Chapter 9: Rational Numbers

# Question: 1

List five rational numbers between:

(i)

(ii)

(iii)

(iv)

## Solution

Rational number is any number that can be expressed in the form of p/q, where p and q are integers and q 0.

(i)             We need to find five rational number between

So, we can write  as rational numbers with denominator $6$.

Therefore, five rational numbers between  would be

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

(ii)          We need to find five rational number between

Let us write  as rational numbers with denominator $10$.

Therefore, five rational numbers between  would be

$\frac{-19}{10},\frac{-9}{5},\frac{-17}{10},\frac{-8}{5},\frac{-3}{2}$

(iii)

Let us write  as rational numbers with the same denominators.

Therefore, five rational numbers between  would be

$\frac{-7}{9},\frac{-34}{45},\frac{-11}{15},\frac{-32}{45},\frac{-31}{45}$

(iv)

Let us write  as rational numbers with the same denominators.

Therefore, five rational numbers between  would be

$\frac{-1}{3},\frac{-1}{6},0,\frac{1}{6},\frac{1}{3}.$

# Question: 2

Write four more rational numbers in each of the following patterns:

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

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

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

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

## Solution

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

We observe that numerator is multiple of 3 and denominator is multiple of 5.

Therefore, the next four rational numbers of this pattern would be

$\begin{array}{l}\frac{-3×5}{5×5},\frac{-3×6}{5×6},\frac{-3×7}{5×7},\frac{-3×8}{5×8}\\ =\frac{-15}{25},\frac{-18}{30},\frac{-21}{35},\frac{-24}{40}\end{array}$

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

Therefore, the next four rational numbers of this pattern would be

$\begin{array}{l}\frac{-1×4}{4×4},\frac{-1×5}{4×5},\frac{-1×6}{4×6},\frac{-1×7}{4×7}\\ =\frac{-4}{16},\frac{-5}{20},\frac{-6}{24},\frac{-7}{28}\end{array}$

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

Therefore, the next four rational numbers of this pattern would be

$\begin{array}{l}\frac{1×5}{-6×5},\frac{1×6}{-6×6},\frac{1×7}{-6×7},\frac{1×8}{-6×8}\\ =\frac{5}{-30},\frac{6}{-36},\frac{7}{-42},\frac{8}{-48}\end{array}$

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

Therefore, the next four rational numbers of this pattern would be

$\begin{array}{l}\frac{-2×-4}{3×-4},\frac{-2×-5}{3×-5},\frac{-2×-6}{3×-6},\frac{-2×-7}{3×-7}\\ =\frac{8}{-12},\frac{10}{-15},\frac{12}{-18},\frac{14}{-21}\end{array}$

# Question: 3

Give four rational numbers equivalent to:

(i)             $\frac{-2}{7}$

(ii)          $\frac{5}{-3}$

(iii)      $\frac{4}{9}$

## Solution

(i)             $\frac{-2}{7}$

Next four

$\begin{array}{l}\frac{-2×2}{7×2}=\frac{-4}{14},\\ \frac{-2×3}{7×3}=\frac{-6}{21},\\ \frac{-2×4}{7×4}=\frac{-8}{28},\\ \frac{-2×5}{7×5}=\frac{-10}{35}\end{array}$

Therefore, four equivalent rational numbers are $\frac{-4}{14},\frac{-6}{21},\frac{-8}{28},\frac{-10}{35}.$

(ii)          $\frac{5}{-3}$

$\begin{array}{l}\frac{5×2}{-3×2}=\frac{10}{-6},\\ \frac{5×3}{-3×3}=\frac{15}{-9},\\ \frac{5×4}{-3×4}=\frac{20}{-12},\\ \frac{5×5}{-3×5}=\frac{25}{-15}\end{array}$

Therefore, four equivalent rational numbers are $\frac{10}{-6},\frac{15}{-9},\frac{20}{-12},\frac{25}{-15}.$

(iii)      $\frac{4}{9}$

$\begin{array}{l}\frac{4×2}{9×2}=\frac{8}{18},\\ \frac{4×3}{9×3}=\frac{12}{27},\\ \frac{4×4}{9×4}=\frac{16}{36},\\ \frac{4×5}{9×5}=\frac{20}{45}\end{array}$

Therefore, four equivalent rational numbers are $\frac{8}{18},\frac{12}{27},\frac{16}{36},\frac{20}{45}.$

# Question: 4

Draw the number line and represent the following rational numbers on it:

(i)             $\frac{3}{4}$

(ii)          $\frac{-5}{8}$

(iii)      $\frac{-7}{4}$

(iv)       $\frac{7}{8}$

## Solution

(i)             $\frac{3}{4}$

Given fraction represent 3 parts out of 4 equal parts. Hence, the space between the integers 0 and 1 on the number line must be divided into 4 parts.

(ii)          $\frac{-5}{8}$

Given fraction represent 5 parts out of 8 equal parts. Negative sign represents that it is on the negative side of the number line. Hence, the space between the integers -1 and 0 must be divided into 8 equal parts on the number line.

(iii)      $\frac{-7}{4}$

We can write given fraction as

$\frac{-7}{4}=-1\frac{3}{4}$

This fraction represents 1 full part and 3 part out of 4 equal parts. Therefore, each space between the integers 0 and -1 and also between -2 and -1 must be divided into 4 equal parts on the number line.

(iv)       $\frac{7}{8}$

Given fraction represents 7 parts out of 8 equal parts. Therefore, space between the  integers 0 and 1 and also between 1 and 2 must be divided into 8 equal parts on the number line.

# Question: 5

The points P, Q, R, S, T, U, A and B on the number line are such that, TR $=$ RS $=$ SU and AP $=$ PQ $=$ QB. Name the rational numbers represented by P, Q, R and S.

## Solution

In the given number line, each part which is between the two numbers is divided into $3$ parts.

Distance between U and T = 1 Unit

It is divided into 3 equal parts.

$\text{TR=RS=}\text{​}\text{\hspace{0.17em}}\text{SU}=\frac{1}{3}$

$R\text{\hspace{0.17em}}=\text{\hspace{0.17em}}-1-\frac{1}{3}=\frac{-3-1}{3}=\frac{-4}{3}$

$S\text{\hspace{0.17em}}=\text{\hspace{0.17em}}-1-\frac{2}{3}=\frac{-3-2}{3}=\frac{-5}{3}$

Similarly,

AB = 1 unit, and it is divided into 3 equal parts.

Therefore,

$\text{P}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}2+\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}$

$\text{Q}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}2+\frac{2}{3}=\frac{6+2}{3}=\frac{8}{3}$

Thus, the rational numbers represented P, Q, R and S are  respectively.

# Question: 6

Which of the following pairs represent the same rational number?

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

## Solution

(i)

[Converting into lowest term]

As  , therefore $\frac{-7}{21}\ne \frac{3}{9}$ and does not represent same rational number.

(ii)

[Converting into lowest term]

$\text{As}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{-4}{5}=\frac{-4}{5}$ therefore $\frac{-16}{20}=\frac{20}{-25}$ and represent same rational numbers.

(iii)

[Converting into lowest term]

$\text{As}\text{\hspace{0.17em}}\frac{2}{3}=\frac{2}{3}$ therefore , and represent same rational number.

(iv)

[Converting into lowest term]

(v)

[Converting into lowest term]

(vi)

[Converting into lowest term]

(vii)

[Converting into lowest term]

# Question: 7

Rewrite the following rational numbers in the simplest form:

(i)             $\frac{-8}{6}$

(ii)          $\frac{25}{45}$

(iii)      $\frac{-44}{72}$

(iv)       $\frac{-8}{10}$

## Solution

(i)             $\begin{array}{l}\text{Given,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{-8}{6}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-8÷2}{6÷2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-4}{3}\end{array}$           [H.C.F. of $8$ and $6$ is $2$ ]

(ii)          $\begin{array}{l}\text{Given}=\frac{25}{45}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{25÷5}{45÷5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{5}{9}\end{array}$          [H.C.F. of $25$ and $45$ is $5$ ]

(iii)      $\begin{array}{l}\text{Given,}\frac{-44}{72}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-44÷4}{72÷4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-11}{18}\end{array}$        [H.C.F. of $44$ and $72$ is $4$ ]

(iv)       $\begin{array}{l}\text{Given,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{-8}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-8÷2}{10÷2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-4}{5}\end{array}$          [H.C.F. of $8$ and $10$ is $2$ ]

# Question: 8

Fill in the boxes with the correct symbol out of $>,\text{\hspace{0.17em}}<,$ and $=$.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

## Solution

(i)

Since, the positive number is always greater than the negative number.

Hence,

(ii)

Making denominator same for both the rational number.

As, -25 > than -28

(iii)

Making denominator same for both the rational number.

(iv)

Making denominator same for both the rational number.

(v)

Making denominator same for both the rational numbers.

(vi)

(vii)

Since, 0 is greater than every negative number.

# Question: 9

Which is greater in each of the following:

(i)             $\frac{2}{3},\frac{5}{2}$

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

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

(iv)       $\frac{-1}{4},\frac{1}{4}$

(v)          $-3\frac{2}{7},-3\frac{4}{5}$

## Solution

(i)             $\frac{2}{3},\frac{5}{2}$

Making denominator same for the rational numbers.

Since

Therefore

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

Making denominator same for both the rational numbers.

Since

Therefore

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

Making denominator same for both the rational numbers.

Since

Therefore

(iv)        Since positive number is always greater than negative number.

(v)          $-3\frac{2}{7},-3\frac{4}{5}$

Since

Therefore

# Question: 10

Write the following rational numbers in ascending order:

(i)             $\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}$

(ii)          $\frac{-1}{3},\frac{-2}{9},\frac{-4}{3}$

(iii)      $\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}$

## Solution

(i)             $\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}$

$\text{As,}-3<-2<-1$

(ii)          $\frac{-1}{3},\frac{-2}{9},\frac{-4}{3}$

$\frac{-1×3}{3×3},\frac{-2}{9},\frac{-4×3}{3×3}$ [Converting these into like fractions]

$\text{As,}\text{\hspace{0.17em}}-12<-3<-2$

$\therefore \frac{-4}{3}<\frac{-1}{3}<\frac{-2}{9}$

(iii)      $\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}$

[Converting these into like fractions]

$\frac{-3×4}{7×4},\frac{-3×14}{2×14},\frac{-3×7}{4×7}$

$\begin{array}{l}\frac{-12}{28},\frac{-42}{28},\frac{-21}{28}\\ \text{As,}\text{\hspace{0.17em}}\text{-42}<-21<-12\end{array}$

# Question: 1

Find the sum:

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

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

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

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

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

(vi)       $\frac{-2}{3}+0$

(vii)    $-2\frac{1}{3}+4\frac{3}{5}$

## Solution

(i)

$\begin{array}{c}\frac{5}{4}+\left(\frac{-11}{4}\right)\\ =\frac{5-11}{4}\\ =\frac{-6}{4}\\ =\frac{-3}{2}\end{array}$

(ii)

$\begin{array}{c}\frac{5}{3}+\frac{3}{5}=\frac{5×5}{3×5}+\frac{3×3}{5×3}\\ =\frac{25}{15}+\frac{9}{15}\end{array}$             [L.C.M. of  ]

$\begin{array}{c}=\frac{25+9}{15}\\ =\frac{34}{15}\\ =2\frac{4}{15}\end{array}$

(iii)

$\begin{array}{c}\frac{-9}{10}+\frac{22}{15}\\ =\frac{-9×3}{10×3}+\frac{22×2}{15×2}\\ =\frac{-27}{30}+\frac{44}{30}\end{array}$  [L.C.M. of  ]

$\begin{array}{l}=\frac{-27+44}{30}\\ =\frac{17}{30}\end{array}$

(iv)

$\begin{array}{c}\frac{-3}{-11}+\frac{5}{9}\\ =\frac{-3×9}{-11×9}+\frac{5×11}{9×11}\\ =\frac{27}{99}+\frac{55}{99}\end{array}$    [L.C.M. of  ]

$\begin{array}{l}=\frac{27+55}{99}\\ =\frac{82}{99}\end{array}$

(v)

$\begin{array}{l}\frac{-8}{19}+\frac{\left(-2\right)}{57}\\ =\frac{-8×3}{19×3}+\frac{\left(-2\right)×1}{57×1}\end{array}$   [L.C.M. of  ]

$\begin{array}{c}=\frac{-24}{57}+\frac{\left(-2\right)}{57}\\ =\frac{-24-2}{57}\\ =\frac{-26}{57}\end{array}$

(vi)

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

(vii)

$\begin{array}{l}-2\frac{1}{3}+4\frac{3}{5}\\ =\frac{-7}{3}+\frac{23}{5}\end{array}$            [L.C.M. of  ]

$\begin{array}{l}=\frac{-7×5}{3×5}+\frac{23×3}{5×3}\\ =\frac{-35}{15}+\frac{69}{15}\end{array}$

$\begin{array}{l}=\frac{-35+69}{15}\\ =\frac{34}{15}\\ =2\frac{4}{15}\end{array}$

# Question: 2

Find

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

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

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

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

(v)          $-2\frac{1}{9}-6$

## Solution

(i)             $\begin{array}{l}\frac{7}{24}-\frac{17}{36}\\ =\frac{7×3}{24×3}-\frac{17×2}{36×2}\end{array}$

[L.C.M. of  ]

$\begin{array}{c}=\frac{21}{72}-\frac{34}{72}\\ =\frac{21-34}{72}\\ =\frac{-13}{72}\end{array}$

(ii)          $\begin{array}{l}\frac{5}{63}-\left(\frac{-6}{21}\right)\\ =\frac{5×1}{63×1}-\left(\frac{-6×3}{21×3}\right)\end{array}$

[L.C.M. of  ]

$\begin{array}{l}=\frac{5}{63}-\frac{-18}{63}\\ =\frac{5-\left(-18\right)}{63}\\ =\frac{5+18}{63}\\ =\frac{23}{63}\end{array}$

(iii)      $\begin{array}{l}\frac{-6}{13}-\left(\frac{-7}{15}\right)\\ =\frac{-6×15}{13×15}-\left(\frac{-7×13}{15×13}\right)\end{array}$

[L.C.M. of  ]

$\begin{array}{l}=\frac{-90}{195}-\left(\frac{-91}{195}\right)\\ =\frac{-90-\left(-91\right)}{195}\\ =\frac{-90+91}{195}\\ =\frac{1}{195}\end{array}$

(iv)       $\begin{array}{l}\frac{-3}{8}-\frac{7}{11}\\ =\frac{-3×11}{8×11}-\frac{7×8}{11×8}\end{array}$

[L.C.M. of  ]

$\begin{array}{l}=\frac{-33}{88}-\frac{56}{88}\\ =\frac{-33-56}{88}\\ =\frac{-89}{88}\\ =-1\frac{1}{88}\end{array}$

(v)          $-2\frac{1}{9}-6=\frac{-19}{9}-\frac{6}{1}$

[L.C.M. of  ]

$\begin{array}{l}=\frac{-19×1}{9×1}-\frac{6×9}{1×9}\\ =\frac{-19}{9}-\frac{54}{9}\end{array}$

$\begin{array}{l}=\frac{-19-54}{9}\\ =\frac{-73}{9}\\ =-8\frac{1}{9}\end{array}$

# Question: 3

Find the product:

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

(ii)          $\frac{3}{10}×\left(-9\right)$

(iii)      $\frac{-6}{5}×\frac{9}{11}$

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

(v)          $\frac{3}{11}×\frac{2}{5}$

(vi)       $\frac{3}{-5}×\frac{-5}{3}$

## Solution

(i)             $\begin{array}{l}\frac{9}{2}×\left(\frac{-7}{4}\right)\\ =\frac{9×\left(-7\right)}{2×4}\end{array}$

$\begin{array}{l}=\frac{-63}{8}\\ =-7\frac{7}{8}\end{array}$

(ii)          $\begin{array}{l}\frac{3}{10}×\left(-9\right)\\ =\frac{3×\left(-9\right)}{10}\end{array}$

$\begin{array}{l}=\frac{-27}{10}\\ =-2\frac{7}{10}\end{array}$

(iii)      $\begin{array}{l}\frac{-6}{5}×\frac{9}{11}\\ =\frac{\left(-6\right)×9}{5×11}\end{array}$

$=\frac{-54}{55}$

(iv)       $\begin{array}{l}\frac{3}{7}×\left(\frac{-2}{5}\right)\\ =\frac{3×\left(-2\right)}{7×5}\end{array}$

$=\frac{-6}{35}$

(v)          $\begin{array}{l}\frac{3}{11}×\frac{2}{5}\\ =\frac{3×2}{11×5}\end{array}$

$=\frac{6}{55}$

(vi)       $\begin{array}{l}\frac{3}{-5}×\left(\frac{-5}{3}\right)\\ =\frac{3×\left(-5\right)}{-5×3}=1\end{array}$

# Question: 4

Find the value of:

(i)             $\left(-4\right)÷\frac{2}{3}$

(ii)          $\frac{-3}{5}÷2$

(iii)      $\frac{-4}{5}÷\left(-3\right)$

(iv)       $\frac{-1}{8}÷\frac{3}{4}$

(v)          $\frac{-2}{13}÷\frac{1}{7}$

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

(vii)    $\frac{3}{13}÷\frac{-4}{65}$

## Solution

(i)             $\begin{array}{l}\left(-4\right)÷\frac{2}{3}\\ =\left(-4\right)×\frac{3}{2}\\ =\left(-2\right)×3\\ =-6\end{array}$

(ii)

$\begin{array}{l}\frac{-3}{5}÷2\\ =\frac{-3}{5}×\frac{1}{2}\\ =\frac{\left(-3\right)×1}{5×2}\\ =\frac{-3}{10}\end{array}$

(iii)

$\frac{-4}{5}÷\left(-3\right)$

$=\frac{\left(-4\right)}{5}×\frac{1}{\left(-3\right)}$

$\begin{array}{l}=\frac{\left(-4\right)×1}{5×\left(-3\right)}\\ =\frac{4}{15}\end{array}$

(iv)

$\begin{array}{l}\frac{-1}{8}÷\frac{3}{4}\\ =\frac{-1}{8}×\frac{4}{3}\\ =\frac{\left(-1\right)×1}{2×3}\\ =\frac{-1}{6}\end{array}$

(v)

$\begin{array}{l}\frac{-2}{13}÷\frac{1}{7}\\ =\frac{-2}{13}×\frac{7}{1}\\ =\frac{\left(-2\right)×7}{13×1}\\ =\frac{-14}{13}\\ =-1\frac{1}{13}\end{array}$

(vi)

$\begin{array}{l}\frac{-7}{12}÷\frac{-2}{13}=\frac{-7}{12}×\frac{13}{\left(-2\right)}\\ =\frac{\left(-7\right)×13}{12×\left(-2\right)}\\ =\frac{-91}{-24}\\ =3\frac{19}{24}\end{array}$

(vii)

$\begin{array}{l}\frac{3}{13}÷\frac{-4}{65}\\ =\frac{3}{13}×\frac{65}{\left(-4\right)}\\ =\frac{3×\left(-5\right)}{1×4}\\ =\frac{-15}{4}\\ =-3\frac{3}{4}\end{array}$