Class 8 Mathematics Chapter 1 Rational Numbers

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

b) All integers are rational numbers.

[Solution Description]

Rational numbers are those numbers that can be expressed as a ratio of two integers, where the denominator is not zero. They include all integers, fractions, and terminating or repeating decimals. Option A is incorrect because irrational numbers like $\sqrt{2}$ cannot be expressed as a ratio of integers. Option B is correct as rational numbers include all integers. Option C is incorrect as rational numbers can indeed be negative. Option D is incorrect because some rational numbers are integers.

Key Concept: Rational Numbers, Equations

b) $x = \frac{2}{7}$

[Solution Description]

To solve the equation $-\frac{7}{8}x + \frac{3}{4} = \frac{1}{2}$, follow these steps:

Step 1: Subtract $\frac{3}{4}$ from both sides of the equation to isolate the term with $x$:

$-\frac{7}{8}x + \frac{3}{4} – \frac{3}{4} = \frac{1}{2} – \frac{3}{4}$

This simplifies to:

$-\frac{7}{8}x = \frac{2}{4} – \frac{3}{4} = -\frac{1}{4}$

Step 2: Multiply both sides by $-\frac{8}{7}$ to solve for $x$:

$x = -\frac{1}{4} \times -\frac{8}{7} = \frac{8}{28} = \frac{2}{7}$

The solution $x = \frac{2}{7}$ 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: Definition of Rational Numbers

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

[Solution Description]

The assertion states that $-7$ is a rational number. According to the definition of rational numbers, any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, is a rational number. Here, $-7$ can be written as $\frac{-7}{1}$, which satisfies the condition for being a rational number. Therefore, the assertion is true. The reason provided correctly defines a rational number, and it accurately explains why $-7$ is a rational number. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Rational Numbers, Equations

c) $2x = 3$

[Solution Description]

To determine which equation requires rational numbers for its solution, we analyze each option:

For $3x + 6 = 9$, solving gives $x = 1$, which is an integer.

For $4x – 8 = 0$, solving gives $x = 2$, which is an integer.

For $2x = 3$, solving gives $x = \frac{3}{2}$, which is a rational number.

For $7x + 14 = 21$, solving gives $x = 1$, which is an integer.

The equation $2x = 3$ requires rational numbers for its solution.

Key Concept: Need for Rational Numbers

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

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

Key Concept: Rational Numbers

b) $-3$

[Solution Description]

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Let’s evaluate each option:

$\sqrt{2}$ cannot be expressed as $\frac{p}{q}$, so it is not a rational number.

$-3$ can be written as $\frac{-3}{1}$, which is of the form $\frac{p}{q}$, so it is a rational number.

$\pi$ is an irrational number and cannot be expressed as $\frac{p}{q}$.

$e$ is also an irrational number and cannot be expressed as $\frac{p}{q}$.

Thus, the correct answer is $-3$.

Key Concept: Rational Numbers

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

[Solution Description]

The assertion states that $\frac{3}{2}$ is a rational number. This is true because $\frac{3}{2}$ can be expressed as the ratio of two integers, 3 and 2, with the denominator not equal to zero. The reason provided is the definition of a rational number: any number that can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Since this definition applies correctly to $\frac{3}{2}$, the reason explains the assertion accurately.

Key Concept: Representation of Rational Numbers on a Number Line

c) Assertion is true, but Reason is false.

[Solution Description]

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

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

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

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

Key Concept: Rational Numbers, Number Line

b) $\frac{p}{q} < \frac{r}{s}$

[Solution Description]

If point $A$ representing $\frac{p}{q}$ is to the left of point $B$ representing $\frac{r}{s}$ on the number line, it means that $\frac{p}{q}$ is less than $\frac{r}{s}$. In other words, $\frac{p}{q} < \frac{r}{s}$. This is because, on a number line, smaller numbers are placed to the left of larger numbers.

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, Commutativity, Associativity

a) $\frac{59}{24}$

[Solution Description]

To solve this, we use the associativity property of addition for rational numbers. First, compute $a + b$:

$a + b = \frac{3}{4} + \frac{5}{6} = \frac{9}{12} + \frac{10}{12} = \frac{19}{12}$

Now, add $c$ to the result:

$(a + b) + c = \frac{19}{12} + \frac{7}{8} = \frac{38}{24} + \frac{21}{24} = \frac{59}{24}$

So, the expression $(a + b) + c$ equals $\frac{59}{24}$.

Key Concept: Closure under Subtraction

a) $\frac{1}{8}$

[Solution Description]

To subtract two rational numbers, we use a common denominator. Here, the denominators are 8 and 4. The least common denominator (LCD) is 8. Convert each fraction to have the LCD:

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

$b = \frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}$

Now, subtract the two fractions:

$a – b = \frac{7}{8} – \frac{6}{8} = \frac{7 – 6}{8} = \frac{1}{8}$

Key Concept: Closure Property

c) The product of two rational numbers is again a rational number

[Solution Description]

Rational numbers are closed under multiplication, meaning that the product of any two rational numbers is also a rational number. For example, $\frac{-2}{4} \times \frac{3}{5} = \frac{-3}{10}$, which is a rational number.

Key Concept: Closure Property of Rational Numbers

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

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

Key Concept: Closure under 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

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

[Solution Description]

The assertion states that the difference of two rational numbers is always a rational number. This is true because, by definition, a rational number can be expressed as the quotient of two integers with a non-zero denominator. When subtracting two rational numbers, say $\frac{a}{b}$ and $\frac{c}{d}$, the result is $\frac{ad – bc}{bd}$, which is also a rational number since both numerator and denominator are integers and the denominator is non-zero. The reason given is that rational numbers are closed under subtraction, which is also true. Moreover, the reason correctly explains the assertion because closure under subtraction directly implies that the difference of any two rational numbers will also be a rational number. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Closure under Multiplication

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

[Solution Description]

Multiply the numerators and denominators of the two fractions.

The numerator of the product is $7 \times 9 = 63$.

The denominator of the product is $12 \times 14 = 168$.

So, the product is $\frac{63}{168}$. Simplifying $\frac{63}{168}$ by dividing both numerator and denominator by 21 gives $\frac{3}{8}$.

Key Concept: Closure under Multiplication

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

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

Key Concept: Associative Property of Division

b) No, because the result of grouping changes the outcome

[Solution Description]

The associative property means:

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

This property does not hold for division of rational numbers.

Let’s take an example:

Let
$a=12$,
$b=3$,
$c=2$

Now compute both sides:

Left Side:

$(12 \div 3) \div 2 = 4 \div 2 = 2$

Right Side:

$12 \div (3 \div 2) = 12 \div 1.5 = 8$

Key Concept: Division by Zero

c) It is undefined

[Solution Description]

Division of any rational number by zero is undefined. This is because there is no number that can be multiplied by zero to get the original numerator.

Key Concept: Subtraction, Commutativity

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

[Solution Description]

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

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

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

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

Key Concept: Commutativity of Addition

c) Commutative

[Solution Description]

Commutative property states that for any two rational numbers $a$ and $b$, $a + b = b + a$. Therefore, addition is commutative for rational numbers.

Key Concept: Commutative Property of Addition

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 the definition of the commutative property of addition for rational numbers. The reason given is that addition is commutative for rational numbers, which directly explains why the assertion is true. Therefore, both the assertion and the reason are true, and the reason correctly explains 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) $\frac{3}{4} \times \frac{2}{5}$

[Solution Description]

The commutative property of multiplication states that for any two rational numbers $a$ and $b$, $a \times b = b \times a$. This means the order in which we multiply the numbers does not change the product. Let’s verify each option:

1. For $\frac{3}{4} \times \frac{2}{5}$, the product is $\frac{6}{20}$. When reversed, $\frac{2}{5} \times \frac{3}{4}$ also equals $\frac{6}{20}$. This satisfies the commutative property.

2. For $-\frac{1}{2} \times -\frac{3}{4}$, the product is $\frac{3}{8}$. When reversed, $-\frac{3}{4} \times -\frac{1}{2}$ also equals $\frac{3}{8}$. This satisfies the commutative property.

3. For $\frac{5}{6} \times -\frac{7}{8}$, the product is $-\frac{35}{48}$. When reversed, $-\frac{7}{8} \times \frac{5}{6}$ also equals $-\frac{35}{48}$. This satisfies the commutative property.

4. For $\frac{9}{10} \times 0$, the product is $0$. When reversed, $0 \times \frac{9}{10}$ also equals $0$. This satisfies the commutative property.

Since all options satisfy the commutative property, the correct answer is any one of them. Here, option 1 is chosen as the correct answer.

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) $p – q$ is greater than $q – p$

[Solution Description]

Calculate $p – q$:

$p – q = \frac{7}{8} – \frac{1}{4} = \frac{7}{8} – \frac{2}{8} = \frac{5}{8}$

Then, calculate $q – p$:

$q – p = \frac{1}{4} – \frac{7}{8} = \frac{2}{8} – \frac{7}{8} = -\frac{5}{8}$

The result shows that $p – q$ is positive while $q – p$ is negative.

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) $\frac{15}{8}$

[Solution Description] To find $a \div b$, we divide $\frac{3}{4}$ by $\frac{2}{5}$. Dividing by a fraction is the same as multiplying by its reciprocal. So, $\frac{3}{4} \div \frac{2}{5} = \frac{3}{4} \times \frac{5}{2} = \frac{15}{8}$.

Key Concept: Non-commutativity of Division

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

[Solution Description]

The assertion states that division is not commutative for rational numbers, which is true. The reason given is that for any two rational numbers $a$ and $b$, $a \div b \neq b \div a$. This is also true because when we divide one rational number by another, the result changes if we swap the numerator and the denominator. For example, $\frac{5}{4} \div \frac{3}{7} = \frac{35}{12}$, whereas $\frac{3}{7} \div \frac{5}{4} = \frac{12}{35}$. Since these results are not equal, the reason correctly explains why division is not commutative for rational numbers.

Key Concept: Associativity of Rational Numbers

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

[Solution Description]

The assertion states that for any three rational numbers $a$, $b$, and $c$, the expression $a + (b + c)$ is equal to $(a + b) + c$. This is a direct statement of the associativity property of addition for rational numbers. The reason provided is that addition is associative for rational numbers, which is a true statement and directly explains why the assertion holds. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Associativity of Rational Numbers

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

[Solution Description]

The assertion states that addition is associative for rational numbers, meaning that the way in which numbers are grouped during addition does not change the result. This is a fundamental property of rational numbers. The reason explains that the associativity property holds for addition due to the fact that the grouping of numbers does not affect the sum. Both the assertion and the reason are true, and the reason correctly explains why the assertion is valid. Therefore, the correct answer is that both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Associative Property of Addition

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

[Solution Description]

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

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

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

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

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

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

Key Concept: Associative Property of Addition

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 Multiplication, Rational Numbers

a) $\frac{-1}{8}$

[Solution Description]

First, calculate $p \times q$:

$\frac{5}{6} \times \frac{-3}{8} = \frac{5 \times -3}{6 \times 8} = \frac{-15}{48} = \frac{-5}{16}$

Next, multiply the result by $r$:

$\frac{-5}{16} \times \frac{2}{5} = \frac{-5 \times 2}{16 \times 5} = \frac{-10}{80} = \frac{-1}{8}$

The final value of $(p \times q) \times r$ is $\frac{-1}{8}$.

Key Concept: Associative Property of Multiplication

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

[Solution Description]

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

Key Concept: Non-associativity of Subtraction

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

[Solution Description]

To verify the assertion and reason, let’s take specific values of $a$, $b$, and $c$. Let $a = \frac{2}{3}$, $b = \frac{4}{5}$, and $c = \frac{1}{2}$.

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

Step 2: Compute $\frac{2}{3} – \frac{4}{5} = \frac{10}{15} – \frac{12}{15} = -\frac{2}{15}$.

Step 3: Subtract $\frac{1}{2}$ from the result: $-\frac{2}{15} – \frac{1}{2} = -\frac{4}{30} – \frac{15}{30} = -\frac{19}{30}$.

Step 4: Now calculate $\frac{2}{3} – \left(\frac{4}{5} – \frac{1}{2}\right)$.

Step 5: Compute $\frac{4}{5} – \frac{1}{2} = \frac{8}{10} – \frac{5}{10} = \frac{3}{10}$.

Step 6: Subtract this result from $\frac{2}{3}$: $\frac{2}{3} – \frac{3}{10} = \frac{20}{30} – \frac{9}{30} = \frac{11}{30}$.

Conclusion: Since $-\frac{19}{30} \neq \frac{11}{30}$, the subtraction is not associative for these rational numbers. Hence, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

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 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: Non-associativity of Division

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

[Solution Description]

To verify the assertion, consider three rational numbers: $a = 1$, $b = \frac{-1}{3}$, and $c = \frac{2}{5}$. First, calculate $(a \div b) \div c$:

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

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

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

Since $-\frac{15}{2} \neq -\frac{6}{5}$, the division operation is not associative for these rational numbers. The reason correctly explains the assertion by stating that $(a \div b) \div c$ is not equal to $a \div (b \div c)$ for any three rational numbers $a$, $b$, and $c$.

Key Concept: Additive Identity, Multiplicative Identity, Rational Numbers

c) $z$ can be any rational number

[Solution Description]

The additive identity property states that $z + 0 = z$. The multiplicative identity property states that $z \times 1 = z$. Therefore, $z + 0 = z \times 1$ simplifies to $z = z$, which is always true for any rational number $z$. This means that $z$ can be any rational number.

Key Concept: Additive Identity

b) $-\frac{3}{5}$

[Solution Description]

According to the additive identity property, adding 0 to any rational number does not change its value. Therefore, $-\frac{3}{5} + 0 = -\frac{3}{5}$.

Key Concept: Additive Identity of Whole Numbers

b) $15$

[Solution Description]

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

Key Concept: Additive Identity

b) -3

[Solution Description]

By the additive identity property, adding 0 to any integer results in the same integer. Therefore, $-3 + 0 = -3$.

Key Concept: Additive Identity

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

[Solution Description]

Adding zero to any rational number does not change its value. This is because zero is the additive identity for rational numbers. Therefore, $\frac{3}{5} + 0 = \frac{3}{5}$.

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: Rational Numbers, Zero as Additive Identity

c) $d$ is the same as $c$

[Solution Description]

The additive identity property of zero ensures that adding zero to any rational number leaves it unchanged. Therefore, $c + 0 = c$. This means that $d$ must be equal to $c$. Hence, the inference is that $d$ is the same as $c$.

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: Zero in Multiplication

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

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

Key Concept: Zero in Multiplication

c) Zero

[Solution Description]

Regardless of the sign of the number, when it is multiplied by zero, the result is always zero. Therefore, the product of zero and a negative number is zero.

Key Concept: Rational numbers, Closure under division

d) Division

[Solution Description]

The set of rational numbers is closed under division if the division of any two non-zero rational numbers results in another rational number. For example, $\frac{5}{2} \div \frac{25}{3} = \frac{5}{6}$, which is a rational number.

Key Concept: Division in Rational Numbers, Zero in Division

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

[Solution Description]

The assertion states that for any rational number $a$, $a \div 0$ is not defined. This is true because division by zero is undefined in mathematics. The reason given is that division by zero leads to an infinite or indeterminate value, which is also true. Moreover, the reason correctly explains why the assertion holds. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

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: Multiplicative Identity

b) Multiplicative identity for whole numbers is 1

[Solution Description]

Whole numbers are a subset of integers, and from the syllabus, we know that $a \times 1 = 1 \times a = a$ holds for rational numbers, including integers and whole numbers. Given that $b \times 1 = b$ for any whole number $b$, it can be concluded that the multiplicative identity for whole 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: Multiplicative Identity, Complex Numbers

a) No solution

[Solution Description]

Using the multiplicative identity property, multiplying by 1 does not change the value of $(z + i)$. So, the equation simplifies to $z + i = z + 3i$. Subtracting $z$ from both sides gives $i = 3i$, which implies $0 = 2i$. This is a contradiction, so there is no solution.

Key Concept: Multiplicative Identity

a) $\frac{5}{6}$

[Solution Description]

Multiplying any rational number by 1 gives the same rational number. So, $\frac{5}{6} \times 1 = \frac{5}{6}$.

Key Concept: Multiplicative Identity

a) The same rational number

[Solution Description]

The multiplicative identity property states that any number multiplied by 1 remains unchanged. For a rational number $a$, $a \times 1 = a$. Therefore, multiplying any rational number by 1 will always give the same rational number as the result.

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 $\frac{5}{8} \times \left( \frac{3}{4} – \frac{1}{2} \right)$ can be simplified using the distributive property of multiplication over subtraction. This is correct because the distributive property allows us to rewrite the expression as $\frac{5}{8} \times \frac{3}{4} – \frac{5}{8} \times \frac{1}{2}$.

The reason provided is also true because it correctly states the distributive property: $a(b – c) = ab – ac$ for any rational numbers $a$, $b$, and $c$.

Furthermore, the reason is the correct explanation of the assertion because it justifies why the expression can be simplified using the distributive property.

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

Key Concept: Distributive Property of Multiplication Over Addition

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

[Solution Description]

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

Key Concept: Distributive Property of Multiplication Over Subtraction

b) $\frac{4}{25}$

[Solution Description]

First, apply the distributive property: $a (b – c) = ab – ac$. Here, $a = \frac{2}{5}$, $b = \frac{7}{10}$, and $c = \frac{3}{10}$. So,

$\frac{2}{5} \times \left( \frac{7}{10} – \frac{3}{10} \right) = \frac{2}{5} \times \frac{7}{10} – \frac{2}{5} \times \frac{3}{10}$

Next, multiply the fractions:

$\frac{2}{5} \times \frac{7}{10} = \frac{14}{50} = \frac{7}{25}$

and

$\frac{2}{5} \times \frac{3}{10} = \frac{6}{50} = \frac{3}{25}$

Now, subtract the second result from the first:

$\frac{7}{25} – \frac{3}{25} = \frac{4}{25}$

Therefore, the simplified form is $\frac{4}{25}$.

Key Concept: Distributive Property of Multiplication Over Subtraction

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

[Solution Description]

The distributive property of multiplication over subtraction states that for any rational numbers $a$, $b$, and $c$, the expression $a(b – c)$ can be simplified to $ab – ac$. This is because multiplication distributes over subtraction, meaning the operation is consistent and valid for all rational numbers. Therefore, the assertion is true, and the reason correctly explains why the assertion holds.

Key Concept: Distributive Property of Multiplication Over Subtraction

b) 2

[Solution Description]

First, subtract the fractions inside the parentheses: $\frac{14}{9} – \frac{7}{9} = \frac{7}{9}$. Next, multiply by $\frac{18}{7}$: $\frac{18}{7} \times \frac{7}{9} = \frac{126}{63} = 2$.

Key Concept: Rational numbers between two rational numbers

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

[Solution Description]

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

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

Now, calculate the average:

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

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

Key Concept: Rational Numbers

c) $1$

[Solution Description]

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

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

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

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

Key Concept: Rational Numbers on a Number Line

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

[Solution Description]

The assertion states that $\frac{-3}{4}$ lies to the left of 0 on the number line. This is true because $\frac{-3}{4}$ is a negative rational number, and all negative numbers are positioned to the left of 0 on the number line. The reason states that all negative rational numbers lie to the left of 0 on the number line, which is also true. Furthermore, the reason correctly explains why the assertion is true, as $\frac{-3}{4}$ is a negative rational number and thus must lie to the left of 0.

Key Concept: Rational Numbers on a Number Line

c) $\frac{1}{8}$

[Solution Description]

To find the midpoint between two rational numbers, we add them and divide by 2.

Step 1: Add $-\frac{1}{2}$ and $\frac{3}{4}$. First, convert $-\frac{1}{2}$ to a denominator of 4: $-\frac{1}{2} = -\frac{2}{4}$.

Step 2: Add $-\frac{2}{4} + \frac{3}{4} = \frac{1}{4}$.

Step 3: Divide $\frac{1}{4}$ by 2: $\frac{1}{4} \div 2 = \frac{1}{8}$.

So, the midpoint is $\frac{1}{8}$.

Key Concept: Finding Rational Numbers Between Two Rational Numbers

c) $-\frac{1}{24}$

[Solution Description] To find a rational number between $-\frac{1}{3}$ and $\frac{1}{4}$, we can again use the mean method. The mean of two numbers $a$ and $b$ is given by $\frac{a + b}{2}$. Here, $a = -\frac{1}{3}$ and $b = \frac{1}{4}$. So, the mean is $\frac{-\frac{1}{3} + \frac{1}{4}}{2} = \frac{-\frac{4}{12} + \frac{3}{12}}{2} = \frac{-\frac{1}{12}}{2} = -\frac{1}{24}$. Therefore, $-\frac{1}{24}$ is a rational number between $-\frac{1}{3}$ and $\frac{1}{4}$.

Key Concept: Finding Rational Numbers Between Two Rational Numbers

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

[Solution Description]

The Assertion states that there are infinitely many rational numbers between any two given rational numbers, which is true as per the syllabus. The Reason mentions that the idea of mean helps us to find rational numbers between two rational numbers, which is also true. Moreover, the Reason correctly explains why the Assertion is true since using the mean is a method to find such rational numbers. Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Rational Numbers

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

[Solution Description]

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

Key Concept: Standard Form of Rational Number

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

[Solution Description]

To find the standard form of a rational number, we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor (GCD). For $\frac{-12}{18}$, the GCD of 12 and 18 is 6. Dividing both numerator and denominator by 6, we get $\frac{-12 \div 6}{18 \div 6} = \frac{-2}{3}$. Thus, the standard form is $\frac{-2}{3}$.

Key Concept: Standard Form of a Rational Number

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

[Solution Description]

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

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

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

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

Key Concept: Standard Form of a Rational Number

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

[Solution Description]

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

Key Concept: Standard Form of a Rational Number

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

[Solution Description]

To convert $\frac{12}{18}$ 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 12 and 18 is 6. Dividing both numerator and denominator by 6 gives us $\frac{12 \div 6}{18 \div 6} = \frac{2}{3}$. Therefore, the standard form of $\frac{12}{18}$ 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 simplify the rational number $\frac{12}{18}$, we find the greatest common divisor (GCD) of the numerator and denominator. The GCD of 12 and 18 is 6. Dividing both the numerator and the denominator by 6, we get $\frac{12 \div 6}{18 \div 6} = \frac{2}{3}$. Therefore, the standard form of $\frac{12}{18}$ is $\frac{2}{3}$. The Assertion states that $\frac{12}{18}$ can be simplified to its standard form, which is true. The Reason explains 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. Furthermore, the Reason correctly explains why the Assertion is true because simplifying a rational number involves removing common factors from the numerator and denominator.

Key Concept: Closure Property, Multiplication

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

[Solution Description]

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

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

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

Key Concept: 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 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 Numbers, Commutativity, Addition

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

[Solution Description]

To determine the correctness of the assertion and reason, consider the properties of rational numbers. Rational numbers are defined as numbers that can be expressed as the quotient of two integers, where the denominator is not zero. The closure property of rational numbers under addition states that the sum of any two rational numbers is also a rational number. This directly supports the assertion. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Comparison of Rational Numbers

d) $\frac{11}{12}$

[Solution Description]

To find which rational number is closer to 1, we calculate the difference between each fraction and 1.

– $1 – \frac{5}{6} = \frac{1}{6}$

– $1 – \frac{7}{8} = \frac{1}{8}$

– $1 – \frac{9}{10} = \frac{1}{10}$

– $1 – \frac{11}{12} = \frac{1}{12}$

Among these differences, $\frac{1}{12}$ is the smallest, indicating that $\frac{11}{12}$ is the closest to 1.

Key Concept: Comparison of Rational Numbers

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

[Solution Description]

To compare the rational numbers $\frac{3}{4}$, $\frac{5}{8}$, $\frac{7}{12}$, and $\frac{1}{2}$, we convert them to have a common denominator. The least common denominator (LCD) for 4, 8, 12, and 2 is 24.

– $\frac{3}{4} = \frac{18}{24}$

– $\frac{5}{8} = \frac{15}{24}$

– $\frac{7}{12} = \frac{14}{24}$

– $\frac{1}{2} = \frac{12}{24}$

Now, comparing the numerators, $\frac{18}{24}$ is the largest.

Key Concept: Converting to Same Denominator

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

[Solution Description]

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

Key Concept: Division of Rational Numbers, 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

a) $-1\frac{1}{2}$ is to the right of $-2\frac{1}{4}$

[Solution Description]

Convert $-1\frac{1}{2}$ and $-2\frac{1}{4}$ to decimals: $-1\frac{1}{2} = -1.5$ and $-2\frac{1}{4} = -2.25$. On a number line, $-1.5$ is to the right of $-2.25$ because $-1.5 > -2.25$.

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: Additive Identity

b) 0

[Solution Description]

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

Key Concept: Commutativity of Rational Numbers under Addition

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. This means that adding any two rational numbers will result in another rational number. The reason provided correctly explains the assertion as it directly relates to the closure property of rational numbers under addition.

Key Concept: Identity for Addition

c) $0$

[Solution Description]

The additive identity for rational numbers is 0. Adding 0 to any rational number does not change its value.

Key Concept: Addition of Rational Numbers, Associativity

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

[Solution Description]

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

Key Concept: Subtraction of Rational Numbers, Closure Property

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

[Solution Description]

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

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

Now, subtracting the two fractions:

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

So, $a – b = \frac{23}{12}$.

Key Concept: Closure Property of Rational Numbers

b) $\frac{11}{20}$

[Solution Description]

To find $\frac{3}{4} – \frac{1}{5}$, we first find a common denominator, which is 20. Then, convert both fractions:

$\frac{3}{4} = \frac{15}{20}$ and $\frac{1}{5} = \frac{4}{20}$.

Subtract the numerators: $\frac{15}{20} – \frac{4}{20} = \frac{11}{20}$.

Therefore, the result is $\frac{11}{20}$.

Key Concept: Closure Property of Rational Numbers

b) The product is always a rational number.

[Solution Description]

The closure property of rational numbers states that the product of any two rational numbers is also a rational number. This means that if $a$ and $b$ are rational numbers, then $a \times b$ is also a rational number.

Key Concept: Multiplication of Rational Numbers, Closure Property

a) $\frac{14}{27}$

[Solution Description]

Multiply $\frac{7}{9}$ and $\frac{2}{3}$ as follows:

$\frac{7}{9} \times \frac{2}{3} = \frac{7 \times 2}{9 \times 3} = \frac{14}{27}$

The result $\frac{14}{27}$ is a rational number, confirming that the product of two rational numbers is also a rational number.

Key Concept: Division of Rational Numbers

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

[Solution Description] The assertion states that division is not associative for rational numbers, which is true. The reason provides an explanation by stating that the grouping of divisions affects the result, which is also true. Furthermore, the reason correctly explains why division is not associative for rational numbers. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: 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 between any two rational numbers, there are countless rational numbers. This is true because the set of rational numbers is dense in the real numbers. The reason states that the mean of two rational numbers is always a rational number lying between them. This is also true because the mean of two rational numbers $a$ and $b$, given by $\frac{a + b}{2}$, is always a rational number that lies between $a$ and $b$. Moreover, the reason correctly explains the assertion, as finding the mean is one way to demonstrate the existence of countless rational numbers between any two rational numbers. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Concept of Mean Method

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

[Solution Description] To find a rational number between $\frac{1}{3}$ and $\frac{1}{2}$, we use the mean method. The mean of two numbers $a$ and $b$ is given by $\text{Mean} = \frac{a + b}{2}$. Substituting $a = \frac{1}{3}$ and $b = \frac{1}{2}$, we get:

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

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

Key Concept: Finding Rational Numbers Between Two Numbers

b) $\frac{31}{40}$

[Solution Description]

To find a rational number between $\frac{3}{4}$ and $\frac{4}{5}$, we use the mean method. First, add the two fractions:

$\frac{3}{4} + \frac{4}{5} = \frac{15}{20} + \frac{16}{20} = \frac{31}{20}$

Next, divide the sum by 2 to get the mean:

$\frac{\frac{31}{20}}{2} = \frac{31}{40}$

Thus, $\frac{31}{40}$ is a rational number between $\frac{3}{4}$ and $\frac{4}{5}$.