[5 marks sum]

Question 15 Marks

Insert seven rational numbers between $2$ and $3.$

Answer

As, 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 $2 \frac{7}{8}$

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Question 25 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 $=\frac{106}{126}, \frac{107}{126} \cdot \frac{108}{126} \cdot \frac{109}{126}, \frac{110}{126}, \frac{111}{126}$
$=\frac{53}{63}, \frac{107}{126}, \frac{6}{7}, \frac{109}{126}, \frac{55}{63}, \frac{37}{42}$

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Question 35 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 $\frac{59}{90}$

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Question 45 Marks

Divide the sum of $\frac{5}{8}$ and $\frac{-11}{12}$ by the difference of $\frac{3}{7}$ and $\frac{5}{14}$.

Answer

Sum 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 $4 \frac{1}{12}$

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Question 55 Marks

For each set of rational number, given below, verify the associative property of addition of rational number : $(i) -1, \frac{5}{6}$ and $\frac{-2}{3}$

Answer

Show 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}$

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Question 65 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 $\frac{-5}{18}$

Answer

Show 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.

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Question 75 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 $\frac{-7}{10}$

Answer

Show 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, $\left(\frac{-2}{5}+\frac{4}{15}\right)+\frac{-7}{10}$

$3$

$5, 15$

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Question 85 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.

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