RS Aggarwal 2019,2020 solution class 7 chapter 4 Rational Numbers Test Paper 4

Test Paper 4

Page-82

Question 1:

Express each of the following rational numbers in standard form:

(i) $\frac{-209}{247}$
(ii) $\frac{-46}{115}$
(iii) $\frac{84}{-147}$

Answer 1:

$\begin{array}{l} \left(i\right) \frac{-209}{247}\\ \mathrm{H}.\mathrm{C}.\mathrm{F}. of 209 and 247 is 19.\\ \begin{array}{l}\mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{the} \mathrm{denominator} \mathrm{by} 19\\ \begin{array}{l}=\frac{-209 ÷ 19}{247 ÷ 19}=\frac{-11}{13}\\ \end{array}\end{array}\end{array}$                                         


$$\begin{gathered} \begin{array}{l}\left(ii\right) \frac{-46}{115} \\ \begin{array}{l}\mathrm{H}.\mathrm{C}.\mathrm{F}. of 46 \mathrm{and} 115 \mathrm{is} 23\\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{the} \mathrm{denominator} \mathrm{by} 23\end{array} \end{array} \\ =\frac{-46 ÷ 23}{115 ÷ 23}=\frac{-2}{5} \end{gathered}$$                                                      

$$\begin{gathered} \left(\mathrm{iii}\right)\frac{84}{-147} \\ \mathrm{Converting} \mathrm{the} \mathrm{number} \mathrm{to} \mathrm{a} \mathrm{positive} \mathrm{denominator}: \\ \frac{84 \times (-1)}{-147 \times (-1)}=\frac{-84}{147} \\ \mathrm{H}.\mathrm{C}.\mathrm{F}. of 84 \mathrm{and} 147 is 21 \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{the} \mathrm{denominator} \mathrm{by} 21 \\ =\frac{-84 ÷ \left(21\right)}{147 ÷ \left(21\right)}=\frac{-4}{7} \end{gathered}$$                                                      

Question 2:

List five rational numbers between −2 and −1.

Answer 2:

$\begin{array}{l}-2=\frac{-2\times 6}{1\times 6}=\frac{-12}{6}\\ \begin{array}{l}-1=\frac{-1\times 6}{1\times 6}=\frac{-6}{6}\\ \begin{array}{l} \mathrm{The} \mathrm{integers} \mathrm{between} -12 \mathrm{and} -6 \mathrm{are} -11,-10,-9,-8,-7.\\ \begin{array}{l}\\ \text{Hence, five rational numbers between }-\text{2 and }-\mathrm{1 are} \frac{-11}{6},\frac{-10}{6},\frac{-9}{6},\frac{-8}{6},\frac{-7}{6}.\end{array}\end{array}\end{array}\end{array}$

Question 3:

The sum of two rational numbers is −4. If one of them is $\frac{-11}{6}$, find the other.

Answer 3:

$$\begin{gathered} \begin{array}{l}\mathrm{Let} \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{be} x.\\ x + \left(\frac{-11}{6}\right)=-4\\ =>x=(-\text{4}) - \left(\frac{-11}{6}\right)\\ \begin{array}{l} =-4 + \frac{11}{6}\\ =\frac{-24 + 11}{6}\\ =\frac{-13}{6}\end{array}\end{array} \\ \mathrm{Hence}, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{-13}{6}. \end{gathered}$$

Question 4:

What should be added to $\frac{-7}{8}$ to get $\frac{5}{9}$?

Answer 4:

$\begin{array}{l}\mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{be} x.\\ x + \left(\frac{-7}{8}\right)=\frac{5}{9}\\ \text{=>x =}\frac{5}{9}- \left(\frac{-7}{8}\right)\\ \begin{array}{l}=\frac{5}{9} + \frac{7}{8}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 9 \mathrm{and} 8 \mathrm{is} 72.\\ \begin{array}{l}=\frac{40 + 63}{72}\\ =\frac{103}{72} \end{array}\end{array}\end{array}$

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

Question 5:

A car is moving at an average speed of $56\frac{3}{5}$ km per hour. How much distance will it cover in $7\frac{1}{2}$ hours?

Answer 5:

$$\begin{gathered} \begin{array}{l}\text{Average speed }=\text{56}\frac{3}{5}\text{ km/h }\\ =\frac{283}{5} \text{km/h}\\ \text{Time }=7\frac{1}{2} \mathrm{h}\\ =\frac{15}{2} \text{h}\\ \text{Distance }= \text{Speed }\times \text{Time}\\ \text{ =}\frac{283}{\sout{5}}\times \frac{\sout{1}{\sout{5}}^3}{2}\\ =\frac{849}{2}\\ \text{ = 424}\frac{1}{2}\text{ km}\end{array} \\ \mathrm{The} \mathrm{car} \mathrm{will} \mathrm{cover} 424\frac{1}{2} \text{km in }7\frac{1}{2} \mathrm{h}. \end{gathered}$$

 

Question 6:

By what number should $-4\frac{3}{8}$ be divided to obtain $-3\frac{1}{2}$?

Answer 6:

$\begin{array}{l}\mathrm{Let} \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{be} x.\\ -\text{4}\frac{3}{8}÷ x=-\text{3}\frac{1}{2}\\ \begin{array}{l}\text{=>x }=\frac{-35}{8} \times \frac{2}{-7}\\ \begin{array}{l}\text{ = }\frac{5}{4}\\ \text{Hence, the required number is }\frac{5}{4}\text{ . }\end{array}\\ \end{array}\\ \end{array}$

Question 7:

How many pieces, each of length $3\frac{3}{4}$ m, can be cut from a rope of length 45 m?

Answer 7:

$\begin{array}{l}\mathrm{Length} \mathrm{of} \mathrm{rope} = 45 \mathrm{m}\\ \text{Number of pieces }=45 ÷ 3\frac{3}{4}\\ \begin{array}{l}\text{ =45 }÷ \frac{15}{4}\\ =\text{45 }\times \frac{4}{15}\end{array}\\ =\text{12}\end{array}$

Question 8:

Find the cost of $3\frac{1}{3}$ m of cloth at Rs $121\frac{1}{2}$ per metre.

Answer 8:

$\begin{array}{l}\text{Cost of 1 m cloth = Rs 121}\frac{1}{2}\\ \text{ Cost of 3}\frac{1}{3}\text{ m cloth = 121}\frac{1}{2}\times 3\frac{1}{3}\\ \text{ =}\frac{{\sout{243}}^{81}}{\sout{2}}\times \frac{\sout{1}{\sout{0}}^5}{\sout{3}}\\ \text{ =Rs 405}\\ \end{array}$

Question 9:

Mark (✓) against the correct answer
$\frac{55}{-66}$ in standard form is

(a) $\frac{5}{-6}$
(b) $\frac{-5}{6}$
(c) $\frac{-55}{66}$
(d) none of these

Answer 9:

$$\begin{gathered} \begin{array}{l}\left(\mathrm{b}\right) \frac{-5}{6}\\ \begin{array}{l} \frac{55}{-66} = \frac{55 \times (-1)}{-66 \times (-1)}=\frac{-55}{66}\\ \end{array}\end{array} \\ \mathrm{H}.\mathrm{C}.\mathrm{F}. \mathrm{of} 55 \mathrm{and} 66 \mathrm{is} 11. \\ \frac{-55 ÷ 11}{66 ÷11}=\frac{-5}{6} \\ \frac{-5}{6} \mathrm{is} \mathrm{the} \mathrm{standard} \mathrm{form}. \end{gathered}$$                                                                

Question 10:

Mark (✓) against the correct answer
What should be subtracted from $\frac{-2}{3}$ to get $\frac{3}{4}$?
(a) $\frac{-17}{12}$
(b) $\frac{17}{12}$
(c) $\frac{-12}{17}$
(d) $\frac{12}{17}$

Answer 10:

$\begin{array}{l}\left(\text{a}\right)\frac{-17}{12}\\ \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{to} \mathrm{be} \mathrm{subtracted} \mathrm{be} x.\\ \frac{-2}{3} - x =\frac{3}{4}\\ \begin{array}{l}\text{=> x = }\frac{-2}{3}- \frac{3}{4}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 3 \mathrm{and} 4 \mathrm{is} 12.\\ \begin{array}{l}=\frac{-8 - 9}{12}\\ =\frac{-17}{12}\end{array}\end{array}\end{array}$

Question 11:

Mark (✓) against the correct answer
The product of two numbers is $\frac{-1}{6}$. If one of them is $\frac{-5}{8}$, the other number is
(a) $\frac{-4}{15}$
(b) $\frac{4}{15}$
(c) $\frac{15}{4}$
(d) $\frac{-15}{4}$

Answer 11:

$$\begin{gathered} \begin{array}{l}\left(\text{b}\right) \frac{4}{15}\\ \mathrm{Let} \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{be} x.\\ \frac{-5}{8} \times x=\frac{-1}{6}\\ \begin{array}{l}\text{ =>x }=\frac{-1}{6}÷\left(\frac{-5}{8}\right)\\ =\frac{-1}{6}\times \left(\frac{8}{-5}\right)\\ =\frac{-4}{-15}\\ =\frac{4}{15}\end{array}\\ \end{array} \\ \mathrm{Hence} , \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{is} \frac{4}{15}. \end{gathered}$$

Question 12:

Mark (✓) against the correct answer
The multiplicative inverse of $\frac{-3}{4}$ is
(a) $\frac{3}{4}$
(b) $\frac{4}{3}$
(c) $\frac{-4}{3}$
(d) none of these

Answer 12:

$\begin{array}{l}\left(\text{c}\right) \frac{-4}{3}\\ \text{Multiplicative iInverse of }\frac{-3}{4}\text{ is }\frac{-4}{3}\text{.}\end{array}$

Question 13:

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 13:

$\begin{array}{l}\left(\text{b}\right)\frac{-5}{14}\\ \text{Missing number}=(-\text{1}) - \left(\frac{-9}{14}\right)\\ =(-1) + \frac{9}{14}\\ =\frac{-14 + 9}{14}\\ \text{ =}\frac{-5}{14}\end{array}$

Question 14:

Mark (✓) against the correct answer
$78\frac{3}{4}÷2\frac{1}{2}=?$
(a) $31\frac{1}{2}$
(b) $39\frac{3}{8}$
(c) $40\frac{1}{3}$
(d) none of these

Answer 14:

$\begin{array}{l}\left(\text{a}\right)\text{ 31}\frac{1}{2}\\ \text{ 78}\frac{3}{4}÷ 2\frac{1}{2}=\frac{315}{4} ÷ \frac{5}{2}\\ =\frac{{\sout{315}}^{63}}{{\sout{4}}_2}\times \frac{\sout{2}}{\sout{5}}\\ =\frac{63}{2}\\ =31\frac{1}{2}\\ \end{array}$

Question 15:

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

Answer 15:

$\begin{array}{l}\left(\text{a}\right)\frac{-5}{6}\\ \text{ Since L.C.M. of 6 and 12 is 12}\text{.}\\ \therefore \frac{-5\times \text{2}}{6\times 2}=\frac{-10}{12}\\ \frac{-7\times 1}{12\times 1}=\frac{-7}{12}\\ \therefore \frac{-5}{6}<\frac{-7}{12}\end{array}$

Page-83

Question 16:

Fill in the blanks.

(i)$\left(......\right)÷\left(\frac{-7}{5}\right)=\frac{-2}{3}$
(ii) $\left(\frac{-65}{14}\right)÷\left(......\right)=2\frac{1}{2}$
(iii) $\left(\frac{-3}{8}\right)+\left(......\right)=\frac{5}{12}$
(iv) Multiplicative inverse of $-1\frac{3}{4}$is ...... .

Answer 16:

$\begin{array}{l}\\ \left(\text{i}\right)(\dots ?.\mathrm{..})÷\frac{-7}{5}=\frac{-2}{3}\\ (\dots ?\dots )=\frac{-2}{3}\times \frac{-7}{5}\\ =\frac{14}{15}\\ \\ \text{(ii})\frac{-65}{14}÷(\dots ?.\mathrm{..})=2\frac{1}{2}\\ (\dots ?.\mathrm{..})=\frac{-65}{14}÷2\frac{1}{2}\\ \text{ =}\frac{-65}{14}÷\frac{5}{2}\\ \text{ =}\frac{-65}{14}\times \frac{2}{5}\\ \text{ =}\frac{-13}{7}\\ \left(\text{iii}\right)(\dots ?...)=\frac{5}{12}-\left(\frac{-3}{8}\right)\\ =\frac{5}{12}+\frac{3}{8}\\ L.C.M. \mathrm{of} 12 \mathrm{and} 8 \mathrm{is} 24.\\ \begin{array}{l}\text{ =}\frac{10+9}{24}\\ =\frac{19}{24}\end{array}\\ \left(\text{iv}\right)\text{Multiplicative inverse of -1 }\frac{3}{4}\text{, i.e. }\frac{-7}{4}\text{, is }\frac{-4}{7}\text{.}\end{array}$

Question 17:

Write 'T' for true and 'F' for false for each of the following:

(i) $\frac{-15}{-11}$ lies to the left of 0 on the number line.
(ii) $\frac{1}{3} \mathrm{and} \frac{-3}{2}$ lie on opposite side of 0 on the number line.
(iii) $\frac{-8}{13}$ lies to the left of 0 on the number line.
(iv) $\frac{-4}{5}>\frac{-2}{3}$.
(v) $\frac{-3}{5}$ is the largest among $\frac{-3}{7},\frac{-7}{10} \mathrm{and} \frac{-5}{6}$.

Answer 17:

$\begin{array}{l}\left(\text{i}\right)\text{False. This is because }\frac{-15}{-11}=\frac{15}{11}, \text{which lies to the right of 0}\text{.}\\ \left(\text{ii}\right)\text{True }\\ \left(\text{iii}\right)\text{True}\\ \left(\text{iv}\right)\text{False }\\ L.C.M. \mathrm{of} 5 \mathrm{and} 3 \mathrm{is} 15.\\ \begin{array}{l}\text{ We know:}\frac{-4\times 3}{5\times 3}=\frac{-12}{15}\text{ and }\frac{-2\times 5}{5\times 3}=\frac{-10}{15}\\ \therefore \frac{-4}{5}<\frac{-2}{5}\\ \left(\text{v}\right)\text{True}\\ \text{L.C.M. of 5}, \text{1}0 and \text{6 is 3}0.\\ \therefore \frac{-3\times 6}{5\times 6}=\frac{-18}{30},\frac{-7\times 3}{10\times 3}=\frac{-21}{30}\text{ and }\frac{-5\times 5}{6\times 5}=\frac{-25}{30}\end{array}\\ \therefore \frac{-3}{5}\text{ is the largest among the given fractions. }\end{array}$