 12}{72 \div 12} = \frac{-1}{6}$

The product of $m$ and $n$ is $\frac{-1}{6}$.

Key Concept: Closure under Division

b) No, only if zero is excluded

[Solution Description]

Rational numbers are not closed under division because division by zero is not defined. However, if we exclude zero, the collection of all other rational numbers is closed under division.

Key Concept: Division of Rational Numbers, Closure under Division

a) $x = 7$, $y = 2$, $z = 5$, $w = 1$

[Solution Description]

Given the equation $\frac{x}{y} \div \frac{z}{w} = \frac{7}{10}$, we can rewrite it as:

$\frac{x}{y} \times \frac{w}{z} = \frac{7}{10}$

Simplifying, we get:

$\frac{xw}{yz} = \frac{7}{10}$

To satisfy this equation, we need to find integers $x, y, z, w$ such that $xw = 7k$ and $yz = 10k$ for some integer $k$. For example, let $k = 1$, then $xw = 7$ and $yz = 10$. One possible solution is $x = 7$, $w = 1$, $y = 2$, and $z = 5$, since:

$\frac{7}{2} \div \frac{5}{1} = \frac{7}{2} \times \frac{1}{5} = \frac{7}{10}$

Key Concept: Commutativity, Rational Numbers

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

[Solution Description]

The assertion states that addition is commutative for rational numbers, which is true as per the syllabus. The reason states that addition is commutative for all types of numbers, including integers, whole numbers, and natural numbers. This is also true, but it does not directly explain why addition is commutative specifically for rational numbers. Therefore, while both the assertion and reason are true, the reason is not the correct explanation of the assertion.

Key Concept: Commutativity of Addition

a) $\( \frac{5}{6} + \frac{1}{3} = \frac{1}{3} + \frac{5}{6} \)$

[Solution Description]

The \textbf{commutative property of addition} states that:
\[
a + b = b + a
\]

We need to find a pair of rational numbers where the order of addition does not affect the sum.

Let’s check each option:

\begin{itemize}
\item[(A)] \( \frac{2}{3} + \frac{1}{4} = \frac{3}{7} \) — This is incorrect. The actual sum is:
\[
\frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12} \ne \frac{3}{7}
\]
Therefore, this does not demonstrate the commutative property.

\item[(B)] \( \frac{5}{6} + \frac{1}{3} = \frac{1}{3} + \frac{5}{6} \) — This is true. The equation shows the same two numbers added in reverse order, which is the essence of the commutative property of addition.

\item[(C)] \( \frac{1}{2} + \frac{2}{5} = \frac{1}{2} \) — This is false. The sum on the left-hand side is:
\[
\frac{1}{2} + \frac{2}{5} = \frac{5}{10} + \frac{4}{10} = \frac{9}{10} \ne \frac{1}{2}
\]

\item[(D)] \( \frac{3}{4} + \frac{1}{4} = 1 \) — While this is a valid sum, it does not show the reversal of terms, so it does not demonstrate the commutative property.

\end{itemize}

\textbf{Final Answer:}
\[
\boxed{\text{B) } \frac{5}{6} + \frac{1}{3} = \frac{1}{3} + \frac{5}{6}}
\]

\end{document}

Key Concept: Commutative Property of Addition

b) $\frac{2}{3} + \frac{7}{9}$

[Solution Description]

The commutative property of addition states that for any two rational numbers, $a + b = b + a$. Here, we have $\frac{7}{9} + \frac{2}{3}$.

Using the commutative property, $\frac{7}{9} + \frac{2}{3}$ is equivalent to $\frac{2}{3} + \frac{7}{9}$.

Therefore, the expression equivalent to $\frac{7}{9} + \frac{2}{3}$ using the commutative property of addition is $\frac{2}{3} + \frac{7}{9}$.

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 for Rational Numbers

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

[Solution Description]

The Commutative Property of Multiplication states that the order in which two rational numbers are multiplied does not affect the product. Mathematically, for any two rational numbers $a$ and $b$, $a \times b = b \times a$. This property holds true for rational numbers as demonstrated by examples such as $-\frac{7}{3} \times \frac{6}{5} = -\frac{42}{15}$ and $\frac{6}{5} \times -\frac{7}{3} = -\frac{42}{15}$, which are equal. Therefore, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Non-commutativity of Subtraction

b) No

[Solution Description]

Subtracting two rational numbers does not yield the same result if their order is reversed. For example, $\frac{2}{5} – \frac{5}{2}$ is not equal to $\frac{5}{2} – \frac{2}{5}$. Therefore, subtraction is not commutative for rational numbers.

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: Reciprocal in Division

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

[Solution Description]

The reciprocal of a fraction $\frac{a}{b}$ is $\frac{b}{a}$. Therefore, the reciprocal of $\frac{2}{5}$ is $\frac{5}{2}$.

Key Concept: Non-commutativity of Division, Rational Numbers

D) No, Expression 1 = $\dfrac{15}{4}$, Expression 2 = $\dfrac{15}{16}$

[Solution Description]

To determine whether the two expressions are equal, we must evaluate each one step-by-step using the rules of fraction division.

\textbf{Expression 1:}
\[
\left( \dfrac{3}{4} \div \dfrac{2}{5} \right) \div \dfrac{1}{2}
\]
Start by evaluating the expression inside the first parentheses:
\[
\dfrac{3}{4} \div \dfrac{2}{5} = \dfrac{3}{4} \times \dfrac{5}{2} = \dfrac{15}{8}
\]
Now divide this result by $\dfrac{1}{2}$:
\[
\dfrac{15}{8} \div \dfrac{1}{2} = \dfrac{15}{8} \times \dfrac{2}{1} = \dfrac{30}{8} = \dfrac{15}{4}
\]

\textbf{Expression 2:}
\[
\dfrac{3}{4} \div \left( \dfrac{2}{5} \div \dfrac{1}{2} \right)
\]
Evaluate the expression inside the parentheses first:
\[
\dfrac{2}{5} \div \dfrac{1}{2} = \dfrac{2}{5} \times \dfrac{2}{1} = \dfrac{4}{5}
\]
Now divide $\dfrac{3}{4}$ by the result:
\[
\dfrac{3}{4} \div \dfrac{4}{5} = \dfrac{3}{4} \times \dfrac{5}{4} = \dfrac{15}{16}
\]

\textbf{Conclusion:}

The two expressions are not equal. Expression 1 evaluates to $\dfrac{15}{4}$, while Expression 2 evaluates to $\dfrac{15}{16}$. Therefore, the correct answer is:

\textbf{Option (D): No, Expression 1 = $\dfrac{15}{4}$, Expression 2 = $\dfrac{15}{16}$.}

\end{document}

Key Concept: Associativity of Addition

b) $\frac{9}{4}$

[Solution Description]

First, solve the inner bracket: $\frac{5}{6} + \frac{2}{3}$. To add these fractions, convert them to a common denominator (LCM of 6 and 3 is 6). So, $\frac{2}{3}$ becomes $\frac{4}{6}$. Now, $\frac{5}{6} + \frac{4}{6} = \frac{9}{6}$. Simplify $\frac{9}{6}$ to $\frac{3}{2}$.

Next, add $\frac{3}{4}$ to $\frac{3}{2}$. Convert $\frac{3}{2}$ to $\frac{6}{4}$ for a common denominator (LCM of 4 and 2 is 4). Now, $\frac{3}{4} + \frac{6}{4} = \frac{9}{4}$. The final answer is $\frac{9}{4}$.

Key Concept: Associativity of Addition

c) Associative Property

[Solution Description]

The equation shows that the grouping of rational numbers does not affect the result when adding them. This is known as the associative property of addition for rational numbers. Thus, the correct answer is “Associative Property.”

Key Concept: Associative Property of Addition

a) $\frac{95}{42}$

[Solution Description]

First, add $\frac{1}{2}$ and $\frac{3}{7}$. The least common denominator (LCD) for 2 and 7 is 14.

$\frac{1}{2} = \frac{7}{14}, \quad \frac{3}{7} = \frac{6}{14}$

Adding these fractions:

$\frac{7}{14} + \frac{6}{14} = \frac{13}{14}$

Now, add $\frac{13}{14}$ to $\frac{4}{3}$. The LCD for 14 and 3 is 42.

$\frac{13}{14} = \frac{39}{42}, \quad \frac{4}{3} = \frac{56}{42}$

Adding these fractions:

$\frac{39}{42} + \frac{56}{42} = \frac{95}{42}$

Therefore, $\left( \frac{1}{2} + \frac{3}{7} \right) + \frac{4}{3} = \frac{95}{42}$.

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, Rational Numbers

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

[Solution Description]

First, calculate $y \times z$:

$\frac{3}{4} \times \frac{2}{3} = \frac{3 \times 2}{4 \times 3} = \frac{6}{12} = \frac{1}{2}$

Next, multiply the result by $x$:

$\frac{-1}{2} \times \frac{1}{2} = \frac{-1 \times 1}{2 \times 2} = \frac{-1}{4}$

The final value of $x \times (y \times z)$ is $\frac{-1}{4}$.

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: Non-associativity of Subtraction

b) $(x – y) – z \neq x – (y – z)$

[Solution Description]

To verify the non-associativity of subtraction, let’s compute $(x – y) – z$ and $x – (y – z)$ using $x = \frac{7}{8}$, $y = \frac{1}{4}$, and $z = \frac{1}{2}$. First, compute $(x – y) – z$: Subtract $y$ from $x$: $\frac{7}{8} – \frac{1}{4} = \frac{7}{8} – \frac{2}{8} = \frac{5}{8}$. Then subtract $z$: $\frac{5}{8} – \frac{1}{2} = \frac{5}{8} – \frac{4}{8} = \frac{1}{8}$. Next, compute $x – (y – z)$: Subtract $z$ from $y$: $\frac{1}{4} – \frac{1}{2} = \frac{1}{4} – \frac{2}{4} = \frac{-1}{4}$. Then subtract this result from $x$: $\frac{7}{8} – (\frac{-1}{4}) = \frac{7}{8} + \frac{1}{4} = \frac{7}{8} + \frac{2}{8} = \frac{9}{8}$. Since $\frac{1}{8} \neq \frac{9}{8}$, subtraction is not associative for rational numbers.

Key Concept: Non-associativity of Subtraction

b) $\left(\frac{3}{4} – \frac{1}{2}\right) – \frac{5}{6} \neq \frac{3}{4} – \left(\frac{1}{2} – \frac{5}{6}\right)$

[Solution Description]

To demonstrate that subtraction is not associative, we need to show that $(a – b) – c \neq a – (b – c)$. Let’s calculate both sides with $a = \frac{3}{4}$, $b = \frac{1}{2}$, and $c = \frac{5}{6}$. First, compute $(a – b) – c$: Subtract $b$ from $a$: $\frac{3}{4} – \frac{1}{2} = \frac{3}{4} – \frac{2}{4} = \frac{1}{4}$. Then subtract $c$: $\frac{1}{4} – \frac{5}{6} = \frac{3}{12} – \frac{10}{12} = \frac{-7}{12}$. Next, compute $a – (b – c)$: Subtract $c$ from $b$: $\frac{1}{2} – \frac{5}{6} = \frac{3}{6} – \frac{5}{6} = \frac{-2}{6} = \frac{-1}{3}$. Then subtract this result from $a$: $\frac{3}{4} – (\frac{-1}{3}) = \frac{3}{4} + \frac{1}{3} = \frac{9}{12} + \frac{4}{12} = \frac{13}{12}$. Since $\frac{-7}{12} \neq \frac{13}{12}$, subtraction 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) $\frac{7}{12}$

[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 Zero and One

b) $y \times 1 = y$

[Solution Description]

For integers, adding 0 results in the same integer, i.e., $y + 0 = y$. Similarly, multiplying by 1 results in the same integer, i.e., $y \times 1 = y$. Both zero and one play their respective roles as additive and multiplicative identities.

Key Concept: Additive Identity

a) $x$

[Solution Description]

The additive identity property states that adding 0 to any whole number leaves it unchanged. Therefore, $x + 0 = x$.

Key Concept: Additive Identity

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

[Solution Description]

According to the syllabus, when zero is added to a rational number, the result is the same rational number. This property is known as the additive identity property. For example, $-\frac{2}{7} + 0 = -\frac{2}{7}$. Here, both the Assertion and Reason are true, and the Reason correctly explains why the Assertion holds.

Key Concept: Whole Numbers, Additive Identity

a) $5$

[Solution Description]

The additive identity property states that for any whole number $z$, $0 + z = z$. Given $0 + z = 5$, it follows that $z = 5$.

Key Concept: Additive Identity, Rational Numbers

a) $\frac{3}{5}$

[Solution Description]

Given the equation $x + 0 = \frac{3}{5}$. Since zero is the additive identity for rational numbers, adding zero to any rational number does not change its value. Therefore, $x$ must be equal to $\frac{3}{5}$.

Key Concept: Zero in Addition

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 $c$, adding zero to $c$ will result in $c$. This is a fundamental property of zero in addition, known as the additive identity property. The reason states that zero is called the identity for the addition of rational numbers. Since the assertion and reason both correctly describe the role of zero in addition and the reason explains why the assertion is true, they are both correct, and the reason is the correct explanation of the assertion.

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: Additive Identity

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

[Solution Description]

The additive identity property states that adding zero to any rational number leaves the number unchanged. Here, $x = \frac{3}{4}$ and $y = 0$. Therefore, $x + y = \frac{3}{4} + 0 = \frac{3}{4}$.

Key Concept: Zero Product Property, Multiplicative Inverse

a) -2

[Solution Description]

The equation $(x^2 – 9)(x + 2) = 0$ can be solved by setting each factor equal to zero:

For $x^2 – 9 = 0$, we get $x = 3$ and $x = -3$.

For $x + 2 = 0$, we get $x = -2$.

The distinct real solutions are $-3$, $-2$, and $3$. Their sum is $-3 + (-2) + 3 = -2$.

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 the product of any number and zero is always zero. This is a fundamental property of multiplication in mathematics. Multiplying any number by zero results in zero because zero represents the absence of quantity. The reason provided is that zero is the additive identity. While it is true that zero is the additive identity (adding zero to any number does not change its value), this property is unrelated to multiplication. The reason does not explain why the product of any number and zero is zero; it discusses addition instead. Therefore, while both the assertion and reason are true, the reason does not correctly explain the assertion.

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: Rational Numbers Closed under Division

c) Rational numbers are closed under division when zero is excluded.

[Solution Description]

Rational numbers are closed under division if zero is excluded. Hence, the statement excluding zero is correct.

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: Multiplicative Identity

b) 1

[Solution Description]

The multiplicative identity for rational numbers is the number that, when multiplied by any rational number, gives back the same rational number. From the given syllabus, we know that $a \times 1 = 1 \times a = a$ for any rational number $a$. Therefore, the multiplicative identity for rational numbers is 1.

Key Concept: One in Multiplication

a) 5

[Solution Description]

Since multiplying any number by 1 gives the number itself, $1 \times y$ where $y = 5$ results in $5$.

Key Concept: Basic Concept

a) The number remains the same

[Solution Description]

When you multiply any number by one, the result is the number itself. This is because one is known as the multiplicative identity in mathematics.

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: Multiplicative Identity, Rational Numbers

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

[Solution Description]

The multiplicative identity for rational numbers is 1. This means that when any rational number is multiplied by 1, the result is the same rational number. Here, $a = \frac{5}{12}$ and $b = 1$. Therefore, the product of $a$ and $b$ is:

$a \times b = \frac{5}{12} \times 1 = \frac{5}{12}$

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

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

[Solution Description]

The assertion states that the expression $3 \times (4 + 5)$ can be simplified using the distributive property as $3 \times 4 + 3 \times 5$. This is correct because according to the distributive property, $a (b + c) = ab + ac$. Therefore, applying this property to the given expression, we get $3 \times (4 + 5) = 3 \times 4 + 3 \times 5$.

The reason provided is also correct as it accurately states the distributive property for multiplication over addition: $a (b + c) = ab + ac$.

Since both the assertion and the reason are true, and the reason correctly explains the assertion, the correct answer is:

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

c) $\frac{49}{64}$

[Solution Description]

Apply the distributive property $a(b + c) = ab + ac$, where $a = \frac{7}{8}$, $b = \frac{3}{4}$, and $c = \frac{1}{8}$.

So,

$\frac{7}{8} \times \left( \frac{3}{4} + \frac{1}{8} \right) = \frac{7}{8} \times \frac{3}{4} + \frac{7}{8} \times \frac{1}{8}$

Calculate each part:

$\frac{7}{8} \times \frac{3}{4} = \frac{21}{32}$

and

$\frac{7}{8} \times \frac{1}{8} = \frac{7}{64}$

Add the two parts:

$\frac{21}{32} + \frac{7}{64}$

Convert $\frac{21}{32}$ to sixty-fourths:

$\frac{21}{32} = \frac{42}{64}$

Now add:

$\frac{42}{64} + \frac{7}{64} = \frac{49}{64}$

Therefore, the correct answer is $\frac{49}{64}$.

Key Concept: Distributive Property of Multiplication Over Subtraction

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

[Solution Description]

First, apply the distributive property:

$\frac{4}{5} \times \left( \frac{7}{8} – \frac{3}{8} \right) = \frac{4}{5} \times \frac{7}{8} – \frac{4}{5} \times \frac{3}{8}$.

Next, perform the multiplication:

$\frac{4}{5} \times \frac{7}{8} = \frac{28}{40}$ and $\frac{4}{5} \times \frac{3}{8} = \frac{12}{40}$.

Now, subtract the two results:

$\frac{28}{40} – \frac{12}{40} = \frac{16}{40}$.

Simplify by dividing numerator and denominator by 8:

$\frac{16}{40} = \frac{2}{5}$.

So, the simplified form is $\frac{2}{5}$.

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: 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 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 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

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

[Solution Description]

The number closest to 0 will have the smallest absolute value. Let’s calculate the absolute values:

Step 1: Find the absolute value of each option:

Option 1: $|\frac{-1}{3}| = \frac{1}{3} \approx 0.333$

Option 2: $|\frac{1}{5}| = \frac{1}{5} = 0.2$

Option 3: $|\frac{2}{7}| \approx 0.286$

Option 4: $|\frac{-1}{4}| = \frac{1}{4} = 0.25$

Step 2: Compare the absolute values. The smallest value is 0.2, corresponding to $\frac{1}{5}$.

So, $\frac{1}{5}$ is closest to 0.

Key Concept: Finding Rational Numbers Between Two Rational Numbers

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

[Solution Description] To find a rational number between $\frac{2}{5}$ and $\frac{3}{5}$, we can use the mean method. The mean of two numbers $a$ and $b$ is given by $\frac{a + b}{2}$. Here, $a = \frac{2}{5}$ and $b = \frac{3}{5}$. So, the mean is $\frac{\frac{2}{5} + \frac{3}{5}}{2} = \frac{\frac{5}{5}}{2} = \frac{1}{2}$. Therefore, $\frac{1}{2}$ is a rational number between $\frac{2}{5}$ and $\frac{3}{5}$.

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: Standard Form of a Rational Number

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

[Solution Description]

To find the standard form of $\frac{108}{-144}$, we first determine the GCD of 108 and 144. The GCD is 36. Dividing both numerator and denominator by 36, we get $\frac{108 \div 36}{-144 \div 36} = \frac{3}{-4}$. Since the denominator should be positive in the standard form, we multiply both numerator and denominator by -1: $\frac{3 \times -1}{-4 \times -1} = \frac{-3}{4}$. Therefore, the standard form is $\frac{-3}{4}$.

Key Concept: Standard Form of a Rational Number

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

[Solution Description]

To convert $\frac{-72}{96}$ into its standard form, we first find the GCD of 72 and 96. The GCD of 72 and 96 is 24. Dividing both numerator and denominator by 24, we get $\frac{-72 \div 24}{96 \div 24} = \frac{-3}{4}$. Since the denominator is already positive, this is the standard form of the 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 a rational number $\frac{p}{q}$ is in its standard form when $p$ and $q$ are integers with no common factors other than 1, and $q > 0$. This is true as it defines the standard form of a rational number. The reason states that the standard form ensures that the representation of the rational number is unique. This is also true because by reducing a rational number to its simplest form, where the numerator and denominator have no common factors other than 1, and the denominator is positive, we ensure a unique representation. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Standard Form of Rational Number, Negative Numerator

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

[Solution Description]

To find the standard form of $\frac{-15}{20}$, we first simplify it by dividing both the numerator and the denominator by their greatest common divisor (GCD). The GCD of 15 and 20 is 5. Dividing both numerator and denominator by 5, we get $\frac{-15 \div 5}{20 \div 5} = \frac{-3}{4}$. Therefore, the standard form of $\frac{-15}{20}$ is $\frac{-3}{4}$.

Key Concept: Converting to Standard Form

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

[Solution Description]

To convert $\frac{144}{216}$ to 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 144 and 216 is 72. Dividing both the numerator and the denominator by 72 gives us $\frac{2}{3}$. Therefore, the standard form of $\frac{144}{216}$ is $\frac{2}{3}$.

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: Rational Numbers, Distributivity

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

[Solution Description] To verify the assertion, let’s simplify both sides of the equation step by step. First, calculate the left-hand side (LHS):

LHS = $\left(-\frac{3}{4} \times \frac{2}{3}\right) + \left(-\frac{3}{4} \times \frac{5}{6}\right)$

LHS = $\left(-\frac{6}{12}\right) + \left(-\frac{15}{24}\right)$

Simplify the fractions:

LHS = $-\frac{1}{2} – \frac{5}{8}$

Find a common denominator and subtract:

LHS = $-\frac{4}{8} – \frac{5}{8} = -\frac{9}{8}$

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

RHS = $-\frac{3}{4} \times \left(\frac{2}{3} + \frac{5}{6}\right)$

Simplify inside the parentheses first:

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

Now multiply:

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

Since LHS = RHS, the assertion is true. The reason given is also correct because the simplification is indeed based on the distributive property of multiplication over addition for rational numbers.

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: 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, 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: Comparison of Rational Numbers

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

[Solution Description]

The assertion is true because comparing rational numbers is easier when they share the same denominator. The reason is also true since having the same denominator simplifies the process by only requiring a comparison of the numerators. Furthermore, the reason correctly explains the assertion because converting to the same denominator is done specifically to facilitate this comparison. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Converting to Same Denominator

b) $\frac{3}{4} < \frac{5}{6}$

[Solution Description]

To compare $\frac{3}{4}$ and $\frac{5}{6}$, we need to convert them to the same denominator. The least common denominator (LCD) of 4 and 6 is 12. We convert each fraction as follows:

1. Convert $\frac{3}{4}$ to a denominator of 12:

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

2. Convert $\frac{5}{6}$ to a denominator of 12:

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

Now, compare $\frac{9}{12}$ and $\frac{10}{12}$. Since 9 < 10, $\frac{9}{12} < \frac{10}{12}$, which means $\frac{3}{4} < \frac{5}{6}$.

Thus, the correct comparison is $\frac{3}{4} < \frac{5}{6}$.

Key Concept: Converting to Same Denominator

a) $\frac{6}{15}$

[Solution Description]

To convert $\frac{2}{5}$ to an equivalent fraction with a denominator of 15, we multiply both the numerator and the denominator by 3 because $5 \times 3 = 15$. Thus, $\frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15}$. Therefore, the equivalent fraction is $\frac{6}{15}$.

Key Concept: Comparing Rational Numbers Using a Number Line

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

[Solution Description]

The assertion states that $\frac{-1}{2}$ lies to the left of 0 on a number line. This is true because negative numbers are positioned to the left of 0 on a number line. The reason given is that negative rational numbers are always located to the left of 0 on a number line, which is also true. Furthermore, the reason correctly explains why $\frac{-1}{2}$ lies to the left of 0. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

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: Commutativity of Addition

b) $\frac{-18}{7}$

[Solution Description]

To add $\frac{3}{7}$ and $\frac{-21}{7}$, we first ensure that both fractions have the same denominator, which they do (7). We then add the numerators: $3 + (-21) = -18$. The result is $\frac{-18}{7}$.

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: 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.

**Assertion**: Calculate $\frac{3}{4} + \frac{-1}{4}$: $\frac{3}{4} + \frac{-1}{4} = \frac{3 – 1}{4} = \frac{2}{4} = \frac{1}{2}$ The result $\frac{1}{2}$ is indeed a rational number, so the Assertion is true.

**Reason**: According to the syllabus, the sum of any two rational numbers is always a rational number. This property is known as closure under addition for rational numbers. Hence, the Reason is also true.

**Correct Explanation**: The Reason correctly explains why the Assertion holds true because the sum of two rational numbers (here, $\frac{3}{4}$ and $\frac{-1}{4}$) will always be a rational number due to the closure property.

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

b) $\frac{7}{4}$

[Solution Description]

Subtracting a negative number is the same as adding its positive counterpart. So, $\frac{11}{12} – \left(-\frac{5}{6}\right) = \frac{11}{12} + \frac{5}{6}$. First, convert $\frac{5}{6}$ to have the same denominator as $\frac{11}{12}$. The least common denominator (LCD) is 12:

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

Now add the two fractions:

$\frac{11}{12} + \frac{10}{12} = \frac{21}{12}.$

Simplify the fraction by dividing numerator and denominator by 3:

$\frac{21}{12} = \frac{7}{4}.$

Key Concept: Subtraction of Rational Numbers

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

[Solution Description]

To find $\frac{7}{6} – \frac{2}{3}$, we first find a common denominator, which is 6. Convert $\frac{2}{3}$ to $\frac{4}{6}$.

Now subtract the numerators: $\frac{7}{6} – \frac{4}{6} = \frac{3}{6} = \frac{1}{2}$.

Therefore, the value is $\frac{1}{2}$.

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: Closure Property of Rational Numbers under Multiplication

a) $\frac{21}{32}$

[Solution Description]

To find the product of $\frac{7}{8}$ and $\frac{3}{4}$, we multiply the numerators together and the denominators together.

Step 1: Multiply the numerators: $7 \times 3 = 21$.

Step 2: Multiply the denominators: $8 \times 4 = 32$.

Step 3: The product is $\frac{21}{32}$. Since 21 and 32 have no common divisors other than 1, the fraction cannot be simplified further.

Key Concept: Division of Rational Numbers, Multiplicative Inverse

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

[Solution Description]

Step 1: Find the multiplicative inverse of $y$.

$y^{-1} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$

Step 2: Divide $x$ by $y^{-1}$.

$\frac{3}{4} \div \frac{3}{2} = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$

Key Concept: Finding Rational Numbers Between Given Numbers

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

[Solution Description]

The assertion states that there are infinitely many rational numbers between any two given rational numbers. This is true because, for any two rational numbers, you can always find their mean, which is also a rational number. By repeating this process, you can generate an infinite sequence of rational numbers between the two given numbers. The reason states that the mean of two rational numbers is always a rational number. This is also true because the sum of two rational numbers is rational, and dividing by 2 (a rational number) preserves rationality. Additionally, the reason correctly explains why the assertion is true, as finding the mean is one method to generate rational numbers between two given rational numbers.

Key Concept: Mean Method

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

[Solution Description]

To find a rational number between $\dfrac{2}{5}$ and $\dfrac{3}{5}$, we use the mean method. The mean of two numbers $a$ and $b$ is given by $\dfrac{a + b}{2}$. Here, $a = \dfrac{2}{5}$ and $b = \dfrac{3}{5}$. So, the mean is calculated as:

$\text{Mean} = \dfrac{\dfrac{2}{5} + \dfrac{3}{5}}{2} = \dfrac{\dfrac{5}{5}}{2} = \dfrac{1}{2}$

Therefore, the rational number between $\dfrac{2}{5}$ and $\dfrac{3}{5}$ is $\dfrac{1}{2}$.

Key Concept: Finding Rational Numbers Between Two Numbers

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

[Solution Description]

To find a rational number between $\frac{1}{2}$ and $\frac{3}{4}$, we can calculate the mean of the two numbers. The mean is given by:

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

First, add $\frac{1}{2}$ and $\frac{3}{4}$. To add these fractions, convert them to have the same denominator:

$\frac{1}{2} = \frac{2}{4}, \quad \frac{3}{4} = \frac{3}{4}$

Now, add the fractions:

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

Next, divide the sum by 2:

$\frac{\frac{5}{4}}{2} = \frac{5}{8}$

Therefore, $\frac{5}{8}$ is a rational number between $\frac{1}{2}$ and $\frac{3}{4}$.