Question 14 Marks
Find $(m+n) \div(m-n)$, if: $m =\frac{4}{5}$ and $n =-\frac{3}{10}$
AnswerUsing formula $( m + n ) \div( m - n )$
$=\left[\frac{4}{5}+\frac{(-3)}{10}\right] \div\left[\frac{4}{5}-\frac{(-3)}{10}\right]$
$=\left[\frac{8+(-3)}{10}\right] \div\left[\frac{8-(-3)}{10}\right]$
$=\left(\frac{8-3}{10}\right) \div\left(\frac{8+3}{10}\right)$
$=\frac{5}{10} \div \frac{11}{10}$
$=\frac{5}{10} \times \frac{10}{11}$
$=\frac{5}{1} \times \frac{1}{11}=\frac{5}{11}$
View full question & answer→Question 24 Marks
Find $(m+n) \div(m-n)$, if: $m =\frac{3}{4}$ and $n =\frac{4}{3}$
AnswerUsing formula $(m+n) \div(m-n)$
$=\left(\frac{3}{4}+\frac{4}{3}\right) \div\left(\frac{3}{4}-\frac{4}{3}\right)$
$ =\left(\frac{3 \times 3}{4 \times 3}+\frac{4 \times 4}{3 \times 4}\right) \div\left(\frac{3 \times 3}{4 \times 3}-\frac{4 \times 4}{3 \times 4}\right)$
$(\text{LCM}$ of $3$ and $4=12)$
$=\left(\frac{9+16}{12}\right) \div\left(\frac{9-16}{12}\right)$
$ =\frac{25}{12} \div-\frac{7}{12}$
$ =\frac{25}{12} \times-\frac{12}{7}$
$=-\frac{25}{7}$
View full question & answer→Question 34 Marks
Find $(m+n) \div(m-n)$, if: $m =\frac{2}{3}$ and $n =\frac{3}{2} $
AnswerUsing formula $(m+n) \div(m-n)$
$=\left(\frac{2}{3}+\frac{3}{2}\right) \div\left(\frac{2}{3}-\frac{3}{2}\right)$
$ =\left(\frac{2 \times 2}{3 \times 2}+\frac{3 \times 3}{2 \times 3}\right) \div\left(\frac{2 \times 2}{3 \times 2}-\frac{3 \times 3}{2 \times 3}\right)$
$( \text{LCM}$ of $3$ and $2=6$ )
$=\left(\frac{4+9}{6}\right) \div\left(\frac{4-9}{6}\right)$
$ =\frac{13}{6} \div\left(\frac{-5}{6}\right)$
$ =\frac{13}{6} \times \frac{6}{-5}=-\frac{13}{5}$
View full question & answer→Question 44 Marks
Verify that $x =\frac{3}{4}, y =\frac{8}{9}$ and $z =-5$
Answer$x =\frac{3}{4}, y =\frac{8}{9}$ and $z =-5$
Using, $x \times( y - z )= x \times y - x \times z$
$\frac{3}{4} \times\left(\frac{8}{9}-(-5)\right)=\frac{3}{4} \times \frac{8}{9}-\frac{3}{4} \times(-5)$
$ \Rightarrow \frac{3}{4} \times\left(\frac{8 \times 1}{9 \times 1}+\frac{5 \times 9}{1 \times 9}\right)=\frac{2}{3}+\frac{15}{4}$
$ \Rightarrow \frac{3}{4} \times\left(\frac{8+45}{9}\right)=\frac{2 \times 4}{3 \times 4}+\frac{15 \times 3}{4 \times 3}$
$ \Rightarrow \frac{3}{4} \times \frac{53}{9}=\frac{8+45}{12}$
$ \Rightarrow \frac{53}{12}=\frac{53}{12}$
View full question & answer→Question 54 Marks
Verify that $x \times( y - z )= x \times y - x \times z$, if $x =\frac{4}{5}, y =-\frac{7}{4}$ and $z =3$
Answer$x =\frac{4}{5}, y =-\frac{7}{4}$ and $z =3$
Using, $x \times( y - z )= x \times y - x \times z$
$\Rightarrow \frac{4}{5} \times\left(\frac{-7}{4}-3\right)=\frac{4}{5} \times \frac{-7}{4}-\frac{4}{5} \times 3$
$ \Rightarrow \frac{4}{5} \times\left(\frac{-7 \times 1-3 \times 4}{4}\right)=\frac{-7}{5}-\frac{12}{5}$
$ \Rightarrow \frac{4}{5} \times\left(\frac{-7-12}{4}\right)=\frac{-7-12}{5}$
$ \Rightarrow \frac{4}{5} \times \frac{-19}{4} \Rightarrow \frac{-19}{5}=\frac{-19}{5}$
View full question & answer→Question 64 Marks
Verify that $( x + y ) \times z = x \times z \times y \times z$, if $x =2, y =\frac{4}{5}$ and $z =\frac{3}{-10}$
Answer$x =2, y =\frac{4}{5}$ and $z =\frac{3}{-10}$
Using, $(x+y) \times z=x \times z+y \times z$
$\Rightarrow\left(\frac{2}{1}+\frac{4}{5}\right) \times \frac{3}{-10}=2 \times \frac{3}{-10}+\frac{4}{5} \times \frac{3}{-10}$
$ \Rightarrow\left(\frac{2 \times 5}{1 \times 5}+\frac{4 \times 1}{5 \times 1}\right) \times \frac{3}{-10}=\frac{3}{-5}+\frac{6}{-25}$
$ \Rightarrow\left(\frac{10+4}{5}\right) \times \frac{3}{-10}=\frac{-3 \times 5}{5 \times 5}+\frac{-6 \times 1}{5 \times 5}$
$ \Rightarrow \frac{14}{5} \times \frac{3}{-10}=\frac{-15-6}{25}$
$ \Rightarrow \frac{-21}{25}=\frac{-21}{25}$
View full question & answer→Question 74 Marks
Verify That $(x+y) \times z=x \times z+y \times z$, If $x=\frac{4}{5}, y=\frac{-2}{3}$ and $z=-4$
Answer$ x=\frac{4}{5}, y=\frac{-2}{3}$ and $z=-4 $
Using, $(x+y) \times z=x \times z \times y \times z$
$\Rightarrow\left(\frac{4}{5}+\frac{-2}{3}\right) \times-4=\frac{4}{5} \times-4+\frac{-2}{3} \times-4$
$\Rightarrow\left(\frac{4 \times 3}{5 \times 3}-\frac{2 \times 5}{3 \times 5}\right) \times-4=\frac{-16}{5}+\frac{8}{3}$
$\Rightarrow \frac{12-10}{15} \times-4=\frac{-48+40}{15}$
$\Rightarrow \frac{2}{15} \times(-4)$
$=\frac{-8}{15}=\frac{-8}{15} $
View full question & answer→Question 84 Marks
Evaluate: $\left(\frac{8}{5} \times \frac{-3}{2}\right)+\left(\frac{-3}{10} \times \frac{9}{16}\right)$
Answer$ =\left(\frac{8 \times(-3)}{5 \times 2}\right)+\left(\frac{(-3) \times 9}{10 \times 16}\right)$
$ =\left(4 \times \frac{-3}{5 \times 1}\right)+\left(\frac{-3 \times 9}{10 \times 16}\right)$
$ =-\frac{12}{5}+\left(\frac{-27}{160}\right)$
| $2$ |
$160, 5$ |
| $2$ |
$80, 5$ |
| $2$ |
$40, 5$ |
| $2$ |
$20, 5$ |
| $2$ |
$10, 5$ |
| $5$ |
$5, 5$ |
|
$1$ |
$\text{LCM}$ of $5$ and $160=160$
$=\frac{(-12) \times 32}{5 \times 32}+\frac{(-27) \times 1}{160 \times 1}$
$ =\frac{-384-27}{160}=\frac{-411}{160}$ View full question & answer→Question 94 Marks
Which rational numbers should be added to $\frac{-7}{8}$ to get $\frac{5}{9}$ ?
AnswerRequired rationa number $=\frac{5}{9}-\left(\frac{-7}{8}\right)$
$=\frac{5}{9}+\frac{7}{8}$
| $2$ |
$9, 8$ |
| $2$ |
$9, 4$ |
| $2$ |
$9, 2$ |
| $3$ |
$9, 1$ |
| $3$ |
$3, 1$ |
|
$1, 1$ |
$\therefore \text{LCM}$ of $9$ and $8=2 \times 2 \times 2 \times 3 \times 3=72$
$=\frac{5 \times 8}{9 \times 8}+\frac{7 \times 9}{8 \times 9}$
$\text{LCM}$ of $9$ and $8=72$
$=\frac{40}{72}+\frac{63}{72}$
$ =\frac{40+63}{72}=\frac{103}{72}=1 \frac{31}{72}$ View full question & answer→Question 104 Marks
The sum of the two rational numbers is $\frac{-2}{3}$. If one of them is $\frac{-8}{5}$ Find the other.
AnswerThe sum of two rational number $=\frac{-2}{3}$
And, one of the numbers $=\frac{-8}{15}$
The other rational number
$=\frac{-2}{3}-\frac{-8}{15}$
| $3$ |
$3, 15$ |
| $5$ |
$1, 5$ |
|
$1, 1$ |
$(\text{LCM}$ of $3$ and $15=15)$
$=\frac{-2 \times 5}{3 \times 5}+\frac{8 \times 1}{15 \times 1}$
$(\text{LCM}$ of $3$ and $15=15$ )
$=\frac{-10+8}{15}=\frac{-2}{15}$ View full question & answer→Question 114 Marks
The sum of two rational numbers is $\frac{9}{20}$. If one of them is $\frac{2}{5}$, find the other. $C$
AnswerThe sum of two rational number $=\frac{9}{20}$
And, one of the numbers $=\frac{2}{5}$
The other rational number
$=\frac{9}{20}-\frac{2}{5}$
| $2$ |
$20, 5$ |
| $2$ |
$10, 5$ |
| $5$ |
$5, 5$ |
|
$1, 1$ |
$\text{LCM}$ of $20$ and $5=20$
$=\frac{9 \times 1}{20 \times 1}-\frac{2 \times 4}{5 \times 4}$
$\text{LCM}$ of $20$ and $5=20$
$=\frac{9}{20}-\frac{8}{20}$
$ =\frac{9-8}{20}=\frac{1}{20}$ View full question & answer→Question 124 Marks
Subtract: $\frac{4}{7}-\frac{-8}{9}-\frac{-13}{7}+\frac{17}{9}$
Answer$ \frac{4}{7}-\frac{-8}{9}-\frac{-13}{7}+\frac{17}{9}$
$ \Rightarrow\left(\frac{4}{7}-\frac{-13}{7}\right)-\left(\frac{-8}{9}-\frac{17}{9}\right)$
$ \Rightarrow\left(\frac{4}{7}+\frac{13}{7}\right)-\left(\frac{-8}{9}-\frac{17}{9}\right)$
$ \Rightarrow \frac{17}{7}-\left(\frac{-25}{9}\right)$
$ \Rightarrow \frac{17}{7}+\frac{25}{9} \quad(\therefore \text { LCM of } 7$ and $9=63)$
$ \Rightarrow \frac{17 \times 9+25 \times 7}{63}$
$ \Rightarrow \frac{153+175}{63}$
$ \Rightarrow \frac{328}{63}$
$ \Rightarrow 5 \frac{13}{63}$
View full question & answer→Question 134 Marks
For the pair of rational numbers, verify commutative property of addition of rational numbers: $(-2)$ and $\frac{3}{-5}$
AnswerShow that: $\frac{-2}{1}+\frac{-3}{5}=\frac{-3}{5}+\frac{-2}{1}$
$=\frac{-2}{1}+\frac{-3}{5} \quad( \text{LCM}$ of $1$ and $5=5)$
$ =\frac{-2 \times 5}{1 \times 5}+\frac{-3 \times 1}{5 \times 1}$
$ =\frac{-10-3}{5}=\frac{-13}{5}$
And, $\frac{-3}{5}+\frac{-2}{1}$
$=\frac{-3 \times 1}{5 \times 1}+\frac{-2 \times 5}{1 \times 5}$
$ =\frac{-3-10}{5}=\frac{-13}{5}$
$ \therefore \frac{-2}{1}+\frac{-3}{5}=\frac{-3}{5}+\frac{-2}{1}$
This verifies that commutative property for the addition of rational numbers.
View full question & answer→Question 144 Marks
For the pair of rational numbers, verify commutative property of addition of rational numbers: $3$ and $\frac{-2}{7}$
AnswerShow that:
$\frac{3}{1}+\frac{-2}{7}=\frac{-2}{7}+\frac{3}{1}$
$ =\frac{3}{1}+\frac{-2}{7} \quad(\text{LCM}$ of $1$ and $7=7)$
$ =\frac{3 \times 7}{1 \times 7}-\frac{2 \times 1}{7 \times 1}$
$ =\frac{21-2}{7}=\frac{19}{7}$
And, $\frac{-2}{7}+\frac{3}{1}$
$=\frac{-2 \times 1}{7 \times 1}+\frac{3 \times 7}{1 \times 7}$
$ =\frac{-2+21}{7}=\frac{19}{7}$
$ \therefore \frac{3}{1}+\frac{-2}{7}=\frac{-2}{7}+\frac{3}{1}$
This verifies that commutative property for the addition of rational numbers.
View full question & answer→Question 154 Marks
For the pair of rational numbers, verify commutative property of addition of rational numbers: $\frac{2}{-5} \text { and } \frac{11}{-15}$
AnswerShow that: $\frac{2}{-5}+\frac{11}{-15}=\frac{11}{-15}+\frac{2}{-5}$
$=\frac{2}{-5}+\frac{11}{-15}$
| $3$ |
$5, 15$ |
| $5$ |
$5, 5$ |
|
$1, 1$ |
$(\therefore \text{LCM}$ of $5$ and $15=15)$
$=\frac{-2 \times 3}{5 \times 3}-\frac{11 \times 1}{15 \times 1}$
$ =\frac{-6-11}{15}=\frac{-17}{15}$
And, $\frac{11}{-15}+\frac{2}{-5}$
$=\frac{-11 \times 1}{15 \times 1}-\frac{2 \times 3}{5 \times 3}=\frac{-11-6}{15}=\frac{-17}{15}$
$ \therefore \frac{2}{-5}+\frac{11}{-15}=\frac{11}{-15}+\frac{2}{-5}$ View full question & answer→Question 164 Marks
For the pair of rational numbers, verify the commutative property of the addition of rational numbers:$\frac{-4}{5}$ and $\frac{-13}{-15}$
Answer$ \frac{-4}{5}$ and $\frac{-13}{-15} $
To verify commutative property, the addition of rational numbers is shown.
$ \begin{array}{l} \frac{-4}{5}+\left(\frac{-13}{-15}\right)=\left(\frac{-13}{-15}\right)+\left(\frac{-4}{5}\right) \\ \begin{array}{l|l} 5 & 5,15 \\ \hline 3 & 1,3 \\ \hline & 1,1 \end{array} \\ \end{array} $
$ 5 \times 3=15 $
$\text{LCM}$ of $5 \& 15=15$
$ \frac{4}{5} \times 15=12 \\ \text { LHS }=\frac{-4}{5}+\left(\frac{-13}{-15}\right)=\frac{-4}{5}+\frac{13}{15}=\frac{-12+13}{15}=\frac{1}{15} $
$ \text { RHS }=\left(\frac{-13}{-15}\right)+\left(\frac{-4}{5}\right)=\frac{13}{15}+\left(\frac{-4}{5}\right)=\frac{13-12}{15}=\frac{1}{15} $
Since $\text{LHS = RHS}$, this verifies the commutative property for the addition of rational numbers.
View full question & answer→Question 174 Marks
For the pair of rational numbers, verify commutative property of addition of rational numbers:$\frac{5}{9}$ and $\frac{5}{-12}$
AnswerTo show that: $\frac{5}{9}+\frac{5}{-12}=\frac{5}{-12}+\frac{5}{9}$
$\therefore \frac{5}{9}+\frac{5}{-12}$
| $2$ |
$9, 12$ |
| $2$ |
$9, 6$ |
| $3$ |
$9, 3$ |
| $3$ |
$3, 1$ |
|
$1, 1$ |
$\text{LCM}$ of $9$ and $2 \times 2 \times 3 \times 3=36$
$=\frac{5 \times 4}{9 \times 4}-\frac{5 \times 3}{12 \times 3}$
$ =\frac{20-15}{36}=\frac{5}{36}$
And, $\frac{5}{-12}+\frac{5}{9}$
$=\frac{5 \times 3}{-12 \times 3}+\frac{5 \times 4}{9 \times 4}$
$ =\frac{-15+20}{36}=\frac{5}{36}$
$ \therefore \frac{5}{9}+\frac{5}{-12}=\frac{5}{-12}+\frac{5}{9}$
This verifies that commutative property for the addition of rational numbers. View full question & answer→Question 184 Marks
For the pair of rational numbers, verify commutative property of addition of rational numbers: $\frac{-8}{7}$ and $\frac{5}{14}$
AnswerTo show that : $\frac{-8}{7}+\frac{5}{14}=\frac{5}{14}+\frac{-8}{7}$
$\therefore \frac{-8}{7}+\frac{5}{14}$
| $2$ |
$7, 14$ |
| $7$ |
$7, 7$ |
|
$1, 1$ |
$\therefore \text{LCM}$ of $2$ and $7=14$
$=\frac{-8 \times 2}{7 \times 2}+\frac{5 \times 1}{14 \times 1}$
$ =\frac{-16+5}{14}=\frac{-11}{14}$
And, $\frac{5}{14}+\frac{-8}{7}$
$=\frac{5 \times 1}{14 \times 1}+\frac{-8 \times 2}{7 \times 2}$
$ =\frac{5-16}{14}=\frac{-11}{14}$
$ \therefore \frac{-8}{7}+\frac{5}{14}=\frac{5}{14}+\frac{-8}{7}$
This verifies that commutative property for the addition of rational numbers. View full question & answer→Question 194 Marks
Evaluate: $\frac{3}{8}+\frac{-5}{12}+\frac{3}{7}+\frac{3}{12}+\frac{-5}{8}+\frac{-2}{7}$
Answer$ =\left(\frac{3}{8}-\frac{5}{8}\right)+\left(\frac{-5}{12}+\frac{3}{12}\right)+\left(\frac{3}{7}-\frac{2}{7}\right)$
$ =\frac{-2}{8}-\frac{2}{12}+\frac{1}{7}$
$ =\frac{-1}{4}-\frac{1}{6}+\frac{1}{7}$
| $2$ |
$4, 6, 7$ |
| $2$ |
$2, 3, 7$ |
| $3$ |
$1, 3, 7$ |
| $7$ |
$1, 1, 7$ |
|
$1, 1, 1$ |
$(\text{LCM}$ of $4,6$ and $7=2 \times 2 \times 3 \times 7=84)$
$=\frac{-1 \times 21}{4 \times 21}-\frac{1 \times 14}{6 \times 14}+\frac{1 \times 12}{7 \times 12}$
$(\text{LCM}$ of $4, 6$ and $7=84)$
$=\frac{-21-14+12}{84}$
$ =\frac{-35+12}{84}=\frac{-23}{84}$ View full question & answer→Question 204 Marks
Insert seven rational numbers between $2$ and $3$.
AnswerAs, we have to find $7$ rational numbers between $2$ and $3$ , we multiply the numbers by $\frac{8}{8}$
$\therefore 2=2 \times \frac{8}{8}=\frac{16}{8}$
and $3=3 \times \frac{8}{8}=\frac{24}{8}$
Thus, $7$ rational numbers between $2$ and $3$
i. e, $(\frac{16}{8}$ and $\frac{24}{8})$ are
$= \frac{17}{8}, \frac{18}{8}, \frac{19}{8}, \frac{20}{8}, \frac{21}{8}, \frac{22}{8}, \frac{23}{8}$
$ =\frac{17}{8}, \frac{9}{4}, \frac{19}{8}, \frac{5}{2}, \frac{21}{8}, \frac{11}{4}, \frac{23}{8}$
$ =2 \frac{1}{8}, 2 \frac{1}{4}, 2 \frac{3}{8}, 2 \frac{1}{2}, 2 \frac{5}{8}, 2 \frac{3}{4}$ and $2 \frac{7}{8}$
View full question & answer→Question 214 Marks
Insert six rational numbers between $\frac{5}{6}$ and $\frac{8}{9}$
Answer$\text{LCM}$ of denominator $6$ and $9$ is $18$
Make, the denominator of each given rational number equal to $18$ i.e the $\text{LCM}$
$\frac{5}{6}=\frac{5 \times 3}{6 \times 3}=\frac{15}{18}$ and
$\frac{8}{9}=\frac{8 \times 2}{9 \times 2}=\frac{16}{18}$
Since six rational numbers are required, multiply the numerator and denominator of each rational number by $6+1=7$
$\therefore \frac{15}{18}=\frac{15 \times 7}{18 \times 7}=\frac{105}{126}$ and
$\frac{16}{18}=\frac{16 \times 7}{18 \times 7}=\frac{112}{126}$
$\therefore$ Required rational numbers between $\frac{5}{6}$ and $\frac{8}{9}$ are
$=\frac{106}{126}, \frac{107}{126} \cdot \frac{108}{126} \cdot \frac{109}{126}, \frac{110}{126}, \frac{111}{126}$
$=\frac{53}{63}, \frac{107}{126}, \frac{6}{7}, \frac{109}{126}, \frac{55}{63}, \frac{37}{42}$
View full question & answer→Question 224 Marks
Insert five rational numbers between $\frac{3}{5}$ and $\frac{2}{3}$
Answer$\text{LCM} $of denominators $5$ and $3$ is $15.$
make the denominator of each given rational number equal to $15$ i.e., the $\text{LCM}.$
$\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}$ and
Since five rational numbers are required, multiply the numerator and denominator of each rational number by $5+1=6$.
$\therefore \frac{9}{15}=\frac{9 \times 6}{15 \times 6}=\frac{54}{90}$ and
$\frac{10}{15}=\frac{10 \times 6}{15 \times 6}=\frac{60}{90}$
$\therefore$ Required rational numbers between $\frac{3}{5}$ and $\frac{2}{3}$ are $\frac{55}{90}, \frac{56}{90}, \frac{57}{90}, \frac{58}{90}$, and $\frac{59}{90}$.
$=\frac{11}{18}, \frac{28}{45}, \frac{19}{35}, \frac{29}{45}$ and $\frac{59}{90}$
View full question & answer→Question 234 Marks
Divide the sum of $\frac{5}{8}$ and $\frac{-11}{12}$ by the difference of $\frac{3}{7}$ and $\frac{5}{14}$.
AnswerSum of $\frac{5}{8}$ and $\frac{-11}{12}=\frac{5}{8}+\left(\frac{-11}{12}\right)$
$=\frac{5}{8}-\frac{11}{12}$
$ =\frac{(5 \times 3)-(11 \times 2)}{24}$
$(\text{LCM}$ of $8$ and $12$ is $24 )$
$=\frac{15-22}{24}=\frac{-7}{24}$
Now, difference of $\frac{3}{7}$ and $\frac{5}{14}$
$=\frac{(3 \times 2)-(5 \times 1)}{14}$ or $\frac{5-(3 \times 2)}{14}$
$(\text{ LCM}$ of $7$ and $13=14)$
$=\frac{6-5}{14}$ or $\frac{5-6}{14}=\frac{1}{14}$ or $\frac{-1}{14}$
Now, divide $\frac{-7}{24}$ by $\frac{1}{14}$ or $\frac{-1}{14}$
$=\frac{\frac{-7}{24}}{\frac{1}{14}}$ or $\frac{\frac{-7}{24}}{\frac{-1}{14}}$
$ =\frac{-7}{24} \times \frac{14}{1}$ or $\frac{-7}{24} \times \frac{-14}{1}$
$ =\frac{-49}{12}$ or $\frac{49}{12}=-4 \frac{1}{12}$ or $4 \frac{1}{12}$
View full question & answer→Question 244 Marks
For each set of rational number, given below, verify the associative property of addition of rational number: $(i)$ $-1, \frac{5}{6}$ and $\frac{-2}{3}$
AnswerShow that :
This verifies associative property of the addition of rational numbers.
$\frac{-1}{1}+\left(\frac{5}{6}+\frac{-2}{3}\right)=\left(\frac{-1}{1}+\frac{5}{6}\right)+\frac{-2}{3}$
$ \therefore \frac{-1}{1}+\left(\frac{5}{6}+\frac{-2}{3}\right)$
| $2$ |
$6, 3$ |
| $3$ |
$3, 3$ |
|
$1, 1$ |
$\therefore \text{LCM}$ of $6$ and $3=6$
$=\frac{-1}{1}+\left(\frac{5 \times 1}{6 \times 1}+\frac{-2 \times 2}{3 \times 2}\right) \quad(\therefore \text{LCM}$ of $6$ and $3=6$ )
$=\frac{-1}{1}+\left(\frac{5-4}{6}\right)$
$ =\frac{-1}{1}+\frac{1}{6}$
$ =\frac{-1 \times 6}{1 \times 6}+\frac{1 \times 1}{6 \times 1} \quad(\therefore \text {LCM}$ of $1$ and $6=1)$
$ =\frac{-6+1}{6}=\frac{-5}{6}$
And, $\left(\frac{-1}{1}+\frac{5}{6}\right)+\frac{-2}{3}$
$=\left(\frac{-1 \times 6}{1 \times 6}+\frac{5 \times 1}{6 \times 1}\right)+\frac{-2}{3} \quad(\therefore \text { LCM}$ of $1$ and $6=6)$
$ =\left(\frac{-6+5}{6}\right)+\frac{-2}{3}$
$ =\frac{-1}{6}+\frac{-2}{3} \quad(\therefore \text {LCM}$ of $1$ and $6=6)$
$ =\frac{-1 \times 1}{6 \times 1}+\frac{-2 \times 2}{3 \times 2}$
$ =\frac{-1-4}{6}=\frac{-5}{6} \quad(\therefore \text { LCM}$ of $6$ and $3=6)$
$ \therefore \frac{-1}{1}+\left(\frac{5}{6}+\frac{-2}{3}\right)=\left(\frac{-1}{1}+\frac{5}{6}\right)+\frac{-2}{3}$ View full question & answer→Question 254 Marks
For each set of rational number, given below, verify the associative property of addition of rational number : $(i) \frac{-7}{9}, \frac{2}{3}$ and $\frac{-5}{18}$
AnswerShow that:
$\frac{-7}{9}+\left(\frac{2}{-3}+\frac{-5}{18}\right)=\left(\frac{-7}{9}+\frac{2}{-3}\right)+\frac{-5}{18}$
$ \therefore \frac{-7}{9}+\left(\frac{2}{-3}+\frac{-5}{18}\right)$
| $2$ |
$3, 18$ |
| $3$ |
$3, 9$ |
| $5$ |
$3, 3$ |
|
$1, 1$ |
$\therefore\text{LCM}$ of $3$ and $18=2 \times 3 \times 3=18$
$=\frac{-7}{9}+\left(\frac{-2 \times 6}{3 \times 6}+\frac{-5 \times 1}{18 \times 1}\right)$
$=\frac{-7}{9}+\frac{-12-5}{18}(\therefore$ $\text{LCM}$ of $3$ and $18=18)$
$=\frac{-7}{9}+\frac{-17}{18}$
$=\frac{-7 \times 2}{9 \times 2}-\frac{17 \times 1}{18 \times 1} \quad(\therefore \text{LCM}$ of $9$ and $18=18)$
$=\frac{-14-17}{18}=\frac{-31}{18}$
And, $\left(\frac{-7}{9}+\frac{2}{3}\right)+\frac{-5}{18}$
| $3$ |
$3, 9$ |
| $3$ |
$3, 3$ |
|
$1, 1$ |
$\therefore \text{LCM}$ of $3$ and $9=9$
$=\left(\frac{-7 \times 1}{9 \times 1}+\frac{-2 \times 3}{3 \times 3}\right)+\frac{-5}{18} \quad(\therefore \text { LCM }=9$ and $3=9)$
$=\frac{-7-6}{9}+\frac{-5}{18}$
$ =\frac{-13}{9}+\frac{-5}{18}$
$ =\frac{-13 \times 2}{9 \times 2}+\frac{-5 \times 1}{18 \times 1}=\frac{-26 \times-5}{18}=\frac{-31}{18}$
$\frac{-7}{9}+\left(\frac{2}{-3}+\frac{-5}{18}\right)=\left(\frac{-7}{9}+\frac{2}{-3}\right)+\frac{-5}{18}$
This verifies associtive property of the addition of rational numbers. View full question & answer→Question 264 Marks
For each set of rational number, given below, verify the associative property of addition of rational number : $(i) \frac{-2}{5}, \frac{4}{15}$ and $\frac{-7}{10}$
AnswerShow that:
$\frac{-2}{5}+\left(\frac{4}{15}+\frac{-7}{10}\right)=\left(\frac{-2}{5}+\frac{4}{15}\right)+\frac{-7}{10}$
$ \therefore \frac{-2}{5}+\left(\frac{4}{15}+\frac{-7}{10}\right)$
| $2$ |
$15 ,10$ |
| $3$ |
$15 ,5$ |
| $5$ |
$5 ,5$ |
|
$1 ,1$ |
$(\therefore \text{LCM}$ of $15,10=2 \times 3 \times 5=30)$
$=\frac{-2}{5}+\left(\frac{4 \times 2}{15 \times 2}+\frac{-7 \times 3}{10 \times 3}\right)$
( $\therefore \text{LCM}$ of $15$ and $10=30$ )
$=\frac{-2}{5}+\left(\frac{8-21}{30}\right)$
$ =\frac{-2}{5}-\frac{13}{30}=\frac{-2 \times 6}{5 \times 6}-\frac{13 \times 1}{30 \times 1}$
$ =\frac{-12-13}{30}=\frac{-25}{30}=\frac{-5}{6}$
And, $\left(\frac{-2}{5}+\frac{4}{15}\right)+\frac{-7}{10}$
View full question & answer→Question 274 Marks
For each set of rational number, given below, verify the associative property of addition of rational number: $(i) \frac{1}{2}, \frac{2}{3}$ and $-\frac{1}{6}$
Answer$\frac{1}{2}, \frac{2}{3}$ and $-\frac{1}{6}$
Show that:
$\frac{1}{2}+\left(\frac{2}{3}+\frac{-1}{6}\right)=\left(\frac{1}{2}+\frac{2}{3}\right)+\frac{-1}{6}$
$ \therefore \frac{1}{2}+\left(\frac{2}{3}+\frac{-1}{6}\right)$
| $2$ |
$3, 6$ |
| $3$ |
$3, 3$ |
|
$1, 1$ |
$\therefore \text{LCM}$ of $3$ and $6=6$
$=\frac{1}{2}+\left(\frac{2 \times 2}{3 \times 2}+\frac{-1 \times 1}{6 \times 1}\right)$
$ =\frac{1}{2}+\left(\frac{4}{6}-\frac{1}{6}\right)$
$ =\frac{1}{2}+\left(\frac{4-1}{6}\right)$
$ =\frac{1}{2}+\left(\frac{3}{6}\right)$
$ =\frac{1 \times 3}{2 \times 3}+\frac{3 \times 1}{6 \times 1} \quad(\therefore \text { LCM}$ of $2$ and $6=3)$
$ =\frac{3+3}{6}=\frac{6}{6}=1$
And, $\left(\frac{1}{2}+\frac{2}{3}\right)+\frac{-1}{6}$
| $2$ |
$2, 3$ |
| $3$ |
$1, 3$ |
|
$1, 1$ |
$(\therefore \text{LCM}$ of $2$ and $3=6)$
$=\left(\frac{1 \times 3}{2 \times 3}+\frac{2 \times 2}{3 \times 2}\right)+\frac{-1}{6}$
$ =\frac{3+4}{6}+\frac{-1}{6}$
$ =\frac{7-1}{6}=\frac{6}{6}=1$
$ \therefore \frac{1}{2}+\left(\frac{2}{3}+\frac{-1}{6}\right)=\left(\frac{1}{2}+\frac{2}{3}\right)+\frac{-1}{6}$
This verifies associative property of the addition of rational numbers. View full question & answer→