:
Each part: 2 Marks
(a) First write the rational numbers with positive denominator = $\frac{-7}{10}$, $\frac{5}{-8}$, $\frac{2}{-3}$.
LCM of denominators (LCM of 10, 8, 3 is 120)
$\frac{-7}{10}$ = $\frac{(-7\times 12)}{(10\times 12)}=\frac{-84}{120}$
$\frac{-5}{8}$ = $\frac{(-5\times 15)}{(8\times 15)}=\frac{-75}{120}$
$\frac{-2}{3}$ = $\frac{(2\times 40)}{(3\times 40)}=\frac{-80}{120}$
Comparing the numerators of these numbers, we get:
- 84 < - 80 < - 75
∴ $\frac{-84}{120}$ < $\frac{-80}{120}$ < $\frac{-75}{120}$;
$\frac{-7}{10}$ < $\frac{-2}{3}$ < $\frac{-5}{8}$.
(b) $\frac{4}{3}$, $\frac{17}{2}$, $\frac{5}{18}$
LCM of denominators (LCM of 3, 2, 18 is 18)
$\frac{4}{3}$ = $\frac{(4\times 6)}{(3\times 6)}=\frac{24}{18}$
$\frac{17}{2}$ = $\frac{(17\times 9)}{(2\times 9)}=\frac{153}{18}$
$\frac{5}{18}$ = $\frac{(5\times 1)}{(18\times 1)}=\frac{5}{18}$
Comparing the numerators of these numbers, we get:
5 < 24 < 153
∴ $\frac{5}{18}$ < $\frac{24}{18}$ < $\frac{153}{18}$;
$\frac{5}{18}$ < $\frac{4}{3}$ < $\frac{17}{2}$.

:
Each part: 1 Mark
(a) Four irrational numbers are $\sqrt{2},\sqrt{3},\sqrt{5}$ and $\pi$
(b) Let the other number be $x$.
$\Rightarrow \frac{3}{4}=\frac{5}{6}\times x$
$\Rightarrow x=\frac{3\times 6}{4\times 5}=\frac{9}{10}$

:
a) Steps: 1 Mark
Answer: 1 Mark
b) Each sub-part: 0.5 Marks
(a) Let the total number of students $=x$.
Then, number of candy bars each student has $=\frac{x}{8}$
Now, no. of student has been half $=\frac{x}{2}$,
Numberof candies each student would have got = 16
$\frac{x}{2}\times 16=x\times \frac{x}{8}$
Or No. of initial students = 64; $\frac{x}{2}=32.$
So, number of candies distributed = 32 x 16 = 512
(b)
(i) $\frac{6}{3}+\frac{-1}{6}=2-\frac{1}{6}=\frac{11}{6}$
(ii)$\frac{-3}{2}-\frac{3}{8}=\frac{-12-3}{8}=\frac{-15}{8}$
(iii)$\frac{-9}{6}+\frac{-3}{5}=\frac{-3}{2}-\frac{3}{5}$
$\frac{-15-6}{10}=\frac{-21}{10}$
(iv)$\frac{-1}{3}-\left(\frac{-3}{5}\right)=\frac{-1}{3}+\frac{3}{5}$
$\frac{-5+9}{15}=\frac{4}{15}$

:
(a) Steps: 1 Mark
Result: 1 Mark
(b) Answer: 1 Mark
(a) Simplified form of the numbers:
$\frac{18}{27}=\frac{2}{3}$
$\frac{7}{3}=2\frac{1}{3}$
$\frac{10}{3}=3\frac{1}{3}$
$\frac{5}{2}=2\frac{1}{2}$
The rational numbers between 2 and 3 are$\frac{7}{3}$ and $\frac{5}{2}$.
(b)Rational number that will lie between $\frac{7}{8}$ and $\frac{1}{2}$ is their average.
$\frac{\frac{7}{8}+\frac{1}{2}}{2}=\frac{\frac{11}{8}}{2}=\frac{11}{16}$
Its reciprocal is $\frac{16}{11}$.

:
Each part: 1 Mark
(a) Converting both numbers with common denominator:$\frac{7\times 8}{13\times 8}and\frac{3\times 13}{8\times 13}$
The numbers are $\frac{56}{104}$ and $\frac{39}{104}$.
Hence, the expression is false.
(b)$\frac{-18}{-13}=\frac{18}{13}$, which is a positive number. Therefore the number lies to the right of 0 on the number line.
Hence the given statement was not true.

:
(a) Solution: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark
(a) Firstly we simplify the fraction $\frac{9}{24}$ = $\frac{3}{8}$
Now the LCM of denominator 8 and 16 is 16.
$\frac{9}{16}+\frac{3}{8}=\frac{9+6}{16}=\frac{15}{16}$
(b)

Arun walks $\frac{1}{3}$km towards East from place A
Then he walks$1\frac{2}{3}=\frac{5}{3}$km towards West, ie. in the opposite direction.
Arun's new position is $\frac{5}{3}-\frac{1}{3}=\frac{4}{3}=1\frac{1}{3}$
Arun is $1\frac{1}{3}$km towards West from place A.

:
(a) Explanation: 2 Marks
(b) Answer: 1 Mark
(a) $\frac{22}{7}$ is a rational number because we can write it in a form of $\frac{p}{q}$.
The actual value of $\pi$ is 3.141516........which implies that there is no end to the values after the decimal, the decimal goes on forever without repeating.
Hence we are notable to find the actual value of $\pi$ therefore we can't represent it on a number line and moreover, we can't represent it in the form of $\frac{p}{q}$, that's why it is an irrational number.
(b) Yes, the numbers $\frac{1}{3}$ and $\frac{-5}{2}$ lie on the opposite sides of 0 on the number line as one of them is positive and the other is negative.

:
Each part: 2 Marks
(a) The LCM of denominator -5 and 2 = -10. Converting these rational numbers to equivalent rational numbers having common denominator.
$\frac{-2\times 2}{5\times 2}=\frac{-4}{10}=\frac{-4\times 2}{10\times 2}=\frac{-8}{20}$
$\frac{1\times 5}{2\times 5}=\frac{5}{10}=\frac{5\times 2}{10\times 2}=\frac{10}{20}$
Clearly -7, -6, -5.............8, 9 are integers between numerators -8 and 10 of these equivalent rational number. Thus we can take any 10 rational numbers and the numbers are $\frac{-7}{20}$, $\frac{-6}{20}$, $\frac{-5}{20}$,$\frac{-3}{20}$,$\frac{-1}{20}$, 0,$\frac{1}{20}$, $\frac{3}{20}$, $\frac{4}{20}$, $\frac{5}{20}$.
(b)$\sqrt{2}$​ = 1.414 (approx)
$\sqrt{3}$​​​​ = 1.732 (approx)
5 rational numbers between $\sqrt{2}$​ and $\sqrt{3}$​​​​ are 1.42, 1.43, 1.44, 1.45, 1.50. (there are many more)
2 irrational numbers between $\sqrt{2}$​ and $\sqrt{3}$​​​​ are $\sqrt{2}.1$ and $\sqrt{2}.9$ (there are many more)

:
Each part: 2 Marks
(a) Distance of trip = 3000km
Rate at which car consumes gas = 6 litres per 100 km
Total amount of gas consumed in the trip $=\frac{6}{100}\times 3000km=180litres$
Cost of gas = 1.20 Rs per litre.
Total cost = 1.20 $\times$ 180 = 216 Rs
$\therefore$ Contribution from each $=\frac{216}{4}=Rs54$
(b) Upper limit $=\frac{6}{9}$
Lower limit $=\frac{-1}{3}$ or$\frac{-3}{9}$
Rational numbers between these are:
$\frac{-2}{9},\frac{-1}{9},0,\frac{1}{9},\frac{2}{9},\frac{3}{9},\frac{4}{9},\frac{5}{9}$

:
Each part: 1 Mark
(a) $\frac{1}{7}+(\frac{2}{3}\times \frac{5}{2})+\frac{-5}{12}$
$\frac{1}{7}+\left(\frac{5}{3}\right)-\frac{5}{12}$
$\frac{12+140-35}{84}$ (LCM of 7, 3, 12 = 84)
$\frac{117}{84}$
(b)$\frac{-5}{7}=\frac{x}{28}$
$\frac{x}{28}=\frac{-5}{7}$
$x=\frac{28\times (-5)}{7}=4\times (-5)=-20$
(c)$\frac{3}{7}÷\frac{21}{-55}$
$\Rightarrow \frac{3}{7}\times \frac{-55}{21}$
$\Rightarrow \frac{1}{7}\times \frac{-55}{7}$
$\Rightarrow \frac{-55}{49}$