RS Aggarwal 2019,2020 solution class 7 chapter 4 Rational Numbers Exercise 4G

Exercise 4G

Page-79

Question 1:

Mark (✓) against the correct answer
$\frac{33}{-55}$ in standard form is

(a) $\frac{3}{-5}$
(b) $\frac{-3}{5}$
(c) $\frac{-33}{55}$
(d) none of these

Answer 1:

$\begin{array}{l}\\ \left(\text{b}\right) \frac{-3}{5}\end{array}$                                                                                 


$\begin{array}{l}\mathrm{H}.\mathrm{C}.\mathrm{F}. \mathrm{of} 33 \mathrm{and} 55 \mathrm{is} 11\\ \frac{-33 ÷ 11}{55 ÷ 11}=\frac{-3}{5}\end{array}$        

Question 2:

Mark (✓) against the correct answer
$\frac{-102}{119}$ in standard form is

(a) $\frac{-4}{7}$
(b) $\frac{-6}{7}$
(c) $\frac{-6}{17}$
(d) none of these

Answer 2:

$\begin{array}{l}\\ \left(\text{b}\right) \frac{-6}{7}\end{array}$



$\begin{array}{l}\mathrm{H}.\mathrm{C}.\mathrm{F}. \mathrm{of} 102 \mathrm{and} 119 \mathrm{is} 17\\ \begin{array}{l}\frac{-102 ÷ 17}{119 ÷ 17}=\frac{-6}{7}\\ \begin{array}{l}\text{ The standard form of }\frac{-102}{119}\text{ is }\frac{-6}{7}\\ \end{array}\end{array}\end{array}$

Question 3:

Mark (✓) against the correct answer
If $\frac{x}{6}=\frac{7}{-3},$  then the value of x is

(a) −14
(b) 14
(c) 21
(d) −21

Answer 3:

$\begin{array}{l}\\ \mathrm{The} \mathrm{c}\text{orrect option is }\left(\text{a}\right).\\ \text{ The value of }x \text{ is }-\text{14}\text{.}\\ [x=\frac{7\times 6}{-3}=\frac{\sout{4}{\sout{2}}^{14}}{-{\sout{3}}_1}=-14]\\ \end{array}$

Question 4:

What should be added to $\frac{-5}{9}$ to get 1?
(a) $\frac{4}{9}$
(b) $\frac{-4}{9}$
(c) $\frac{14}{9}$
(d) $\frac{-14}{9}$

Answer 4:

$$\begin{gathered} \begin{array}{l}\\ \mathrm{The} \mathrm{c}\text{orrect option is }\left(\text{c}\right).\\ \frac{14}{9}\text{ should be added to }\frac{-5}{9}\text{ to get 1.}\\ \mathrm{Let} \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{be} x.\end{array} \\ x + \left(\frac{-5}{9}\right)=1 \\ x = 1 - \frac{(-5)}{9}=\frac{9 + 5}{9}=\frac{14}{9} \end{gathered}$$

Question 5:

What should be subtracted from $\frac{-3}{4}$ to get $\frac{5}{6}$?
(a) $\frac{19}{12}$
(b) $\frac{-19}{12}$
(c) $\frac{1}{12}$
(d) $\frac{-1}{12}$

Answer 5:

$$\begin{gathered} \begin{array}{l}\\ \mathrm{The} \mathrm{c}\text{orrect option is }\left(\text{b}\right).\\ \text{Let the number that is to be subtracted be }x.\\ \\ \frac{-3}{4}-x=\frac{5}{6}\\ =>-x=\frac{5}{6} - \left(\frac{-3}{4}\right)\\ =>-x=\frac{5}{6} + \frac{3}{4}\\ =>-x=\frac{(5 \times 2) + (3 \times 3)}{12}\\ =>x=-\frac{19}{12}\end{array} \\ \mathrm{Hence}, -\frac{19}{12}\text{ should be subtracted from} \frac{-3}{4}\text{ to get }\frac{5}{6}\text{.} \end{gathered}$$

Question 6:

Mark (✓) against the correct answer
Which is smaller out of $\frac{5}{-6} \mathrm{and} \frac{-7}{12}$?
(a) $\frac{5}{-6}$
(b) $\frac{-7}{12}$
(c) cannot be compared

Answer 6:

$\begin{array}{l}\\ \mathrm{The} \mathrm{c}\text{orrect option is }\left(\text{a}\right).\\ \frac{5 \times -1}{-6 \times -1}=\frac{-5}{6}\\ \\ \text{L.C.M.}\text{ of 6 and 12 is 12}\text{.}\\ \begin{array}{l}\therefore \frac{-5\times 2}{6\times 2}=\frac{-10}{12}\text{ and }\frac{-7\times 1}{12\times 1}=\frac{-7}{12}\\ \mathrm{Hence}, \frac{5}{-6}\text{ is smaller than }\frac{-7}{12}\text{.}\end{array}\end{array}$

Question 7:

Mark (✓) against the correct answer
Which is larger out of $\frac{2}{-3}$ and $\frac{-4}{5}$?
(a) $\frac{2}{-3}$
(b) $\frac{-4}{5}$
(c) cannot be compared

Answer 7:

$\begin{array}{l}\\ \mathrm{The} \mathrm{c}\text{orrect option is }\left(\text{a}\right).\\ \frac{2\times -1}{-3\times -1}=\frac{-2}{3}\\ \begin{array}{l}\text{L.C.M. }\text{}\text{of 3 and 5 is 15}\text{.}\\ \therefore \frac{-2\times 5}{3\times 5}=\frac{-10}{15}\text{ and }\frac{-4\times 3}{5\times 3}=\frac{-12}{15}\\ \mathrm{Thus}, \frac{2}{-3}\text{ is greater than} \frac{-4}{5} .\end{array}\\ \end{array}$

Question 8:

Mark (✓) against the correct answer
Reciprocal of −6 is

(a) 6
(b) $\frac{1}{6}$
(c) $\frac{-1}{6}$
(d) none of these

Answer 8:

$\begin{array}{l}\\ \mathrm{The} \mathrm{c}\text{orrect option is }\left(\text{c}\right).\\ \text{Reciprocal of }-\text{6 is }\frac{-1}{6}\text{.}\end{array}$

Question 9:

Mark (✓) against the correct answer
Multiplicative inverse of $\frac{-2}{3}$ is
(a) $\frac{2}{3}$
(b) $\frac{-3}{2}$
(c) $\frac{3}{2}$
(d) none of these

Answer 9:

$\begin{array}{l}\\ \text{ The correct option is }\left(\text{b}\right).\\ \text{Multiplicative inverse of }\frac{-2}{3}\text{ is }\frac{-3}{2}\text{.}\end{array}$

Question 10:

Mark (✓) against the correct answer
$-2\frac{1}{9}-6=?$

(a) $-8\frac{1}{9}$
(b) $8\frac{1}{9}$
(c) $4\frac{1}{9}$
(d) $-4\frac{1}{9}$

Answer 10:

$\begin{array}{l}\\ \text{ The correct option is }\left(\text{a}\right).\\ -2\frac{1}{9} - 6\\ =\frac{-19}{9} - 6 = \frac{-19 - 54}{9}\\ =\frac{-73}{9}=-8\frac{1}{9}\\ \end{array}$

Question 11:

Mark (✓) against the correct answer
$\frac{-6}{13}-\left(\frac{-7}{15}\right)=?$

(a) $\frac{-181}{195}$
(b) $\frac{181}{195}$
(c) $\frac{1}{195}$
(d) $\frac{-1}{195}$

Answer 11:

$\begin{array}{l}\\ \text{ The correct option is }\left(\text{c}\right).\\ \frac{-6}{13}- \frac{[-7]}{15}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 13 \mathrm{and} 15 \mathrm{is} 195.\\ \begin{array}{l}=\frac{-6}{13} - \frac{[-7]}{15}\\ =\frac{-90 + 91}{195}\\ =\frac{1}{195}\end{array}\\ \end{array}$

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Question 12:

Mark (✓) against the correct answer
$-2\frac{1}{3}+4\frac{3}{5}=?$

(a) $-2\frac{4}{15}$
(b)  $2\frac{4}{15}$
(c) $-2\frac{1}{5}$
(d) $2\frac{2}{15}$

Answer 12:

$\begin{array}{l}\\ \text{ The correct option is }\left(\text{b}\right).\\ -2\frac{1}{3} + 4\frac{3}{5}\\ =\frac{-7}{3} + \frac{23}{5}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 5 \mathrm{and} 5 \mathrm{is} 15.\\ \begin{array}{l}=\frac{-35 + 69}{15}\\ =\frac{34}{15}\\ =2\frac{4}{15}\end{array}\\ \end{array}$

Question 13:

Mark (✓) against the correct answer
$\frac{2}{3}-1\frac{5}{7}=?$

(a) $1\frac{1}{21}$
(b)  $-1\frac{1}{21}$
(c) $\frac{5}{21}$
(d) $\frac{-5}{21}$

Answer 13:

$\begin{array}{l}\\ \text{ The correct option is }\left(\text{b}\right).\\ \frac{2}{3}-1\frac{5}{7}\\ =\frac{2}{3} - \frac{12}{7}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 3 \mathrm{and} 7 \mathrm{is} 21.\\ \begin{array}{l}=\frac{14 - 36}{21}\\ =\frac{-22}{21}\\ =-1\frac{1}{21}\end{array}\\ \end{array}$

Question 14:

Which is greater between $\frac{-4}{9}$ and $\frac{-5}{12}$?
(a) $\frac{-4}{9}$
(b) $\frac{-5}{12}$
(c) both are equal

Answer 14:

The correct option is (b).
$$\begin{gathered} \begin{array}{l}\\ \frac{-5}{12}\text{ is greater than }\frac{-4}{9}\text{.}\end{array} \\ \begin{array}{l}\text{L}\text{.C}\text{.M}\text{. of 9 and 12 is 36.}\\ \frac{-5\times 3}{12\times 3}=\frac{-15}{36}\\ \frac{-4\times 4}{12\times 4}=\frac{-16}{36}\\ (-15)>(-16)\\ \frac{-5}{12}>\frac{-4}{9}\end{array} \end{gathered}$$

Question 15:

Mark (✓) against the correct answer
$\frac{-9}{14}+?=-1$

(a) $\frac{5}{14}$
(b) $\frac{-5}{14}$
(c) $\frac{1}{7}$
(d) $\frac{-1}{7}$

Answer 15:

$\begin{array}{l}\\ \text{ The correct option is }\left(\text{b}\right).\\ \frac{-9}{14}+?=-\text{1}\\ \therefore ?=-1-\frac{(-9)}{14}\\ ?=\frac{-14 + 9}{14}\\ ?=\frac{-5}{14}\end{array}$

Question 16:

Mark (✓) against the correct answer
$\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}=?$

(a) $\frac{3}{4}$
(b) $\frac{-3}{4}$
(c) $\frac{-7}{12}$
(d) $\frac{7}{12}$

Answer 16:

$\begin{array}{l}\\ \left(\text{a}\right) \frac{3}{4}\\ \frac{5}{4}-\frac{7}{6}-\frac{(-2)}{3}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 4, 6 \mathrm{and} 3 \mathrm{is} 12.\\ \begin{array}{l}=\frac{15 - 14 + 8}{12}\\ =\frac{23 - 14}{12}\\ =\frac{{\sout{9}}^3}{\sout{1}{\sout{2}}_4}=\frac{3}{4}\end{array}\\ \end{array}$

Question 17:

Mark (✓) against the correct answer
$1÷\frac{1}{2}=?$

(a) $\frac{1}{2}$
(b) 2
(c) $2\frac{1}{2}$
(d) $1\frac{1}{2}$

Answer 17:

$\begin{array}{l}\\ \left(\text{b}\right) 2\\ 1÷\frac{1}{2}\\ =1\times \frac{2}{1}\\ =2\end{array}$

Question 18:

Mark (✓) against the correct answer
$\frac{-3}{14}\times ?=\frac{5}{12}$

(a) $\frac{-35}{18}$
(b) $\frac{35}{18}$
(c) $\frac{7}{3}$
(d) $\frac{-7}{3}$

Answer 18:

$\begin{array}{l}\\ \left(\text{a}\right) \frac{-35}{18}\\ \\ \begin{array}{l}?=\frac{5}{12}÷\frac{(-3)}{14}\\ =\frac{5}{12}\times \frac{14}{(-3)}\\ =\frac{70}{-36}\\ =\frac{35\times -1}{-18\times -1}\\ ?=\frac{-35}{18}\end{array}\\ \\ \\ \end{array}$

Question 19:

Mark (✓) against the correct answer
$0÷\frac{-7}{5}=?$

(a) not defined
(b) $\frac{-5}{7}$
(c) 0
(d) $\frac{5}{7}$

Answer 19:

$\begin{array}{l}\\ \left(\text{c}\right)\text{ 0}\\ \text{ 0 }÷ \frac{-7}{5} = 0\end{array}$

Question 20:

Mark (✓) against the correct answer
$\frac{-3}{8}÷0=?$

(a) $\frac{-3}{8}$
(b) 0
(c) $\frac{-8}{3}$
(d) not defined

Answer 20:

$\begin{array}{l}\\ \left(\text{d}\right) \mathrm{Not} \mathrm{defined}\\ \text{This is because }\frac{-3}{8}÷ 0 \mathrm{is} \text{not defined}\text{.}\end{array}$