a) Each rational number: 0.5 Marks
b) Each operation: 0.5 Marks
(a) There are infinite numbers of rational numbers that lie between two points on a number line.

Consider the rational numbers $\frac{2}{5}$ and $\frac{6}{5}$.
Take the average of two numbers $\frac{2+6}{5\times 2}$ = $\frac{8}{10}$

$\frac{8}{10}$ lies in the middle of $\frac{2}{5}$ and $\frac{6}{5}$.

Now take the average of $\frac{2}{5}$ and $\frac{8}{10}$. It is $(\frac{2}{5}+\frac{8}{10})\times \frac{1}{2}$ = $\frac{4+8}{10}\times \frac{1}{2}$ = $\frac{12}{20}$

So, as we proceed in this, we will get an infinite number of rational number between two points.
(b) (i) $x+y=\frac{1}{10}+\frac{-3}{8}$
$\Rightarrow \frac{1}{10}-\frac{3}{8}=\frac{8-30}{80}$
$\Rightarrow \frac{-22}{80}=\frac{-11}{40}$
(ii) $x-y=\frac{1}{10}-\frac{-3}{8}$
$\Rightarrow \frac{1}{10}+\frac{3}{8}=\frac{8+30}{80}$
$\Rightarrow \frac{38}{80}=\frac{19}{40}$
(iii) $x\times y=\frac{1}{10}\times \frac{-3}{8}=\frac{-3}{80}$
(iv) $x÷y=\frac{1}{10}÷\frac{-3}{8}$
$\Rightarrow \frac{1}{10}\times \frac{8}{-3}=\frac{8}{-30}$
$\Rightarrow \frac{-8}{30}=\frac{-4}{15}$

Each part: 1 Mark
(a) Amit played $4\frac{1}{2}=\frac{9}{2}$ times of his brother.
He cut down his play by $1\frac{1}{3}=\frac{4}{3}$
The duration Amit plays now
= $\frac{9}{2}-\frac{4}{3}=\frac{19}{6}=3\frac{1}{6}$.

So, now Amit plays $3\frac{1}{6}$ times as his brother.
(b) Let the number be x.
$\frac{-6}{11}+x=\frac{-5}{8}$
$x=\frac{-5}{8}-\frac{-6}{11}$
$x=\frac{-5}{8}+\frac{6}{11}$
$x=\frac{-7}{88}$
(c)  For $\frac{-3}{8}$ and $\frac{x}{-24}$ to be equivalent.
$\Rightarrow \frac{-3}{8}=\frac{x}{-24}$
$\Rightarrow x=\frac{-3\times -24}{8}$ 
$\Rightarrow x=\frac{-3\times -3}{1}=9$

(a) Answer: 2 Marks
(b) Answer: 1 Mark
(a) When we multiply by 13 in denominator and the numerator of $\frac{19}{26}$ we get $\frac{247}{338}$, therefore $\frac{19}{26}=\frac{247}{338}$.
By inspection, when the numerator is same and the denominator is different then the fraction having a large value of denominator is smaller than the fraction having less value of denominator.
Hence, $\frac{19}{26}$ < $\frac{19}{21}$.

For comparing $\frac{19}{26}$ and $\frac{13}{16}$ we have to make the denominator of both the fraction by taking LCM of 26 and 16 = 416.
$\frac{(19\times 16)}{(26\times 16)}=\frac{304}{416}$, $\frac{(13\times 26)}{(16\times 26)}=\frac{338}{416}$
338>304. Hence, $\frac{26}{32}$ >$\frac{19}{26}$.
(b)  A rational number which lies between $\frac{1}{2}$ and $\frac{1}{4}$ is:
$\Rightarrow \frac{\frac{1}{2}+\frac{1}{4}}{2}$
$\Rightarrow \frac{\frac{3}{4}}{2}$
$\Rightarrow \frac{3}{8}$

(a)  Formula: 1 Mark
      Steps: 1 Mark
      Answer: 1 Mark
(b) Answer: 1 Mark
(a) To find the length of the diagonal, d, of each rectangle, use Pythagoras Theorem.
For A, d $=\sqrt{(}{1}^2+2^2)=\sqrt{(}1+4)=\sqrt{(}5)$
For B, d $=\sqrt{(}{2}^2+3^2)=\sqrt{(}4+9)=\sqrt{(}13)$
For C, d $=\sqrt{(}{3}^2+4^2)=\sqrt{(}9+16)=\sqrt{(}25)$
For D, d $=\sqrt{(}{4}^2+5^2)=\sqrt{(}16+25)=\sqrt{(}41)$
Of these, only $\sqrt{(}25)=5$ is rational.
The length of the diagonal of rectangle C is rational.
The length of the diagonal of rectangle A, B and D are irrational
(b)  $\frac{5}{13}-\frac{-8}{26}$
$\Rightarrow \frac{5}{13}+\frac{8}{26}$
$\Rightarrow \frac{10+8}{26}$
$\Rightarrow \frac{18}{26}=\frac{9}{13}$

Each part: 2 Marks
(i) Find the LCM of denominators, i.e. the LCM of 3, 2, 9 and 4 = 36.

Now make the denominators common to the above fraction.

So, $\frac{4\times 12}{3\times 12}$ = $\frac{48}{36}$ ; $\frac{3\times 18}{2\times 18}$ = $\frac{54}{36}$ ; $\frac{7\times 4}{9\times 4}$ = $\frac{28}{36}$ ; $\frac{5\times 9}{4\times 9}$ = $\frac{45}{36}$;

Now arrange the fractions according to their numerator, fraction having large magnitude of numerator will be of greater magnitude. 

So, correct sequence is $\frac{28}{36}$, $\frac{45}{36}$, $\frac{48}{36}$, $\frac{54}{36}$.
$\Rightarrow \frac{7}{9}<\frac{5}{4}<\frac{4}{3}<\frac{3}{2}$
(ii)  Find the LCM of denominators, i.e. the LCM of 5, 7, 2 and 20 = 140.

Now make the denominators common to the above fraction.

So, $\frac{8\times 28}{5\times 28}$ = $\frac{224}{140}$; 
$\frac{12\times 20}{7\times 20}$ = $\frac{240}{140}$; 
$\frac{9\times 70}{2\times 70}$ = $\frac{630}{140}$; 
$\frac{23\times 7}{20\times 7}$ = $\frac{161}{140}$;

Now arrange the fractions according to their numerator, fraction having large magnitude of numerator will be of greater magnitude. 

So, correct sequence is $\frac{161}{140}$, $\frac{224}{140}$, $\frac{240}{140}$, $\frac{630}{140}$.
$\Rightarrow \frac{23}{20}<\frac{8}{5}<\frac{12}{7}<\frac{9}{2}$

Each part: 2 Marks
(a) Total money invested = 20 cr

Amount allocated to advertisement team = $20\times \frac{1}{4}$ = 5 cr.

Remaining amount: 20 - 5 = 15cr; Two third of 15 cr is allocated for development team $=15\times \frac{2}{3}=10cr$

Let graphics team got $x$ ,then content team got $\frac{3}{2}$$x$.

By solving, $x$ + $\frac{3}{2}$$x$ =10; $x$=4

Hence, amount allocated for graphics and content team were 4cr and 6cr respectively.
(b) Area of the walkway = (60 $\times$ 2)feet${}^2$ =120feet${}^2$
Area of 1 tile = (2 $\times$ 2)feet${}^2$ = 4feet${}^2$
Total tiles required = $\frac{120}{4}=30$
1 box contains 6 tiles.
$\therefore$ No. of boxes for 30 tiles
$=\frac{30}{6}=5$

Each part: 2 Marks
(a) After striking to the surface, the height gained by ball is $30m\times \frac{2}{3}=20m$.

Now it attained the height of 20m, height attained by ball after $2^{nd}$ strike is $20\times \frac{2}{5}=8m$.

After $2^{nd}$ strike, height attained by ball is $8\times \frac{1}{2}=4$.

So for $3^{rd}$ strike, it falls from a height of 4m.

Again he attained half of the height of previous so this time it covers 2m.

So total distance covered by ball = 30+ 20+8+4+2 = 64.
(b) (i) $\frac{-1}{5}<\frac{-3}{5}$ is incorrect
 The following inequalities are correct:
(ii) $\frac{-3}{5}<\frac{-1}{5}$
(iii) $\frac{3}{5}>\frac{1}{5}$