Direct and inverse proportions; compound proportions; rates of work problems.
📖 4 min read · 3 worked examples · 8 practice questions
Today we'll focus on three key goals: defining direct and inverse proportion, understanding how to combine ratios into compound proportions, and applying rates of work to everyday problems. First, a direct proportion means when one quantity doubles, the other also doubles. An inverse proportion is the opposite: when one quantity doubles, the other halves. Next, compound proportion lets us link several ratios together, like converting units step‑by‑step, and we'll see how to combine them efficiently. Finally, rates of work let us solve real‑world tasks—like figuring out how many days a team needs to finish a field irrigation project when each person works at a different speed. We'll keep checking for understanding, so feel free to raise your hand if anything feels unclear.
Class, let's dive into the theory of direct and inverse proportion. These relationships are the backbone of many real‑world situations we'll encounter. First, direct proportion means that as one quantity x increases, y increases at a constant rate. Mathematically we write y = k·x, where k is the constant of proportionality. You'll see this expressed on the slide as 'y ∝ x → y = k·x'. Think of a farmer planting mango trees: if each tree yields 20 kg of fruit, then 5 trees give 100 kg, 10 trees give 200 kg, and so on—straight‑line growth. For inverse proportion: when one variable grows, the other shrinks so that their product stays constant. The formula is y = k⁄x. A Kenyan example: the time it takes to travel a fixed distance at constant speed is inversely proportional to the speed. Double the speed, halve the travel time. Let's compare the two relationships in this table: direct proportion has a linear graph, constant ratio y/x, and both variables increase together; inverse proportion has a hyperbolic graph, constant product xy, and one variable rises while the other falls. Any questions so far? If you're comfortable, try writing a quick example on your notebook: choose a constant k and plot both y = kx and y = k⁄x for x = 1 to 5.
Everyone, let's dive into the idea of a compound proportion. This is when we link two or more simple ratios together to solve a multi‑step problem. Notice the chain of ratios on the slide: a⁄b = c⁄d = e⁄f. Each equals sign tells us the relationships are all the same, so we can move from one pair to the next. Step 1: Identify the ratios involved in the problem. Ask yourself, which quantities correspond to a, b, c, d, and so on. Step 2: Multiply or divide successive ratios. For example, if we have a⁄b = 3⁄4 and c⁄d = 2⁄3, we can combine them by (a⁄b)·(c⁄d) = (3⁄4)·(2⁄3) = ½. Step 3: Use the resulting relationship to solve the original question. Plug the combined ratio back into the story problem and calculate the unknown value. Here, the line highlights the flow from one step to the next—just like following a recipe.
The problem states: ten kilograms of maize costs Ksh 800. We need to find the cost of twenty‑five kilograms. First we set up the proportion. Cost over weight equals 800 divided by 10, which we write as (\frac{cost}{weight}=\frac{800}{10}=\frac{x}{25}). Solving for x gives us x equals Ksh 2000, so a 25 kg bag of maize costs two thousand shillings.
Let's work through Example 2, which is about inverse proportion. The question says: If four workers finish a field task in six days, how many days would it take six workers to finish the same task? Because time and number of workers are inversely proportional, we set time × workers = constant. Here, 6 days × 4 workers = t × 6 workers, so solving gives t = 4 days. With more workers the time decreases: six workers complete the job in four days. Any questions before we move on?
Let's work through Worked Example 3, which deals with a compound proportion. First, remember the original recommendation: 2 liters of fertilizer for every 500 square metres. We set up the first proportion for the area increase: ( \frac{2\text{ L}}{500\text{ m}^2} = \frac{x}{1200\text{ m}^2} ). Solving for x gives the amount needed if concentration stayed the same. That calculation yields 4.8 L, but our new field also requires a 1.5‑times higher concentration, so we must multiply 4.8 L by 1.5. Multiplying 4.8 L by 1.5 gives a final answer of 7.2 L of fertilizer for the larger, more concentrated field. The key steps are: adjust for the larger area, then adjust for the higher concentration. Any questions before we move on?
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