EDMA102: MEASUREMENT SUPPLEMENTARY ASSIGNMENT, QUESTION 1: Ratio and Proportion, QUESTION 2: ALGEBRAIC WORD PROBLEM-AGE PROBLEM, QUESTION 3: PERCENTAGE

Family A: $2 \times \text{K}1,000 = \text{K}2,000$
Family B: $3 \times \text{K}1,000 = \text{K}3,000$
Family C: $4 \times \text{K}1,000 = \text{K}4,000$
The shares are: Family A: $\boxed{\text{K}2,000}$, Family B: $\boxed{\text{K}3,000}$, Family C: $\boxed{\text{K}4,000}$.

b) If 12 workers can clear a garden in 9 days, how long will it take 18 workers working at the same rate?
This is an inverse proportion problem. Let $W$ be the number of workers and $D$ be the number of days.

$$W_1 D_1 = W_2 D_2$$ $$12 \text{ workers} \times 9 \text{ days} = 18 \text{ workers} \times D_2$$

$$108 = 18 D_2$$ $$D_2 = \frac{108}{18}$$ $$D_2 = 6 \text{ days}$$

It will take 18 workers $\boxed{6 \text{ days}}$.

QUESTION 2: ALGEBRAIC WORD PROBLEM - AGE PROBLEM

a) Define the variables for present ages.
Let $s$ be the son's present age in years.
Let $f$ be the father's present age in years.

b) Form expressions for their ages in 20 years.
Son's age in 20 years: $s + 20$
Father's age in 20 years: $f + 20$

c) Form the correct equation.
From the problem:

1. Father is now 4 times as old as his son: $f = 4s$
2. In 20 years, father will be twice as old as his son: $f + 20 = 2(s + 20)$

d) Solve the equation.
Substitute the first equation ($f = 4s$) into the second equation:

$$4s + 20 = 2(s + 20)$$ $$4s + 20 = 2s + 40$$

Subtract $2s$ from both sides:

$$2s + 20 = 40$$

Subtract 20 from both sides:

$$2s = 20$$

Divide by 2:

$$s = 10$$

Now find the father's age using $f = 4s$:

$$f = 4 \times 10$$ $$f = 40$$

e) State both present ages with check.
The son's present age is $\boxed{10 \text{ years}}$.
The father's present age is $\boxed{40 \text{ years}}$.

Check:
Present: Father (40) is 4 times Son (10) ($40 = 4 \times 10$). Correct.
In 20 years: Son will be $10 + 20 = 30$ years old. Father will be $40 + 20 = 60$ years old.
Father (60) will be twice Son (30) ($60 = 2 \times 30$). Correct.

QUESTION 3: PERCENTAGE

a) Find the marked price.
Original price = $\text{K}2,000$. Markup = 40%.

$$\text{Marked price} = \text{Original price} + \text{Markup amount}$$ $$\text{Marked price} = \text{K}2,000 + \text{K}800 = \text{K}2,800$$

The marked price is $\boxed{\text{K}2,800}$.