Find the values of the following fractions of given numbers: a) 1/2 of 16, b) 1/5 of 20, c) 1/7 of 42, d) 1/10 of 40, e) 4/5 of 35, f) 2/3 of 36, g) 9/10 of 100, h) 6/7 of 84.

$$\frac{1}{2} \times 16 = \frac{16}{2} = 8$$

The value is $\boxed{\textbf{8}}$.

b) $\frac{1}{5}$ of 20

$$\frac{1}{5} \times 20 = \frac{20}{5} = 4$$

The value is $\boxed{\textbf{4}}$.

c) $\frac{1}{7}$ of 42

$$\frac{1}{7} \times 42 = \frac{42}{7} = 6$$

The value is $\boxed{\textbf{6}}$.

d) $\frac{1}{10}$ of 40

$$\frac{1}{10} \times 40 = \frac{40}{10} = 4$$

The value is $\boxed{\textbf{4}}$.

e) $\frac{4}{5}$ of 35

$$\frac{4}{5} \times 35 = \frac{4 \times 35}{5} = \frac{140}{5} = 28$$

The value is $\boxed{\textbf{28}}$.

f) $\frac{2}{3}$ of 36

$$\frac{2}{3} \times 36 = \frac{2 \times 36}{3} = \frac{72}{3} = 24$$

The value is $\boxed{\textbf{24}}$.

g) $\frac{9}{10}$ of 100

$$\frac{9}{10} \times 100 = \frac{9 \times 100}{10} = \frac{900}{10} = 90$$

The value is $\boxed{\textbf{90}}$.

h) $\frac{6}{7}$ of 84

$$\frac{6}{7} \times 84 = \frac{6 \times 84}{7} = \frac{504}{7} = 72$$

The value is $\boxed{\textbf{72}}$.

2. Work out the following:

a) $\frac{1}{3} \times 12$

$$\frac{1}{3} \times 12 = \frac{12}{3} = 4$$

The product is $\boxed{\textbf{4}}$.

b) $\frac{1}{4} \times 8$

$$\frac{1}{4} \times 8 = \frac{8}{4} = 2$$

The product is $\boxed{\textbf{2}}$.

c) $\frac{3}{4} \times 8$

$$\frac{3}{4} \times 8 = \frac{3 \times 8}{4} = \frac{24}{4} = 6$$

The product is $\boxed{\textbf{6}}$.

d) $\frac{4}{5} \times 20$

$$\frac{4}{5} \times 20 = \frac{4 \times 20}{5} = \frac{80}{5} = 16$$

The product is $\boxed{\textbf{16}}$.

e) $\frac{2}{3} \times 9$

$$\frac{2}{3} \times 9 = \frac{2 \times 9}{3} = \frac{18}{3} = 6$$

The product is $\boxed{\textbf{6}}$.

f) $\frac{1}{6} \times 36$

$$\frac{1}{6} \times 36 = \frac{36}{6} = 6$$

The product is $\boxed{\textbf{6}}$.

g) $\frac{5}{6} \times 24$

$$\frac{5}{6} \times 24 = \frac{5 \times 24}{6} = \frac{120}{6} = 20$$

The product is $\boxed{\textbf{20}}$.

h) $\frac{4}{9} \times 45$

$$\frac{4}{9} \times 45 = \frac{4 \times 45}{9} = \frac{180}{9} = 20$$

The product is $\boxed{\textbf{20}}$.

Word Problems:

: At 10 years Kgomotso was $\frac{1}{3}$ the weight of her father. If her father was 108 kg, what was her weight?

$$\text{Kgomotso's weight} = \frac{1}{3} \times 108 \text{ kg} = \frac{108}{3} \text{ kg} = 36 \text{ kg}$$

Kgomotso's weight was $\boxed{\textbf{36 kg}}$.

: In a certain month in winter, BMC killed $\frac{3}{4}$ of its normal number. If it normally killed 6600 cattle, how many did it kill in this month?

$$\text{Cattle killed} = \frac{3}{4} \times 6600 = \frac{3 \times 6600}{4} = \frac{19800}{4} = 4950$$

BMC killed $\boxed{\textbf{4950 cattle}}$ in that month.

: A bag contains counters. $\frac{3}{5}$ of the counters are blue. How many counters are blue if there are 60 counters in the bag?

$$\text{Blue counters} = \frac{3}{5} \times 60 = \frac{3 \times 60}{5} = \frac{180}{5} = 36$$

There are $\boxed{\textbf{36 blue counters}}$ in the bag.

: Of the 9380 people tested for HIV, $\frac{3}{20}$ of them were HIV positive. How many people were HIV positive?

$$\text{HIV positive people} = \frac{3}{20} \times 9380 = \frac{3 \times 9380}{20} = \frac{28140}{20} = 1407$$

$\boxed{\textbf{1407 people}}$ were HIV positive.