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Compound Proportions and Rates of Work

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

Grade 09 Pathway: N/A

First Principles

Objective: Master the principles of compound proportion (multivariable relationships) and combined work/flow rates using first principles.

Interactive Tank Filling & Work Rate Simulator

Adjust the flow rates of Tap A (Inlet), Tap B (Inlet), and Drain C (Outlet) to see how individual rates combine into a net rate and determine total filling time.




Net Rate: 0.33 tank/hr
Time to Fill: 3.0 hours
1 Job = 1 Fully Filled Water Tank

1. The Core Principle: Why Times Don't Add, but Rates Do

Suppose Amina takes 3 hours to weed a shamba, and Brian takes 6 hours to weed the same shamba. If they work together, will it take \(3 + 6 = 9\) hours? Absolutely not! Working together must take less time than the fastest worker.

Because time is inversely proportional to work output, we cannot add hours directly. Instead, we convert each person's effort into a rate (fraction of the job completed per unit of time):

  • Amina weeds at a rate of \(R_A = \frac{1}{3}\) shamba per hour.
  • Brian weeds at a rate of \(R_B = \frac{1}{6}\) shamba per hour.
  • Together, their combined rate is: \[R_{\text{total}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ shamba per hour}\]
  • The total time is the reciprocal of the total rate: \[T = \frac{1}{R_{\text{total}}} = \frac{1}{\frac{1}{2}} = 2\text{ hours}\]

2. Compound Proportion: Multivariable Scaling

In many real-world tasks—such as building classrooms or harvesting coffee—more than two variables change at once (e.g., number of workers, hours worked per day, and total output). The fundamental invariance rule states:

\[\frac{\text{Men} \times \text{Time}}{\text{Work Done}} = \text{Constant}\]

If \(M_1\) men working \(D_1\) days for \(H_1\) hours/day produce \(W_1\) units, then for any other setup:

\[\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}\]

Key Formulas

1. Work-Rate Relationship: \[\text{Work} = \text{Rate} \times \text{Time} \quad \Longleftrightarrow \quad R = \frac{W}{T} \quad \Longleftrightarrow \quad T = \frac{W}{R}\] When completing 1 whole job, \(W = 1\), so \(R = \frac{1}{T}\) and \(T = \frac{1}{R}\).
2. Combined Rates (Cooperative Work & Pipes): \[R_{\text{net}} = R_1 + R_2 + R_3 - R_{\text{drain}}\] \[T_{\text{combined}} = \frac{1}{R_{\text{net}}} = \frac{1}{\frac{1}{T_1} + \frac{1}{T_2} - \frac{1}{T_{\text{drain}}}}\]
3. Two-Worker Shortcut Formula: \[T_{\text{together}} = \frac{T_1 \times T_2}{T_1 + T_2}\]
4. General Compound Proportion (Man-Hours to Work): \[\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}\]
  • \(M\) = Number of workers / machines
  • \(D\) = Number of days
  • \(H\) = Hours per day
  • \(W\) = Output / work completed / length of road / bags of harvest

Worked Examples

Example 1 (Easy — Work Rates):
Pipe A can fill an irrigation reservoir in 6 hours, while Pipe B can fill it in 3 hours. How long will it take to fill the reservoir if both pipes run simultaneously?

Step-by-step Solution:
  1. Identify individual rates: \[R_A = \frac{1}{6}\text{ reservoir/hr}, \quad R_B = \frac{1}{3}\text{ reservoir/hr}\]
  2. Calculate combined rate: \[R_{\text{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}\text{ reservoir/hr}\]
  3. Find time: \[T = \frac{1}{R_{\text{total}}} = \frac{1}{\frac{1}{2}} = 2\text{ hours}\]
Answer: \(2\) hours (or 120 minutes).
Example 2 (Medium — Compound Proportion):
A dairy cooperative in Eldoret uses 6 automatic milking machines to milk 180 cows in 4 hours. How many cows can 9 machines milk in 5 hours, assuming all machines operate at the same constant rate?

Step-by-step Solution:
  1. Set up the compound proportion ratio: \[\frac{M_1 \times T_1}{W_1} = \frac{M_2 \times T_2}{W_2}\]
  2. Substitute known values: \[\frac{6 \times 4}{180} = \frac{9 \times 5}{W_2}\]
  3. Simplify the left side: \[\frac{24}{180} = \frac{2}{15}\]
  4. Solve for \(W_2\): \[\frac{2}{15} = \frac{45}{W_2} \implies 2 W_2 = 45 \times 15 = 675 \implies W_2 = \frac{675}{2} = 337.5\]
Answer: \(337.5\) (or \(337\) whole cows).
Example 3 (Hard — Combined Rate with Outlet & Interruption):
An inlet tap fills a community water kiosk tank in 4 hours. A leakage drain empties the full tank in 12 hours. The tap is opened at 8:00 AM with the drain mistakenly left open. At 10:00 AM, the caretaker notices and closes the drain. At what exact time will the tank be completely full?

Step-by-step Solution:
  1. Phase 1 (8:00 AM to 10:00 AM = 2 hours): Both tap and drain are open. \[R_{\text{net}} = R_{\text{tap}} - R_{\text{drain}} = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}\text{ tank/hr}\] Work done in 2 hours: \[W_1 = 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\text{ of the tank}\]
  2. Phase 2 (From 10:00 AM onwards): Only tap is open. Remaining work: \[W_{\text{rem}} = 1 - \frac{1}{3} = \frac{2}{3}\text{ of the tank}\] Rate of tap alone \(= \frac{1}{4}\) tank/hr. Time required: \[T_2 = \frac{W_{\text{rem}}}{R_{\text{tap}}} = \frac{\frac{2}{3}}{\frac{1}{4}} = \frac{2}{3} \times 4 = \frac{8}{3}\text{ hours} = 2\text{ hours } 40\text{ minutes}\]
  3. Total Time Calculation: 10:00 AM + 2 hours 40 minutes = 12:40 PM (Total time = 4 hours 40 minutes).
Answer: \(12:40\text{ PM}\).

Common Mistakes

Mistake 1: Adding times directly instead of adding rates.
Incorrect: If Builder Juma takes 4 days and Builder Otieno takes 6 days, together they take \(4 + 6 = 10\) days.
Correction Two people working together must take less time than the fastest individual. Always sum their unit rates: \[R_{\text{net}} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \implies T = \frac{12}{5} = 2.4\text{ days}\]
Why it feels right We naturally equate "combining people" with the addition operator \((+)\), forgetting that time is inversely proportional to productivity.
Mistake 2: Treating inverse proportions as direct proportions.
Incorrect: If 4 workers build a fence in 6 days, then 8 workers will take \(\frac{8 \times 6}{4} = 12\) days.
Correction Doubling the workforce halves the required duration (assuming equal productivity): \[M_1 \times D_1 = M_2 \times D_2 \implies 4 \times 6 = 8 \times D_2 \implies D_2 = \frac{24}{8} = 3\text{ days}\]
Why it feels right Direct cross-multiplication \(\frac{a}{b} = \frac{c}{d}\) is over-applied to scenarios where more input results in less required time.
Mistake 3: Forgetting to subtract drain/outflow rates.
Incorrect: An inlet pipe fills at \(\frac{1}{4}\) tank/hr and a drain leaks at \(\frac{1}{10}\) tank/hr, so net rate is \(\frac{1}{4} + \frac{1}{10}\).
Correction Outflows and opposing forces work against progress, so their rates must be subtracted: \[R_{\text{net}} = \frac{1}{4} - \frac{1}{10} = \frac{5 - 2}{20} = \frac{3}{20}\text{ tank/hr}\]
Why it feels right All numbers in the question text are given as positive values, which masks the opposing direction of action.

Real World

Solar Pump 1 (Primary): Delivers water at a rate capable of filling the tank alone in 6 hours.
Electric Booster Pump 2 (Secondary): Delivers water at a rate capable of filling the tank alone in 4 hours.
Field Irrigation Demand (Drain): While filling, field irrigation lines draw water continuously, capable of emptying a full tank in 12 hours.
Determine individual rates in tanks per hour: \[R_1 = +\frac{1}{6}, \quad R_2 = +\frac{1}{4}, \quad R_{\text{draw}} = -\frac{1}{12}\]
Calculate combined net inflow rate: \[R_{\text{net}} = \frac{1}{6} + \frac{1}{4} - \frac{1}{12}\] Finding the common denominator \(12\): \[R_{\text{net}} = \frac{2 + 3 - 1}{12} = \frac{4}{12} = \frac{1}{3}\text{ tank/hour}\]
Calculate time required: \[T = \frac{1}{R_{\text{net}}} = \frac{1}{\frac{1}{3}} = 3\text{ hours}\]
Volumetric Verification: \[\text{Flow Rate} = \frac{18{,}000\text{ L}}{3\text{ hrs}} = 6{,}000\text{ litres per hour}\] Even with active field watering, both pumps working together will completely charge the header tank in exactly 3 hours!

Practice

A water tank at a primary school can be filled by Pipe A in 3 hours and by Pipe B in 5 hours. If both pipes are opened together, how many minutes will it take to fill the tank completely? (Type only the number, e.g., 112.5)
Review the concepts above.
At Mombasa market, 5 kg of maize costs Ksh 600. At this constant rate, how much would 12 kg of the same maize cost in Ksh? (Type only the number, e.g., 1440)
Review the concepts above.
A small factory uses machines to produce mobile phone chargers. When 2 machines operate for 5 days, they produce a total of 150 chargers. Assuming each machine works at a constant rate, how many chargers will be produced if 4 machines operate for 7 days? (Type only the number, e.g., 420)
Review the concepts above.
A matatu travels 120 km in 2.5 hours. If the driver increases the speed by 20% and drives for 3 hours, how far in km will the matatu travel? (Type only the number, e.g., 172.8)
Review the concepts above.
A matatu driver and his assistant together can transport 240 passengers from the central bus station to the market in 6 hours. The driver alone can transport 10 passengers per hour more than his assistant. How many hours would it take the driver alone to transport all 240 passengers? (Type only the number, e.g., 9.6)
Review the concepts above.
An inlet pipe can fill a community borehole tank in 6 hours. Due to an open irrigation tap at the bottom, it takes 10 hours to fill the tank when both are open. How many hours would it take the irrigation tap alone to completely empty the full tank? (Type only the number, e.g., 15)
Review the concepts above.