Hello, and welcome to today's mathematics lesson. Today, we are going to explore direct and inverse variations, including time and work problems. By the end of this lesson, you will understand how quantities relate to each other, how to identify direct and inverse variation, and how to solve practical problems involving work and time.
Let us begin with the fundamental idea of variation. Two quantities are said to be in variation if a change in one quantity causes a corresponding change in the other quantity. We encounter this everywhere in daily life. For instance, the more articles you buy, the more you pay. The more money you deposit in a bank, the more interest you earn. Or consider speed and time: the faster a car travels, the less time it takes to cover a fixed distance.
There are two main types of variation that we need to master. First, direct variation: two quantities have direct variation when an increase in one causes an increase in the other, and a decrease in one causes a decrease in the other. They move in the same direction. Second, inverse variation: here, an increase in one quantity causes a decrease in the other, and vice versa. They move in opposite directions.
Let me illustrate with a simple comparison. Suppose 7 bags weigh 560 kilograms. Then one bag weighs 80 kilograms, and 14 bags would weigh 1120 kilograms. As the number of bags doubles, the weight doubles too. This is direct variation.
Now consider work and workers. If 7 men complete a job in 560 days, then one man would take 3920 days, but 14 men would finish in just 280 days. As workers increase, time decreases. This is inverse variation.
Let us examine direct variation more formally. When two quantities x and y are in direct variation, the ratio x/y is always constant. This means x₁/y₁ = x₂/y₂ = x₃/y₃ = ... all equal the same constant value.
Here is a practical example. Suppose 8 articles cost 40 rupees, 10 articles cost 50 rupees, and 15 articles cost 75 rupees. Check the ratios: 8/40 = 1/5, 10/50 = 1/5, 15/75 = 1/5. The ratio remains constant, confirming direct variation.
Let us work through a problem together. Given that x and y vary directly, with x equal to 5 when y equals 10, x equal to a when y equals 16, x equal to b when y equals 48, and x equal to 42 when y equals c. We write 5/10 = a/16 = b/48 = 42/c. From 5/10 = a/16, we get a equals 8. From 5/10 = b/48, we get b equals 24. And from 5/10 = 42/c, we find c equals 84.
Now we turn to inverse variation. Two quantities x and y are in inverse variation when their product is constant. That is, xy = k for some fixed value k. Equivalently, x₁y₁ = x₂y₂ = x₃y₃ = ... all equal the same constant.
Consider this table: when x is 2, y is 32; when x is 4, y is 16; when x is 8, y is 8. The products are 2 × 32 = 64, 4 × 16 = 64, and 8 × 8 = 64. Since the product stays constant, x and y vary inversely.
Here is a worked example. If p and q vary inversely, with p equal to 8 when q equals 2.5, we can find other values. The constant product is 8 × 2.5 = 20. So when p is 2, we have 2q = 20, giving q equals 10. When q is 5, we have 5p = 20, giving p equals 4. And when p is 10, we have 10q = 20, giving q equals 2.
There is a helpful technique called the arrow method for solving variation problems quickly. For direct variation, draw arrows in the same direction in both columns. For inverse variation, draw arrows in opposite directions. Then set up your proportion accordingly: the ratio of values at the arrow heads equals the ratio of values at the arrow tails.
For example, if 15 pens cost 375 rupees, how many pens cost 800 rupees? Since this is direct variation, both arrows point the same way. We get x/15 = 800/375, which gives x equals 32 pens.
For inverse variation, suppose 50 boys each get 75 rupees. If 60 boys share the same amount, how much does each get? The arrows point opposite ways. We get x/75 = 50/60, so x equals 62.5 rupees each.
Now we move to the final section: Time and Work. This builds directly on inverse variation. The fundamental principle is simple: if a person can complete work in n days, their one day's work is 1/n of the total work. Conversely, if one day's work is 1/m, the work is completed in m days.
When people work together, we add their daily work rates. Suppose A completes work in 80 days and B in 100 days. A's daily work is 1/80, B's is 1/100. Together, they complete 1/80 + 1/100 = 9/400 per day.
If they work together for 20 days, they complete 9/400 × 20 = 9/20 of the work. The remaining 11/20 is done by A alone at 1/80 per day, taking 11/20 ÷ 1/80 = 44 days.
Here is another common pattern. Suppose A and B together take 15 days, B and C take 20 days, and C and A take 30 days. We write: A plus B equals 1/15, B plus C equals 1/20, C plus A equals 1/30. Adding all three: twice of A plus B plus C equals (4+3+2)/60 = 3/20. So A plus B plus C equals 3/40 per day, meaning together they take 40/3 = 13 1/3 days.
To find individual times: A alone equals 3/40 − 1/20 = 1/40, so 40 days. B alone equals 3/40 − 1/30 = 1/24, so 24 days. C alone equals 3/40 − 1/15 = 1/120, so 120 days.
Let us recap the key takeaways from this lesson.
First, in direct variation, the ratio of quantities is constant: x/y = k. Second, in inverse variation, the product of quantities is constant: xy = k. Third, for time and work problems, one day's work equals 1/n, and combined work rates add together. Fourth, the arrow method provides a quick visual approach to setting up proportions correctly. Fifth, when solving combined work problems, look for patterns where pairing information helps you find individual rates. Sixth, always verify whether your answer makes practical sense: more workers should mean less time, and faster speeds should mean shorter journeys.
You have now learned how quantities vary together, how to solve problems involving direct and inverse proportion, and how to tackle time and work scenarios. Practice identifying the type of variation first, then apply the appropriate method. With consistent practice, these problems will become straightforward. Thank you for listening, and keep exploring the beautiful patterns in mathematics.