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#### Page No 60:

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).

#### Page No 60:

(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

#### Page No 60:

(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

#### Page No 60:

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.

#### Page No 60:

Negative rational numbers:

(iii) $\frac{-5}{7}$

(iv) $\frac{4}{-9}$

(v)  -6

(vi) $\frac{1}{-2}$

#### Page No 60:

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

#### Page No 61:

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

#### Page No 61:

(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:

#### Page No 61:

(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:

#### Page No 61:

(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:

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

#### Page No 61:

(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:

#### Page No 61:

(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:

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

$\frac{14}{-5}$=​$\frac{-70}{25}$

#### Page No 61:

(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:

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

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

#### Page No 61:

(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}$

#### Page No 61:

(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:

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

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

#### Page No 61:

(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:

#### Page No 61:

(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)

#### Page No 61:

(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.

#### Page No 61:

(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)

#### Page No 61:

(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.

#### Page No 61:

(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.

#### Page No 66:

(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.

#### Page No 66:

#### Page No 66:

#### Page No 66:

#### Page No 69:

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

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

#### Page No 70:

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

#### Page No 70:

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

#### Page No 70:

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

#### Page No 72:

(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)

.

#### Page No 72:

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

#### Page No 72:

Let the required number be x.

#### Page No 72:

Let the number that is to be added be x.

#### Page No 72:

Let the number that is to be added be x.

#### Page No 72:

Let the number that is to be added be x.

#### Page No 75:

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

#### Page No 75:

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

#### Page No 78:

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

#### Page No 79:

#### Page No 80:

The correct option is (b).

#### Page No 82:

#### Page No 82:

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

#### Page No 82: