Chapter 1 Rational Numbers (Class 8 - Maths NCERT Exemplar Solutions)

Solution:

We check each option based on the properties of rational numbers:

(a) $\frac{2}{3} + \frac{5}{4} = \frac{5}{4} + \frac{2}{3}$: This is True as addition is commutative for rational numbers.

(b) $\frac{2}{3} - \frac{5}{4} = \frac{5}{4} - \frac{2}{3}$: This is False because subtraction is not commutative for rational numbers. Let's verify:

L.H.S. $= \frac{2}{3} - \frac{5}{4} = \frac{8 - 15}{12} = \frac{-7}{12}$

R.H.S. $= \frac{5}{4} - \frac{2}{3} = \frac{15 - 8}{12} = \frac{7}{12}$

Since $\frac{-7}{12} \neq \frac{7}{12}$, the statement is not true.

(c) $\frac{2}{3} \times \frac{5}{4} = \frac{5}{4} \times \frac{2}{3}$: This is True as multiplication is commutative for rational numbers.

(d) $\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5}$: This is True by the definition of division of rational numbers (multiplying by the reciprocal).

The correct option is (b).

Solution:

The given number is $\frac{0}{1}$, which is equal to $0$.

The multiplicative inverse of a rational number $a$ is $\frac{1}{a}$ such that $a \times \frac{1}{a} = 1$.

For $0$, the multiplicative inverse would be $\frac{1}{0}$.

Since division by zero is not defined in mathematics, the multiplicative inverse of $0$ does not exist.

The correct option is (d).

Solution:

First, let's convert the given numbers to have a common denominator or decimal form for easier comparison:

$\frac{-3}{4} = -0.75$

$\frac{1}{2} = 0.50$

We check the values in option (c):

$\frac{-1}{4} = -0.25$

$0 = 0$

$\frac{1}{4} = 0.25$

Comparing these with the range $(-0.75, 0.50)$:

$-0.75 < -0.25 < 0.50$ (True)

$-0.75 < 0 < 0.50$ (True)

$-0.75 < 0.25 < 0.50$ (True)

In other options, values like $\frac{3}{4} (0.75)$ or $\frac{-5}{4} (-1.25)$ fall outside the range.

The correct option is (c).

Solution:

Let the non-zero rational number be $x = \frac{a}{b}$ where $a, b \neq 0$.

Its reciprocal is $\frac{1}{x} = \frac{b}{a}$.

The product is:

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

The correct answer is 1.

Solution:

Given: $x = \frac{1}{3}$ and $y = \frac{6}{7}$

First, calculate $xy$:

$xy = \frac{1}{3} \times \frac{6}{7} = \frac{1 \times 6}{3 \times 7} = \frac{2}{7}$

Next, calculate $\frac{y}{x}$:

$\frac{y}{x} = \frac{6}{7} \div \frac{1}{3} = \frac{6}{7} \times 3 = \frac{18}{7}$

Now, find the value of $xy - \frac{y}{x}$:

$xy - \frac{y}{x} = \frac{2}{7} - \frac{18}{7} = \frac{2 - 18}{7} = \frac{-16}{7}$

The correct answer is $-\frac{16}{7}$.

Solution:

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

The number $0$ is a rational number (can be written as $\frac{0}{1}$).

The reciprocal of $0$ would be $\frac{1}{0}$, which is not defined.

Therefore, not every rational number has a reciprocal.

The statement is False.

Solution:

Let's convert the fractions to decimals for comparison:

$\frac{-4}{5} = -0.8$

$\frac{-5}{4} = -1.25$

On a number line, $-0.8$ lies to the right of $-1.25$.

Since a number to the right is always larger than a number to the left on a number line:

$-0.8 > -1.25$

Therefore, $\frac{-4}{5}$ is larger than $\frac{-5}{4}$.

The statement is True.

Solution:

To Find: The value of the expression $\frac{4}{7} \times \frac{14}{3} \div \frac{2}{3}$

Using the rule of division for rational numbers (multiplication by reciprocal):

$\frac{4}{7} \times \frac{14}{3} \div \frac{2}{3} = \frac{4}{7} \times \frac{14}{3} \times \frac{3}{2}$

Now, perform the multiplication and simplify by canceling terms:

$\frac{4 \times \cancel{14}^{\cancel{2}^{1}} \times \cancel{3}^{1}}{\cancel{7}_{1} \times \cancel{3}_{1} \times \cancel{2}_{1}}$

$= 4 \times 1$

$= 4$

The final value is 4.

Solution:

The given expression is:

$\frac{2}{3} \times \frac{-5}{7} + \frac{7}{3} + \frac{2}{3} \times \frac{-2}{7}$

Rearranging the terms using Commutative Property of addition to group terms with the common factor $\frac{2}{3}$:

$= \frac{2}{3} \times \left(\frac{-5}{7}\right) + \frac{2}{3} \times \left(\frac{-2}{7}\right) + \frac{7}{3}$

Using the Distributive Property [$a \times b + a \times c = a(b + c)$]:

$= \frac{2}{3} \times \left[\frac{-5}{7} + \left(\frac{-2}{7}\right)\right] + \frac{7}{3}$

Simplifying the terms inside the bracket:

$= \frac{2}{3} \times \left[\frac{-5 - 2}{7}\right] + \frac{7}{3}$

$= \frac{2}{3} \times \left(\frac{-7}{7}\right) + \frac{7}{3}$

$= \frac{2}{3} \times (-1) + \frac{7}{3}$

Now, calculating the final sum:

$= -\frac{2}{3} + \frac{7}{3}$

$= \frac{-2 + 7}{3}$

$= \frac{5}{3}$

Therefore, the value of the expression is $\frac{5}{3}$.

Given:

Point O represents the number 0.

Point P represents the number 3.

Point Z represents the number -5.

OQ = QR = RS = SP (segment OP is divided into 4 equal parts).

ZT = TO (T is the midpoint of segment ZO).

To Find:

Rational numbers represented by points Q, R, S, and T.

Number Line Representation:

Solution:

Step 1: Finding the values of Q, R, and S

The total length of the segment OP is calculated as:

$OP = P - O = 3 - 0 = 3$ units.

Since the segment OP is divided into 4 equal parts, the length of each part is $\frac{3}{4}$ units.

The points can be calculated by adding the part length successively starting from O:

Point Q $= O + \frac{3}{4} = 0 + \frac{3}{4} = \frac{3}{4}$

Point R $= Q + \frac{3}{4} = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2}$

Point S $= R + \frac{3}{4} = \frac{6}{4} + \frac{3}{4} = \frac{9}{4}$

Step 2: Finding the value of T

Point T lies between Z $(-5)$ and O $(0)$ such that $ZT = TO$. This implies that T is the midpoint of the segment ZO.

The value of T is the average of the values at Z and O:

$T = -\frac{5}{2} = -2.5$

Final Answer:

The rational numbers represented by the points are:

Q = $\frac{3}{4}$

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

S = $\frac{9}{4}$

T = $-\frac{5}{2}$

Given:

Total area of the field $= 49\frac{4}{5}$ ha.

Total number of people dividing the field $= 1 \text{ (son)} + 2 \text{ (daughters)} = 3 \text{ people}$.

To Find:

Area of each one's share.

Solution:

First, convert the mixed fraction of the area into an improper fraction:

$49\frac{4}{5} = \frac{49 \times 5 + 4}{5} = \frac{245 + 4}{5} = \frac{249}{5}$ ha.

Now, divide this area equally among 3 children:

Share of each child $= \frac{249}{5} \div 3$

Share of each child $= \frac{249}{5} \times \frac{1}{3}$

Share of each child $= \frac{\cancel{249}^{83}}{5} \times \frac{1}{\cancel{3}_{1}}$

Share of each child $= \frac{83}{5}$ ha.

Converting back to a mixed fraction:

$\frac{83}{5} = 16\frac{3}{5}$ ha.

Therefore, each child's share is $16\frac{3}{5}$ ha.

Given:

Three rational numbers are:

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

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

$c = -\frac{5}{6}$

Verification (i): Associative property of addition

To verify: $a + (b + c) = (a + b) + c$

L.H.S. $= a + (b + c)$

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

$= \frac{2}{3} + \left[\frac{4 \times 6 - 5 \times 5}{30}\right]$

$= \frac{2}{3} + \left[\frac{24 - 25}{30}\right]$

$= \frac{2}{3} + \left(-\frac{1}{30}\right) = \frac{20 - 1}{30}$

R.H.S. $= (a + b) + c$

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

$= \left(\frac{2 \times 5 + 4 \times 3}{15}\right) - \frac{5}{6}$

$= \frac{10 + 12}{15} - \frac{5}{6} = \frac{22}{15} - \frac{5}{6}$

$= \frac{22 \times 2 - 5 \times 5}{30} = \frac{44 - 25}{30}$

Since L.H.S. = R.H.S., the associative property of addition is verified.

Verification (ii): Associative property of multiplication

To verify: $a \times (b \times c) = (a \times b) \times c$

L.H.S. $= a \times (b \times c)$

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

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

$= \frac{2}{3} \times \left[\frac{-20}{30}\right] = \frac{2}{3} \times \left(-\frac{2}{3}\right)$

R.H.S. $= (a \times b) \times c$

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

$= \frac{8}{15} \times \left(-\frac{5}{6}\right)$

$= \frac{8 \times (-5)}{15 \times 6} = \frac{-40}{90}$

Since L.H.S. = R.H.S., the associative property of multiplication is verified.

Solutions:

(i) $\frac{5}{3} + 0 = \frac{5}{3}$

(ii) $\frac{-2}{5} + 0 = \frac{-2}{5}$

(iii) $\frac{3}{7} + 0 = \frac{3}{7}$

(iv) $\frac{2}{3} \times 1 = \frac{2}{3}$

(v) $\frac{-6}{7} \times 1 = \frac{-6}{7}$

(vi) $\frac{9}{8} \times 1 = \frac{9}{8}$

Observations:

1. In cases (i), (ii), and (iii), adding zero to any rational number results in the same rational number. Therefore, 0 is the additive identity for rational numbers.

2. In cases (iv), (v), and (vi), multiplying any rational number by one results in the same rational number. Therefore, 1 is the multiplicative identity for rational numbers.

Solution:

To find rational numbers between $\frac{-5}{6}$ and $\frac{7}{8}$, we first find a common denominator.

The L.C.M. of 6 and 8 is 24.

Converting the fractions:

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

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

Now, we can choose any five rational numbers between $\frac{-20}{24}$ and $\frac{21}{24}$.

Some examples are: $\frac{-19}{24}, \frac{-10}{24}, 0, \frac{5}{24}, \frac{20}{24}$.

After simplifying some of these, we get $-\frac{19}{24}, -\frac{5}{12}, 0, \frac{5}{24}, \frac{5}{6}$.

Solution:

The given rational numbers are $\frac{2}{3}, \frac{-4}{5}, \frac{1}{2}, \text{ and } \frac{1}{3}$.

Observation:

1. $\frac{2}{3}$ is a positive rational number.

2. $\frac{1}{2}$ is a positive rational number.

3. $\frac{1}{3}$ is a positive rational number.

4. $\frac{-4}{5}$ is a negative rational number.

Conclusion:

The rational number $\frac{-4}{5}$ is different from the others because it is the only negative rational number in the set, while all others are positive.

Given:

Product of two rational numbers $= -7$

One of the numbers $= -10$

To Find:

The other rational number.

Solution:

Let the other rational number be $x$.

According to the problem:

To find $x$, we divide the product by the given number:

$x = \frac{-7}{-10}$

$x = \frac{7}{10}$

Verification:

$\frac{7}{10} \times (-10) = -7$ (Correct)

Therefore, the other rational number is $\frac{7}{10}$.

Solution:

By definition, a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, is called a rational number.

While natural numbers, whole numbers, and integers can also be expressed in this form (by taking $q = 1$), the most comprehensive and specific term for the given definition is a rational number.

The correct option is (d).

Solution:

The standard definition of a rational number requires two conditions:

1. Both the numerator ($p$) and the denominator ($q$) must be integers.

2. The denominator ($q$) must not be zero ($q \neq 0$), because division by zero is undefined.

The numerator $p$ can be zero; therefore, $p \neq 0$ is not a requirement.

The correct option is (b).

Solution:

The expression shows the addition of two rational numbers, $\frac{3}{8}$ and $\frac{-5}{7}$, resulting in another rational number, $\frac{-19}{56}$.

The property stating that the sum of any two rational numbers is always a rational number is known as the closure property under addition.

The correct option is (a).

Solution:

Let us examine the closure property for rational numbers across different operations:

1. Addition: Sum of two rational numbers is always rational. (True)

2. Subtraction: Difference of two rational numbers is always rational. (True)

3. Multiplication: Product of two rational numbers is always rational. (True)

4. Division: If we divide a rational number by another rational number $q$, the result is rational except when $q = 0$. Since $0$ is a rational number and division by $0$ is undefined, the set of rational numbers is not closed under division.

The correct option is (d).

Solution:

The expression shows that changing the order of the addends does not change the sum:

$a + b = b + a$

This property is called the commutative property of addition.

The correct option is (a).

Solution:

The associative property of multiplication for rational numbers states that the grouping of numbers does not affect the product:

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

In option (a), we see exactly this structure:

$\frac{2}{3} \times \left( \frac{-6}{7} \times \frac{3}{5} \right) = \left( \frac{2}{3} \times \frac{-6}{7} \right) \times \frac{3}{5}$

Other options demonstrate commutativity or a mix of properties.

The correct option is (a).

Solution:

When zero is added to any rational number, the value of the rational number remains unchanged:

$a + 0 = a = 0 + a$

Therefore, zero is called the additive identity (or identity for addition) for rational numbers.

The correct option is (a).

Solution:

When any rational number is multiplied by one, the value of the rational number remains the same:

$a \times 1 = a = 1 \times a$

Therefore, one is called the multiplicative identity (or identity for multiplication) for rational numbers.

The correct option is (c).

The additive inverse of a rational number $a$ is $-a$ such that their sum is zero ($a + (-a) = 0$).

Additive inverse of $\frac{-7}{19} = - \left( \frac{-7}{19} \right) = \frac{7}{19}$

The correct option is (b).

The multiplicative inverse of a rational number $\frac{p}{q}$ is $\frac{q}{p}$.

The sign of the number does not change when taking its reciprocal. If the original number is negative, its reciprocal will also be negative.

Therefore, the multiplicative inverse of a negative rational number is a negative rational number.

The correct option is (b).

The property $x + 0 = 0 + x = x$ states that when zero is added to any rational number, the number remains unchanged.

This defines $0$ as the additive identity (or identity for addition) of rational numbers.

The correct option is (a).

To get a product of $1$, we must multiply a number by its multiplicative inverse (reciprocal).

The multiplicative inverse of $\frac{8}{21}$ is $\frac{21}{8}$.

$\frac{8}{21} \times \frac{21}{8} = 1$

The correct option is (c).

The expression $-(-x)$ represents the negative of a negative number.

According to the rules of signs in mathematics, the product of two negative signs results in a positive sign.

$-(-x) = x$

The correct option is (b).

First, convert the mixed fraction $-1\frac{1}{7}$ into an improper fraction:

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

The multiplicative inverse of $-\frac{8}{7}$ is obtained by interchanging the numerator and denominator while keeping the sign.

Multiplicative inverse $= -\frac{7}{8}$ or $\frac{7}{-8}$

The correct option is (d).

Zero is the additive identity for the set of rational numbers.

Adding zero to any rational number $x$ does not change its value.

$x + 0 = x$

The correct option is (a).

The reciprocal of a number $n$ is $\frac{1}{n}$.

Reciprocal of $1 = \frac{1}{1} = 1$

The correct option is (a).

Solution:

The reciprocal of a number $x$ is $1/x$.

The reciprocal of $-1$ is $\frac{1}{-1}$, which is equal to $-1$.

This is because the product of a number and its reciprocal must be $1$. Here, $(-1) \times (-1) = 1$.

The correct option is (b).

Solution:

The reciprocal of a number $x$ is defined as $\frac{1}{x}$.

If we try to find the reciprocal of $0$, we get $\frac{1}{0}$.

In mathematics, division by zero is not defined.

Therefore, the reciprocal of $0$ is not defined.

The correct option is (d).

Solution:

The reciprocal (or multiplicative inverse) of a non-zero rational number $\frac{p}{q}$ is the number which when multiplied by $\frac{p}{q}$ gives the product $1$.

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

Thus, the reciprocal of $\frac{p}{q}$ is $\frac{q}{p}$.

The correct option is (d).

Given:

$y$ is the reciprocal of $x$. This means $y = \frac{1}{x}$.

Solution:

We need to find the reciprocal of $y$.

The reciprocal of $y$ is $\frac{1}{y}$.

Substituting the value of $y$:

$\text{Reciprocal of } y = \frac{1}{\left(\frac{1}{x}\right)} = x$

The correct option is (a).

Solution:

First, let's find the product of the given expression:

$\frac{-3}{8} \times \left( \frac{-7}{13} \right) = \frac{(-3) \times (-7)}{8 \times 13} = \frac{21}{104}$

Now, we find the reciprocal of the resulting product $\frac{21}{104}$.

Reciprocal of $\frac{21}{104} = \frac{104}{21}$

The correct option is (a).

Solution:

The distributive property of multiplication over addition is expressed as:

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

Looking at option (a):

L.H.S. is $-\frac{1}{4} \times \left\{ \frac{2}{3} + \left( \frac{-4}{7} \right) \right\}$

R.H.S. is $\left[ -\frac{1}{4} \times \frac{2}{3} \right] + \left[ -\frac{1}{4} \times \left( \frac{-4}{7} \right) \right]$

This matches the identity $a \times (b + c) = ab + ac$.

The correct option is (a).

Solution:

Rational numbers are dense. This means that no matter how close two rational numbers are, there is always another rational number between them.

By repeatedly finding the mean (average) or changing the denominator, we can generate an endless sequence of numbers.

Therefore, there are infinitely many rational numbers between any two given rational numbers.

The correct option is (d).

Solution:

The expression $\frac{x + y}{2}$ represents the mean or average of two numbers $x$ and $y$.

On a number line, the average of two numbers always lies exactly at the midpoint between them.

Thus, $\frac{x + y}{2}$ is a rational number that lies between $x$ and $y$.

The correct option is (a).

Solution:

To find a rational number between two given rational numbers $x$ and $y$, we find their arithmetic mean.

The arithmetic mean of $x$ and $y$ is given by $\frac{x + y}{2}$.

This value is guaranteed to be greater than the smaller number and smaller than the larger number.

Therefore, statement (b) is always true.

The correct option is (b).

Given:

Rational number $= \frac{5}{7}$

Required numerator $= 45$

Solution:

To find the equivalent rational number, we need to determine the factor by which the original numerator must be multiplied to get 45.

$45 \div 5 = 9$

Now, multiply both the numerator and the denominator of the given fraction by 9:

$\frac{5 \times 9}{7 \times 9} = \frac{45}{63}$

The equivalent rational number is $\frac{45}{63}$.

Given:

Rational number $= \frac{7}{9}$

Required denominator $= 45$

Solution:

To find the equivalent rational number, we need to determine the factor by which the original denominator must be multiplied to get 45.

$45 \div 9 = 5$

Now, multiply both the numerator and the denominator of the given fraction by 5:

$\frac{7 \times 5}{9 \times 5} = \frac{35}{45}$

The equivalent rational number is $\frac{35}{45}$.

Given:

Rational numbers: $\frac{15}{20}$ and $\frac{35}{40}$

Solution:

To compare the two rational numbers, we first make their denominators equal by finding the L.C.M. of 20 and 40.

L.C.M. of 20 and 40 is 40.

Now, convert $\frac{15}{20}$ into an equivalent fraction with denominator 40:

$\frac{15 \times 2}{20 \times 2} = \frac{30}{40}$

Now, comparing $\frac{30}{40}$ and $\frac{35}{40}$:

Since $35 > 30$, it follows that $\frac{35}{40} > \frac{30}{40}$.

The greater number is $\frac{35}{40}$.

Solution:

The reciprocal of a rational number $\frac{a}{b}$ is $\frac{b}{a}$.

If a rational number is positive, both its numerator and denominator have the same sign (either both positive or both negative). When we take the reciprocal, the sign remains the same.

For example, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$, which is also positive.

Therefore, the reciprocal of a positive rational number is positive.

Solution:

The reciprocal of a rational number $\frac{a}{b}$ is $\frac{b}{a}$.

If a rational number is negative, one of its terms (either numerator or denominator) is negative. Taking the reciprocal does not change the sign of the overall value.

For example, the reciprocal of $-\frac{4}{5}$ is $-\frac{5}{4}$, which is also negative.

Therefore, the reciprocal of a negative rational number is negative.

Solution:

The reciprocal of a number $x$ is defined as $\frac{1}{x}$.

For zero, the reciprocal would be $\frac{1}{0}$. However, division by zero is not defined in mathematics.

Therefore, zero has no reciprocal.

Solution:

A number $x$ is its own reciprocal if $x = \frac{1}{x}$, which implies $x^{2} = 1$.

Solving for $x$, we find:

1. If $x = 1$, then $\frac{1}{1} = 1$.

2. If $x = -1$, then $\frac{1}{-1} = -1$.

The numbers 1 and -1 are their own reciprocal.

Given:

$y$ is the reciprocal of $x$.

Solution:

We need to find the reciprocal of $y^{2}$ in terms of $x$.

First, find $y^{2}$ by squaring equation (i):

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

Now, the reciprocal of $y^{2}$ is $\frac{1}{y^{2}}$.

Reciprocal of $y^{2} = \frac{1}{\left( \frac{1}{x^{2}} \right)} = x^{2}$

Therefore, the reciprocal of $y^{2}$ in terms of $x$ is $x^{2}$.

Solution:

First, we find the product of the given rational numbers:

$\frac{2}{5} \times \left( \frac{-4}{9} \right) = \frac{2 \times (-4)}{5 \times 9} = \frac{-8}{45}$

The reciprocal of a rational number $\frac{p}{q}$ is $\frac{q}{p}$.

Therefore, the reciprocal of $\frac{-8}{45}$ is $\frac{45}{-8}$ or $-\frac{45}{8}$.

Solution:

According to the laws of exponents and properties of rational numbers, the reciprocal of a product is the product of the reciprocals.

$(a \times b)^{-1} = a^{-1} \times b^{-1}$

Applying this property to the given expression:

$(213 \times 657)^{-1} = 213^{-1} \times 657^{-1}$

The blank should be filled with $657^{-1}$.

Solution:

The negative of a number is its additive inverse. For any number $a$, its negative is $-a$ such that $a + (-a) = 0$.

The negative of $1$ is $-1$ because $1 + (-1) = 0$.

The blank should be filled with $-1$.

Solution:

According to the distributive property of multiplication over addition for rational numbers:

$\frac{a}{b} \times \left( \frac{c}{d} + \frac{e}{f} \right) = \left( \frac{a}{b} \times \frac{c}{d} \right) + \left( \frac{a}{b} \times \frac{e}{f} \right)$

The blanks should be filled with $\left( \frac{a}{b} \times \frac{c}{d} \right)$ and $\left( \frac{a}{b} \times \frac{e}{f} \right)$.

Solution:

To compare $\frac{-5}{7}$ and $-3$, let us convert them to a common denominator or decimal form.

$\frac{-5}{7} \approx -0.71$

The other number is $-3$.

On a number line, $-0.71$ lies to the right of $-3$. Numbers to the right are always larger than numbers to the left.

Therefore, $\frac{-5}{7} > -3$.

The blank should be filled with greater.

Solution:

Rational numbers are dense on the number line. Between any two distinct rational numbers, we can find another rational number by taking their average or mean.

This process can be repeated an infinite number of times.

Therefore, there are infinitely many rational numbers between any two rational numbers.

Solution:

On a number line, zero is the center point. Positive numbers are represented to the right of zero, and negative numbers are represented to the left of zero.

Since $\frac{1}{3}$ is a positive rational number and $\frac{-1}{3}$ is a negative rational number, they lie on opposite sides of zero.

Solution:

Let a negative rational number be $-x$, where $x$ is a positive rational number.

The negative of this number is given by $-(-x)$.

In mathematics, the negative of a negative value results in a positive value:

$-(-x) = x$

Therefore, the negative of a negative rational number is always a positive rational number.

Solution:

Rational numbers follow the Commutative Property for both addition and multiplication.

This property states that for any two rational numbers $a$ and $b$:

This means the result remains the same regardless of the sequence in which the numbers are placed.

Therefore, the blank should be filled with order.

Solution:

The reciprocal of a rational number $\frac{a}{b}$ is defined as $\frac{b}{a}$. To find the reciprocal, we interchange the numerator and the denominator while keeping the sign of the number unchanged.

Given rational number $= \frac{-5}{7}$

Its reciprocal $= \frac{7}{-5}$

This can also be written as $-\frac{7}{5}$ or $\frac{-7}{5}$.

Solution:

The multiplicative inverse of a rational number is another name for its reciprocal. A number multiplied by its multiplicative inverse always results in the multiplicative identity, which is $1$.

Given rational number $= \frac{4}{3}$

Let its multiplicative inverse be $x$.

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

Therefore, the multiplicative inverse is $\frac{3}{4}$.

Solution:

To convert a decimal number into the $\frac{p}{q}$ form, we count the number of decimal places and divide the number by the corresponding power of $10$.

The number $10.11$ has two decimal places.

Since $1011$ and $100$ do not have any common factors other than $1$, the fraction is already in its simplest form.

Therefore, the blank should be filled with $\frac{1011}{100}$.

Solution:

This statement represents the Distributive Property of Multiplication over Addition for rational numbers. The property states:

Applying this to the given expression where $a = \frac{1}{5}$, $b = \frac{2}{7}$, and $c = \frac{3}{8}$:

$\frac{1}{5} \times \left[ \frac{2}{7} + \frac{3}{8} \right] = \left[ \frac{1}{5} \times \frac{2}{7} \right] + \left[ \frac{1}{5} \times \frac{3}{8} \right]$

Therefore, the blank should be filled with $\left[ \frac{1}{5} \times \frac{3}{8} \right]$.

Solution:

Rational numbers with a denominator of $1$ are equivalent to integers ($n = \frac{n}{1}$).

We need to find the integers that lie between $-2$ and $-5$ on the number line.

Looking at the number line between $-2$ and $-5$:

$-5 < -4 < -3 < -2$

The integers are $-3$ and $-4$. Written as rational numbers with denominator $1$, they are $\frac{-3}{1}$ and $\frac{-4}{1}$.

Therefore, the two numbers are $-3$ and $-4$.

By definition, a rational number is in the form $\frac{x}{y}$ where $x$ and $y$ are integers and $y \neq 0$.

Integers include negative numbers (e.g., $-1, -2, -3, ...$), whereas whole numbers consist only of $\{0, 1, 2, 3, ...\}$. Thus, $y$ can be a negative integer.

The statement is False.

A rational number is defined as $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

There is no restriction on the numerator $p$ being zero. In fact, $0$ is a rational number because it can be written as $\frac{0}{1}$, where $p = 0$.

The statement is False.

In a rational number $\frac{r}{s}$, the division is by $s$. Division by zero is undefined in mathematics.

Therefore, the condition $s \neq 0$ is essential for the number to be a valid rational number.

The statement is True.

To check if $\frac{5}{6}$ lies between $\frac{2}{3}$ and $1$, we convert them to a common denominator.

L.C.M. of $3, 6, 1$ is $6$.

$\frac{2}{3} = \frac{4}{6}$

$1 = \frac{6}{6}$

Now, comparing the numerators: $4 < 5 < 6$.

Since $\frac{4}{6} < \frac{5}{6} < \frac{6}{6}$, the number lies in between.

The statement is True.

Let's simplify the given fraction:

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

The question asks if $\frac{5}{10}$ lies between $\frac{1}{2}$ and $1$. For a number to lie "between" two values, it must be strictly greater than the smaller value and strictly less than the larger value.

Since $\frac{1}{2}$ is equal to $\frac{1}{2}$, it does not lie between them.

The statement is False.

Converting the rational number to decimal form:

$\frac{-7}{2} = -3.5$

On a number line, $-3.5$ is located exactly halfway between $-3$ and $-4$.

Therefore, it lies between them.

The statement is True.

Simplifying the fraction:

$\frac{9}{6} = \frac{3}{2} = 1.5$

Since $1 < 1.5 < 2$, the value $1.5$ lies between $1$ and $2$.

The statement is True.

The multiplicative inverse (reciprocal) of a rational number is a number such that their product is $1$.

Multiply $\frac{a}{b}$ by $\frac{b}{a}$:

$\frac{a}{b} \times \frac{b}{a} = \frac{ab}{ab} = 1$

Since the product is $1$ and it is given that $a \neq 0$ (and $b$ must be non-zero for the fraction to exist), the inverse is correct.

The statement is True.

The multiplicative inverse of a rational number $a$ is the number which, when multiplied by $a$, gives the product $1$. Let us check:

$\frac{-3}{5} \times \frac{5}{3} = -1$

Since the product is not $1$, the multiplicative inverse of $\frac{-3}{5}$ should be $\frac{-5}{3}$.

The statement is False (F).

The additive inverse of a rational number is the number which, when added to the original number, gives a sum of $0$. Let us check:

$\frac{1}{2} + (-2) = \frac{1 - 4}{2} = \frac{-3}{2} \neq 0$

The additive inverse of $\frac{1}{2}$ is $-\frac{1}{2}$.

The statement is False (F).

By definition, if $\frac{x}{y}$ is the additive inverse of $\frac{c}{d}$, then their sum must be equal to the additive identity, which is $0$.

The statement is True (T).

For any rational number $x$, adding $1$ increases its value by $1$ unit. The equation $x + 1 = x$ would imply that $1 = 0$, which is impossible.

The additive identity is $0$, meaning $x + 0 = x$.

The statement is False (F).

If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d}$, it means $\frac{x}{y} = -\frac{c}{d}$.

Substituting this into the given expression:

$\left( -\frac{c}{d} \right) - \frac{c}{d} = -\frac{2c}{d}$

This is not equal to $0$ unless $c = 0$. Since the property must hold for all rational numbers, this statement is incorrect.

The statement is False (F).

The reciprocal of a non-zero rational number $\frac{q}{p}$ is found by interchanging the numerator and the denominator, which results in $\frac{p}{q}$.

The statement incorrectly claims the reciprocal is the number itself.

The statement is False (F).

If $x + y = 0$, then $y$ is the additive inverse (negative) of $x$. Thus, $y = -x$.

Multiplying both sides by $-1$, we get $-y = x$.

The negative of $x$ is $y$. The statement claims $-y$ is the negative of $x$. Since $-y = x$, it would mean $x$ is the negative of $x$, which is only true for $0$.

The statement is False (F).

The negative of any rational number $x$ is $-x$. The negative of $(-x)$ is $-(-x)$.

In mathematics, the negative of a negative results in the original positive value:

$-(-x) = x$

The statement is True (T).

In the context of rational numbers, every number is categorized as either positive, negative, or zero. Zero is a neutral number, meaning it is neither positive nor negative.

While other rational numbers have a distinct negative counterpart (for example, the negative of $5$ is $-5$), zero is its own additive inverse ($0 + 0 = 0$). This means there is no separate or distinct "negative" value for zero that exists outside of itself.

Therefore, as a distinct signed value, the negative of zero does not exist.

The statement is True (T).

The negative (additive inverse) of $1$ is $-1$ because $1 + (-1) = 0$.

Since $1 \neq -1$, the negative of $1$ is not itself.

The statement is False (F).

Subtraction is not commutative for rational numbers. If we take $x = \frac{1}{2}$ and $y = \frac{1}{4}$:

$x - y = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

$y - x = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$

Since $\frac{1}{4} \neq -\frac{1}{4}$, the equality $x - y = y - x$ does not hold for all rational numbers.

The statement is False (F).

Multiplication is commutative for rational numbers. This means for any two rational numbers $x$ and $y$, the product remains the same regardless of the order.

The statement is True (T).

The product of any rational number $x$ and zero is always zero.

$x \times 0 = 0$

The multiplicative identity is $1$, which means $x \times 1 = x$.

The statement is False (F).

In rational numbers, multiplication distributes over addition, but addition does not distribute over multiplication.

$x + (y \times z)$ is not equal to $(x + y) \times (x + z)$.

For example, if $x = 1, y = 2, z = 3$:

$1 + (2 \times 3) = 1 + 6 = 7$

$(1 + 2) \times (1 + 3) = 3 \times 4 = 12$

The statement is False (F).

The distributive property of multiplication over addition states:

The statement provides $ab + bc$ instead of $ab + ac$. These are not equivalent in general.

The statement is False (F).

The reciprocal of a number $n$ is $1/n$. A number is its own reciprocal if $n = 1/n$.

For $n = 1$, the reciprocal is $1/1 = 1$.

For $n = -1$, the reciprocal is $1/-1 = -1$.

Since both $1$ and $-1$ are their own reciprocals, $1$ is not the only such number.

The statement is False (F).

The reciprocal of a rational number $x$ is $1/x$.

If we take the rational number $-1$, its reciprocal is $\frac{1}{-1} = -1$.

Since $-1$ is indeed the reciprocal of the rational number $-1$, the statement that it is not the reciprocal of any rational number is incorrect.

The statement is False (F).

Let us test this with a rational number, say $x = 5$.

L.H.S. $= x + (-1) = 5 - 1 = 4$

R.H.S. $= -x = -5$

Since $4 \neq -5$, the equation does not hold for all rational numbers.

The statement is False (F).

If $x < y$, then subtracting $y$ from $x$ will always result in a negative value.

For example, let $x = 2$ and $y = 5$. Here $x < y$.

$x - y = 2 - 5 = -3$

The result $-3$ is a negative rational number, not a positive one.

The statement is False (F).

When two negative rational numbers are added, their absolute values are added and the common negative sign is placed before the sum.

For example, $\left( -\frac{1}{2} \right) + \left( -\frac{1}{3} \right) = \frac{-3 - 2}{6} = -\frac{5}{6}$, which is also a negative rational number.

The statement is True (T).

Rational numbers possess the property of density, which states that between any two distinct rational numbers, there exist infinitely many rational numbers.

The statement claiming there are exactly ten is incorrect.

The statement is False (F).

Rational numbers are closed under addition, multiplication, and subtraction. This means the result of subtracting one rational number from another is always a rational number.

For example, $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$, which is a rational number.

The statement is False (F).

Subtraction is not commutative for rational numbers because $a - b$ is not equal to $b - a$ in general.

For example, $\frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$, but $\frac{1}{4} - \frac{3}{4} = -\frac{2}{4} = -\frac{1}{2}$.

Since $\frac{1}{2} \neq -\frac{1}{2}$, subtraction is not commutative.

The statement is False (F).

To compare $\frac{-3}{4}$ and $-2$, let us convert them to decimals:

$\frac{-3}{4} = -0.75$

On a number line, $-0.75$ lies to the right of $-2$. Any number to the right of another on the number line is greater.

Therefore, $\frac{-3}{4} > -2$, which means it is larger, not smaller.

The statement is False (F).

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

The number $0$ can be written as $\frac{0}{1}$ (or $\frac{0}{2}, \frac{0}{3}$, etc.), where $0$ and $1$ are integers and the denominator is not zero.

The statement is True (T).

Positive rational numbers extend infinitely in the positive direction on the number line ($0$ to $+\infty$).

Numbers like $1001, 2500.5$, or $\frac{5000}{2}$ are positive rational numbers, but they do not lie between $0$ and $1000$.

The statement is False (F).

The population of a country is always represented by a whole number (counting number).

Every whole number $n$ is a rational number because it can be expressed in the form $\frac{n}{1}$. For example, if the population was $1,027,015,247$, it is $\frac{1027015247}{1}$.

The statement is True (T).

According to the density property of rational numbers, between any two distinct rational numbers, there are infinitely many (countless) rational numbers.

No matter how close $\frac{5}{6}$ and $\frac{8}{9}$ are, we can always find more numbers between them by using the mean method or by increasing the denominator.

The statement is True (T).

By definition, $x^{-1}$ means $\frac{1}{x}$.

The reciprocal of a number $\frac{1}{x}$ is found by interchanging the numerator and the denominator, which gives $\frac{x}{1} = x$.

The statement incorrectly identifies the reciprocal as $\frac{1}{x}$.

The statement is False (F).

On a number line, positive rational numbers lie to the right of zero, and negative rational numbers lie to the left of zero.

The number $\frac{57}{23}$ is a positive rational number; therefore, it must lie to the right of zero.

The statement is False (F).

The rational number $\frac{7}{-4}$ is a negative rational number because it has one negative sign ($-\frac{7}{4}$).

Negative rational numbers always lie to the left of zero on the number line.

The statement is False (F).

The rational number $\frac{-8}{-3}$ is equivalent to $\frac{8}{3}$ because the negative signs in the numerator and denominator cancel each other out.

Since $\frac{8}{3}$ is a positive rational number, it lies to the right of zero on the number line.

Only zero lies neither to the right nor to the left of itself.

The statement is False (F).

On a number line, zero is the reference point. Rational numbers with a positive sign lie to the right side of zero, while rational numbers with a negative sign lie to the left side of zero.

Since $\frac{1}{2}$ is a positive rational number and $-1$ is a negative rational number, they are positioned on different sides of zero.

The statement is True (T).

A fraction is usually defined as a number in the form $\frac{a}{b}$, where $a$ and $b$ are whole numbers and $b \neq 0$.

A rational number is a number in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Since all whole numbers are also integers, every fraction satisfy the definition of a rational number.

The statement is True (T).

Any integer $n$ can be expressed in the form $\frac{n}{1}$, where both $n$ and $1$ are integers and the denominator is not zero.

For example, the integer $-5$ can be written as $\frac{-5}{1}$. This fits the definition of a rational number.

The statement is True (T).

Just like integers and fractions, all rational numbers have a unique position on the number line. Positive rational numbers are marked to the right of zero, and negative rational numbers are marked to the left of zero.

The statement is True (T).

Let a negative rational number be represented as $-x$ (where $x > 0$). The "negative" of this number is its additive inverse, which is given by $-(-x)$.

In algebra, the negative of a negative value always results in a positive value:

$-(-x) = x$

The statement is True (T).

If $x$ and $y$ are two numbers such that $x > y$, then subtracting the smaller number from the larger number always results in a value greater than zero.

Since the result is greater than zero, $x - y$ is a positive rational number.

The statement is True (T).

The set of rational numbers includes infinitely many negative numbers (like $-1, -100, -1000000, ...$).

Since negative rational numbers are smaller than zero, zero cannot be the smallest rational number. In fact, there is no "smallest" rational number as they extend to $-\infty$.

The statement is False (F).

Whole numbers are the set $\{0, 1, 2, 3, ...\}$. Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

As we can see, every element in the set of whole numbers is also present in the set of integers.

The statement is True (T).

Whole numbers are the set $\{0, 1, 2, 3, ...\}$. A rational number is any number that can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

Any whole number $n$ can be written as $\frac{n}{1}$. For example, $5 = \frac{5}{1}$ and $0 = \frac{0}{1}$. Since these satisfy the definition of rational numbers, every whole number is indeed a rational number.

The statement is True (T).

While $0$ is a whole number, it can also be expressed as $\frac{0}{1}, \frac{0}{2}$, etc. In these expressions, the numerator and denominator are integers, and the denominator is not zero.

Therefore, $0$ satisfies the definition of a rational number.

The statement is False (F).

On a number line, zero acts as the origin. All positive numbers are located to the right of zero, and all negative numbers are located to the left of zero.

Since $\frac{1}{2}$ is a positive rational number and $-\frac{5}{2}$ is a negative rational number, they are positioned on opposite sides of zero.

The statement is True (T).

This statement refers to the Commutative Property. For any two rational numbers $a$ and $b$, $a + b = b + a$ and $a \times b = b \times a$.

In the given example:

L.H.S. $= \frac{-4}{5} \times \frac{-6}{5} = \frac{24}{25}$

R.H.S. $= \frac{-6}{5} \times \frac{-4}{5} = \frac{24}{25}$

Since the results are identical, it confirms that multiplication of rational numbers is commutative.

The statement is True (T).

To Find:

The rational numbers from the provided list that are also integers.

Solution:

A rational number $\frac{p}{q}$ is an integer if the numerator $p$ is exactly divisible by the denominator $q$. Let us simplify each number in the list:

1. $\frac{9}{4} = 2.25$ (Not an integer)

2. $\frac{8}{4} = 2$ (Integer)

3. $\frac{7}{4} = 1.75$ (Not an integer)

4. $\frac{6}{4} = 1.5$ (Not an integer)

5. $\frac{9}{3} = 3$ (Integer)

6. $\frac{8}{3} = 2.66...$ (Not an integer)

7. $\frac{7}{3} = 2.33...$ (Not an integer)

8. $\frac{6}{3} = 2$ (Integer)

9. $\frac{5}{2} = 2.5$ (Not an integer)

10. $\frac{4}{2} = 2$ (Integer)

11. $\frac{3}{1} = 3$ (Integer)

12. $\frac{3}{2} = 1.5$ (Not an integer)

13. $\frac{1}{1} = 1$ (Integer)

14. $\frac{0}{1} = 0$ (Integer)

15. $\frac{-1}{1} = -1$ (Integer)

16. $\frac{-2}{1} = -2$ (Integer)

17. $\frac{-3}{2} = -1.5$ (Not an integer)

18. $\frac{-4}{2} = -2$ (Integer)

19. $\frac{-5}{2} = -2.5$ (Not an integer)

20. $\frac{-6}{2} = -3$ (Integer)

Final Answer:

The rational numbers which are also integers are: $\frac{8}{4}, \frac{9}{3}, \frac{6}{3}, \frac{4}{2}, \frac{3}{1}, \frac{1}{1}, \frac{0}{1}, \frac{-1}{1}, \frac{-2}{1}, \frac{-4}{2}, \frac{-6}{2}$.

Solution:

To solve this, we must simplify each given rational number to its lowest (standard) form and check if the denominator is 4.

1. $\frac{7}{8}$: The numerator and denominator have no common factors other than 1. It is already in its lowest form. The denominator is 8.

2. $\frac{64}{16}$:

$\frac{\cancel{64}^{4}}{\cancel{16}_{1}} = \frac{4}{1}$

In its lowest form, the denominator is 1.

3. $\frac{36}{-12}$:

$\frac{\cancel{36}^{3}}{\cancel{-12}_{-1}} = \frac{3}{-1} = \frac{-3}{1}$

In its lowest form, the denominator is 1.

4. $\frac{-16}{17}$: It is already in its lowest form. The denominator is 17.

5. $\frac{5}{-4}$:

$\frac{5}{-4} = \frac{-5}{4}$

It is in its lowest form, and the denominator is 4.

6. $\frac{140}{28}$:

$\frac{\cancel{140}^{5}}{\cancel{28}_{1}} = \frac{5}{1}$

In its lowest form, the denominator is 1.

Conclusion:

Only $\frac{5}{-4}$ can be written as a rational number with denominator 4 in its lowest form.

Solution (a):

We rearrange the terms to group those with common denominators:

$\frac{4}{7} + \left( \frac{-4}{9} \right) + \frac{3}{7} + \left( \frac{-13}{9} \right)$

$= \left( \frac{4}{7} + \frac{3}{7} \right) + \left( \frac{-4}{9} + \frac{-13}{9} \right)$

$= \left( \frac{4 + 3}{7} \right) + \left( \frac{-4 - 13}{9} \right)$

$= \frac{7}{7} + \left( \frac{-17}{9} \right)$

$= 1 - \frac{17}{9}$

$= \frac{9 - 17}{9} = \mathbf{\frac{-8}{9}}$

Solution (b):

$-5 + \frac{7}{10} + \frac{3}{7} + (-3) + \frac{5}{14} + \frac{-4}{5}$

Rearranging integers together and fractions with related denominators:

$= \{-5 + (-3)\} + \left( \frac{7}{10} - \frac{4}{5} \right) + \left( \frac{3}{7} + \frac{5}{14} \right)$

$= -8 + \left( \frac{7 - 8}{10} \right) + \left( \frac{6 + 5}{14} \right)$

$= -8 + \left( \frac{-1}{10} \right) + \frac{11}{14}$

Taking L.C.M. of 10 and 14, which is 70:

$= \frac{-8 \times 70 - 1 \times 7 + 11 \times 5}{70}$

$= \frac{-560 - 7 + 55}{70}$

$= \frac{-567 + 55}{70} = \frac{-512}{70}$

Simplifying the fraction:

$\frac{\cancel{-512}^{256}}{\cancel{70}_{35}} = \mathbf{\frac{-256}{35}}$

Verification (i): $x = \frac{3}{5}$

L.H.S. $= -(-x) = -\left( -\frac{3}{5} \right) = \frac{3}{5}$

R.H.S. $= x = \frac{3}{5}$

Since L.H.S. = R.H.S., it is verified.

Verification (ii): $x = \frac{-7}{9}$

L.H.S. $= -(-x) = -\left[ -\left( \frac{-7}{9} \right) \right] = -\left[ \frac{7}{9} \right] = -\frac{7}{9}$

R.H.S. $= x = -\frac{7}{9}$

Since L.H.S. = R.H.S., it is verified.

Verification (iii): $x = \frac{13}{-15} = -\frac{13}{15}$

L.H.S. $= -(-x) = -\left[ -\left( -\frac{13}{15} \right) \right] = -\left[ \frac{13}{15} \right] = -\frac{13}{15}$

R.H.S. $= x = -\frac{13}{15}$

Since L.H.S. = R.H.S., it is verified.

Examples of Closure Property:

1. Addition: Let $\frac{1}{2}$ and $\frac{1}{4}$ be two rational numbers.

$\frac{1}{2} + \frac{1}{4} = \frac{2 + 1}{4} = \frac{3}{4}$, which is a rational number. Hence, closed under addition.

2. Subtraction: Let $\frac{3}{5}$ and $\frac{1}{5}$ be two rational numbers.

$\frac{3}{5} - \frac{1}{5} = \frac{2}{5}$, which is a rational number. Hence, closed under subtraction.

3. Multiplication: Let $\frac{-2}{3}$ and $\frac{4}{5}$ be two rational numbers.

$\frac{-2}{3} \times \frac{4}{5} = \frac{-8}{15}$, which is a rational number. Hence, closed under multiplication.

Closure under Division:

No, rational numbers are not closed under division. This is because division by zero (which is a rational number) is not defined.

Example 1: $\frac{2}{3} \div \frac{5}{7} = \frac{2}{3} \times \frac{7}{5} = \frac{14}{15}$ (Result is a rational number).

Example 2: $\frac{2}{3} \div 0$. Since zero is a rational number but division by it is undefined, the result is not a rational number. Thus, closure property fails.

Verification (a): $x = \frac{1}{2}, y = \frac{1}{2}$

L.H.S. $= x + y = \frac{1}{2} + \frac{1}{2} = 1$

R.H.S. $= y + x = \frac{1}{2} + \frac{1}{2} = 1$

L.H.S. = R.H.S. (Verified)

Verification (b): $x = \frac{-2}{3}, y = \frac{-5}{6}$

L.H.S. $= x + y = \frac{-2}{3} + \left( \frac{-5}{6} \right) = \frac{-4 - 5}{6} = \frac{-9}{6} = -\frac{3}{2}$

R.H.S. $= y + x = \frac{-5}{6} + \left( \frac{-2}{3} \right) = \frac{-5 - 4}{6} = \frac{-9}{6} = -\frac{3}{2}$

L.H.S. = R.H.S. (Verified)

Verification (c): $x = \frac{-3}{7}, y = \frac{20}{21}$

L.H.S. $= x + y = \frac{-3}{7} + \frac{20}{21} = \frac{-9 + 20}{21} = \frac{11}{21}$

R.H.S. $= y + x = \frac{20}{21} + \left( \frac{-3}{7} \right) = \frac{20 - 9}{21} = \frac{11}{21}$

L.H.S. = R.H.S. (Verified)

Verification (d): $x = \frac{-2}{5}, y = \frac{-9}{10}$

L.H.S. $= x + y = \frac{-2}{5} + \left( \frac{-9}{10} \right) = \frac{-4 - 9}{10} = \frac{-13}{10}$

R.H.S. $= y + x = \frac{-9}{10} + \left( \frac{-2}{5} \right) = \frac{-9 - 4}{10} = \frac{-13}{10}$

L.H.S. = R.H.S. (Verified)

(a) Solution:

The given expression is $\left[ \frac{1}{2} \times \frac{1}{4} \right] + \left[ \frac{1}{2} \times 6 \right]$

We can see that $\frac{1}{2}$ is a common factor in both terms. We use the Distributive Property of Multiplication over Addition:

Applying the property where $a = \frac{1}{2}$, $b = \frac{1}{4}$ and $c = 6$:

$= \frac{1}{2} \times \left( \frac{1}{4} + 6 \right)$

$= \frac{1}{2} \times \left( \frac{1 + 24}{4} \right)$

$= \frac{1}{2} \times \frac{25}{4}$

$= \frac{25}{8}$

Converting to a mixed fraction:

$= 3\frac{1}{8}$

(b) Solution:

The given expression is $\left[ \frac{1}{5} \times \frac{2}{15} \right] - \left[ \frac{1}{5} \times \frac{2}{5} \right]$

We use the Distributive Property of Multiplication over Subtraction:

Applying the property where $a = \frac{1}{5}$, $b = \frac{2}{15}$ and $c = \frac{2}{5}$:

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

Taking the L.C.M. of 15 and 5, which is 15:

$= \frac{1}{5} \times \left( \frac{2 - 6}{15} \right)$

$= \frac{1}{5} \times \left( \frac{-4}{15} \right)$

$= \frac{1 \times (-4)}{5 \times 15}$

$= -\frac{4}{75}$

(c) Solution:

The given expression is $\frac{-3}{5} \times \left\{ \frac{3}{7} + \left( \frac{-5}{6} \right) \right\}$

We use the Distributive Property of Multiplication over Addition:

Applying the property:

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

$= \left( \frac{-9}{35} \right) + \left( \frac{\cancel{-3}^{1}}{\cancel{5}_{1}} \times \frac{-\cancel{5}^{1}}{\cancel{6}_{2}} \right)$

$= \frac{-9}{35} + \frac{1}{2}$

Taking the L.C.M. of 35 and 2, which is 70:

$= \frac{-18 + 35}{70}$

$= \frac{17}{70}$

Solution:

The expression shows that the way in which three rational numbers are grouped for multiplication does not change the final product.

General form: $a \times (b \times c) = (a \times b) \times c$

This is the Associative Property of Multiplication.

Verification (a):

L.H.S. $= x \times y = 7 \times \frac{1}{2} = \frac{7}{2}$

R.H.S. $= y \times x = \frac{1}{2} \times 7 = \frac{7}{2}$

L.H.S. = R.H.S. (Verified)

Verification (b):

L.H.S. $= \frac{2}{3} \times \frac{9}{4} = \frac{18}{12} = \frac{3}{2}$

R.H.S. $= \frac{9}{4} \times \frac{2}{3} = \frac{18}{12} = \frac{3}{2}$

L.H.S. = R.H.S. (Verified)

Verification (c):

L.H.S. $= \frac{-5}{7} \times \frac{14}{15} = \frac{-5 \times 14}{7 \times 15} = \frac{-70}{105} = \frac{-2}{3}$

R.H.S. $= \frac{14}{15} \times \frac{-5}{7} = \frac{14 \times (-5)}{15 \times 7} = \frac{-70}{105} = \frac{-2}{3}$

L.H.S. = R.H.S. (Verified)

Verification (d):

L.H.S. $= \frac{-3}{8} \times \frac{-4}{9} = \frac{12}{72} = \frac{1}{6}$

R.H.S. $= \frac{-4}{9} \times \frac{-3}{8} = \frac{12}{72} = \frac{1}{6}$

L.H.S. = R.H.S. (Verified)

Property Name: Associative Property of Multiplication.

Verification (a):

L.H.S. $= 1 \times \left( \frac{-1}{2} \times \frac{1}{4} \right) = 1 \times \left( \frac{-1}{8} \right) = -\frac{1}{8}$

R.H.S. $= \left( 1 \times \frac{-1}{2} \right) \times \frac{1}{4} = \left( -\frac{1}{2} \right) \times \frac{1}{4} = -\frac{1}{8}$

L.H.S. = R.H.S. (Verified)

Verification (b):

L.H.S. $= \frac{2}{3} \times \left( \frac{-3}{7} \times \frac{1}{2} \right) = \frac{2}{3} \times \left( \frac{-3}{14} \right) = \frac{-6}{42} = -\frac{1}{7}$

R.H.S. $= \left( \frac{2}{3} \times \frac{-3}{7} \right) \times \frac{1}{2} = \left( \frac{-2}{7} \right) \times \frac{1}{2} = -\frac{1}{7}$

L.H.S. = R.H.S. (Verified)

Verification (c):

L.H.S. $= \frac{-2}{7} \times \left( \frac{-5}{6} \times \frac{1}{4} \right) = \frac{-2}{7} \times \left( \frac{-5}{24} \right) = \frac{10}{168} = \frac{5}{84}$

R.H.S. $= \left( \frac{-2}{7} \times \frac{-5}{6} \right) \times \frac{1}{4} = \left( \frac{10}{42} \right) \times \frac{1}{4} = \frac{10}{168} = \frac{5}{84}$

L.H.S. = R.H.S. (Verified)

Verification (d):

Taking $z = 3$ (since $z$ was missing in the question):

L.H.S. $= 0 \times \left( \frac{1}{2} \times 3 \right) = 0 \times \frac{3}{2} = 0$

R.H.S. $= \left( 0 \times \frac{1}{2} \right) \times 3 = 0 \times 3 = 0$

L.H.S. = R.H.S. (Verified)

Verification (a):

L.H.S. $= \frac{-1}{2} \times \left( \frac{3}{4} + \frac{1}{4} \right) = \frac{-1}{2} \times \left( \frac{4}{4} \right) = -\frac{1}{2}$

R.H.S. $= \left( \frac{-1}{2} \times \frac{3}{4} \right) + \left( \frac{-1}{2} \times \frac{1}{4} \right) = \frac{-3}{8} + \frac{-1}{8} = \frac{-4}{8} = -\frac{1}{2}$

L.H.S. = R.H.S. (Verified)

Verification (b):

L.H.S. $= \frac{-1}{2} \times \left( \frac{2}{3} + \frac{3}{4} \right) = \frac{-1}{2} \times \left( \frac{8 + 9}{12} \right) = \frac{-1}{2} \times \frac{17}{12} = -\frac{17}{24}$

R.H.S. $= \left( \frac{-1}{2} \times \frac{2}{3} \right) + \left( \frac{-1}{2} \times \frac{3}{4} \right) = -\frac{2}{6} - \frac{3}{8} = \frac{-8 - 9}{24} = -\frac{17}{24}$

L.H.S. = R.H.S. (Verified)

Verification (c):

L.H.S. $= \frac{-2}{3} \times \left( \frac{-4}{6} + \frac{-7}{9} \right) = \frac{-2}{3} \times \left( \frac{-12 - 14}{18} \right) = \frac{-2}{3} \times \frac{-26}{18} = \frac{52}{54} = \frac{26}{27}$

R.H.S. $= \left( \frac{-2}{3} \times \frac{-4}{6} \right) + \left( \frac{-2}{3} \times \frac{-7}{9} \right) = \frac{8}{18} + \frac{14}{27} = \frac{24 + 28}{54} = \frac{52}{54} = \frac{26}{27}$

L.H.S. = R.H.S. (Verified)

Verification (d):

L.H.S. $= \frac{-1}{5} \times \left( \frac{2}{15} - \frac{3}{10} \right) = \frac{-1}{5} \times \left( \frac{4 - 9}{30} \right) = \frac{-1}{5} \times \frac{-5}{30} = \frac{5}{150} = \frac{1}{30}$

R.H.S. $= \left( \frac{-1}{5} \times \frac{2}{15} \right) + \left( \frac{-1}{5} \times \frac{-3}{10} \right) = -\frac{2}{75} + \frac{3}{50} = \frac{-4 + 9}{150} = \frac{5}{150} = \frac{1}{30}$

L.H.S. = R.H.S. (Verified)

Solution:

According to the Distributive Property: $a \times (b + c) = (a \times b) + (a \times c)$

(a) $\frac{3}{5} \times \left[ \frac{35}{24} + 10 \right]$

$= \left( \frac{3}{5} \times \frac{35}{24} \right) + \left( \frac{3}{5} \times 10 \right)$

$= \left( \frac{\cancel{3}^{1}}{\cancel{5}_{1}} \times \frac{\cancel{35}^{7}}{\cancel{24}_{8}} \right) + \left( \frac{3}{\cancel{5}_{1}} \times \cancel{10}^{2} \right)$

$= \frac{7}{8} + 6 = \frac{7 + 48}{8} = \mathbf{\frac{55}{8}}$

(b) $\frac{-5}{4} \times \left[ \frac{8}{5} + \frac{16}{15} \right]$

$= \left( \frac{-5}{4} \times \frac{8}{5} \right) + \left( \frac{-5}{4} \times \frac{16}{15} \right)$

$= \left( \frac{-\cancel{5}^{1}}{\cancel{4}_{1}} \times \frac{\cancel{8}^{2}}{\cancel{5}_{1}} \right) + \left( \frac{-\cancel{5}^{1}}{\cancel{4}_{1}} \times \frac{\cancel{16}^{4}}{\cancel{15}_{3}} \right)$

$= -2 + \left( -\frac{4}{3} \right) = \frac{-6 - 4}{3} = \mathbf{-\frac{10}{3}}$

(c) $\frac{2}{7} \times \left[ \frac{7}{16} - \frac{21}{4} \right]$

$= \left( \frac{2}{7} \times \frac{7}{16} \right) - \left( \frac{2}{7} \times \frac{21}{4} \right)$

$= \left( \frac{\cancel{2}^{1}}{\cancel{7}_{1}} \times \frac{\cancel{7}^{1}}{\cancel{16}_{8}} \right) - \left( \frac{\cancel{2}^{1}}{\cancel{7}_{1}} \times \frac{\cancel{21}^{3}}{\cancel{4}_{2}} \right)$

$= \frac{1}{8} - \frac{3}{2} = \frac{1 - 12}{8} = \mathbf{-\frac{11}{8}}$

(d) $\frac{3}{4} \times \left[ \frac{8}{9} - 40 \right]$

$= \left( \frac{3}{4} \times \frac{8}{9} \right) - \left( \frac{3}{4} \times 40 \right)$

$= \left( \frac{\cancel{3}^{1}}{\cancel{4}_{1}} \times \frac{\cancel{8}^{2}}{\cancel{9}_{3}} \right) - \left( \frac{3}{\cancel{4}_{1}} \times \cancel{40}^{10} \right)$

$= \frac{2}{3} - 30 = \frac{2 - 90}{3} = \mathbf{-\frac{88}{3}}$

(a) Solution:

The given expression is $\frac{32}{5} + \frac{23}{11} \times \frac{22}{15}$

According to the BODMAS rule, we must perform the multiplication first.

$= \frac{32}{5} + \left( \frac{23}{\cancel{11}_{1}} \times \frac{\cancel{22}^{2}}{15} \right)$

$= \frac{32}{5} + \frac{46}{15}$

To add these rational numbers, we find the L.C.M. of the denominators 5 and 15, which is 15.

$= \frac{32 \times 3 + 46 \times 1}{15}$

$= \frac{96 + 46}{15}$

$= \mathbf{\frac{142}{15}}$

(b) Solution:

The given expression is $\frac{3}{7} \times \frac{28}{15} \div \frac{14}{5}$

First, we convert the division into multiplication by taking the reciprocal of the divisor.

$= \frac{3}{7} \times \frac{28}{15} \times \frac{5}{14}$

Now, we simplify the expression by canceling common factors in the numerators and denominators.

$= \frac{\cancel{3}^{1}}{\cancel{7}_{1}} \times \frac{\cancel{28}^{4}}{\cancel{15}_{5}} \times \frac{5}{14}$

$= \frac{1}{1} \times \frac{4}{\cancel{5}_{1}} \times \frac{\cancel{5}^{1}}{14}$

$= \frac{\cancel{4}^{2}}{\cancel{14}_{7}}$

$= \mathbf{\frac{2}{7}}$

(c) Solution:

The given expression is $\frac{3}{7} + \frac{-2}{21} \times \frac{-5}{6}$

We perform the multiplication operation first.

$= \frac{3}{7} + \left( \frac{-\cancel{2}^{1}}{21} \times \frac{-5}{\cancel{6}_{3}} \right)$

$= \frac{3}{7} + \left( \frac{-1 \times -5}{21 \times 3} \right)$

$= \frac{3}{7} + \frac{5}{63}$

The L.C.M. of 7 and 63 is 63.

$= \frac{3 \times 9 + 5}{63}$

$= \frac{27 + 5}{63}$

$= \mathbf{\frac{32}{63}}$

(d) Solution:

The given expression is $\frac{7}{8} + \frac{1}{16} - \frac{1}{12}$

To simplify this, we find the L.C.M. of the denominators 8, 16, and 12.

L.C.M. (8, 16, 12) = 48

$= \frac{7 \times 6 + 1 \times 3 - 1 \times 4}{48}$

$= \frac{42 + 3 - 4}{48}$

$= \frac{45 - 4}{48}$

$= \mathbf{\frac{41}{48}}$

Solution:

The rational number that does not belong with the others is $-\frac{7}{3}$.

Reasoning:

1. The rational numbers $-\frac{5}{11}, -\frac{1}{2}, \text{ and } -\frac{4}{9}$ are all proper fractions (their numerators are smaller than their denominators in absolute value). On a number line, they all lie between 0 and -1.

2. The rational number $-\frac{7}{3}$ is an improper fraction ($7 > 3$). When converted to a mixed number, it is $-2\frac{1}{3}$. On a number line, it lies between -2 and -3, which is beyond the range of the other three numbers.

Given:

Total length of wire $= \frac{19}{4}$ metres

Total cost $= \textsf{₹} \frac{171}{2}$

To Find:

Cost of one metre of wire.

Solution:

Cost of 1 metre $= \text{Total Cost} \div \text{Total Length}$

Cost $= \frac{171}{2} \div \frac{19}{4}$

Cost $= \frac{171}{2} \times \frac{4}{19}$

Cost $= \frac{\cancel{171}^{9}}{\cancel{2}_{1}} \times \frac{\cancel{4}^{2}}{\cancel{19}_{1}}$

Cost $= 9 \times 2 = 18$

Final Answer: The cost of one metre of wire is $\textsf{₹} 18$.

Given:

Total Distance $= \frac{1445}{2}$ km

Total Time $= \frac{17}{2}$ hours

To Find:

Speed of the train in km/h.

Solution:

We know that, $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$

Speed $= \frac{1445}{2} \div \frac{17}{2}$

Speed $= \frac{1445}{2} \times \frac{2}{17}$

Speed $= \frac{1445}{\cancel{2}} \times \frac{\cancel{2}}{17} = \frac{1445}{17}$

By division:

$1445 \div 17 = 85$

Final Answer: The speed of the train is 85 km/h.

Given:

Total length of cloth $= 24$ m

Number of shirts that can be made $= 16$

To Find:

Cloth needed for making one shirt.

Solution:

To find the cloth required for one shirt, we divide the total length of the cloth by the total number of shirts.

Cloth for one shirt $= \frac{\text{Total length of cloth}}{\text{Number of shirts}}$

Cloth for one shirt $= \frac{24}{16}$ m

Simplifying the fraction by dividing both numerator and denominator by their highest common factor, 8:

Cloth for one shirt $= \frac{\cancel{24}^{3}}{\cancel{16}_{2}}$ m

Cloth for one shirt $= 1.5$ m

Therefore, $1.5$ m of cloth is needed to make one shirt.

Given:

Fraction of money $= \frac{7}{11}$

Value of this fraction $= \textsf{₹} 77,000$

To Find:

Total money in Hamid's bank account.

Solution:

Let the total money in Hamid's bank account be $x$.

According to the question:

To find $x$, we transpose $\frac{7}{11}$ to the other side:

$x = 77,000 \times \frac{11}{7}$

$x = \frac{\cancel{77000}^{11000} \times 11}{\cancel{7}_{1}}$

$x = 11,000 \times 11$

$x = 1,21,000$

Therefore, Hamid has $\textsf{₹} 1,21,000$ in his bank account.

Given:

Total length of the rope $= 117\frac{1}{3}$ m

Length of each small piece $= 7\frac{1}{3}$ m

Solution:

First, convert the mixed fractions into improper fractions:

Total length $= 117\frac{1}{3} = \frac{117 \times 3 + 1}{3} = \frac{351 + 1}{3} = \frac{352}{3}$ m

Length of each piece $= 7\frac{1}{3} = \frac{7 \times 3 + 1}{3} = \frac{21 + 1}{3} = \frac{22}{3}$ m

Now, the number of pieces is found by dividing the total length by the length of one piece:

Number of pieces $= \frac{352}{3} \div \frac{22}{3}$

Number of pieces $= \frac{352}{3} \times \frac{3}{22}$

Number of pieces $= \frac{352}{\cancel{3}} \times \frac{\cancel{3}}{22} = \frac{352}{22}$

Simplifying the fraction:

Number of pieces $= \frac{\cancel{352}^{16}}{\cancel{22}_{1}} = 16$

Therefore, there are 16 such small pieces.

Given:

Total number of students $= 48$

Students above average $= \frac{1}{6}$ of total

Average students $= \frac{1}{4}$ of total

Solution:

Step 1: Calculate the number of students who are above average.

Above average students $= \frac{1}{6} \times 48 = 8$

Step 2: Calculate the number of average students.

Average students $= \frac{1}{4} \times 48 = 12$

Step 3: Calculate the number of below average students.

Below average students $= \text{Total students} - (\text{Above average} + \text{Average})$

Below average students $= 48 - (8 + 12)$

Below average students $= 48 - 20 = 28$

Therefore, there are 28 below average students in the class.

To Find:

Total number of students in the school.

Solution:

Let the total number of students in the school be $x$.

Number of students coming by car $= \frac{2}{5}x$

Number of students coming by bus $= \frac{1}{4}x$

First, find the total fraction of students who come by vehicle:

Total by vehicle $= \frac{2}{5}x + \frac{1}{4}x = \left( \frac{8 + 5}{20} \right)x = \frac{13}{20}x$

Remaining students (who walk) $= x - \frac{13}{20}x = \frac{20x - 13x}{20} = \frac{7}{20}x$

Now, it is given that $\frac{1}{3}$ of these walkers walk on their own.

Students walking on their own $= \frac{1}{3} \times \left( \frac{7}{20}x \right) = \frac{7}{60}x$

According to the given condition, this number is equal to 224.

$x = \frac{224 \times 60}{7}$

$x = \frac{\cancel{224}^{32} \times 60}{\cancel{7}_{1}}$

$x = 32 \times 60$

$x = 1920$

Therefore, 1920 students study in that school.

Given:

Total monthly allowance $= \textsf{₹} 2,016$

Seema's share $= \frac{1}{2}$ of Huma's share

Hubna's share $= 1\frac{2}{3}$ times Seema's share

Solution:

Let Huma's share be $x$.

Then, Seema's share $= \frac{1}{2}x$

Hubna's share $= 1\frac{2}{3} \times (\text{Seema's share}) = \frac{5}{3} \times \frac{x}{2} = \frac{5x}{6}$

The total sum of their shares is equal to the total allowance:

$x + \frac{x}{2} + \frac{5x}{6} = 2016$

Taking the L.C.M. of 1, 2, and 6, which is 6:

$\frac{6x + 3x + 5x}{6} = 2016$

$\frac{14x}{6} = 2016$

$14x = 2016 \times 6$

$14x = 12096$

$x = \frac{12096}{14} = 864$

Now, calculating individual shares:

Huma's share $= x = \mathbf{\textsf{₹} 864}$

Seema's share $= \frac{x}{2} = \frac{864}{2} = \mathbf{\textsf{₹} 432}$

Hubna's share $= \frac{5x}{6} = \frac{5 \times 864}{6} = 5 \times 144 = \mathbf{\textsf{₹} 720}$

Given:

Total cost of constructing the room $= \textsf{₹} 60,000$.

Contribution of the elder daughter $= \frac{3}{8}$ of the mother's contribution.

Contribution of the younger daughter $= \frac{1}{2}$ of the mother's contribution.

To Find:

The individual contributions of the mother, the elder daughter, and the younger daughter.

Solution:

Let the mother's contribution be $\textsf{₹} x$.

Then, the elder daughter's contribution $= \frac{3}{8} \times x = \frac{3x}{8}$

And, the younger daughter's contribution $= \frac{1}{2} \times x = \frac{x}{2}$

The total contribution is the sum of the individual contributions of all three members:

$\text{Mother's share} + \text{Elder daughter's share} + \text{Younger daughter's share} $$ = \text{Total Cost}$

$x + \frac{3x}{8} + \frac{x}{2} = 60000$

To solve for $x$, find the L.C.M. of the denominators (1, 8, and 2), which is 8:

$\frac{8x + 3x + 4x}{8} = 60000$

$\frac{15x}{8} = 60000$

$15x = 60000 \times 8$

$15x = 480000$

$x = \frac{480000}{15}$

$x = 32000$

Individual Contributions:

1. Mother's contribution $= x = \mathbf{\textsf{₹} 32,000}$

2. Elder daughter's contribution $= \frac{3}{8} \times 32000 = 3 \times 4000 = \mathbf{\textsf{₹} 12,000}$

3. Younger daughter's contribution $= \frac{1}{2} \times 32000 = \mathbf{\textsf{₹} 16,000}$

Verification:

$\textsf{₹} 32,000 + \textsf{₹} 12,000 + \textsf{₹} 16,000 = \textsf{₹} 60,000$. (Verified)

Solution:

The first expression is in the form $a \times (b \times c)$.

The second expression is in the form $(a \times c) \times b$.

To reach from the first to the second, we change the order of multiplication ($b$ and $c$) and change the grouping.

The properties that allow this are the Commutative property and the Associative property of multiplication for rational numbers.

Properties Used:

(i) Commutative property of multiplication: $a \times b = b \times a$.

(ii) Distributive property of multiplication over addition: $a \times (b + c) = ab + ac$.

(iii) Associative property of addition: $a + (b + c) = (a + b) + c$.

(iv) Role of Zero / Additive identity: $a + 0 = 0 + a = a$.

(v) Role of One / Multiplicative identity: $a \times 1 = 1 \times a = a$.

(i) Solution for $-1\frac{1}{8}$:

First, convert the mixed fraction into an improper fraction:

$-1\frac{1}{8} = -\frac{1 \times 8 + 1}{8} = -\frac{9}{8}$

The multiplicative inverse of $-\frac{9}{8}$ is its reciprocal.

Multiplicative inverse $= \mathbf{-\frac{8}{9}}$

(ii) Solution for $3\frac{1}{3}$:

First, convert the mixed fraction into an improper fraction:

$3\frac{1}{3} = \frac{3 \times 3 + 1}{3} = \frac{10}{3}$

The multiplicative inverse of $\frac{10}{3}$ is its reciprocal.

Multiplicative inverse $= \mathbf{\frac{3}{10}}$

Given:

The rational numbers are $\frac{1}{4}$, $\frac{13}{16}$, and $\frac{5}{8}$.

Solution:

To arrange the rational numbers in descending order, we must first make their denominators equal by finding the L.C.M. (Least Common Multiple) of the denominators 4, 16, and 8.

L.C.M. of 4, 16, and 8 is 16.

Now, convert each fraction into an equivalent fraction with denominator 16:

$\frac{1}{4} = \frac{1 \times 4}{4 \times 4} = \frac{4}{16}$

$\frac{13}{16}$ is already with denominator 16.

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

Now, comparing the numerators of the equivalent fractions: $13, 4, \text{ and } 10$.

Since $13 > 10 > 4$, we have:

$\frac{13}{16} > \frac{10}{16} > \frac{4}{16}$

Replacing them with the original fractions:

$\frac{13}{16} > \frac{5}{8} > \frac{1}{4}$

Final Answer:

The descending order is $\frac{13}{16}, \frac{5}{8}, \frac{1}{4}$.

Given:

Product of two numbers $= \frac{-14}{27}$

One rational number $= \frac{7}{9}$

To Find:

The other rational number.

Solution:

Let the other rational number be $x$.

According to the question:

$x \times \frac{7}{9} = \frac{-14}{27}$

To find $x$, divide the product by the given number:

$x = \frac{-14}{27} \div \frac{7}{9}$

$x = \frac{-14}{27} \times \frac{9}{7}$

Simplify the expression by canceling common factors:

$x = \frac{\cancel{-14}^{-2}}{\cancel{27}_{3}} \times \frac{\cancel{9}^{1}}{\cancel{7}_{1}}$

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

Final Answer:

The other rational number is $-\frac{2}{3}$.

Given:

Rational number $= \frac{-15}{20}$

Required product $= \frac{-5}{7}$

Solution:

First, let's simplify the given number to its lowest form:

$\frac{-15}{20} = \frac{\cancel{-15}^{-3}}{\cancel{20}_{4}} = -\frac{3}{4}$

Let the number to be multiplied be $x$.

$x \times \left( -\frac{3}{4} \right) = \frac{-5}{7}$

$x = \frac{-5}{7} \div \left( -\frac{3}{4} \right)$

$x = \frac{-5}{7} \times \left( -\frac{4}{3} \right)$

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

$x = \frac{20}{21}$

Final Answer:

We should multiply by $\frac{20}{21}$.

Given:

Rational number $= \frac{-8}{13}$

Product $= 24$

Solution:

Let the required number be $x$.

$x \times \left( \frac{-8}{13} \right) = 24$

$x = 24 \div \left( \frac{-8}{13} \right)$

$x = 24 \times \left( \frac{13}{-8} \right)$

Simplify the expression:

$x = \frac{\cancel{24}^{3} \times 13}{\cancel{-8}_{-1}}$

$x = 3 \times (-13)$

$x = -39$

Final Answer:

The required number is $-39$.

Given:

Product of two rational numbers $= -7$

One number $= -5$

To Find:

The other rational number.

Solution:

Let the other rational number be $x$.

$x \times (-5) = -7$

$x = \frac{-7}{-5}$

Since both the numerator and the denominator are negative, the quotient is positive:

$x = \frac{7}{5}$

Final Answer:

The other rational number is $\frac{7}{5}$.

Solution:

The multiplicative inverse of a rational number $x$ is given by $\frac{1}{x}$. We are looking for a number $x$ such that its multiplicative inverse is $-1$.

Let the rational number be $x$.

By cross-multiplication, we get:

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

$x = -1$

Verification: The product of a number and its multiplicative inverse must be $1$. Here, $(-1) \times (-1) = 1$.

Therefore, the rational number whose multiplicative inverse is $-1$ is $-1$.

Solution:

To find five rational numbers between $0$ and $1$, we can express $0$ and $1$ as rational numbers with a denominator greater than $5$. Let us take the denominator as $6$.

$0 = \frac{0}{6}$

$1 = \frac{6}{6}$

The rational numbers between $\frac{0}{6}$ and $\frac{6}{6}$ are:

$\frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \text{ and } \frac{5}{6}$

On simplifying, we get:

$\frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \text{ and } \frac{5}{6}$.

Solution:

The absolute value of a rational number $x$, denoted as $|x|$, is its numerical value regardless of its sign.

We need to find two numbers such that $|x| = \frac{1}{5}$.

1. If $x$ is a positive rational number, then $| \frac{1}{5} | = \frac{1}{5}$.

2. If $x$ is a negative rational number, then $| -\frac{1}{5} | = \frac{1}{5}$.

Therefore, the two rational numbers are $\frac{1}{5}$ and $-\frac{1}{5}$.

Given:

Total length of the rope $= 40$ metres.

Length of each piece $= \frac{10}{3}$ metre.

To Find:

The number of pieces.

Solution:

The number of pieces is obtained by dividing the total length of the rope by the length of one piece.

Number of pieces $= 40 \div \frac{10}{3}$

Number of pieces $= 40 \times \frac{3}{10}$

Number of pieces $= \frac{\cancel{40}^{4}}{1} \times \frac{3}{\cancel{10}_{1}}$

Number of pieces $= 4 \times 3$

Number of pieces $= 12$

Therefore, there are 12 such pieces.

Given:

Total length of the rope $= 5\frac{1}{2}$ metres.

Number of equal pieces $= 12$.

Solution:

First, we convert the mixed fraction of the total length into an improper fraction:

$5\frac{1}{2} = \frac{5 \times 2 + 1}{2} = \frac{11}{2}$ metres.

The length of each piece is found by dividing the total length by the number of pieces:

Length of each piece $= \frac{11}{2} \div 12$

Length of each piece $= \frac{11}{2} \times \frac{1}{12}$

Length of each piece $= \frac{11 \times 1}{2 \times 12}$

Length of each piece $= \frac{11}{24}$ metre.

Therefore, the length of each piece is $\frac{11}{24}$ metre.

Given:

Rational numbers: $\frac{8}{7}$, $\frac{-9}{8}$, $\frac{-3}{2}$, $0$, and $\frac{2}{5}$.

Solution:

To arrange the numbers in descending order (largest to smallest), we first categorize them into positive, zero, and negative groups.

Positive numbers: $\frac{8}{7}$ and $\frac{2}{5}$

Zero: $0$

Negative numbers: $\frac{-9}{8}$ and $\frac{-3}{2}$

Step 1: Compare positive numbers.

$\frac{8}{7} \approx 1.14$ and $\frac{2}{5} = 0.4$.

Since $1.14 > 0.4$, we have $\frac{8}{7} > \frac{2}{5}$.

Step 2: Compare negative numbers.

$\frac{-9}{8} = -1.125$ and $\frac{-3}{2} = -1.5$.

On the number line, $-1.125$ is to the right of $-1.5$, so $\frac{-9}{8} > \frac{-3}{2}$.

Step 3: Arrange all numbers.

Positive numbers $> 0 >$ Negative numbers.

$\frac{8}{7} > \frac{2}{5} > 0 > \frac{-9}{8} > \frac{-3}{2}$

Final Answer: The descending order is $\frac{8}{7}, \frac{2}{5}, 0, \frac{-9}{8}, \frac{-3}{2}$.

(i) Solution for $0 \div \frac{2}{3}$:

We know that zero divided by any non-zero rational number is always zero.

$0 \div \frac{2}{3} = 0 \times \frac{3}{2} = \mathbf{0}$

(ii) Solution for $\frac{1}{3} \times \frac{-5}{7} \times \frac{-21}{10}$:

$= \frac{1 \times (-5) \times (-21)}{3 \times 7 \times 10}$

$= \frac{105}{210}$

Dividing both numerator and denominator by 105:

$= \frac{\cancel{105}^{1}}{\cancel{210}_{2}} = \mathbf{\frac{1}{2}}$

Given:

Temperature in Celsius ($C$) $= -16^\circ\text{C}$

Formula: $\frac{C}{5} = \frac{F - 32}{9}$

Solution:

Substituting $C = -16$ in the formula:

$\frac{-16}{5} = \frac{F - 32}{9}$

$-3.2 = \frac{F - 32}{9}$

Multiply both sides by 9:

$-3.2 \times 9 = F - 32$

$-28.8 = F - 32$

$F = 32 - 28.8$

$F = 3.2$

Final Answer: The temperature is $3.2^\circ\text{F}$.

Given:

The number is $7$.

Solution:

Step 1: Find the additive inverse of 7.

The additive inverse is $-7$ (because $7 + (-7) = 0$).

Step 2: Find the multiplicative inverse of 7.

The multiplicative inverse is $\frac{1}{7}$ (because $7 \times \frac{1}{7} = 1$).

Step 3: Find their sum.

Sum $= -7 + \frac{1}{7}$

Sum $= \frac{-7 \times 7 + 1}{7}$

Sum $= \frac{-49 + 1}{7} = -\frac{48}{7}$

Final Answer: The sum is $-\frac{48}{7}$.

Given:

The number is $-\frac{1}{3}$.

Solution:

Step 1: Find the additive inverse of $-\frac{1}{3}$.

The additive inverse is $\frac{1}{3}$.

Step 2: Find the multiplicative inverse of $-\frac{1}{3}$.

The multiplicative inverse is $-3$.

Step 3: Find their product.

Product $= \frac{1}{3} \times (-3)$

Product $= \frac{1}{\cancel{3}} \times (-\cancel{3})$

Product $= -1$

Final Answer: The product is $-1$.

Given (from the diagram):

Wingspan of Albatross $= 3\frac{3}{5}$ m

Wingspan of Sea Gull $= 1\frac{7}{10}$ m

Wingspan of Golden Eagle $= 2\frac{1}{2}$ m

Wingspan of Blue Jay $= \frac{41}{100}$ m

Solution (a):

Difference $= \text{Wingspan of Albatross} - \text{Wingspan of Sea Gull}$

$= 3\frac{3}{5} - 1\frac{7}{10}$

$= \frac{18}{5} - \frac{17}{10}$

$= \frac{36 - 17}{10} = \frac{19}{10} = 1\frac{9}{10}$ m

Therefore, the wingspan of an Albatross is $1\frac{9}{10}$ m longer than that of a Sea Gull.

Solution (b):

Difference $= \text{Wingspan of Golden Eagle} - \text{Wingspan of Blue Jay}$

$= 2\frac{1}{2} - \frac{41}{100}$

$= \frac{5}{2} - \frac{41}{100}$

$= \frac{250 - 41}{100} = \frac{209}{100} = 2\frac{9}{100}$ m

Therefore, the wingspan of a Golden Eagle is $2\frac{9}{100}$ m longer than that of a Blue Jay.

Given:

Length of strip $= 8\frac{3}{4} = \frac{35}{4}$ cm

Width of strip $= 1\frac{1}{4} = \frac{5}{4}$ cm

Diameter of each circle $= 1\frac{1}{4} = \frac{5}{4}$ cm

Solution:

1. Number of full circles:

Since the width of the strip and the diameter of the circle are equal, the circles can only be cut along the length.

Number of circles $= \text{Length of strip} \div \text{Diameter of circle}$

$= \frac{35}{4} \div \frac{5}{4}$

$= \frac{35}{4} \times \frac{4}{5} = 7$ circles

Thus, Shalini can cut 7 full circles.

2. Calculation of Wastage:

Wastage $= \text{Area of strip} - \text{Area of 7 circles}$

Area of strip $= \frac{35}{4} \times \frac{5}{4} = \frac{175}{16}$ cm$^{2}$

Radius ($r$) of circle $= \text{Diameter} \div 2 = \frac{5}{4} \times \frac{1}{2} = \frac{5}{8}$ cm

Area of 7 circles $= 7 \times \pi r^{2} = 7 \times \frac{22}{7} \times \frac{5}{8} \times \frac{5}{8} = \frac{22 \times 25}{64} = \frac{11 \times 25}{32} = \frac{275}{32}$ cm$^{2}$

Wastage $= \frac{175}{16} - \frac{275}{32} = \frac{350 - 275}{32} = \frac{75}{32}$ cm$^{2}$

Therefore, the wastage is $2\frac{11}{32}$ cm$^{2}$.

Given:

Sugar for 1st recipe $= \frac{1}{2}$ cup

Sugar for 2nd recipe $= 2$ tablespoons

Conversion: 1 tablespoon $= \frac{1}{16}$ cup

Solution:

First, convert the 2nd recipe's requirement into cups:

Sugar for 2nd recipe $= 2 \times \frac{1}{16} = \frac{1}{8}$ cup

Now, find the difference:

Difference $= \frac{1}{2} - \frac{1}{8}$

$= \frac{4 - 1}{8} = \frac{3}{8}$ cup

Therefore, the first recipe requires $\frac{3}{8}$ cup more sugar.

Solution (a):

Difference $= \text{Distance of Soni} - \text{Distance of Nancy}$

$= \frac{1}{20} - \frac{1}{32}$

L.C.M. of 20 and 32 is 160.

$= \frac{8 - 5}{160} = \frac{3}{160}$ km

Soni hopped $\frac{3}{160}$ km farther than Nancy.

Solution (b):

Total distance $= \text{Distance of Seema} + \text{Distance of Megha}$

$= \frac{1}{25} + \frac{1}{40}$

L.C.M. of 25 and 40 is 200.

$= \frac{8 + 5}{200} = \frac{13}{200}$ km

Total distance covered is $\frac{13}{200}$ km.

Solution (c):

Comparing Nancy ($\frac{1}{32}$) and Megha ($\frac{1}{40}$):

When the numerators are the same, the fraction with the smaller denominator is larger.

Since $32 < 40$, it implies $\frac{1}{32} > \frac{1}{40}$.

Therefore, Nancy hopped farther.

Step 1: Locate distances from the table.

Distance between Himgaon and Rawalpur $= 98\frac{3}{4}$ km

Distance between Sonapur and Ramgarh $= 40\frac{2}{3}$ km

Distance between Himgaon and Sonapur $= 100\frac{5}{6}$ km

Distance between Sonapur and Rawalpur $= 16\frac{1}{2}$ km

Solution (a):

We need to compare $98\frac{3}{4}$ and $40\frac{2}{3}$.

Clearly, $98 > 40$.

Therefore, the distance between Himgaon and Rawalpur is greater than the distance between Sonapur and Ramgarh.

Solution (b):

Total drive $= (\text{Himgaon to Sonapur}) + (\text{Sonapur to Rawalpur})$

$= 100\frac{5}{6} + 16\frac{1}{2}$

$= \frac{605}{6} + \frac{33}{2}$

$= \frac{605 + 99}{6} = \frac{704}{6}$

Simplifying:

$= \frac{352}{3} = 117\frac{1}{3}$ km

You would drive a total of $117\frac{1}{3}$ km.

Given:

Recycled portions: Paper = $\frac{5}{11}$, Aluminium cans = $\frac{5}{8}$, Glass = $\frac{2}{5}$, Scrap = $\frac{3}{4}$.

Solution:

(a) Paper recycled:

Comparing $\frac{5}{11}$ with $\frac{1}{2}$:

$\frac{5}{11} \approx 0.45$ and $\frac{1}{2} = 0.5$

Since $0.45 < 0.5$, the amount of paper recycled is less than $\frac{1}{2}$.

(b) Items with amount less than $\frac{1}{2}$:

Let's convert all fractions to decimals:

Paper: $\frac{5}{11} \approx 0.45$ (Less than 0.5)

Aluminium cans: $\frac{5}{8} = 0.625$ (More than 0.5)

Glass: $\frac{2}{5} = 0.4$ (Less than 0.5)

Scrap: $\frac{3}{4} = 0.75$ (More than 0.5)

Items with amount less than $\frac{1}{2}$ are Paper and Glass.

(c) Aluminium cans:

As calculated above, $\frac{5}{8} = 0.625$. Since $0.625 > 0.5$, the quantity is more than half.

(d) Arrangement (Greatest to smallest):

Comparing decimals: $0.75 > 0.625 > 0.45 > 0.4$

Order: Scrap, Aluminium cans, Paper, Glass.

Solution:

To express the given widths as rational numbers in the form $\frac{p}{q}$ and arrange them, we first convert each into a fraction and then compare their decimal values.

Step 1: Expressing in $\frac{p}{q}$ form and finding decimal equivalents:

1. $97.28$ cm:

$97.28 = \frac{9728}{100}$

Dividing both numerator and denominator by 4:

$= \frac{2432}{25}$ cm

Decimal value $= 97.28$

2. $98\frac{4}{9}$ cm:

$98\frac{4}{9} = \frac{98 \times 9 + 4}{9} = \frac{882 + 4}{9}$

$= \frac{886}{9}$ cm

Decimal value $\approx 98.44$ (approx)

3. $98\frac{1}{25}$ cm:

$98\frac{1}{25} = \frac{98 \times 25 + 1}{25} = \frac{2450 + 1}{25}$

$= \frac{2451}{25}$ cm

Decimal value $= 98.04$

4. $97.94$ cm:

$97.94 = \frac{9794}{100}$

Dividing both numerator and denominator by 2:

$= \frac{4897}{50}$ cm

Decimal value $= 97.94$

Step 2: Arranging in ascending order:

Comparing the decimal values: $97.28 < 97.94 < 98.04 < 98.44$

Therefore, the widths in ascending order are:

$97.28$ cm, $97.94$ cm, $98\frac{1}{25}$ cm, $98\frac{4}{9}$ cm

In $\frac{p}{q}$ form: $\frac{2432}{25}, \frac{4897}{50}, \frac{2451}{25}, \frac{886}{9}$

Given:

Height of existing coaster $= \frac{2}{3}$ m

Scale of new coaster $= \frac{3}{5}$ times

Solution:

Height of new coaster $= \frac{3}{5} \times \frac{2}{3}$

$= \frac{\cancel{3} \times 2}{5 \times \cancel{3}} = \frac{2}{5}$ m

Height of the new roller coaster will be $0.4$ m (or $\frac{2}{5}$ m).

Solution:

To convert the decimal rainfall values into the rational form $\frac{p}{q}$, we write the number without the decimal point as the numerator and a power of 10 as the denominator, then simplify to the lowest terms.

1. May: $2.6934$

$2.6934 = \frac{26934}{10000}$

Dividing both by 2:

$= \mathbf{\frac{13467}{5000}}$

2. June: $0.6096$

$0.6096 = \frac{6096}{10000}$

Dividing both by 16:

$= \mathbf{\frac{381}{625}}$

3. July: $-6.9088$

$-6.9088 = -\frac{69088}{10000}$

Dividing both by 16:

$= \mathbf{-\frac{4318}{625}}$

4. August: $-8.636$

$-8.636 = -\frac{8636}{1000}$

Dividing both by 4:

$= \mathbf{-\frac{2159}{250}}$

Solution:

To express each decimal in the form $\frac{p}{q}$, we write the given decimal as a fraction with a power of 10 in the denominator and then simplify it to its lowest terms.

Arranging states from the least to the greatest male life expectancy (Ascending Order):

By comparing the decimal values (Male %), the order is as follows:

1. Madhya Pradesh ($56.5$)

2. Assam ($57.1$)

3. Orissa ($57.6$)

4. Uttar Pradesh ($58.9$)

5. Rajasthan ($59.8$)

6. Bihar ($60.7$)

7. Andhra Pradesh ($61.6$)

8. Gujarat ($61.9$)

9. Karnataka ($62.4$)

10. West Bengal ($62.8$)

11. Tamil Nadu ($63.7$)

12. Haryana ($64.1$)

13. Maharashtra ($64.5$)

14. Himachal Pradesh ($65.1$)

15. Punjab ($66.9$)

16. Kerala ($70.6$)

Note: India (Average) at $60.8$ lies between Bihar and Andhra Pradesh.

Given:

Current length of skirt $= 35\frac{7}{8}$ cm

Length of hem $= 3\frac{1}{8}$ cm

Solution:

New length $= 35\frac{7}{8} + 3\frac{1}{8}$

$= \frac{287}{8} + \frac{25}{8}$

$= \frac{312}{8} = 39$ cm

The skirt will be 39 cm long.

Given: Allowance of each $= \textsf{₹} 1260$.

Solution for Manavi:

Money saved $= \frac{1}{2} \times 1260 = \textsf{₹} 630$

Money spent $= \frac{1}{4} \times 1260 = \textsf{₹} 315$

Amount left $= 1260 - (630 + 315) = 1260 - 945 = \mathbf{\textsf{₹} 315}$

Solution for Kuber:

Money saved $= \frac{1}{3} \times 1260 = \textsf{₹} 420$

Money spent $= \frac{3}{5} \times 1260 = 3 \times 252 = \textsf{₹} 756$

Amount left $= 1260 - (420 + 756) = 1260 - 1176 = \mathbf{\textsf{₹} 84}$