Question 12 Marks
Find using distributivity.
(i) $\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times\left(\frac{-3}{9}\right)\right\}$
Answer$ \text {(i) We have, } \begin{array}{c} \left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times\left(\frac{-3}{9}\right)\right\} =\frac{9}{16} \times\left[\frac{4}{12}+\left(\frac{-3}{9}\right)\right] \end{array}\quad$
[by distributivity, taking $\frac{9}{16}$ as common factor]
$\begin{array}{l}=\frac{9}{16} \times\left[\frac{4}{12}-\frac{3}{9}\right] \\ =\frac{9}{16} \times\left[\frac{12-12}{36}\right] \quad[\because \text { LCM of } 12 \text { and } 9=36] \\ =\frac{9}{16} \times \frac{0}{36}=\frac{9}{16} \times 0 \\ =0\quad\left[\because \frac{0}{36}=0\right]\end{array} $
View full question & answer→Question 22 Marks
Find using distributivity.
(i) $\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}$
Answer(i) We have,
$\begin{array}{r}\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\} =\frac{7}{5} \times\left[\frac{-3}{12}+\frac{5}{12}\right]\end{array}\quad$
[by distributivity, taking $\frac{7}{5}$ as common factor]
$=\frac{7}{5} \times\left[\frac{-3+5}{12}\right]=\frac{7}{5} \times \frac{2}{12}=\frac{7}{30}$
View full question & answer→Question 32 Marks
Tell, what property allows you to compute
$ \frac{1}{3} \times\left(6 \times \frac{4}{3}\right) \text { as }\left(\frac{1}{3} \times 6\right) \times \frac{4}{3} ? $
AnswerWe know that for any three rational numbers $a, b$ and $c$, if $a \times(b \times c)=(a \times b) \times c$
Then, multiplication is associative for rational numbers.
Here, $\frac{1}{3} \times\left(6 \times \frac{4}{3}\right)=\frac{1}{3} \times \frac{24}{3}=\frac{8}{3}$
and $\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}=\frac{6}{3} \times \frac{4}{3}=2 \times \frac{4}{3}=\frac{8}{3}$
$\therefore \quad \frac{1}{3} \times\left(6 \times \frac{4}{3}\right)=\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}$
Hence, we used the associative property for rational numbers.
View full question & answer→Question 42 Marks
By what numbers should we multiply $\frac{-15}{20},$so that the product may be $\frac{-5}{7}?$
AnswerWe have, $\frac{-15}{20} \times$ Number $=\frac{-5}{7}$
$ \Rightarrow \text { Number }=\frac{-5}{7}+\left(\frac{-15}{20}\right)=\frac{-5}{7} \times \frac{20}{-15}=\frac{-20}{-21}=\frac{20}{21} $
So, the required number is $\frac{20}{21}$.
View full question & answer→Question 52 Marks
Name the property used in each of the following
(i) $\left(-\frac{7}{11}\right) \times\left(\frac{-3}{5}\right)=\left(\frac{-3}{5}\right) \times\left(\frac{-7}{11}\right)$
(ii) $\frac{2}{3} \times[\frac{3}{4}+(\frac{-1}{2})]=[(\frac{-2}{3})\times\frac{3}{4}] +[(\frac{-2}{3})\times
(\frac{-1}{2})]$
(iii) $\frac{1}{3} +[\frac{4}{9}+(\frac{4}{3})]-[\frac{1}{3}+\frac{4}{9}]+(\frac{-4}{3})$
(iv) $\frac{-2}{7}+0=0+(\frac{-2}{7})=\frac{2}{7}$
(v) $\frac{3}{8} \times1=1\times\frac{3}{8}=\frac{3}{8}$
Answer(i) Commutative property over multiplication
(ii) Distributive property over addition
(iii) Associative property over addition
(iv) Existence of additive identity
(v) Existence of multiplicative identity
View full question & answer→Question 62 Marks
Verify that
$ \frac{5}{6} \times\left(-\frac{4}{5}+\frac{-6}{10}\right)=\left[\frac{5}{6} \times\left(\frac{-4}{5}\right)\right]+\left[\frac{5}{6} \times\left(\frac{-6}{10}\right)\right].$
AnswerWe have,
LHS $=\frac{5}{6} \times\left(-\frac{4}{5}+\frac{(-6)}{10}\right)=\frac{5}{6} \times\left(\frac{-8-6}{10}\right)$
$=\frac{5}{6} \times\left(\frac{-14}{10}\right)=\frac{-70}{60}=\frac{-7}{6}$
Also, RHS $=\left[\frac{5}{6} \times\left(\frac{-4}{5}\right)\right]+\left[\frac{5}{6} \times\left(\frac{-6}{10}\right)\right]$
$=\left(\frac{-20}{30}\right)+\left(\frac{-30}{60}\right)=\frac{-20}{30}-\frac{30}{60}$
$=\frac{-40-30}{60}=\frac{-70}{60}=\frac{-7}{6}$
Thus, LHS = RHS
View full question & answer→Question 72 Marks
Simplify
$ \frac{-8}{9} \times\left(\frac{3}{7}+\frac{7}{4}\right) $
Answer$\frac{-8}{9} \times\left(\frac{3}{7}+\frac{7}{4}\right)=\frac{-8}{9} \times\left(\frac{12+49}{28}\right)$
$=\frac{-8}{9} \times \frac{61}{28}=\frac{-122}{63}$
We can also solve it as
$\frac{-8}{9} \times\left(\frac{3}{7}+\frac{7}{4}\right)=\frac{-8}{9} \times \frac{3}{7}+\left(\frac{-8}{9}\right) \times \frac{7}{4}\quad$ [by distributive property]
$=\frac{-8}{21}-\frac{14}{9}=\frac{-24-98}{63}=\frac{-122}{63}$
Thus, $\frac{-8}{9} \times\left(\frac{3}{7}+\frac{7}{4}\right)=\frac{-122}{63}$
View full question & answer→Question 82 Marks
Find the value of $\frac{x}{y}+x y$, using appropriate property and name it.
(i) $x=\frac{1}{3}, y=\frac{2}{5}$
(ii) $x=\frac{3}{4}, y=2$
Answer(i) $\frac{29}{30};$ Distributive $\quad$ (ii) $\frac{15}{8};$ Distributive
View full question & answer→Question 92 Marks
Solve $-5+\frac{7}{10}+\frac{3}{7}+(-3)+\frac{5}{14}+\frac{-4}{5}.$
View full question & answer→Question 102 Marks
Verify the distributive property under addition for three numbers $\frac{1}{5}, 2, \frac{-6}{9}.$
AnswerLet $x=\frac{1}{5}, y=2$ and $z=\frac{-6}{9}$
Distributive property under addition is
$x \times(y+z)=x \times y+x \times z$
LHS $=x \times(y+z)=\frac{1}{5} \times\left[2+\left(\frac{-6}{9}\right)\right]$
$=\frac{1}{5} \times\left(2-\frac{6}{9}\right)=\frac{1}{5} \times\left(\frac{18-6}{9}\right)$
$=\frac{1}{5} \times \frac{12}{9}=\frac{4}{15}$
RHS $=x \times y+x \times z=\frac{1}{5} \times 2+\frac{1}{5} \times\left(\frac{-6}{9}\right)$
$=\frac{2}{5}-\frac{6}{45}=\frac{18-6}{45}=\frac{12}{45}=\frac{4}{15}$
View full question & answer→Question 112 Marks
If $x=\frac{9}{10}, y=\frac{4}{3}, z=6$, then verify that $(x+y)+z \neq x+(y+z)$
AnswerLet $x=\frac{9}{10}, y=\frac{4}{3}$ and $z=6$
LHS $=(x \div y) \div z=\left(\frac{9}{10} \div \frac{4}{3}\right) \div 6$
$=\left(\frac{9}{10} \times \frac{3}{4}\right) \div 6=\frac{27}{40} \times \frac{1}{6}=\frac{9}{80}$
$\text{RHS} =x \div(y \div z)=\frac{9}{10} \div\left(\frac{4}{3} \div 6\right)$
$=\frac{9}{10}+\left(\frac{4}{3} \times \frac{1}{6}\right)=\frac{9}{10}+\left(\frac{2}{9}\right)=\frac{9}{10} \times \frac{9}{2}=\frac{81}{20}$
$\therefore(x+y)+z \neq x+(y+z)$
Hence, the rational numbers are not associative under division.
View full question & answer→Question 122 Marks
Divide the sum of two numbers $\frac{1}{5}$ and $\frac{6}{5}$ by their product.
AnswerSum of numbers $\frac{1}{5}$ and $\frac{6}{5}=\frac{1}{5}+\frac{6}{5}=\frac{7}{5}$
Product of numbers $\frac{1}{5}$ and $\frac{6}{5}=\frac{1}{5} \times \frac{6}{5}=\frac{6}{25}$
Then, $\frac{7}{5} \div \frac{6}{25}=\frac{7}{5} \times \frac{25}{6}=\frac{35}{6}$
Hence, $\frac{35}{6}$ is the required number.
View full question & answer→Question 132 Marks
Simplify each of the following and write the name of property used for simplifying
(i) $\left[\frac{5}{9} \times\left(\frac{-6}{7}\right)\right] \times\left[\frac{9}{5} \times\left(\frac{21}{12}\right)\right]$
(ii) $\frac{3}{2} \times \frac{15}{11}+\frac{7}{3}-\frac{15}{11} \times\left(\frac{-19}{21}\right)$
(iii) $\frac{-4}{17} \times \frac{21}{33} \times\left(\frac{-51}{8}\right) \times\left(\frac{-11}{7}\right)$
Answer(i) $\frac{-3}{2}$; commutative, associative, existence of inverse
(ii) $\frac{2593}{462}$; commutative, distributive
(iii) $\frac{-3}{2}$; commutative, associative
View full question & answer→Question 142 Marks
Verify the property $x \times y=y \times x$ of rational numbers using
(i) $x=\frac{-5}{7}$ and $y=\frac{14}{15}$ $\quad$ (ii) $x=\frac{2}{3}$ and $y=\frac{9}{4}$
Answer(i) We know that $x \times y=y \times x$
$\Rightarrow \quad \frac{-5}{7} \times \frac{14}{15}=\frac{14}{15} \times \frac{-5}{7}$
$\therefore \quad \frac{-2}{3}=\frac{-2}{3}$
which shows LHS $=$ RHS
Hence, it is verified.
(ii) We know that
$x \times y=y \times x$
$\Rightarrow \quad \frac{2}{3} \times \frac{9}{4}=\frac{9}{4} \times \frac{2}{3}$
$\therefore \quad \frac{18}{12}=\frac{18}{12}$
which shows LHS $=$ RHS
Hence, it is verified.
View full question & answer→Question 152 Marks
Find the multiplicative inverse of the foilowing
(i) $-6 \times\left(\frac{-5}{42}\right)$ $\quad$(ii) $\frac{-33}{119}$
(iii) -1 $\quad$$\quad$$\quad$$\quad$ (iv) $\frac{-6}{17} \times \frac{17}{5}$
Answer(i) $-6 \times \frac{-5}{42}=\frac{5}{7}$
$\therefore$ Multiplicative inverse of $\frac{5}{7}$ is $\frac{7}{5}.$
(ii) Multiplicative inverse of $\frac{-33}{119}$ is $\frac{-119}{33}.$
(iii) Multiplicative inverse of -1 is -1
(iv) We have, $\frac{-6}{17} \times \frac{17}{5}=\frac{-6}{5}$
$\therefore$ Multiplicative inverse of $\frac{-6}{5}$ is $\frac{-5}{6}.$
View full question & answer→Question 162 Marks
Subtract the sum of $\frac{-7}{8}$ and $\frac{5}{16}$ from the sum of $\frac{3}{5}$ and $\frac{2}{9}.$
AnswerSum of $\frac{-7}{8}$ and $\frac{5}{16}=\frac{-7}{8}+\frac{5}{16}=\frac{-14+5}{16}=\frac{-9}{16}$
Sum of $\frac{3}{5}$ and $\frac{2}{9}=\frac{3}{5}+\frac{2}{9}=\frac{27+10}{45}=\frac{37}{45}$
$\therefore \quad$ Required answer $=\frac{37}{45}-\left(\frac{-9}{16}\right)$
$=\frac{37}{45}+\frac{9}{16}=\frac{592+405}{720}$
$=\frac{997}{720}=1\frac{277}{720}$
View full question & answer→Question 172 Marks
Simplify
(I) $\frac{3}{7} \times \frac{28}{15}÷\frac{14}{5}$
(ii) $\frac{3}{7}+\left(\frac{-2}{21}\right) \times\left(\frac{-5}{6}\right)$
Answer(i) We have, $ \begin{aligned} \frac{3}{7} \times \frac{28}{15} ÷\frac{14}{5} =\frac{3}{7} \times \frac{28}{15} \times \frac{5}{14} =\frac{3}{7} \times\left(\frac{28}{15} \times \frac{5}{14}\right)=\frac{3}{7} \times \frac{2}{3}=\frac{2}{7} \end{aligned} $
(ii) We have, $\frac{3}{7}+\left(\frac{-2}{21}\right) \times\left(\frac{-5}{6}\right)=\frac{3}{7}+\left[\left(\frac{-2}{21}\right) \times\left(\frac{-5}{6}\right)\right]$
$=\frac{3}{7}+\left(\frac{-5}{-63}\right)=\frac{3}{7}+\frac{5}{63}=\frac{27+5}{63}=\frac{32}{63}$
View full question & answer→Question 182 Marks
Solve
(i) $\frac{4}{7}+\left(-\frac{4}{9}\right)+\frac{3}{7}+\left(-\frac{13}{9}\right)$
(ii) $\left(\frac{-3}{7}+\frac{9}{14}\right)+\left(\frac{-5}{2}\right)$
(iii) $\frac{18}{19}+\left(\frac{-37}{19}+\frac{74}{57}\right)$
Answer
$\begin{array}{l}\text{(i) We have,} \quad \frac{4}{7}+\left(-\frac{4}{9}\right)+\frac{3}{7}+\left(-\frac{13}{9}\right) \\ =\left(\frac{4}{7}+\frac{3}{7}\right)+\left(-\frac{4}{9}+\frac{-13}{9}\right)=\frac{7}{7}+\left(-\frac{4}{9}-\frac{13}{9}\right) \\ =1+\left(-\frac{17}{9}\right)=1-\frac{17}{9}=\frac{9-17}{9}=-\frac{8}{9} \end{array}$
$\text{(ii) We have, }\left(-\frac{3}{7}÷ \frac{9}{14}\right) ÷\left(\frac{-5}{2}\right)$
$=\left(-\frac{3}{7} \times \frac{14}{9}\right) \times\left(\frac{-2}{5}\right)=\left(-\frac{2}{3}\right) \times\left(\frac{-2}{5}\right)=\frac{4}{15}$
$\text{(iii) We have, }\frac{18}{19} ÷\left(\frac{-37}{19} ÷ \frac{74}{57}\right)$
$=\frac{18}{19}÷(\frac{-37}{19}\times\frac{57}{74})=\frac{18}{19}÷(\frac{-3}{2})$
$=\frac{18}{19}\times(\frac{-2}{3})=\frac{-12}{19}$
View full question & answer→Question 192 Marks
Using commutativity and associativity of addition of rational numbers, express the following as a rational number
$ \frac{5}{2}+\left(\frac{-3}{7}\right)+\frac{1}{2}+\frac{4}{7}$
Answer$\frac{5}{2}+\left(\frac{1}{2}+\frac{-3}{7}\right)+\frac{4}{7}\quad$ [by commutative property of rational numbers]
$ =\left(\frac{5}{2}+\frac{1}{2}\right)+\frac{(-3)}{7}+\frac{4}{7}\quad$ [by associative property of rational numbers]
$\begin{array}{l} =\frac{5+1}{2}+\frac{(-3)+4}{7}=\frac{6}{2}+\frac{1}{7}=\frac{6 \times 7+2 \times 1}{2 \times 7} \\ =\frac{42+2}{14}=\frac{44}{14}=\frac{22}{7} \end{array}$
View full question & answer→Question 202 Marks
Solve $\frac{-3}{13}+\frac{15}{17} \times\left(\frac{-3}{13}\right)+\frac{17}{15}.$
AnswerWe have,
$ \begin{array}{l} \frac{-3}{13}+\frac{15}{17} \times \frac{-3}{13}+\frac{17}{15} \\ =\frac{-3}{13} \times\left[1+\frac{15}{17}\right]+\frac{17}{15}=\frac{-3}{13} \times\left[\frac{17+15}{17}\right]+\frac{17}{15} \\ =\frac{-3}{13} \times\left[\frac{32}{17}\right]+\frac{17}{15}=\frac{-96}{221}+\frac{17}{15} \\ =\frac{-96 \times 15+221 \times 17}{221 \times 15}=\frac{-1440+3757}{3315}=\frac{2317}{3315} \end{array} $
View full question & answer→