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  $=\frac{x}{2}$,

Number of 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) $\frac{-1}{3}-\left(\frac{-3}{5}\right)=\frac{-1}{3}+\frac{3}{5}$
$\frac{-5+9}{15}=\frac{4}{15}$

Each part: 2 Marks
(a) As given in the question AP = PQ = QB = $\frac{1}{3}$, because interval of one between AB is divided by into three equal parts.
Similarly, US = SR = RT = $\frac{1}{3}$, since the interval of one between U and T is divided into three equal parts.  

So, value of P $=2+\frac{1}{3}=\frac{7}{3}$;

Q $=\frac{7}{3}+\frac{1}{3}=\frac{8}{3}$

Similarly for R $=-1-\frac{1}{3}=-\frac{4}{3}$;

S $=-\frac{4}{3}-\frac{1}{3}=-\frac{5}{3}$
(b) Time taken by 1 person to prepare 1 bag = 2 minutes
Time taken by 6 people to prepare 1 bag = $\frac{2}{6}$minutes
Time taken by 6 people (Ashley and her 5 friends) to prepare 900 bags
= $\frac{2}{6}\times 900=300minutes=5hrs$

Each part: 2 Marks
(a) In normal jump frog covers 5m in first jump but this time his jump covered $=5+(5\times \frac{1}{5})=6m$

In the second jump it covered $=4+(4\times \frac{1}{2})=6m$.

So, distance covered in second jump = 6m.
Total distance covered by frog = 6m + 6m = 12m
(b) (i) The statement is false as $\frac{-21}{5}$ and $\frac{7}{-31}$ both are negative numbers and lie to the left side of  0 on the number line i.e. they are at the same side to that of 0 which is left.
(ii) This statement is true as the number $\frac{-3}{-5}$ is actually positive $\frac{3}{5}$  and  $\frac{-4}{7}$ is negative. Therefore, $\frac{3}{5}$ is  is on the right of  $\frac{-4}{7}$ on the number line.

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)

$\frac{-8}{19}+\frac{-4}{57}=\frac{(-8\times 3)}{(19\times 3)}+\frac{-4}{57}$

                                               $=\frac{-24}{57}+\frac{-4}{57}$

                                               $=\frac{-28}{57}$

When we have to compare two rational numbers, we first make their denominators equal and then compare their numerators.
Consider $\frac{24}{40}$ and $\frac{35}{50}.$ LCM of 40 and 50 is 200.
$⟹\frac{(24\times 5)}{(40\times 5)}=\frac{120}{200}$
and    $\frac{(35\times 4)}{(50\times 4)}=\frac{140}{200}\ne \frac{120}{200}$. 
Thus, $\frac{24}{40}$ and $\frac{35}{50}$ are not equivalent.
Considering $\frac{-25}{35}$ and $\frac{55}{-77},$ LCM of 35 and 77 is 385.
$⟹\frac{(-25\times 11)}{(35\times 11)}=\frac{-275}{385}$
and     $\frac{(55\times 5)}{(-77\times 5)}=\frac{-275}{385}.$
Thus,  $\frac{-25}{35}$ and $\frac{55}{-77}$ are equivalent rational numbers.
Similarly, on checking the remaining options we find that $\frac{-25}{35}$ and $\frac{55}{-77}$ is the only pair of equivalent rational numbers.

To compare two rational numbers we make their denominators equal and then compare their numerators.
LCM of the denominator 4 and 5 is 20.
$\frac{2\times 4}{5\times 4}=\frac{8}{20}$
$\frac{3\times 5}{4\times 5}=\frac{15}{20}$
$\text{Comparing}\frac{8}{20}\text{and}\frac{15}{20},$
$\text{we see that}\frac{8}{20}<\frac{15}{20}$
$\text{Hence,}\frac{2}{5}<\frac{3}{4}$.

$-\frac{16}{21}÷-\frac{4}{3}=-\frac{16}{21}\times -\frac{3}{4}$
$=\frac{4}{7}$