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#### Answer:

The numbers that are in the form of  $\frac{p}{q}$, where p and q are integers and q ≠0, are called rational numbers.

For example:

Five positive rational numbers:

$\frac{5}{7},\frac{-3}{-4},\frac{7}{8},\frac{-14}{-15},\frac{5}{9}$

Five negative rational numbers:

$\frac{-3}{7},\frac{-3}{8},\frac{8}{-9},\frac{-19}{25},\frac{8}{-25}$

Yes, there is a rational number that is neither positive nor negative, i.e. zero (0).

#### Answer:

(i) $\frac{8}{19}$
Numerator = 8

Denominator =19

(ii)$\frac{5}{-8}$
Numerator  = 5

Denominator = −8

(iii) $\frac{-13}{5}$

Numerator = −13
Denominator =15

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

Numerator = −8
Denominator = −11

(v) 9
i.e $\frac{9}{1}$
Numerator = 9
Denominator = 1

#### Answer:

(i) 5
The rational number will be $\frac{5}{1}$.
Numerator = 5
Denominator = 1

(ii) -3
The rational number will be $\frac{-3}{1}$.
Numerator   = -3
Denominator = 1

(iii)1
The rational number will be $\frac{1}{1}$.
Numerator = 1
Denominator = 1

(iv) 0
The rational number will be $\frac{0}{1}$.
Numerator =0
Denominator = 1

(v) -23
The rational number will be $\frac{-23}{1}$.
Numerator = -23
Denominator = 1

#### Answer:

Positive rational numbers:
(iii) $\frac{-5}{-8}$

(iv) $\frac{37}{53}$
(vi) 8 because 8 can be written as .

0 is neither positive nor negative.

#### Answer:

Negative rational numbers:

(iii) $\frac{-5}{7}$

(iv) $\frac{4}{-9}$

(v)  -6

(vi) $\frac{1}{-2}$

#### Answer:

(i) Following are the four rational numbers that are equivalent to $\frac{6}{11}$.

(ii) Following are the four rational numbers that are equivalent to $\frac{-3}{8}$.
$\frac{-3×2}{8×2}$,
$\frac{-3×3}{8×3}$$\frac{-3×4}{8×4}$ and $\frac{-3×5}{8×5}$

i.e. $\frac{-6}{16}$$\frac{-9}{24}$$\frac{-12}{32}$ and $\frac{-15}{40}$

(iii) Following are the four rational numbers that are equivalent to $\frac{7}{-15}$.

(iv) Following are the four rational numbers that are equivalent to 8, i.e. $\frac{8}{1}$.

(v) Following are the four rational numbers that are equivalent to ­­-1, i.e. $\frac{1}{1}$.

(vi)
Following are the four rational numbers that are equivalent to ­­-1, i.e. $\frac{-1}{1}$.

#### Answer:

(i) $\frac{12×\left(-1\right)}{\left(-17\right)×\left(-1\right)}=\frac{-12}{17}$

(ii) $\frac{1×\left(-1\right)}{\left(-2\right)×\left(-1\right)}=\frac{-1}{2}$

(iii) $\frac{-8}{-19}=\frac{-8×\left(-1\right)}{\left(-19\right)×\left(-1\right)}=\frac{8}{19}$

(iv) $\frac{11×\left(-1\right)}{-6×\left(-1\right)}=\frac{-11}{6}$

#### Answer:

(i) Numerator of  $\frac{5}{8}$  is 5.
5 should be multiplied by 3 to get 15.
Multiplying both the numerator and the denominator by 3:

(ii)  Numerator of  $\frac{5}{8}$  is 5.
5 should be multiplied by −2 to get −10.
Multiplying both the numerator and the denominator by −2:

#### Answer:

(i) Denominator of $\frac{4}{7}$ is 7.
7 should be multiplied by 3 to get 21.
Multiplying both the numerator and the denominator by 3:

$\frac{4×3}{7×3}$$\frac{12}{21}$

$\frac{4×3}{7×3}$$\frac{4}{7}$

(ii)
Denominator of $\frac{4}{7}$ is 7.
7 should be multiplied by -5 to get −35.
Multiplying both the numerator and the denominator by 5:

#### Answer:

(i) Numerator of $\frac{-12}{13}$ is −12.
−12 should be multiplied by 4 to get 48.
Multiplying both the numerator and the denominator by 4
:

$\begin{array}{l}\frac{-12×4}{13×4}=\frac{-48}{52}\\ \\ \frac{-12}{13}=\frac{-48}{52}\end{array}$

(ii)
Numerator of $\frac{-12}{13}$ is −12.​
12 should be multiplied by 5 to get 60

Multiplying its numerator and denominator by -5:

$\begin{array}{l}\frac{-12×\left(-5\right)}{13×\left(-5\right)}=\frac{60}{-65}\\ \\ \frac{-12}{13}=\frac{60}{-65}\end{array}$

#### Answer:

(i) Denominator of$\frac{-8}{11}$  is 11.
Clearly, 11×2= 22

Multiplying both the numerator and the denominator by 2:

$\begin{array}{l}\frac{-8×2}{11×2}=\frac{-16}{22}\\ \\ \frac{-8}{11}=\frac{-16}{22}\end{array}$

(ii)
Denominator of$\frac{-8}{11}$  is 11.
Clearly, 11×5=55

Multiplying both the numerator and the denominator by 5:

#### Answer:

(i) Numerator of $\frac{14}{-5}$ is 14.
Clearly, 14×4=56

Multiplying both the numerator and the denominator by 4:

$\frac{14×4}{-5×4}$=$\frac{56}{-20}$

$\frac{14}{-5}$=$\frac{56}{-20}$

(ii) −70
Numerator of $\frac{14}{-5}$ is 14.​
Clearly, 14×(−5)=−70
Multiplying both the numerator and the denominator by -5:

$\frac{14×\left(-5\right)}{\left(-5\right)×\left(-5\right)}$=$\frac{-70}{25}$

$\frac{14}{-5}$=​$\frac{-70}{25}$

#### Answer:

(i) Denominator of $\frac{13}{-8}$ is −8.
Clearly, (
−8)×5= −40
Multiplying both the numerator and the denominator by 5:
$\begin{array}{l}\frac{13×5}{-8×5}=\frac{65}{-40}\\ \\ \frac{13}{-8}=\frac{65}{-40}\end{array}$

(ii) Denominator of $\frac{13}{-8}$ is −8.
Clearly, (−8)×(
−4)= 32

Multiplying both the numerator and the denominator by −4:

$\frac{13×\left(-4\right)}{-8×\left(-4\right)}=\frac{-52}{32}$

$\frac{13}{-8}$=$\frac{-52}{32}$

#### Answer:

(i) Numerator of  $\frac{-36}{24}$ is -36.

Clearly, (−36) ÷ 4= (−9)
Dividing both the numerator and the denominator by 4:

$\frac{-36÷4}{24÷4}=\frac{-9}{6}$

(ii)
Numerator of  $\frac{-36}{24}$ is −36.​
Clearly, (−36) ÷ ( −6) = 6
Dividing both the numerator and the denominator by -6:

$\frac{-36÷\left(-6\right)}{24÷\left(-6\right)}=\frac{6}{-4}$

$\frac{-36}{24}$$\frac{6}{-4}$

#### Answer:

(i) Denominator of $\frac{84}{-147}$ is 147.
∴ −147 ÷(−21)=7
Dividing both the numerator and the denominator by -21:

$\begin{array}{l}\frac{84÷\left(-21\right)}{-147÷\left(-21\right)}=\frac{-4}{7}\\ \\ \frac{84}{-147}=\frac{-4}{7}\end{array}$

(ii)
Denominator of
$\frac{84}{-147}$ is 147.
−147÷3=−49
Dividing both the numerator and the denominator by 3:

$\begin{array}{l}\frac{84÷3}{-147÷3}=\frac{28}{-49}\\ \end{array}$

$\frac{84}{-147}=\frac{28}{-49}$

#### Answer:

(i) $\frac{35}{49}$
H.C.F. of 35 and 49 is 7.

Dividing the numerator and the denominator by 7:

$\frac{35÷7}{49÷7}=\frac{5}{7}$
So,
in the standard form.

(ii)$\frac{8}{-36}$
Denominator is -36, which is negative.
Multiplying both the numerator and the denominator by -1:

$\frac{8×\left(-1\right)}{-36×\left(-1\right)}=\frac{-8}{36}$

H.C.F. of 8 and 36 is 4
.
Dividing its numerator and denominator by 4:

$\frac{-8÷4}{36÷4}=\frac{-2}{9}$

So,  in the standard form.

(iii) $\frac{-27}{45}$

H.C.F. of 27 and 45 is 9.

Dividing its numerator and denominator by 9:
$\frac{-27÷9}{45÷9}=\frac{-3}{5}$
Hence,  in the standard form.

H.C.F. of 14 and 49 is 7.
Dividing both the numerator and the denominator by 7.

H.C.F. of 91 and 78 is 13.
Dividing both the numerator and the denominator by 13:

H.C.F. of 68 and 119 is 17.
Dividing both the numerator and the denominator by 17:

H.C.F. of 87 and 116 is 29.
Dividing both the numerator and the denominator by 29:

The denominator is negative.
Multiplying both the numerator and denominator by -1:

H.C.F. of 299 and 161 is 23.
Dividing both the numerator and the denominator by 23:

#### Answer:

(i)

$\frac{-9×4}{5×4}=\frac{-36}{20}\phantom{\rule{0ex}{0ex}}\frac{-9×\left(-3\right)}{5×\left(-3\right)}=\frac{27}{-15}\phantom{\rule{0ex}{0ex}}\frac{-9×5}{5×5}=\frac{-45}{25}\phantom{\rule{0ex}{0ex}}\therefore \frac{-9}{5}=\frac{-36}{20}=\frac{27}{-15}=\frac{-45}{25}$

(ii)

#### Answer:

(i) $\frac{-13}{7},\frac{39}{-21}$
We have:
(−13)×(−21) = 273

And 7×39=273

(ii)
We have:

3×16=48

And (−8) ×(−6) =48

∴ 3×16 =(−8)×(−6)

(iii)

We have:

9×(−16)= −144

And 4×(-36)= −144

9×(−16) = 4×(−36)

$\frac{9}{4}=\frac{-36}{-16}$
Therefore, they are equivalent rational numbers.

(iv)

We have:

7×60 =420
And 15×(-28)= −420

∴ 7×60 ≠15×(−28)
Therefore, the rational numbers are not equivalent.

(v)

We have:
3 ×4=12
And 12×(−1)= −12

12 ≠ −12
Therefore, the rational numbers are not equivalent.

(vi) $\frac{2}{3},\frac{3}{2}$

We have:

2×2=4

And 3×3=9

2×2≠3×3

Therefore, the rational numbers are not equivalent.

#### Answer:

(i)$\frac{-1}{5}=\frac{8}{x}$

=> −x =5×8
=> x= −40

(ii)$\frac{7}{-3}=\frac{x}{6}$
=> (
−3)x=7×6

=>
x=$\frac{\left(7×6\right)}{\left(-3\right)}$
=>  x=−14

(iii) $\frac{3}{5}=\frac{x}{-25}$
=>    5x=3×(−25)

=>   x=$\frac{3×\left(-25\right)}{5}$
=>x  = (−15)

(iv)$\frac{13}{6}=\frac{-65}{x}$

=> 13x=6×(−65)

=>  x=$\frac{6×\left(-65\right)}{13}$

=>  x= 6×(−​5)

=>  x = −​30

(v)$\frac{16}{x}=-4$
=> $\mathrm{x =}\frac{16}{\left(-4\right)}$
=>  x= (−4)

vi)$\frac{-48}{x}=2$
=>
$\frac{-48}{2}=\frac{x}{1}$
=>$2x=\left(-48\right)×1$
=>$x=\frac{-48}{2}$
x= (−24)

#### Answer:

(i)

8×15 =120
And ( −10)×(−12)=120

8×15 =(−10) ×(−12)

Therefore, the rational numbers are equal.

ii)

(−3)×(−21) =63
And 7× 9=63

∴ (−3)×(−21) =7×9

Therefore, the rational numbers are equal.

(iii) $\frac{-8}{-14},\frac{15}{21}$

(−8) × 21 = −168
And 15 ×(
−​14) = − ​210

(−8) × 21 ≠ 15 × 14

Therefore, the rational numbers are not equal.

#### Answer:

(i) False
For example,
1 is smaller than zero and is a rational number.
(ii)True
All integers can be written with the denominator 1.

(iii) False
Though 0 is an integer, when the denominator is 0, it is not a rational number.
For example, $\frac{1}{0}$ is not a rational number.

(iv)True
(v) False
−1 is a rational number but not a fraction.

#### Answer:

(i)

(ii)

(iii) (7/3)=2+(1/3)

(iv)

$\frac{22}{7}$ can be written as . So, we need to move to the right of point 3. Then, we need to move $\frac{1}{7}$ distance more to the right.

(v) $\frac{37}{8}$ can be written as 4+$\frac{5}{8}$.  So, we need to move to the right of point 4. Then, we need to move $\frac{5}{8}$ distance more to the right.

(vi)

(vii)

(viii)

$\frac{-12}{7}$ can be written as $-1-\frac{5}{7}$. So, we need to move to the left of point -1. Then, we need to move $\frac{5}{7}$ distance more to the left.

(ix)

$\frac{36}{-5}$ can be written as $-7-\frac{1}{5}$. So, we need to move to the left of point -7. Then, we need to move $\frac{1}{5}$ distance more to the left.

(x) $\frac{-43}{9}$ can be written as $-4-\frac{7}{9}$. So, we need to move to the left of point -4. Then, we need to move $\frac{7}{9}$ distance more to the left.

#### Answer:

#### Answer:

#### Answer:

#### Answer:

(i)
$\frac{12}{7}+\frac{3}{7}=\frac{12+3}{7}=\frac{15}{7}$

(ii)
$\frac{-2}{5}+\frac{1}{5}=\frac{-2+1}{5}=\frac{-1}{5}$

(iii)

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

$\frac{-3}{8}+\frac{1}{8}=\frac{-3+1}{8}=\frac{-2}{8}$

(iv)

$\frac{7}{-11}×\frac{-1}{-1}=\frac{-7}{11}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{-5}{11}+\frac{-7}{11}=\frac{-5+\left(-7\right)}{11}=\frac{-5-7}{11}=\frac{-12}{11}$

(v)

$\frac{-11}{-13}×\frac{-1}{-1}=\frac{11}{13}$

=$\frac{-9}{13}+\frac{11}{13}=\frac{-9+11}{13}=\frac{2}{13}$

(vi)

$\frac{-2}{9}+\frac{-5}{9}=\frac{-2-5}{9}=\frac{-7}{9}$

(vii)

$\frac{\left(-17\right)}{9}+\frac{\left(-11\right)}{9}=\frac{-17-11}{9}=\frac{-28}{9}$

(viii)
$\frac{5}{-7}×\frac{-1}{-1}=\frac{-5}{7}$

$\frac{-3}{7}+\frac{\left(-5\right)}{7}=\frac{-3-5}{7}=\frac{-8}{7}$

#### Answer:

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

(ii)$\frac{-5}{9}+\frac{2}{3}$

(iii)$-4+\frac{1}{2}$

(iv)
$\frac{-7}{27}+\frac{5}{18}$

The denominators of the given rational numbers are 27 and 18.

L.C.M. of 27 and 18 is 54.

(v)$\frac{-5}{36}+\left(\frac{-7}{12}\right)$

The denominators of the given rational numbers are 36 and 12.
L.C.M. of 36 and 12 is 36.

(vi)
$\frac{1}{-9}+\left(\frac{4}{-27}\right)$

(vii)

(viii)$\frac{27}{-4}+\left(\frac{-15}{8}\right)$

#### Answer:

(ii)
$\begin{array}{l}\frac{\begin{array}{l}\\ -12\end{array}}{7}+\frac{3}{7}+\frac{-2}{7}\\ =\frac{\left(-12\right)}{7}+\frac{3}{7}+\frac{\left(-2\right)}{7}\\ =\frac{-12+3-2}{7}\\ =\frac{-14+3}{7}\\ =\frac{-11}{7}\end{array}$

#### Answer:

$\left(v\right)\phantom{\rule{0ex}{0ex}}2+\frac{-1}{2}+\frac{-3}{4}$
2+12+3

#### Answer:

(i) Additive inverse of 5 is −5.

(ii) Additive inverse of −9 is 9.

(iii) Additive inverse of .

(iv) Additive inverse of .

(v) Additive inverse of $\frac{15}{-4}=\frac{15×\left(-1\right)}{\left(-4\right)×\left(-1\right)}$

(vi) Additive inverse of $\frac{-18}{-13}=\frac{-18×\left(-1\right)}{\left(-13\right)×\left(-1\right)}$

(vii) Additive inverse of​ 0 is 0.

(viii) Additive inverse of   $\frac{1}{-6}=\frac{1×\left(-1\right)}{\left(-6\right)×\left(-1\right)}$

(vi)

.

#### Answer:

[L.C.M. of 8 and 4 is 8.]

#### Answer:

Let the required number be x.

#### Answer:

Let the number that is to be added be x.

#### Answer:

Let the number that is to be added be x.

#### Answer:

Let the number that is to be added be x.

#### Answer:

$\begin{array}{l}\left(\mathrm{vi}\right)\frac{\overline{)2}{\overline{)5}}^{5}}{-{\overline{)9}}_{3}}×\frac{{\overline{)3}}^{1}}{-\overline{)1}{\overline{)0}}_{2}}=\frac{5}{3}×\frac{1}{2}=\frac{5}{6}\\ \left(\mathrm{vii}\right)\frac{-{\overline{)7}}^{1}}{\overline{)1}{\overline{)0}}_{1}}×\frac{-\overline{)4}{\overline{)0}}^{4}}{\overline{)2}{\overline{)1}}_{3}}=\frac{4}{3}\\ \left(\mathrm{viii}\right)\frac{-\overline{)3}{\overline{)6}}^{12}}{{\overline{)5}}_{1}}×\frac{\overline{)2}{\overline{)0}}^{4}}{-{\overline{)3}}_{1}}=12×4=48\\ \left(\mathrm{ix}\right)\frac{-\overline{)1}{\overline{)3}}^{1}}{\overline{)1}{\overline{)5}}_{3}}×\frac{-\overline{)2}{\overline{)5}}^{5}}{\overline{)2}{\overline{)6}}_{2}}=\frac{-1}{3}×\frac{-5}{2}=\frac{5}{6}\end{array}$

#### Answer:

$\begin{array}{l}\text{(i)}\\ \frac{7}{\overline{)2}{\overline{)4}}_{1}}×\left(-\overline{)4}{\overline{)8}}^{2}\right)\\ =7×\left(-2\right)\\ =-14\\ \left(ii\right)\\ \frac{-19}{\overline{)3}{\overline{)6}}_{9}}×\overline{)1}{\overline{)6}}^{4}\\ =\frac{-19}{9}×4\\ =\frac{-76}{9}\\ \\ \left(iii\right)\\ \frac{-{\overline{)3}}^{1}}{{\overline{)4}}_{1}}×\frac{{\overline{)4}}^{1}}{{\overline{)3}}_{1}}\\ =-1\\ \\ \left(iv\right)\\ -13×\frac{17}{26}\\ =\frac{-\overline{)1}{\overline{)3}}^{1}×17}{\overline{)2}{\overline{)6}}^{2}}\\ =\frac{-17}{2}\\ \\ \left(v\right)\\ \frac{-13}{{\overline{)5}}_{1}}×\left(-\overline{)1}{\overline{)0}}^{2}\right)\\ =26\\ \\ \left(vi\right)\\ \frac{\left(-{\overline{)9}}^{1}\right)}{\overline{)1}{\overline{)6}}_{1}}×\frac{\left(-\overline{)6}{\overline{)4}}^{4}\right)}{\overline{)2}{\overline{)7}}_{3}}\\ =\frac{4}{3}\\ \end{array}$

#### Answer:

$\begin{array}{l}\text{(i}\right)\frac{4}{9}÷\left(\frac{-5}{12}\right)\\ =\frac{4}{{\overline{)9}}_{3}}×\frac{\overline{)1}{\overline{)2}}^{4}}{\left(-5\right)}\\ =\frac{4×4}{3×\left(-5\right)}\\ =\frac{-16}{15}\\ \text{(ii}\right)-8÷\left(\frac{-5}{16}\right)\\ =-8×\frac{-16}{5}\\ =\frac{128}{5}\\ \text{(iii}\right)\frac{-12}{7}÷\left(-18\right)\\ =\frac{-\overline{)1}{\overline{)2}}^{2}}{7}×\left(\frac{-1}{\overline{)1}{\overline{)8}}_{3}}\right)\\ =\frac{2}{21}\\ \text{(iv}\right)\frac{-1}{10}÷\left(\frac{-8}{5}\right)\\ =\frac{-1}{\overline{)1}{\overline{)0}}^{2}}×\left(\frac{\overline{)5}}{-8}\right)\\ =\frac{-1}{2}×\frac{1}{\left(-8\right)}\\ =\frac{-1}{-16}\\ =\frac{1}{16}\\ \text{(v}\right)\frac{-16}{35}÷\left(\frac{-15}{14}\right)\\ =\frac{-16}{\overline{)3}{\overline{)5}}_{5}}×\frac{\overline{)1}{\overline{)4}}^{2}}{\left(-15\right)}\\ =\frac{-32}{-75}\\ =\frac{32}{75}\\ \text{(vi}\right)\left(\frac{-65}{14}\right)÷\left(\frac{13}{-7}\right)\\ =\left(\frac{-\overline{)6}{\overline{)5}}^{5}}{\overline{)1}{\overline{)4}}_{2}}\right)×\frac{\left(-\overline{)7}\right)}{\overline{)1}\overline{)3}}\\ =\left(\frac{-5}{2}\right)×\left(\frac{-1}{1}\right)\\ =\frac{5}{2}\end{array}$

#### Answer:

#### Answer:

The correct option is (b).

#### Answer:

#### Answer:

Hence , the other number is $\frac{103}{72}$

#### Answer:

#### Answer:

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