RS Aggarwal 2019 Solutions for Class 7 Math Chapter 4 - Rational Numbers

Question 1:

Question 2:

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).   

Question 3:

$\begin{array}{l}\text{i) }\frac{5}{-8}\text{ is a rational number because it is in the form of }\frac{p}{q}\text{, where }\mathit{\text{p}}\text{ and q are integers and}\mathit{\text{ q}}\text{≠0. }\\ \text{ii)}\frac{-6}{11}\text{ is a rational number because it is in the form of }\frac{p}{q}\text{, where }\mathit{\text{p}}\text{ and q are integers and}\mathit{\text{ q}}\text{≠0. }\\ \text{iii)}\frac{-13}{15}\text{is a rational number because it is in the form of }\frac{p}{q}, w\text{here p and q are integers and}\mathit{\text{ q}}\text{≠0. }\\ \text{iv)}\frac{-8}{-12}\text{is a rational number because it is in the form of }\frac{p}{q}, w\text{here p and q are integers and}\mathit{\text{ q}}\ne 0.\\ \text{v) 6 is a rational number because it is in the form of }\frac{p}{q}, w\text{here p and q are integers and }\mathit{\text{q}}\ne 0.\\ \text{vi) -3 is a rational number because it is in the form of }\frac{p}{q}, w\text{here}\mathit{\text{ p}}\text{ and }\mathit{\text{q}}\text{ are integers and }\mathit{\text{q}}\ne 0.\\ \text{vii) 0 = }\frac{0}{1}\text{is a rational number because it is in the form of }\frac{p}{q}, w\text{here}\mathit{\text{ p}}\text{ and }\mathit{\text{q}}\text{ are integers and }\mathit{\text{q}}\ne 0.\\ \text{viii)}\frac{0}{1}\text{ is a rational number because it is in the form of }\frac{p}{q}, w\text{here}\mathit{\text{ p}}\text{ and }\mathit{\text{q}}\text{ are integers and }\mathit{\text{q}}\ne 0.\\ \text{ix)}\frac{1}{0}\text{is not a rational number because, here,}\mathit{\text{ q}}\text{ = 0.}\\ \text{x)}\frac{0}{0}\text{ is not a rational number because, here, }\mathit{\text{q}}\text{= 0}\text{.}\end{array}\begin{array}{l}\\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}$

Question 4:

(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

Question 5:

(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

Question 6:

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

(iv) $\frac{37}{53}$
(vi) 8 because 8 can be written as $\frac{8}{1}, \mathrm{where} 1\ne 0$.

0 is neither positive nor negative.

Question 7:

Negative rational numbers:

(iii) $\frac{-5}{7}$

(iv) $\frac{4}{-9}$

(v)  -6

(vi) $\frac{1}{-2}$

Question 8:

(i) Following are the four rational numbers that are equivalent to $\frac{6}{11}$.
$\begin{array}{l}\frac{6\times 2}{11\times 2},\frac{6\times 3}{11\times 3},\frac{6\times 4}{11\times 4}\text{ and }\frac{6\times 5}{11\times 5}\\ \\ \mathrm{i}.\mathrm{e}. \frac{12}{22},\frac{18}{33},\frac{24}{44}\text{ and }\frac{30}{55}\end{array}$

(ii) Following are the four rational numbers that are equivalent to $\frac{-3}{8}$.
$\frac{-3\times 2}{8\times 2}$,$\frac{-3\times 3}{8\times 3}$, $\frac{-3\times 4}{8\times 4}$ and $\frac{-3\times 5}{8\times 5}$
 
  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 $\frac{7}{-15}$.
$\begin{array}{l}\frac{\text{7}\times 2}{-15\times 2},\frac{\text{7}\times 3}{-15\times 3},\frac{\text{7}\times 4}{-15\times 4}\text{ and }\frac{\text{7}\times 5}{-15\times 5}\\ \\ \text{i}\text{.e}\text{. }\frac{14}{-30}\text{, }\frac{21}{-45},\frac{28}{-60}\text{ and }\frac{35}{-75}\end{array}$

(iv) Following are the four rational numbers that are equivalent to 8, i.e. $\frac{8}{1}$.
$\begin{array}{l}\frac{8\times 2}{1\times 2},\frac{8\times 3}{1\times 3},\frac{8\times 4}{1\times 4}\text{ and }\frac{8\times 5}{1\times 5}\\ \\ \mathrm{i}.\mathrm{e}. \frac{16}{2},\frac{24}{3},\frac{32}{4}\text{ and }\frac{40}{5}\end{array}$

(v) Following are the four rational numbers that are equivalent to ­­-1, i.e. $\frac{1}{1}$.
$\begin{array}{l}\frac{1\times 2}{1\times 2},\frac{1\times 3}{1\times 3},\frac{1\times 4}{1\times 4}\text{ and }\frac{1\times 5}{1\times 5}\\ \\ \mathrm{i}.\mathrm{e}. \frac{2}{2},\frac{3}{3},\frac{4}{4}\text{ and }\frac{5}{5}\end{array}$
(vi) Following are the four rational numbers that are equivalent to ­­-1, i.e. $\frac{-1}{1}$.
$\begin{array}{l}\frac{-1\times 2}{1\times 2},\frac{-1\times 3}{1\times 3},\frac{-1\times 4}{1\times 4}\text{ and }\frac{-1\times 5}{1\times 5}\\ \\ \mathrm{i}.\mathrm{e}. \frac{-2}{2},\frac{-3}{3},\frac{-4}{4}\text{ and }\frac{-5}{5}\end{array}$
 

Question 9:

(i) $\frac{12\times (-1)}{(-17)\times (-1)}=\frac{-12}{17}$

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

(iii) $\frac{-8}{-19}=\frac{-8\times (-1)}{(-19)\times (-1)}=\frac{8}{19}$

(iv) $\frac{11\times (-1)}{-6\times (-1)}=\frac{-11}{6}$

Question 10:

(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:

$\begin{array}{l}\frac{5\times 3}{8\times 3}=\frac{15}{24}\\ \\ \frac{5}{8}=\frac{15}{24}\end{array}$

(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:

$\begin{array}{l}\frac{5\times (-2)}{8\times (-2)}=\frac{-10}{-16}\\ \frac{5}{8}=\frac{-10}{-16}\end{array}$

Question 11:

(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:

$\frac{4\times 3}{7\times 3}$= $\frac{12}{21}$

$\frac{4\times 3}{7\times 3}$= $\frac{4}{7}$ 

(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:

$\begin{array}{l}\frac{4\times (-5)}{7\times (-5)}=\frac{-20}{-35}\\ \frac{4}{7}=\frac{-20}{-35}\end{array}$

Question 12:

(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:

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

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

 Multiplying its numerator and denominator by -5:

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

Question 13:

(i) Denominator of$\frac{-8}{11}$ is 11.
​Clearly, 11×2= ​22
Multiplying both the numerator and the denominator by 2:

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

(ii) Denominator of$\frac{-8}{11}$  is 11.
Clearly, 11×5=55
Multiplying both the numerator and the denominator by 5:

$\begin{array}{l}\frac{-8\times 5}{11\times 5}=\frac{-40}{55}\\ \\ \frac{-8}{11}=\frac{-40}{55}\end{array}$

Question 14:

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

Multiplying both the numerator and the denominator by 4:
$\frac{14\times 4}{-5\times 4}$=$\frac{56}{-20}$


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

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

$\frac{14\times (-5)}{(-5)\times (-5)}$=$\frac{-70}{25}$

$\frac{14}{-5}$=​$\frac{-70}{25}$

Question 15:

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


(ii) Denominator of $\frac{13}{-8}$ is −8.
Clearly, (−8)×(−4)= 32
Multiplying both the numerator and the denominator by −4:

$\frac{13\times (-4)}{-8\times (-4)}=\frac{-52}{32}$

 $\frac{13}{-8}$=$\frac{-52}{32}$

Question 16:

(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:

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


$\frac{-36}{24}$= $\frac{6}{-4}$

Question 17:

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

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

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

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

 $\frac{84}{-147}=\frac{28}{-49}$

Question 18:

(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, $\frac{35}{49} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{5}{7}$ in the standard form.

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

$\frac{8\times (-1)}{-36\times (-1)}=\frac{-8}{36}$


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, $\frac{8}{-36} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{-2}{9}$ 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, $\frac{-27}{45} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{-3}{5}$ in the standard form.

$$\begin{gathered} \left(\mathrm{iv}\right) \frac{-14}{-49} \\ \mathrm{The} \mathrm{denominator} \mathrm{is} \mathrm{negative}. \\ \mathrm{Multiplying} \mathrm{its} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{by} -1: \\ \frac{-14\times (-1)}{-49\times (-1)}=\frac{14}{49} \end{gathered}$$


H.C.F. of 14 and 49 is 7.
Dividing both the numerator and the denominator by 7.
$$\begin{gathered} \frac{14÷7}{49÷7}=\frac{2}{7} \\ \mathrm{Hence}, \frac{-14}{-49 } \mathrm{is} \mathrm{equal} \mathrm{to} \frac{2}{7} \mathrm{in} \mathrm{the} \mathrm{standard} \mathrm{form}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right) \frac{91}{-78} \\ \mathrm{The} \mathrm{denominator} \mathrm{is} \mathrm{negative}. \\ M\mathrm{ultiplying} \mathrm{its} \mathrm{denominator} \mathrm{and} \mathrm{denominator} \mathrm{by} -1: \\ \frac{91\times (-1)}{-78\times (-1)}=\frac{-91}{78} \end{gathered}$$


H.C.F. of 91 and 78 is 13.
Dividing both the numerator and the denominator by 13:
$$\begin{gathered} \frac{-91÷13}{78÷13}=\frac{-7}{6} \\ \mathrm{Hence}, \frac{91}{-78} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{-7}{6} \mathrm{in} \mathrm{the} \mathrm{standard} \mathrm{form}. \end{gathered}$$

 $\left(\mathrm{vi}\right) \frac{-68}{119}$


H.C.F. of 68 and 119 is 17.
Dividing both the numerator and the denominator by 17:
$$\begin{gathered} \frac{-68÷17}{119÷17}=\frac{-4}{7} \\ \mathrm{Hence}, \frac{-68}{119} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{-4}{7} \mathrm{in} \mathrm{the} \mathrm{standard} \mathrm{form}. \end{gathered}$$

$\left(vii\right) \frac{-87}{116}$


H.C.F. of 87 and 116 is 29.
Dividing both the numerator and the denominator by 29:
$$\begin{gathered} \frac{-87÷29}{116÷29}=\frac{-3}{4} \\ \mathrm{Hence}, \frac{-87}{116} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{-3}{4}\mathrm{in} \mathrm{the} \mathrm{standard} \mathrm{form}. \end{gathered}$$

$\left(\mathrm{viii}\right) \frac{299}{-161}$
The denominator is negative.
Multiplying both the numerator and denominator by -1:

$\frac{299\times (-1)}{-161\times (-1) }=\frac{-299}{161}$


H.C.F. of 299 and 161 is 23.
Dividing both the numerator and the denominator by 23:
$$\begin{gathered} \frac{-299÷23}{161÷23}=\frac{-13}{7} \\ \mathrm{Hence}, \frac{299}{-161} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{-13}{7}\mathrm{in} \mathrm{the} \mathrm{standard} \mathrm{form}. \end{gathered}$$

Question 19:

(i)

 $$\begin{gathered} \frac{-9\times 4}{5\times 4}=\frac{-36}{20} \\ \frac{-9\times (-3)}{5\times (-3)}=\frac{27}{-15} \\ \frac{-9\times 5}{5\times 5}=\frac{-45}{25} \\ \therefore \frac{-9}{5}=\frac{-36}{20}=\frac{27}{-15}=\frac{-45}{25} \end{gathered}$$


(ii) 
$$\begin{gathered} \frac{-6\times 3}{11\times 3}=\frac{-18}{33} \\ \frac{-6\times 4}{11\times 4}=\frac{-24}{44} \\ \therefore \frac{-6}{11}=\frac{-18}{33}=\frac{-24}{44} \end{gathered}$$

Question 20:

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

And 7×39=273

$\begin{array}{l}(-\text{13})\times (-\text{21})=\text{7}\times \text{39}\\ \text{or }\frac{-\text{13}}{7}=\frac{\text{39}}{-21}\\ \text{Hence, }\frac{-\text{13}}{7}\mathrm{and} \frac{\text{39}}{21}\text{ are equivalent rational numbers. }\end{array}$

(ii) $\frac{3}{-8}, \frac{-6}{16}$
We have:

3×16=48

And (−8) ×(−6) =48

∴ 3×16 =(−8)×(−6)
$\frac{3}{-8} =\frac{-6}{16}$

(iii)$\frac{9}{4}, \frac{-36}{-16}$
 
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)$\frac{7}{15}, \frac{-28}{60}$ 

We have:

7×60 =420
And 15×(-28)= −420

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

(v) $\frac{3}{12}, \frac{-1}{4}$

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.           

Question 21:

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

=> −x =5×8
=> x= −40
 
(ii)$\frac{7}{-3}=\frac{x}{6}$
=> (−3)x=7×6

=> x=$\frac{(7\times 6)}{(-3)}$
=>  x=−14
 
 
(iii) $\frac{3}{5}=\frac{x}{-25}$
=>    5x=3×(−25)

=>   x=$\frac{\mathrm{3\times (-25)}}{5}$
=>x  = (−15)

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

=> 13x=6×(−65)

=>  x=$\frac{\mathrm{6\times }(-65)}{13}$

=>  x= 6×(−​5)

=>  x = −​30

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

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

Question 22:

(i)$\frac{8}{-12} \mathrm{and}\frac{-10}{15}$


8×15 =120
And ( −10)×(−12)=120

8×15 =(−10) ×(−12)

$\therefore \frac{8}{-12} =\frac{-10}{15}$

Therefore, the rational numbers are equal.

ii)$\frac{-3}{9}, \frac{7}{-21}$

(−3)×(−21) =63
And 7× 9=63

∴ (−3)×(−21) =7×9

$\frac{-3}{9}= \frac{7}{-21}$

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.

Question 1:

(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.

Question 2:

(i)


(ii)
 


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



(iv) 

$\frac{22}{7}$ can be written as $3 + \frac{1}{7}$. 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. 

Question 3:

$\begin{array}{l}\\ \left(\text{i}\right)\frac{5}{6}. \mathrm{This} \mathrm{is} \mathrm{because} 0\text{ can be written as }\frac{0}{6}\text{ and }\frac{0}{6}<\frac{5}{6}.\\ \left(\text{ii}\right)\frac{-3}{5} <0. \mathrm{This} \mathrm{is} \mathrm{because}0\text{ can be written as }\frac{0}{5}\text{ and }-3<0.\\ \left(\text{iii}\right)\frac{5}{8} >\frac{3}{8}. \mathrm{This} \mathrm{is} \mathrm{because} \text{ 5 }>\text{3}.\\ \left(\text{iv}\right)\frac{7}{9} >\frac{5}{9}. \mathrm{This} \mathrm{is} \mathrm{because} \text{7} > \text{5.}\\ \left(\text{v}\right)\frac{-6}{11} <\frac{-5}{11}. \mathrm{This} \mathrm{is} \mathrm{because} -\text{6} < -\text{5.}\\ \left(\text{vi}\right)\frac{-15}{4} > \frac{-17}{4},-\text{15} > -\text{17}\end{array}$