GRADE 9 Mathematics NUMBERS – COMPOUND PROPORTIONS AND RATES OF WORK Notes

Mathematics — Numbers
Subtopic: Compound Proportions and Rates of Work
(Age ~14, Kenyan context)

Learning goals:

• Understand what compound proportion means.
• Solve problems where several ratios affect a quantity.
• Use rates of work (workers and time) to find how long tasks take together.

1. What is Compound Proportion?

Compound proportion is when more than one proportional change affects a quantity. Each factor may be directly proportional (increase leads to increase) or inversely proportional (increase leads to decrease).

Example of compound proportion (items & machines)

"3 machines make 120 bricks in 4 hours. How many machines are needed to make 200 bricks in 5 hours?"

Step-by-step:

1. Find rate per machine per hour:
   3 machines → 120 bricks in 4 h ⇒ 1 machine in 4 h makes 120 ÷ 3 = 40 bricks ⇒ 1 machine in 1 h makes 40 ÷ 4 = 10 bricks
2. Say x machines are needed for 200 bricks in 5 h. Total bricks = (machines × rate per machine per hour × hours):
3. So: x × 10 × 5 = 200 ⇒ x = 200 ÷ 50 = 4 machines.

2. Direct and Inverse Proportion in Compound Problems

- Direct proportion: if A doubles, B doubles (same direction). Use factor = new/old.
- Inverse proportion: if A doubles, B halves (opposite direction). Use factor = old/new.

3. Rates of Work (Workers and Time)

Basic idea: Work = Rate × Time. If a worker does a job in T days, rate = 1/T job per day. When workers work together, rates add.

Example 1 — Two workers

"James can paint a fence in 6 hours. Wanjiru can paint the same fence in 4 hours. How long if they work together?"

Solution:

1. James' rate = 1/6 fence per hour. Wanjiru's rate = 1/4 fence per hour.
2. Combined rate = 1/6 + 1/4 = (2 + 3)/12 = 5/12 fence per hour.
3. Time = 1 ÷ (5/12) = 12/5 = 2.4 hours = 2 hours 24 minutes.

Example 2 — Different start times

"A can finish a job in 8 days. B can finish it in 12 days. A starts and works alone for 3 days, then B joins. How many more days until finish?"

Solution:

1. A's daily rate = 1/8. In 3 days A does 3 × 1/8 = 3/8 of the job.
2. Remaining = 1 − 3/8 = 5/8.
3. Together rate = 1/8 + 1/12 = (3 + 2)/24 = 5/24 per day.
4. Time to finish = (5/8) ÷ (5/24) = (5/8) × (24/5) = 3 days.
5. So total = 3 + 3 = 6 days.

4. More Compound Proportion Examples

Example 3 — Multiple factors

"A contractor finds that with 4 masons, a wall is built in 15 days using 10 sacks of cement per 5 m². If he wants to build a wall twice as big in 10 days, how many masons and how many sacks are needed?"

Approach:

1. Area doubles → twice the work (direct ×2).
2. Time reduces from 15 to 10 days → work per day must increase (inverse ×15/10 = ×3/2).
3. So overall factor to daily workforce = 2 × 15/10 = 3.
4. Needed masons = 4 × 3 = 12 masons.
5. Cement: original used 10 sacks for 5 m². Twice area ⇒ ×2 ⇒ 20 sacks. (Assuming cement needed proportional to area.)

5. Quick Methods and Tips

• Turn each proportional change into a simple multiplying factor.
• For direct: factor = new/old. For inverse: factor = old/new.
• Multiply all factors to get net change. Apply net change to original quantity.
• For work problems, convert times to rates (1/time) before adding.

6. Practice Questions (Try these)

1. Five workers build a hut in 12 days. How many workers are needed to build the same hut in 8 days?
2. A machine fills 500 bottles in 4 hours. How many hours will 3 such machines take to fill 1500 bottles?
3. Jane alone takes 10 days to do a job; Peter takes 15 days. If they start together but Jane leaves after 4 days, how many more days will Peter need?
4. 2 pumps can empty a tank in 6 hours. How long will 5 pumps take (same pump rate)?

7. Answers (check your work)

1. Workers ∝ 1/time (inverse). Factor = 12/8 = 3/2 → workers = 5 × 3/2 = 7.5 ⇒ 8 workers (must be whole). If fractional allowed, 7.5.
2. One machine rate = 500/4 = 125 bottles/hour. Three machines rate = 375 bottles/hour. Time = 1500 ÷ 375 = 4 hours.
3. Combined rate = 1/10 + 1/15 = (3+2)/30 = 1/6 job/day. Jane + Peter do 4 × 1/6 = 2/3 job. Remaining = 1/3. Peter alone rate = 1/15 → time = (1/3) ÷ (1/15) = 5 days.
4. 2 pumps → 6 h, so 1 pump → 12 h (inverse). 5 pumps → 12 ÷ 5 = 2.4 h = 2 h 24 min.

Final tip: practice making a small table of factors (direct and inverse) for each change. This makes compound proportion problems much easier.