Chapter 8 Rational Numbers

Solution:

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To reduce a rational number (fraction) to its standard form, we follow these steps:

1. If the denominator is negative, make it positive by multiplying both the numerator and the denominator by -1.

2. Find the Highest Common Factor (HCF) of the absolute values of the numerator and the denominator.

3. Divide both the numerator and the denominator by their HCF.

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The given rational number is $\frac{-45}{30}$.

The numerator is -45 and the denominator is 30.

The denominator 30 is already positive.

Now, let's find the HCF of the absolute values of the numerator and denominator, which are 45 and 30.

We can find the HCF by prime factorization.

Prime factorization of 45:

$\begin{array}{c|cc} 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $45 = 3 \times 3 \times 5 = 3^2 \times 5^1$.

Prime factorization of 30:

$\begin{array}{c|cc} 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $30 = 2 \times 3 \times 5 = 2^1 \times 3^1 \times 5^1$.

To find the HCF, we take the common prime factors with the lowest power.

The common prime factors are 3 and 5. The lowest power of 3 is $3^1$ and the lowest power of 5 is $5^1$.

HCF(45, 30) = $3^1 \times 5^1 = 3 \times 5 = 15$.

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Now, we divide both the numerator (-45) and the denominator (30) by their HCF, which is 15.

$\frac{-45}{30} = \frac{-45 \div 15}{30 \div 15}$

Performing the division:

$-45 \div 15 = -3$

$30 \div 15 = 2$

So, $\frac{-45}{30} = \frac{-3}{2}$.

The resulting fraction is $\frac{-3}{2}$. The denominator is 2, which is positive, and the HCF of the absolute values of the numerator (3) and the denominator (2) is 1. Thus, it is in standard form.

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Conclusion: The standard form of $\frac{-45}{30}$ is $\frac{-3}{2}$.

Solution:

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To reduce a rational number to its standard form, we ensure the denominator is positive, find the HCF of the absolute values of the numerator and denominator, and then divide both by the HCF.

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(i) $\frac{36}{-24}$

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The given rational number is $\frac{36}{-24}$.

The numerator is 36 and the denominator is -24.

The denominator is negative. To make the denominator positive, multiply both the numerator and the denominator by -1:

$\frac{36}{-24} = \frac{36 \times (-1)}{-24 \times (-1)} = \frac{-36}{24}$.

Now, we find the HCF of the absolute values of the numerator (-36) and the denominator (24), which are 36 and 24.

Prime factorization of 36: $2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$.

Prime factorization of 24: $2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$.

The common prime factors are 2 and 3. The lowest power of 2 is $2^2$ and the lowest power of 3 is $3^1$.

HCF(36, 24) = $2^2 \times 3^1 = 4 \times 3 = 12$.

Now, divide both the numerator (-36) and the denominator (24) by the HCF (12):

$\frac{-36}{24} = \frac{-36 \div 12}{24 \div 12} = \frac{-3}{2}$.

The resulting fraction $\frac{-3}{2}$ has a positive denominator and the HCF of the absolute values of the numerator (3) and denominator (2) is 1. Thus, it is in standard form.

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Conclusion for (i): The standard form of $\frac{36}{-24}$ is $\frac{-3}{2}$.

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(ii) $\frac{-3}{-15}$

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The given rational number is $\frac{-3}{-15}$.

The numerator is -3 and the denominator is -15.

The denominator is negative. To make the denominator positive, multiply both the numerator and the denominator by -1:

$\frac{-3}{-15} = \frac{-3 \times (-1)}{-15 \times (-1)} = \frac{3}{15}$.

Now, we find the HCF of the absolute values of the numerator (3) and the denominator (15), which are 3 and 15.

Prime factorization of 3 is $3^1$.

Prime factorization of 15 is $3 \times 5 = 3^1 \times 5^1$.

The common prime factor is 3. The lowest power of 3 is $3^1$.

HCF(3, 15) = 3.

Now, divide both the numerator (3) and the denominator (15) by the HCF (3):

$\frac{3}{15} = \frac{3 \div 3}{15 \div 3} = \frac{1}{5}$.

The resulting fraction $\frac{1}{5}$ has a positive denominator and the HCF of the absolute values of the numerator (1) and denominator (5) is 1. Thus, it is in standard form.

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Conclusion for (ii): The standard form of $\frac{-3}{-15}$ is $\frac{1}{5}$.

Solution:

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To check if two rational numbers represent the same value, we can reduce both fractions to their standard form and compare them. If the standard forms are identical, the rational numbers are the same.

Let's reduce the first rational number, $\frac{4}{-9}$, to its standard form.

The denominator is negative. To make it positive, we multiply both the numerator and the denominator by -1:

$\frac{4}{-9} = \frac{4 \times (-1)}{-9 \times (-1)} = \frac{-4}{9}$.

Now, we find the HCF of the absolute values of the numerator (-4) and the denominator (9), which are 4 and 9.

Prime factors of 4: $2 \times 2 = 2^2$.

Prime factors of 9: $3 \times 3 = 3^2$.

The HCF(4, 9) is 1, as they have no common prime factors.

Since the HCF of the absolute values is 1 and the denominator is positive, $\frac{-4}{9}$ is the standard form of $\frac{4}{-9}$.

Next, let's reduce the second rational number, $\frac{-16}{36}$, to its standard form.

The denominator (36) is already positive.

We find the HCF of the absolute values of the numerator (-16) and the denominator (36), which are 16 and 36.

Prime factorization of 16: $2 \times 2 \times 2 \times 2 = 2^4$.

Prime factorization of 36: $2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$.

The common prime factor is 2. The lowest power of 2 is $2^2$.

HCF(16, 36) = $2^2 = 4$.

Now, we divide both the numerator (-16) and the denominator (36) by their HCF (4):

$\frac{-16}{36} = \frac{-16 \div 4}{36 \div 4} = \frac{-4}{9}$.

The resulting fraction $\frac{-4}{9}$ has a positive denominator and the HCF of the absolute values of the numerator (4) and denominator (9) is 1. Therefore, $\frac{-4}{9}$ is the standard form of $\frac{-16}{36}$.

Comparing the standard forms:

Standard form of $\frac{4}{-9}$ is $\frac{-4}{9}$.

Standard form of $\frac{-16}{36}$ is $\frac{-4}{9}$.

Since the standard forms are the same, they represent the same rational number.

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Conclusion: Yes, $\frac{4}{-9}$ and $\frac{-16}{36}$ represent the same rational number.

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Alternate Solution (Using Cross-Multiplication):

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Two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ are equal if and only if $a \times d = b \times c$.

Here, the two rational numbers are $\frac{4}{-9}$ and $\frac{-16}{36}$.

We compare $\frac{4}{-9}$ and $\frac{-16}{36}$.

Cross-multiplying gives:

$4 \times 36$ and $(-9) \times (-16)$.

Calculate the products:

$4 \times 36 = 144$.

$(-9) \times (-16) = 9 \times 16 = 144$.

Since the cross-products are equal ($144 = 144$), the two rational numbers are equal.

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Conclusion: Yes, $\frac{4}{-9}$ and $\frac{-16}{36}$ represent the same rational number.

Solution:

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To find rational numbers between two given rational numbers, we can express both numbers as equivalent fractions with a common denominator. If there are not enough integers between the numerators, we can increase the common denominator.

The given numbers are -2 and -1.

We can write these as rational numbers with a denominator of 1:

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

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

To find rational numbers between $\frac{-2}{1}$ and $\frac{-1}{1}$, let's express them with a larger common denominator, say 4.

To convert $\frac{-2}{1}$ to an equivalent fraction with denominator 4, multiply the numerator and denominator by 4:

$\frac{-2}{1} = \frac{-2 \times 4}{1 \times 4} = \frac{-8}{4}$.

To convert $\frac{-1}{1}$ to an equivalent fraction with denominator 4, multiply the numerator and denominator by 4:

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

Now we need to find rational numbers between $\frac{-8}{4}$ and $\frac{-4}{4}$. These fractions have the same denominator.

The integers between the numerators -8 and -4 are -7, -6, and -5.

Using these integers as numerators with the denominator 4, we get the following rational numbers:

$\frac{-7}{4}$, $\frac{-6}{4}$, $\frac{-5}{4}$.

These three rational numbers are between $\frac{-8}{4}$ (which is -2) and $\frac{-4}{4}$ (which is -1).

We can simplify $\frac{-6}{4}$:

$\frac{-6}{4} = \frac{\cancel{-6}^{-3}}{\cancel{4}_{2}} = \frac{-3}{2}$.

Thus, three rational numbers between -2 and -1 are $\frac{-7}{4}$, $\frac{-6}{4}$ (or $\frac{-3}{2}$), and $\frac{-5}{4}$.

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Conclusion: Three rational numbers between -2 and -1 are $\frac{-7}{4}$, $\frac{-3}{2}$, and $\frac{-5}{4}$.

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Alternate Method:

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We can also express -2 and -1 with a denominator of 10:

$-2 = \frac{-2 \times 10}{1 \times 10} = \frac{-20}{10}$

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

The integers between the numerators -20 and -10 are -19, -18, -17, -16, -15, -14, -13, -12, -11.

Any three of these integers can be used as numerators with the denominator 10:

For example, $\frac{-19}{10}$, $\frac{-15}{10}$, $\frac{-11}{10}$.

We can simplify $\frac{-15}{10}$:

$\frac{-15}{10} = \frac{\cancel{-15}^{-3}}{\cancel{10}_{2}} = \frac{-3}{2}$.

Thus, another set of three rational numbers between -2 and -1 could be $\frac{-19}{10}$, $\frac{-3}{2}$, and $\frac{-11}{10}$.

There are infinitely many rational numbers between any two distinct rational numbers.

Solution:

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We are given the sequence of rational numbers:

$\frac{-1}{3}$, $\frac{-2}{6}$, $\frac{-3}{9}$, $\frac{-4}{12}$, ...

Let's examine the pattern of the numerators and denominators.

The numerators are -1, -2, -3, -4, ...

This sequence is obtained by multiplying -1 by consecutive positive integers: $-1 \times 1$, $-1 \times 2$, $-1 \times 3$, $-1 \times 4$, ...

The denominators are 3, 6, 9, 12, ...

This sequence is obtained by multiplying 3 by consecutive positive integers: $3 \times 1$, $3 \times 2$, $3 \times 3$, $3 \times 4$, ...

The pattern shows that each term in the sequence is an equivalent fraction of $\frac{-1}{3}$, obtained by multiplying both the numerator and the denominator of $\frac{-1}{3}$ by the same positive integer (which is the position of the term in the sequence).

The $n$-th term in the sequence is $\frac{-1 \times n}{3 \times n}$.

The first term is for $n=1$: $\frac{-1 \times 1}{3 \times 1} = \frac{-1}{3}$.

The second term is for $n=2$: $\frac{-1 \times 2}{3 \times 2} = \frac{-2}{6}$.

The third term is for $n=3$: $\frac{-1 \times 3}{3 \times 3} = \frac{-3}{9}$.

The fourth term is for $n=4$: $\frac{-1 \times 4}{3 \times 4} = \frac{-4}{12}$.

To find the next four numbers, we need to continue this pattern for $n=5, 6, 7, 8$.

The fifth number (for $n=5$):

$\frac{-1 \times 5}{3 \times 5} = \frac{-5}{15}$.

The sixth number (for $n=6$):

$\frac{-1 \times 6}{3 \times 6} = \frac{-6}{18}$.

The seventh number (for $n=7$):

$\frac{-1 \times 7}{3 \times 7} = \frac{-7}{21}$.

The eighth number (for $n=8$):

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

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Conclusion: The next four numbers in the pattern are $\frac{-5}{15}$, $\frac{-6}{18}$, $\frac{-7}{21}$, and $\frac{-8}{24}$.

Solution:

To find rational numbers between two given rational numbers, we express both numbers as fractions with a common denominator. We can increase the common denominator as needed to find the desired number of rational numbers between them.

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(i) Between –1 and 0

We write -1 and 0 as rational numbers:

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

$0 = \frac{0}{1}$

To find five rational numbers, we can express both numbers with a denominator larger than $5+1=6$. Let's use a denominator of 6.

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

$0 = \frac{0 \times 6}{1 \times 6} = \frac{0}{6}$

Now we look for integers between the numerators -6 and 0. These integers are -5, -4, -3, -2, -1.

Using these integers as numerators with the denominator 6, we get the rational numbers:

$\frac{-5}{6}$, $\frac{-4}{6}$, $\frac{-3}{6}$, $\frac{-2}{6}$, $\frac{-1}{6}$.

These are five rational numbers between -1 and 0. We can simplify some of them:

$\frac{-4}{6} = \frac{\cancel{-4}^{-2}}{\cancel{6}_{3}} = \frac{-2}{3}$

$\frac{-3}{6} = \frac{\cancel{-3}^{-1}}{\cancel{6}_{2}} = \frac{-1}{2}$

$\frac{-2}{6} = \frac{\cancel{-2}^{-1}}{\cancel{6}_{3}} = \frac{-1}{3}$

So, five rational numbers between -1 and 0 are $\frac{-5}{6}$, $\frac{-2}{3}$, $\frac{-1}{2}$, $\frac{-1}{3}$, $\frac{-1}{6}$.

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(ii) Between –2 and –1

We write -2 and -1 as rational numbers:

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

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

To find five rational numbers, we can express both numbers with a denominator larger than 6. Let's use a denominator of 6.

$-2 = \frac{-2 \times 6}{1 \times 6} = \frac{-12}{6}$

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

Now we look for integers between the numerators -12 and -6. These integers are -11, -10, -9, -8, -7.

Using these integers as numerators with the denominator 6, we get the rational numbers:

$\frac{-11}{6}$, $\frac{-10}{6}$, $\frac{-9}{6}$, $\frac{-8}{6}$, $\frac{-7}{6}$.

These are five rational numbers between -2 and -1. We can simplify some of them:

$\frac{-10}{6} = \frac{\cancel{-10}^{-5}}{\cancel{6}_{3}} = \frac{-5}{3}$

$\frac{-9}{6} = \frac{\cancel{-9}^{-3}}{\cancel{6}_{2}} = \frac{-3}{2}$

$\frac{-8}{6} = \frac{\cancel{-8}^{-4}}{\cancel{6}_{3}} = \frac{-4}{3}$

So, five rational numbers between -2 and -1 are $\frac{-11}{6}$, $\frac{-5}{3}$, $\frac{-3}{2}$, $\frac{-4}{3}$, $\frac{-7}{6}$.

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(iii) Between $\frac{-4}{5}$ and $\frac{-2}{3}$

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

First, find a common denominator for 5 and 3. The LCM of 5 and 3 is 15.

Convert the fractions to have denominator 15:

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

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

We need to find five rational numbers between $\frac{-12}{15}$ and $\frac{-10}{15}$. The only integer between -12 and -10 is -11.

We need to increase the denominator to find more numbers. Let's multiply the current denominator (15) by a factor that gives us at least $5+1=6$ integers between the numerators. Multiplying by 6 gives a denominator of $15 \times 6 = 90$.

Convert the fractions to have a denominator of 90:

$\frac{-4}{5} = \frac{-4 \times 18}{5 \times 18} = \frac{-72}{90}$

$\frac{-2}{3} = \frac{-2 \times 30}{3 \times 30} = \frac{-60}{90}$

Now we look for five integers between the numerators -72 and -60. We can choose any five from -71, -70, -69, -68, -67, -66, -65, -64, -63, -62, -61.

Let's choose -71, -70, -69, -68, -67.

Using these integers as numerators with the denominator 90, we get the rational numbers:

$\frac{-71}{90}$, $\frac{-70}{90}$, $\frac{-69}{90}$, $\frac{-68}{90}$, $\frac{-67}{90}$.

These are five rational numbers between $\frac{-4}{5}$ and $\frac{-2}{3}$. We can simplify some of them:

$\frac{-70}{90} = \frac{\cancel{-70}^{-7}}{\cancel{90}_{9}} = \frac{-7}{9}$

$\frac{-69}{90} = \frac{\cancel{-69}^{-23}}{\cancel{90}_{30}} = \frac{-23}{30}$

$\frac{-68}{90} = \frac{\cancel{-68}^{-34}}{\cancel{90}_{45}} = \frac{-34}{45}$

So, five rational numbers between $\frac{-4}{5}$ and $\frac{-2}{3}$ are $\frac{-71}{90}$, $\frac{-7}{9}$, $\frac{-23}{30}$, $\frac{-34}{45}$, $\frac{-67}{90}$.

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(iv) Between $-\frac{1}{2}$ and $\frac{2}{3}$

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

First, find a common denominator for 2 and 3. The LCM of 2 and 3 is 6.

Convert the fractions to have denominator 6:

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

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

Now we need to find five rational numbers between $\frac{-3}{6}$ and $\frac{4}{6}$. We look for integers between the numerators -3 and 4. These integers are -2, -1, 0, 1, 2, 3.

There are more than five integers available. We can choose any five.

Let's choose -2, -1, 0, 1, 2.

Using these integers as numerators with the denominator 6, we get the rational numbers:

$\frac{-2}{6}$, $\frac{-1}{6}$, $\frac{0}{6}$, $\frac{1}{6}$, $\frac{2}{6}$.

These are five rational numbers between $-\frac{1}{2}$ and $\frac{2}{3}$. We can simplify some of them:

$\frac{-2}{6} = \frac{\cancel{-2}^{-1}}{\cancel{6}_{3}} = \frac{-1}{3}$

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

$\frac{2}{6} = \frac{\cancel{2}^{1}}{\cancel{6}_{3}} = \frac{1}{3}$

So, five rational numbers between $-\frac{1}{2}$ and $\frac{2}{3}$ are $\frac{-1}{3}$, $\frac{-1}{6}$, $0$, $\frac{1}{6}$, $\frac{1}{3}$.

Solution:

We will analyse the pattern in each sequence and determine the rule to generate the next four terms.

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(i) Pattern: $\frac{-3}{5}$ , $\frac{-6}{10}$ , $\frac{-9}{15}$ , $\frac{-12}{20}$ ,......

Let's look at the terms:

First term: $\frac{-3}{5} = \frac{-3 \times 1}{5 \times 1}$

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

Third term: $\frac{-9}{15} = \frac{-3 \times 3}{5 \times 3}$

Fourth term: $\frac{-12}{20} = \frac{-3 \times 4}{5 \times 4}$

The pattern shows that each term is obtained by multiplying the numerator and the denominator of the base fraction $\frac{-3}{5}$ by consecutive positive integers $n = 1, 2, 3, 4, \dots$. The $n$-th term is given by $\frac{-3 \times n}{5 \times n}$.

To find the next four numbers, we use $n=5, 6, 7, 8$.

For $n=5$: $\frac{-3 \times 5}{5 \times 5} = \frac{-15}{25}$

For $n=6$: $\frac{-3 \times 6}{5 \times 6} = \frac{-18}{30}$

For $n=7$: $\frac{-3 \times 7}{5 \times 7} = \frac{-21}{35}$

For $n=8$: $\frac{-3 \times 8}{5 \times 8} = \frac{-24}{40}$

The next four rational numbers are $\frac{-15}{25}$, $\frac{-18}{30}$, $\frac{-21}{35}$, $\frac{-24}{40}$.

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(ii) Pattern: $\frac{-1}{4}$ , $\frac{-2}{8}$ , $\frac{-3}{12}$ , ....…

Let's look at the terms:

First term: $\frac{-1}{4} = \frac{-1 \times 1}{4 \times 1}$

Second term: $\frac{-2}{8} = \frac{-1 \times 2}{4 \times 2}$

Third term: $\frac{-3}{12} = \frac{-1 \times 3}{4 \times 3}$

The pattern shows that each term is obtained by multiplying the numerator and the denominator of the base fraction $\frac{-1}{4}$ by consecutive positive integers $n = 1, 2, 3, \dots$. The $n$-th term is given by $\frac{-1 \times n}{4 \times n}$.

The given terms correspond to $n=1, 2, 3$. To find the next four numbers, we use $n=4, 5, 6, 7$.

For $n=4$: $\frac{-1 \times 4}{4 \times 4} = \frac{-4}{16}$

For $n=5$: $\frac{-1 \times 5}{4 \times 5} = \frac{-5}{20}$

For $n=6$: $\frac{-1 \times 6}{4 \times 6} = \frac{-6}{24}$

For $n=7$: $\frac{-1 \times 7}{4 \times 7} = \frac{-7}{28}$

The next four rational numbers are $\frac{-4}{16}$, $\frac{-5}{20}$, $\frac{-6}{24}$, $\frac{-7}{28}$.

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(iii) Pattern: $\frac{-1}{6}$ , $\frac{2}{-12}$ , $\frac{3}{-18}$ , $\frac{4}{-24}$ ,......

Let's look at the terms:

First term: $\frac{-1}{6}$

Second term: $\frac{2}{-12}$. The numerator is 2. The denominator is $-12 = -6 \times 2$.

Third term: $\frac{3}{-18}$. The numerator is 3. The denominator is $-18 = -6 \times 3$.

Fourth term: $\frac{4}{-24}$. The numerator is 4. The denominator is $-24 = -6 \times 4$.

The pattern appears to be:

The first term is $\frac{-1}{6}$.

For $n \ge 2$, the $n$-th term is $\frac{n}{-6 \times n}$.

The given terms correspond to $n=1$ (first term) and $n=2, 3, 4$ (subsequent terms). To find the next four numbers, we use $n=5, 6, 7, 8$ following the rule for $n \ge 2$.

For $n=5$: $\frac{5}{-6 \times 5} = \frac{5}{-30}$

For $n=6$: $\frac{6}{-6 \times 6} = \frac{6}{-36}$

For $n=7$: $\frac{7}{-6 \times 7} = \frac{7}{-42}$

For $n=8$: $\frac{8}{-6 \times 8} = \frac{8}{-48}$

The next four rational numbers are $\frac{5}{-30}$, $\frac{6}{-36}$, $\frac{7}{-42}$, $\frac{8}{-48}$.

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(iv) Pattern: $\frac{-2}{3}$ , $\frac{2}{-3}$ , $\frac{4}{-6}$ , $\frac{6}{-9}$ ,..…

Let's look at the terms:

First term: $\frac{-2}{3}$

Second term: $\frac{2}{-3}$. This is an equivalent fraction to $\frac{-2}{3}$.

Third term: $\frac{4}{-6}$. This is $\frac{2 \times 2}{-3 \times 2}$. It's obtained by multiplying the numerator and denominator of the second term $\frac{2}{-3}$ by 2.

Fourth term: $\frac{6}{-9}$. This is $\frac{2 \times 3}{-3 \times 3}$. It's obtained by multiplying the numerator and denominator of the second term $\frac{2}{-3}$ by 3.

The pattern appears to be:

The first term is $\frac{-2}{3}$.

For $n \ge 2$, the $n$-th term is obtained by multiplying the numerator and denominator of the second term $\frac{2}{-3}$ by $k = n-1$. So the $n$-th term ($n \ge 2$) is $\frac{2 \times (n-1)}{-3 \times (n-1)}$.

The given terms correspond to $n=1$ (first term) and $n=2, 3, 4$ (subsequent terms where $k=1, 2, 3$). To find the next four numbers, we use $n=5, 6, 7, 8$, applying the rule for $n \ge 2$ with $k = n-1 = 4, 5, 6, 7$.

For $n=5$ ($k=4$): $\frac{2 \times 4}{-3 \times 4} = \frac{8}{-12}$

For $n=6$ ($k=5$): $\frac{2 \times 5}{-3 \times 5} = \frac{10}{-15}$

For $n=7$ ($k=6$): $\frac{2 \times 6}{-3 \times 6} = \frac{12}{-18}$

For $n=8$ ($k=7$): $\frac{2 \times 7}{-3 \times 7} = \frac{14}{-21}$

The next four rational numbers are $\frac{8}{-12}$, $\frac{10}{-15}$, $\frac{12}{-18}$, $\frac{14}{-21}$.

Solution:

To find rational numbers equivalent to a given fraction, we multiply both the numerator and the denominator by the same non-zero integer.

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(i) Equivalent rational numbers to $\frac{-2}{7}$:

Multiply the numerator and denominator by consecutive integers, for example, 2, 3, 4, and 5.

Multiplying by 2: $\frac{-2 \times 2}{7 \times 2} = \frac{-4}{14}$

Multiplying by 3: $\frac{-2 \times 3}{7 \times 3} = \frac{-6}{21}$

Multiplying by 4: $\frac{-2 \times 4}{7 \times 4} = \frac{-8}{28}$

Multiplying by 5: $\frac{-2 \times 5}{7 \times 5} = \frac{-10}{35}$

Four rational numbers equivalent to $\frac{-2}{7}$ are $\frac{-4}{14}$, $\frac{-6}{21}$, $\frac{-8}{28}$, and $\frac{-10}{35}$.

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(ii) Equivalent rational numbers to $\frac{5}{-3}$:

Multiply the numerator and denominator by consecutive integers, for example, 2, 3, 4, and 5.

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

Multiplying by 3: $\frac{5 \times 3}{-3 \times 3} = \frac{15}{-9}$

Multiplying by 4: $\frac{5 \times 4}{-3 \times 4} = \frac{20}{-12}$

Multiplying by 5: $\frac{5 \times 5}{-3 \times 5} = \frac{25}{-15}$

Four rational numbers equivalent to $\frac{5}{-3}$ are $\frac{10}{-6}$, $\frac{15}{-9}$, $\frac{20}{-12}$, and $\frac{25}{-15}$.

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(iii) Equivalent rational numbers to $\frac{4}{9}$:

Multiply the numerator and denominator by consecutive integers, for example, 2, 3, 4, and 5.

Multiplying by 2: $\frac{4 \times 2}{9 \times 2} = \frac{8}{18}$

Multiplying by 3: $\frac{4 \times 3}{9 \times 3} = \frac{12}{27}$

Multiplying by 4: $\frac{4 \times 4}{9 \times 4} = \frac{16}{36}$

Multiplying by 5: $\frac{4 \times 5}{9 \times 5} = \frac{20}{45}$

Four rational numbers equivalent to $\frac{4}{9}$ are $\frac{8}{18}$, $\frac{12}{27}$, $\frac{16}{36}$, and $\frac{20}{45}$.

Solution:

To represent rational numbers on a number line, we first draw a straight line and mark integers on it. Then, we divide the segment between the relevant integers into parts equal to the denominator of the fraction. The numerator tells us how many parts to count from the starting integer (usually 0).

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(i) Represent $\frac{3}{4}$ on the number line.

The rational number $\frac{3}{4}$ is positive and less than 1. It lies between 0 and 1.

The denominator is 4, so we divide the segment between 0 and 1 into 4 equal parts.

The numerator is 3, so we start from 0 and count 3 parts towards the positive side (towards 1).

The third mark from 0 represents $\frac{3}{4}$.

To draw this, mark points 0 and 1 on your number line. Divide the distance between 0 and 1 into 4 equal segments. Label the division points from left to right as $\frac{1}{4}$, $\frac{2}{4}$ (or $\frac{1}{2}$), $\frac{3}{4}$. Mark the point at $\frac{3}{4}$.

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(ii) Represent $\frac{-5}{8}$ on the number line.

The rational number $\frac{-5}{8}$ is negative and lies between -1 and 0.

The denominator is 8, so we divide the segment between 0 and -1 into 8 equal parts.

The numerator is -5 (considering the sign with the number), so we start from 0 and count 5 parts towards the negative side (towards -1).

The fifth mark from 0 towards -1 represents $\frac{-5}{8}$.

To draw this, mark points -1 and 0 on your number line. Divide the distance between 0 and -1 into 8 equal segments. Label the division points from right to left (starting from 0) as $\frac{-1}{8}$, $\frac{-2}{8}$, $\frac{-3}{8}$, $\frac{-4}{8}$ (or $\frac{-1}{2}$), $\frac{-5}{8}$, $\frac{-6}{8}$, $\frac{-7}{8}$. Mark the point at $\frac{-5}{8}$.

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(iii) Represent $\frac{-7}{4}$ on the number line.

The rational number $\frac{-7}{4}$ is an improper fraction. Convert it to a mixed number: $\frac{-7}{4} = -1 \frac{3}{4}$.

This means the number is 1 whole unit less than 0 and then another $\frac{3}{4}$ less than that. It lies between -2 and -1.

The denominator is 4, so we divide the segment between -1 and -2 into 4 equal parts.

The fraction part is $\frac{3}{4}$. Starting from -1, we count 3 parts towards the negative side (towards -2).

The third mark from -1 towards -2 represents $-1 \frac{3}{4}$ or $\frac{-7}{4}$.

To draw this, mark points -2, -1, and 0 on your number line. Locate -1. Divide the distance between -1 and -2 into 4 equal segments. Label the division points from right to left (starting from -1) as $-1\frac{1}{4}$ (or $\frac{-5}{4}$), $-1\frac{2}{4}$ (or $\frac{-6}{4}$ / $\frac{-3}{2}$), $-1\frac{3}{4}$ (or $\frac{-7}{4}$). Mark the point at $\frac{-7}{4}$.

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(iv) Represent $\frac{7}{8}$ on the number line.

The rational number $\frac{7}{8}$ is positive and less than 1. It lies between 0 and 1.

The denominator is 8, so we divide the segment between 0 and 1 into 8 equal parts.

The numerator is 7, so we start from 0 and count 7 parts towards the positive side (towards 1).

The seventh mark from 0 represents $\frac{7}{8}$.

To draw this, mark points 0 and 1 on your number line. Divide the distance between 0 and 1 into 8 equal segments. Label the division points from left to right as $\frac{1}{8}$, $\frac{2}{8}$, $\frac{3}{8}$, $\frac{4}{8}$ (or $\frac{1}{2}$), $\frac{5}{8}$, $\frac{6}{8}$, $\frac{7}{8}$. Mark the point at $\frac{7}{8}$.

Solution:

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Given:

Points P, Q, R, S, T, U, A and B are on the number line.

From the number line shown:

Point A is at 2.

Point B is at 3.

Point U is at -2.

Point T is at -1.

The segments are divided such that TR = RS = SU and AP = PQ = QB.

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To Find:

The rational numbers represented by points P, Q, R, and S.

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Solution:

First, consider the segment TU on the number line. Point T is at -1 and Point U is at -2.

The length of the segment TU is $|-2 - (-1)| = |-2+1| = |-1| = 1$ unit.

The condition TR = RS = SU means that the segment from T (-1) to U (-2) is divided into 3 equal parts by the points R and S.

The length of each equal part is $\frac{1}{3}$ of the unit distance.

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Point R is the first division point from T (-1) towards U (-2).

The position of R is $T + (\text{length of TR}) = -1 + (-\frac{1}{3})$ (moving in the negative direction).

R $= -1 - \frac{1}{3} = \frac{-3}{3} - \frac{1}{3} = \frac{-3-1}{3} = \frac{-4}{3}$.

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Point S is the second division point from T (-1) towards U (-2).

The position of S is $T + (\text{length of TR}) + (\text{length of RS}) = -1 + (-\frac{1}{3}) + (-\frac{1}{3})$.

S $= -1 - \frac{1}{3} - \frac{1}{3} = -1 - \frac{2}{3} = \frac{-3}{3} - \frac{2}{3} = \frac{-3-2}{3} = \frac{-5}{3}$.

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Next, consider the segment AB on the number line. Point A is at 2 and Point B is at 3.

The length of the segment AB is $|3 - 2| = 1$ unit.

The condition AP = PQ = QB means that the segment from A (2) to B (3) is divided into 3 equal parts by the points P and Q.

The length of each equal part is $\frac{1}{3}$ of the unit distance.

---

Point P is the first division point from A (2) towards B (3).

The position of P is $A + (\text{length of AP}) = 2 + \frac{1}{3}$.

P $= 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{6+1}{3} = \frac{7}{3}$.

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Point Q is the second division point from A (2) towards B (3).

The position of Q is $A + (\text{length of AP}) + (\text{length of PQ}) = 2 + \frac{1}{3} + \frac{1}{3}$.

Q $= 2 + \frac{2}{3} = \frac{6}{3} + \frac{2}{3} = \frac{6+2}{3} = \frac{8}{3}$.

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The rational numbers represented by the points are:

P represents $\frac{7}{3}$.

Q represents $\frac{8}{3}$.

R represents $\frac{-4}{3}$.

S represents $\frac{-5}{3}$.

Solution:

To check if two rational numbers represent the same value, we can reduce both fractions to their standard form. If the standard forms are identical, the rational numbers are the same.

(i) $\frac{-7}{21}$ and $\frac{3}{9}$

Standard form of $\frac{-7}{21}$:

Divide numerator and denominator by their HCF, which is 7.

$\frac{-7 \div 7}{21 \div 7} = \frac{-1}{3}$.

Standard form of $\frac{3}{9}$:

Divide numerator and denominator by their HCF, which is 3.

$\frac{3 \div 3}{9 \div 3} = \frac{1}{3}$.

Comparing the standard forms: $\frac{-1}{3}$ and $\frac{1}{3}$. They are not the same.

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(ii) $\frac{-16}{20}$ and $\frac{20}{-25}$

Standard form of $\frac{-16}{20}$:

The HCF of 16 and 20 is 4. Divide numerator and denominator by 4.

$\frac{-16 \div 4}{20 \div 4} = \frac{-4}{5}$.

Standard form of $\frac{20}{-25}$:

The denominator is negative. Multiply numerator and denominator by -1: $\frac{20 \times -1}{-25 \times -1} = \frac{-20}{25}$.

The HCF of 20 and 25 is 5. Divide numerator and denominator by 5.

$\frac{-20 \div 5}{25 \div 5} = \frac{-4}{5}$.

Comparing the standard forms: $\frac{-4}{5}$ and $\frac{-4}{5}$. They are the same.

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(iii) $\frac{-2}{-3}$ and $\frac{2}{3}$

Standard form of $\frac{-2}{-3}$:

The denominator is negative. Multiply numerator and denominator by -1.

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

Standard form of $\frac{2}{3}$:

The denominator is positive and the HCF of 2 and 3 is 1. This is already in standard form.

$\frac{2}{3}$.

Comparing the standard forms: $\frac{2}{3}$ and $\frac{2}{3}$. They are the same.

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(iv) $\frac{-3}{5}$ and $\frac{-12}{20}$

Standard form of $\frac{-3}{5}$:

The denominator is positive and the HCF of 3 and 5 is 1. This is already in standard form.

$\frac{-3}{5}$.

Standard form of $\frac{-12}{20}$:

The HCF of 12 and 20 is 4. Divide numerator and denominator by 4.

$\frac{-12 \div 4}{20 \div 4} = \frac{-3}{5}$.

Comparing the standard forms: $\frac{-3}{5}$ and $\frac{-3}{5}$. They are the same.

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(v) $\frac{8}{-5}$ and $\frac{-24}{15}$

Standard form of $\frac{8}{-5}$:

The denominator is negative. Multiply numerator and denominator by -1.

$\frac{8 \times -1}{-5 \times -1} = \frac{-8}{5}$.

Standard form of $\frac{-24}{15}$:

The HCF of 24 and 15 is 3. Divide numerator and denominator by 3.

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

Comparing the standard forms: $\frac{-8}{5}$ and $\frac{-8}{5}$. They are the same.

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(vi) $\frac{1}{3}$ and $\frac{-1}{9}$

Standard form of $\frac{1}{3}$:

The denominator is positive and the HCF of 1 and 3 is 1. This is already in standard form.

$\frac{1}{3}$.

Standard form of $\frac{-1}{9}$:

The denominator is positive and the HCF of 1 and 9 is 1. This is already in standard form.

$\frac{-1}{9}$.

Comparing the standard forms: $\frac{1}{3}$ and $\frac{-1}{9}$. They are not the same.

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(vii) $\frac{-5}{-9}$ and $\frac{5}{-9}$

Standard form of $\frac{-5}{-9}$:

The denominator is negative. Multiply numerator and denominator by -1.

$\frac{-5 \times -1}{-9 \times -1} = \frac{5}{9}$.

Standard form of $\frac{5}{-9}$:

The denominator is negative. Multiply numerator and denominator by -1.

$\frac{5 \times -1}{-9 \times -1} = \frac{-5}{9}$.

Comparing the standard forms: $\frac{5}{9}$ and $\frac{-5}{9}$. They are not the same.

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The pairs that represent the same rational number are (ii), (iii), (iv), and (v).

Solution:

To rewrite a rational number in its simplest form (standard form), we divide both the numerator and the denominator by their HCF and ensure the denominator is positive.

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(i) $\frac{-8}{6}$

The given fraction is $\frac{-8}{6}$. The numerator is -8 and the denominator is 6.

The HCF of the absolute values of the numerator (8) and denominator (6) is HCF(8, 6) = 2.

Divide both the numerator and the denominator by their HCF, 2:

$\frac{-8 \div 2}{6 \div 2} = \frac{-4}{3}$.

The denominator (3) is positive. The HCF of 4 and 3 is 1. So, $\frac{-4}{3}$ is the simplest form.

The simplest form of $\frac{-8}{6}$ is $\frac{-4}{3}$.

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(ii) $\frac{25}{45}$

The given fraction is $\frac{25}{45}$. The numerator is 25 and the denominator is 45.

The HCF of 25 and 45 is HCF(25, 45) = 5.

Divide both the numerator and the denominator by their HCF, 5:

$\frac{25 \div 5}{45 \div 5} = \frac{5}{9}$.

The denominator (9) is positive. The HCF of 5 and 9 is 1. So, $\frac{5}{9}$ is the simplest form.

The simplest form of $\frac{25}{45}$ is $\frac{5}{9}$.

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(iii) $\frac{-44}{72}$

The given fraction is $\frac{-44}{72}$. The numerator is -44 and the denominator is 72.

The HCF of the absolute values of the numerator (44) and denominator (72) is HCF(44, 72).

We find the prime factorization of 44 and 72:

Prime factorization of 44:

$\begin{array}{c|cc} 2 & 44 \\ \hline 2 & 22 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

So, $44 = 2 \times 2 \times 11 = 2^2 \times 11^1$.

Prime factorization of 72:

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.

The common prime factor is 2. The lowest power is $2^2$.

HCF(44, 72) = $2^2 = 4$.

Divide both the numerator and the denominator by their HCF, 4:

$\frac{-44 \div 4}{72 \div 4} = \frac{-11}{18}$.

The denominator (18) is positive. The HCF of 11 and 18 is 1. So, $\frac{-11}{18}$ is the simplest form.

The simplest form of $\frac{-44}{72}$ is $\frac{-11}{18}$.

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(iv) $\frac{-8}{10}$

The given fraction is $\frac{-8}{10}$. The numerator is -8 and the denominator is 10.

The HCF of the absolute values of the numerator (8) and denominator (10) is HCF(8, 10) = 2.

Divide both the numerator and the denominator by their HCF, 2:

$\frac{-8 \div 2}{10 \div 2} = \frac{-4}{5}$.

The denominator (5) is positive. The HCF of 4 and 5 is 1. So, $\frac{-4}{5}$ is the simplest form.

The simplest form of $\frac{-8}{10}$ is $\frac{-4}{5}$.

Solution:

To compare rational numbers, we can express them with a common denominator and then compare their numerators. Alternatively, we can use cross-multiplication ($\frac{a}{b}$ vs $\frac{c}{d}$ is equivalent to comparing $ad$ vs $bc$). We will use the common denominator method.

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(i) $\frac{-5}{7}$ ⬜ $\frac{2}{3}$

Using common denominator (LCM of 7 and 3 is 21):

$\frac{-5}{7} = \frac{-5 \times 3}{7 \times 3} = \frac{-15}{21}$

$\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}$

Comparing numerators, $-15 < 14$.

So, $\frac{-15}{21} < \frac{14}{21}$, which means $\frac{-5}{7} < \frac{2}{3}$.

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(ii) $\frac{-4}{5}$ ⬜ $\frac{-5}{7}$

Using common denominator (LCM of 5 and 7 is 35):

$\frac{-4}{5} = \frac{-4 \times 7}{5 \times 7} = \frac{-28}{35}$

$\frac{-5}{7} = \frac{-5 \times 5}{7 \times 5} = \frac{-25}{35}$

Comparing numerators, $-28 < -25$.

So, $\frac{-28}{35} < \frac{-25}{35}$, which means $\frac{-4}{5} < \frac{-5}{7}$.

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(iii) $\frac{-7}{8}$ ⬜ $\frac{14}{-16}$

First, write $\frac{14}{-16}$ with a positive denominator: $\frac{14 \times -1}{-16 \times -1} = \frac{-14}{16}$.

Now compare $\frac{-7}{8}$ and $\frac{-14}{16}$.

Using common denominator (LCM of 8 and 16 is 16):

$\frac{-7}{8} = \frac{-7 \times 2}{8 \times 2} = \frac{-14}{16}$

$\frac{-14}{16}$ is already in this form.

Comparing numerators, $-14 = -14$.

So, $\frac{-14}{16} = \frac{-14}{16}$, which means $\frac{-7}{8} = \frac{14}{-16}$.

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(iv) $\frac{-8}{5}$ ⬜ $\frac{-7}{4}$

Using common denominator (LCM of 5 and 4 is 20):

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

$\frac{-7}{4} = \frac{-7 \times 5}{4 \times 5} = \frac{-35}{20}$

Comparing numerators, $-32 > -35$.

So, $\frac{-32}{20} > \frac{-35}{20}$, which means $\frac{-8}{5} > \frac{-7}{4}$.

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(v) $\frac{1}{-3}$ ⬜ $\frac{-1}{4}$

First, write $\frac{1}{-3}$ with a positive denominator: $\frac{1 \times -1}{-3 \times -1} = \frac{-1}{3}$.

Now compare $\frac{-1}{3}$ and $\frac{-1}{4}$.

Using common denominator (LCM of 3 and 4 is 12):

$\frac{-1}{3} = \frac{-1 \times 4}{3 \times 4} = \frac{-4}{12}$

$\frac{-1}{4} = \frac{-1 \times 3}{4 \times 3} = \frac{-3}{12}$

Comparing numerators, $-4 < -3$.

So, $\frac{-4}{12} < \frac{-3}{12}$, which means $\frac{1}{-3} < \frac{-1}{4}$.

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(vi) $\frac{5}{-11}$ ⬜ $\frac{-5}{11}$

First, write $\frac{5}{-11}$ with a positive denominator: $\frac{5 \times -1}{-11 \times -1} = \frac{-5}{11}$.

Now compare $\frac{-5}{11}$ and $\frac{-5}{11}$.

They are identical.

So, $\frac{5}{-11} = \frac{-5}{11}$.

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(vii) 0 ⬜ $\frac{-7}{6}$

A negative rational number is always less than 0.

Since $\frac{-7}{6}$ is a negative rational number, $0$ is greater than $\frac{-7}{6}$.

So, $0 > \frac{-7}{6}$.

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Filling in the boxes with the correct symbols:

(i) $\frac{-5}{7} < \frac{2}{3}$

(ii) $\frac{-4}{5} < \frac{-5}{7}$

(iii) $\frac{-7}{8} = \frac{14}{-16}$

(iv) $\frac{-8}{5} > \frac{-7}{4}$

(v) $\frac{1}{-3} < \frac{-1}{4}$

(vi) $\frac{5}{-11} = \frac{-5}{11}$

(vii) $0 > \frac{-7}{6}$

Solution:

To compare two rational numbers, we can write them with a common denominator and then compare their numerators. For negative numbers, the one with the larger numerator (when the denominator is positive) is greater.

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(i) Compare $\frac{2}{3}$ and $\frac{5}{2}$

The denominators are 3 and 2. The LCM of 3 and 2 is 6.

Convert the fractions to have denominator 6:

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

$\frac{5}{2} = \frac{5 \times 3}{2 \times 3} = \frac{15}{6}$

Now compare the numerators 4 and 15. Since $4 < 15$, we have $\frac{4}{6} < \frac{15}{6}$.

Thus, $\frac{2}{3} < \frac{5}{2}$.

Therefore, $\frac{5}{2}$ is greater.

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(ii) Compare $\frac{-5}{6}$ and $\frac{-4}{3}$

The denominators are 6 and 3. The LCM of 6 and 3 is 6.

Convert the fractions to have denominator 6:

$\frac{-5}{6}$ is already in this form.

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

Now compare the numerators -5 and -8. Since $-5 > -8$, we have $\frac{-5}{6} > \frac{-8}{6}$.

Thus, $\frac{-5}{6} > \frac{-4}{3}$.

Therefore, $\frac{-5}{6}$ is greater.

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(iii) Compare $\frac{-3}{4}$ and $\frac{2}{-3}$

First, write $\frac{2}{-3}$ with a positive denominator: $\frac{2}{-3} = \frac{2 \times (-1)}{-3 \times (-1)} = \frac{-2}{3}$.

Now compare $\frac{-3}{4}$ and $\frac{-2}{3}$.

The denominators are 4 and 3. The LCM of 4 and 3 is 12.

Convert the fractions to have denominator 12:

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

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

Now compare the numerators -9 and -8. Since $-9 < -8$, we have $\frac{-9}{12} < \frac{-8}{12}$.

Thus, $\frac{-3}{4} < \frac{-2}{3}$.

Therefore, $\frac{-2}{3}$ (or $\frac{2}{-3}$) is greater.

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(iv) Compare $\frac{-1}{4}$ and $\frac{1}{4}$

One number is negative ($\frac{-1}{4}$) and the other is positive ($\frac{1}{4}$).

A positive number is always greater than a negative number.

So, $\frac{1}{4} > \frac{-1}{4}$.

Therefore, $\frac{1}{4}$ is greater.

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(v) Compare $-3\frac{2}{7}$ and $-3\frac{4}{4}$

First, simplify the second number:

$-3\frac{4}{4} = -3 + (-\frac{4}{4}) = -3 + (-1) = -4$.

Now compare $-3\frac{2}{7}$ and $-4$.

Convert the mixed number to an improper fraction:

$-3\frac{2}{7} = -\frac{(3 \times 7) + 2}{7} = -\frac{21 + 2}{7} = \frac{-23}{7}$.

Write -4 as a fraction with denominator 7: $-4 = \frac{-4 \times 7}{7} = \frac{-28}{7}$.

Now compare $\frac{-23}{7}$ and $\frac{-28}{7}$. Both are negative.

Compare the numerators -23 and -28. Since $-23 > -28$, we have $\frac{-23}{7} > \frac{-28}{7}$.

Thus, $-3\frac{2}{7} > -4$.

Therefore, $-3\frac{2}{7}$ is greater.

Solution:

To arrange rational numbers in ascending order, we first express them with a common denominator. Then, we compare the numerators. For negative numbers, the smaller the numerator, the larger the number.

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(i) Arrange $\frac{-3}{5}$ , $\frac{-2}{5}$ , $\frac{-1}{5}$ in ascending order.

The denominators are already the same (5).

We compare the numerators: -3, -2, -1.

In ascending order, these are -3, -2, -1.

So, the rational numbers in ascending order are $\frac{-3}{5}$, $\frac{-2}{5}$, $\frac{-1}{5}$.

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(ii) Arrange $\frac{-1}{3}$ , $\frac{-2}{9}$ , $\frac{-4}{3}$ in ascending order.

The denominators are 3, 9, and 3. The LCM of 3 and 9 is 9.

Convert the fractions to have denominator 9:

$\frac{-1}{3} = \frac{-1 \times 3}{3 \times 3} = \frac{-3}{9}$

$\frac{-2}{9}$ is already in this form.

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

Now we compare the numerators: -3, -2, -12.

In ascending order, these are -12, -3, -2.

So, the rational numbers in ascending order are $\frac{-12}{9}$, $\frac{-3}{9}$, $\frac{-2}{9}$.

Replacing with the original fractions:

$\frac{-4}{3}$, $\frac{-1}{3}$, $\frac{-2}{9}$.

The ascending order is $\frac{-4}{3} < \frac{-1}{3} < \frac{-2}{9}$.

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(iii) Arrange $\frac{-3}{7}$ , $\frac{-3}{2}$ , $\frac{-3}{4}$ in ascending order.

The numerators are the same (-3). The denominators are 7, 2, and 4.

For negative rational numbers with the same numerator, the one with the smaller denominator (in terms of absolute value) is smaller (more negative).

Let's compare the absolute values of the denominators: 7, 2, 4. In ascending order, these are 2, 4, 7.

So, for negative numerators, the fractions in ascending order will correspond to denominators in descending order of their absolute values. This means the order of fractions will be with denominators 2, then 4, then 7.

Alternatively, using a common denominator (LCM of 7, 2, 4 is 28):

$\frac{-3}{7} = \frac{-3 \times 4}{7 \times 4} = \frac{-12}{28}$

$\frac{-3}{2} = \frac{-3 \times 14}{2 \times 14} = \frac{-42}{28}$

$\frac{-3}{4} = \frac{-3 \times 7}{4 \times 7} = \frac{-21}{28}$

Now we compare the numerators: -12, -42, -21.

In ascending order, these are -42, -21, -12.

So, the rational numbers in ascending order are $\frac{-42}{28}$, $\frac{-21}{28}$, $\frac{-12}{28}$.

Replacing with the original fractions:

$\frac{-3}{2}$, $\frac{-3}{4}$, $\frac{-3}{7}$.

The ascending order is $\frac{-3}{2} < \frac{-3}{4} < \frac{-3}{7}$.

Solution:

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Given:

Starting point: P

First movement: $\frac{2}{3}$ km towards east from P.

Second movement: $1\frac{5}{7}$ km towards west from the position reached after the first movement.

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To Find:

Satpal's final position from P.

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Solution:

Let us represent the place P by the origin, 0, on a number line.

Let us consider the distance covered towards the east as positive and the distance covered towards the west as negative.

Distance covered towards east from P = $+ \frac{2}{3}$ km.

Distance covered towards west from the position reached = $-1\frac{5}{7}$ km.

We first convert the mixed number to an improper fraction:

$1\frac{5}{7} = \frac{(1 \times 7) + 5}{7} = \frac{7 + 5}{7} = \frac{12}{7}$.

So, the movement towards west is $-\frac{12}{7}$ km.

To find the final position of Satpal from P, we add the two displacements:

Final position from P = $\frac{2}{3} + (-\frac{12}{7}) = \frac{2}{3} - \frac{12}{7}$.

To perform the subtraction, we find the common denominator for 3 and 7. The LCM of 3 and 7 is 21.

We convert both fractions to equivalent fractions with denominator 21:

$\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}$

$\frac{12}{7} = \frac{12 \times 3}{7 \times 3} = \frac{36}{21}$

Now, subtract the fractions:

Final position = $\frac{14}{21} - \frac{36}{21} = \frac{14 - 36}{21}$.

Subtracting the numerators: $14 - 36 = -22$.

So, the final position is $\frac{-22}{21}$ km from P.

The negative sign indicates that the final position is towards the west of P.

We can express the answer as a mixed number:

$\frac{22}{21}$ as a mixed number is $1 \frac{1}{21}$.

So, $\frac{-22}{21} = -1 \frac{1}{21}$.

This means Satpal is $1 \frac{1}{21}$ km away from P in the west direction.

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Conclusion: Satpal will be $1 \frac{1}{21}$ km to the west of P.

Solution:

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(i) $\frac{5}{4} + \left( \frac{-11}{4} \right)$

The denominators are the same (4).

Add the numerators and keep the common denominator:

$\frac{5 + (-11)}{4} = \frac{5 - 11}{4} = \frac{-6}{4}$.

Reduce the resulting fraction to its simplest form by dividing the numerator and the denominator by their HCF, which is 2.

$\frac{-6 \div 2}{4 \div 2} = \frac{-3}{2}$.

So, the sum is $\frac{-3}{2}$.

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(ii) $\frac{5}{3} + \frac{3}{5}$

The denominators are 3 and 5. The Least Common Multiple (LCM) of 3 and 5 is 15.

Convert each fraction to an equivalent fraction with the denominator 15:

$\frac{5}{3} = \frac{5 \times 5}{3 \times 5} = \frac{25}{15}$

$\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}$

Add the converted fractions:

$\frac{25}{15} + \frac{9}{15} = \frac{25+9}{15} = \frac{34}{15}$.

The fraction $\frac{34}{15}$ is in its simplest form as the HCF of 34 and 15 is 1.

So, the sum is $\frac{34}{15}$.

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(iii) $\frac{-9}{10} + \frac{22}{15}$

The denominators are 10 and 15. The LCM of 10 and 15 is 30.

Convert each fraction to an equivalent fraction with the denominator 30:

$\frac{-9}{10} = \frac{-9 \times 3}{10 \times 3} = \frac{-27}{30}$

$\frac{22}{15} = \frac{22 \times 2}{15 \times 2} = \frac{44}{30}$

Add the converted fractions:

$\frac{-27}{30} + \frac{44}{30} = \frac{-27+44}{30} = \frac{17}{30}$.

The fraction $\frac{17}{30}$ is in its simplest form as the HCF of 17 and 30 is 1.

So, the sum is $\frac{17}{30}$.

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(iv) $\frac{-3}{-11} + \frac{5}{9}$

First, rewrite the first rational number with a positive denominator:

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

Now, find the sum of $\frac{3}{11}$ and $\frac{5}{9}$. The denominators are 11 and 9. The LCM of 11 and 9 is 99.

Convert each fraction to an equivalent fraction with the denominator 99:

$\frac{3}{11} = \frac{3 \times 9}{11 \times 9} = \frac{27}{99}$

$\frac{5}{9} = \frac{5 \times 11}{9 \times 11} = \frac{55}{99}$

Add the converted fractions:

$\frac{27}{99} + \frac{55}{99} = \frac{27+55}{99} = \frac{82}{99}$.

The fraction $\frac{82}{99}$ is in its simplest form as the HCF of 82 and 99 is 1.

So, the sum is $\frac{82}{99}$.

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(v) $\frac{-8}{19} + \frac{(-2)}{57}$

The denominators are 19 and 57. The LCM of 19 and 57 is 57, since $57 = 3 \times 19$.

Convert the first fraction to have the denominator 57:

$\frac{-8}{19} = \frac{-8 \times 3}{19 \times 3} = \frac{-24}{57}$

The second fraction $\frac{-2}{57}$ is already in this form.

Add the fractions:

$\frac{-24}{57} + \frac{-2}{57} = \frac{-24 + (-2)}{57} = \frac{-24 - 2}{57} = \frac{-26}{57}$.

The fraction $\frac{-26}{57}$ is in its simplest form as the HCF of 26 and 57 is 1.

So, the sum is $\frac{-26}{57}$.

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(vi) $\frac{-2}{3} + 0$

Adding zero to any rational number does not change the number.

$\frac{-2}{3} + 0 = \frac{-2}{3}$.

So, the sum is $\frac{-2}{3}$.

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(vii) $-2\frac{1}{3} + 4\frac{3}{5}$

Convert the mixed numbers into improper fractions:

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

$4\frac{3}{5} = 4 + \frac{3}{5} = \frac{4 \times 5 + 3}{5} = \frac{23}{5}$.

Now, find the sum of $\frac{-7}{3}$ and $\frac{23}{5}$. The denominators are 3 and 5. The LCM of 3 and 5 is 15.

Convert each fraction to an equivalent fraction with the denominator 15:

$\frac{-7}{3} = \frac{-7 \times 5}{3 \times 5} = \frac{-35}{15}$

$\frac{23}{5} = \frac{23 \times 3}{5 \times 3} = \frac{69}{15}$

Add the converted fractions:

$\frac{-35}{15} + \frac{69}{15} = \frac{-35+69}{15} = \frac{34}{15}$.

The fraction $\frac{34}{15}$ is in its simplest form as the HCF of 34 and 15 is 1.

So, the sum is $\frac{34}{15}$.

Solution:

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(i) $\frac{7}{24} - \frac{17}{36}$

To subtract fractions, we find a common denominator. The denominators are 24 and 36.

The LCM of 24 and 36 is 72.

Convert each fraction to an equivalent fraction with the denominator 72:

$\frac{7}{24} = \frac{7 \times 3}{24 \times 3} = \frac{21}{72}$

$\frac{17}{36} = \frac{17 \times 2}{36 \times 2} = \frac{34}{72}$

Now, subtract the equivalent fractions:

$\frac{21}{72} - \frac{34}{72} = \frac{21 - 34}{72} = \frac{-13}{72}$.

The fraction $\frac{-13}{72}$ is in its simplest form.

So, $\frac{7}{24} - \frac{17}{36} = \frac{-13}{72}$.

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(ii) $\frac{5}{63} - \left( \frac{-6}{21} \right)$

Subtracting a negative number is the same as adding the positive number:

$\frac{5}{63} - \left( \frac{-6}{21} \right) = \frac{5}{63} + \frac{6}{21}$.

The denominators are 63 and 21. The LCM of 63 and 21 is 63.

Convert $\frac{6}{21}$ to an equivalent fraction with the denominator 63:

$\frac{6}{21} = \frac{6 \times 3}{21 \times 3} = \frac{18}{63}$.

$\frac{5}{63}$ is already in this form.

Now, add the fractions:

$\frac{5}{63} + \frac{18}{63} = \frac{5 + 18}{63} = \frac{23}{63}$.

The fraction $\frac{23}{63}$ is in its simplest form.

So, $\frac{5}{63} - \left( \frac{-6}{21} \right) = \frac{23}{63}$.

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(iii) $\frac{-6}{13} - \left( \frac{-7}{15} \right)$

Subtracting a negative number is the same as adding the positive number:

$\frac{-6}{13} - \left( \frac{-7}{15} \right) = \frac{-6}{13} + \frac{7}{15}$.

The denominators are 13 and 15. Since 13 and 15 are coprime, the LCM of 13 and 15 is $13 \times 15 = 195$.

Convert each fraction to an equivalent fraction with the denominator 195:

$\frac{-6}{13} = \frac{-6 \times 15}{13 \times 15} = \frac{-90}{195}$

$\frac{7}{15} = \frac{7 \times 13}{15 \times 13} = \frac{91}{195}$

Now, add the fractions:

$\frac{-90}{195} + \frac{91}{195} = \frac{-90 + 91}{195} = \frac{1}{195}$.

The fraction $\frac{1}{195}$ is in its simplest form.

So, $\frac{-6}{13} - \left( \frac{-7}{15} \right) = \frac{1}{195}$.

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(iv) $\frac{-3}{8} - \frac{7}{11}$

The denominators are 8 and 11. Since 8 and 11 are coprime, the LCM of 8 and 11 is $8 \times 11 = 88$.

Convert each fraction to an equivalent fraction with the denominator 88:

$\frac{-3}{8} = \frac{-3 \times 11}{8 \times 11} = \frac{-33}{88}$

$\frac{7}{11} = \frac{7 \times 8}{11 \times 8} = \frac{56}{88}$

Now, subtract the fractions:

$\frac{-33}{88} - \frac{56}{88} = \frac{-33 - 56}{88} = \frac{-89}{88}$.

The fraction $\frac{-89}{88}$ is in its simplest form.

So, $\frac{-3}{8} - \frac{7}{11} = \frac{-89}{88}$.

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(v) $-2\frac{1}{9} - 6$

Convert the mixed number to an improper fraction:

$-2\frac{1}{9} = -\left(2 + \frac{1}{9}\right) = -\left(\frac{2 \times 9 + 1}{9}\right) = -\frac{19}{9}$.

Write 6 as a fraction with the same denominator, 9:

$6 = \frac{6}{1} = \frac{6 \times 9}{1 \times 9} = \frac{54}{9}$.

Now, perform the subtraction:

$\frac{-19}{9} - \frac{54}{9} = \frac{-19 - 54}{9} = \frac{-73}{9}$.

The fraction $\frac{-73}{9}$ is in its simplest form. It can also be written as a mixed number: $-8\frac{1}{9}$.

So, $-2\frac{1}{9} - 6 = \frac{-73}{9}$ or $-8\frac{1}{9}$.

Solution:

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

$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

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(i) $\frac{9}{2} × \left( \frac{-7}{4} \right)$

Multiply the numerators (9 and -7) and the denominators (2 and 4):

$\frac{9 \times (-7)}{2 \times 4} = \frac{-63}{8}$.

The fraction $\frac{-63}{8}$ is in its simplest form.

So, the product is $\frac{-63}{8}$.

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(ii) $\frac{3}{10} × (-9)$

Write -9 as a rational number: $-9 = \frac{-9}{1}$.

Multiply the numerators (3 and -9) and the denominators (10 and 1):

$\frac{3 \times (-9)}{10 \times 1} = \frac{-27}{10}$.

The fraction $\frac{-27}{10}$ is in its simplest form.

So, the product is $\frac{-27}{10}$.

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(iii) $\frac{-6}{5} × \frac{9}{11}$

Multiply the numerators (-6 and 9) and the denominators (5 and 11):

$\frac{-6 \times 9}{5 \times 11} = \frac{-54}{55}$.

The fraction $\frac{-54}{55}$ is in its simplest form.

So, the product is $\frac{-54}{55}$.

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(iv) $\frac{3}{7} × \left( \frac{-2}{5} \right)$

Multiply the numerators (3 and -2) and the denominators (7 and 5):

$\frac{3 \times (-2)}{7 \times 5} = \frac{-6}{35}$.

The fraction $\frac{-6}{35}$ is in its simplest form.

So, the product is $\frac{-6}{35}$.

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(v) $\frac{3}{11} × \frac{2}{5}$

Multiply the numerators (3 and 2) and the denominators (11 and 5):

$\frac{3 \times 2}{11 \times 5} = \frac{6}{55}$.

The fraction $\frac{6}{55}$ is in its simplest form.

So, the product is $\frac{6}{55}$.

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(vi) $\frac{3}{-5} × \frac{-5}{3}$

Multiply the numerators (3 and -5) and the denominators (-5 and 3):

$\frac{3 \times (-5)}{-5 \times 3} = \frac{-15}{-15}$.

Simplify the resulting fraction:

$\frac{-15}{-15} = 1$.

Alternatively, we can cancel out common factors before multiplying:

Cancel the factor 3 from numerator and denominator: $\frac{\cancel{3} \times (-5)}{-5 \times \cancel{3}} = \frac{-5}{-5}$.

Cancel the factor -5 from numerator and denominator: $\frac{\cancel{-5}}{\cancel{-5}} = 1$.

The result is 1.

So, the product is 1.

Solution:

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To divide a rational number by another rational number, we multiply the first rational number by the reciprocal of the second rational number. That is, $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$, where $\frac{c}{d} \neq 0$.

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(i) $(-4) ÷ \frac{2}{3}$

Rewrite -4 as a rational number: $-4 = \frac{-4}{1}$.

The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

Now multiply $\frac{-4}{1}$ by $\frac{3}{2}$:

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

Simplify the resulting fraction:

$\frac{-12}{2} = -6$.

So, $(-4) ÷ \frac{2}{3} = -6$.

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(ii) $\frac{-3}{5} ÷ 2$

Rewrite 2 as a rational number: $2 = \frac{2}{1}$.

The reciprocal of $\frac{2}{1}$ is $\frac{1}{2}$.

Now multiply $\frac{-3}{5}$ by $\frac{1}{2}$:

$\frac{-3}{5} \times \frac{1}{2} = \frac{-3 \times 1}{5 \times 2} = \frac{-3}{10}$.

The fraction $\frac{-3}{10}$ is in its simplest form.

So, $\frac{-3}{5} ÷ 2 = \frac{-3}{10}$.

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(iii) $\frac{-4}{5} ÷ (-3)$

Rewrite -3 as a rational number: $-3 = \frac{-3}{1}$.

The reciprocal of $\frac{-3}{1}$ is $\frac{1}{-3}$.

Now multiply $\frac{-4}{5}$ by $\frac{1}{-3}$:

$\frac{-4}{5} \times \frac{1}{-3} = \frac{-4 \times 1}{5 \times (-3)} = \frac{-4}{-15}$.

Rewrite the fraction with a positive denominator (standard form):

$\frac{-4}{-15} = \frac{-4 \times -1}{-15 \times -1} = \frac{4}{15}$.

The fraction $\frac{4}{15}$ is in its simplest form.

So, $\frac{-4}{5} ÷ (-3) = \frac{4}{15}$.

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(iv) $\frac{-1}{8} ÷ \frac{3}{4}$

The reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$.

Now multiply $\frac{-1}{8}$ by $\frac{4}{3}$:

$\frac{-1}{8} \times \frac{4}{3} = \frac{-1 \times 4}{8 \times 3} = \frac{-4}{24}$.

Simplify the resulting fraction by dividing the numerator and denominator by their HCF, which is 4:

$\frac{-4 \div 4}{24 \div 4} = \frac{-1}{6}$.

So, $\frac{-1}{8} ÷ \frac{3}{4} = \frac{-1}{6}$.

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(v) $\frac{-2}{13} ÷ \frac{1}{7}$

The reciprocal of $\frac{1}{7}$ is $\frac{7}{1}$.

Now multiply $\frac{-2}{13}$ by $\frac{7}{1}$:

$\frac{-2}{13} \times \frac{7}{1} = \frac{-2 \times 7}{13 \times 1} = \frac{-14}{13}$.

The fraction $\frac{-14}{13}$ is in its simplest form.

So, $\frac{-2}{13} ÷ \frac{1}{7} = \frac{-14}{13}$.

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(vi) $\frac{-7}{12} ÷ \left( \frac{-2}{13} \right)$

The reciprocal of $\frac{-2}{13}$ is $\frac{13}{-2}$.

Now multiply $\frac{-7}{12}$ by $\frac{13}{-2}$:

$\frac{-7}{12} \times \frac{13}{-2} = \frac{-7 \times 13}{12 \times (-2)} = \frac{-91}{-24}$.

Rewrite the fraction with a positive denominator (standard form):

$\frac{-91}{-24} = \frac{-91 \times -1}{-24 \times -1} = \frac{91}{24}$.

The fraction $\frac{91}{24}$ is in its simplest form.

So, $\frac{-7}{12} ÷ \left( \frac{-2}{13} \right) = \frac{91}{24}$.

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(vii) $\frac{3}{13} ÷ \left( \frac{-4}{65} \right)$

The reciprocal of $\frac{-4}{65}$ is $\frac{65}{-4}$.

Now multiply $\frac{3}{13}$ by $\frac{65}{-4}$:

$\frac{3}{13} \times \frac{65}{-4} = \frac{3 \times 65}{13 \times (-4)}$.

We can cancel out the common factor 13 from the numerator (65) and the denominator (13):

$\frac{3 \times \cancel{65}^{5}}{\cancel{13}^{1} \times (-4)} = \frac{3 \times 5}{1 \times (-4)} = \frac{15}{-4}$.

Rewrite the fraction with a positive denominator (standard form):

$\frac{15}{-4} = \frac{15 \times -1}{-4 \times -1} = \frac{-15}{4}$.

The fraction $\frac{-15}{4}$ is in its simplest form.

So, $\frac{3}{13} ÷ \left( \frac{-4}{65} \right) = \frac{-15}{4}$.