Hello, and welcome to your mathematics lesson today. We are going to explore the Unitary Method, including its powerful applications in Time and Work problems. By the end of this lesson, you will understand how to find the value of one unit first, and then use it to solve for any required quantity. You will also master the difference between direct and inverse variation, and learn how to tackle work problems with confidence.
Let us begin with the heart of this chapter: what exactly is the unitary method?
Imagine you know the cost of fifteen metres of cloth is three hundred rupees. To find the cost of ten metres, you would first find the cost of just one metre. That would be 300/15, which equals twenty rupees. Then, ten metres would cost 10 × 20, giving you two hundred rupees.
Or consider this: if fifteen men can complete some work in three hundred days, how long would ten men take? First, find how long one man would take: 300 × 15, which equals four thousand five hundred days. Then, ten men would take 4500/10, which equals four hundred fifty days.
The method in which the value of a unit quantity is first calculated to get the value of a required quantity is called the unitary method.
Now, within the unitary method, we encounter two fundamental types of relationships between quantities.
The first is direct variation. Here, when one quantity increases, the other also increases, and when one decreases, the other decreases too. Think about it: with more money, you can buy more articles. With greater speed, you cover more distance in the same time. With fewer workers, less work gets done. These are all direct variations — the two quantities move in the same direction.
The second type is inverse variation. Here, the relationship flips: when one quantity increases, the other decreases, and vice versa. Consider speed and time: the faster you travel, the less time you need for the same distance. Or think about workers and days: fewer men means more days required to finish the same work. These quantities move in opposite directions.
Let us see direct variation in action with a clear example.
A man earns four hundred rupees in ten days. How much will he earn in twenty-eight days?
First, find his daily earnings: 400/10 equals forty rupees per day. Then, for twenty-eight days: 28 × 40 equals one thousand one hundred twenty rupees. Notice how more days meant more earnings — this confirms direct variation.
Here is another example with decimals. Zero point seven five metres of cloth costs forty-five rupees. What is the cost of zero point six metres?
First, find the cost per metre: 45/0.75 equals sixty rupees per metre. Then, zero point six metres costs zero point six multiplied by sixty, which equals thirty-six rupees.
Now, let us turn to inverse variation, where the logic reverses.
Four men can complete a piece of work in five days. How many men would be needed to finish it in four days?
First, find the work in terms of man-days: one man would take 4 × 5, which equals twenty days. So, to finish in four days, you need 20/4, which equals five men. Notice: fewer days required more men — this is inverse variation.
Consider another classic inverse variation problem. With a speed of sixty kilometres per hour, a journey takes four hours. What speed is needed to complete the same journey in three hours?
First, find the total distance: 60 × 4 equals two hundred forty kilometres. Then, for three hours, the speed must be 240/3, which equals eighty kilometres per hour. Higher speed meant less time — inverse variation at work.
The unitary method also handles fractions elegantly.
Suppose 8/15 of a cargo costs two thousand rupees. What is the cost of 3/5 of the same cargo?
First, find the cost of the whole cargo: 2000 ÷ 8/15 equals 2000 × 15/8, which equals three thousand seven hundred fifty rupees. Then, 3/5 of this equals 3750 × 3/5, which equals two thousand two hundred fifty rupees.
Time calculations also follow this method. A watch gains forty-two seconds in three days and eight hours — that is eighty hours total. How long will it take to gain two minutes six seconds, which is one hundred twenty-six seconds?
First, find time per second gained: 80/42 hours per second. Then for one hundred twenty-six seconds: 80/42 × 126 equals two hundred forty hours, or ten days.
Now we arrive at a fascinating extension: Time and Work.
Here is the key insight: if a person completes a work in fifty days, their one day's work equals 1/50 of the total work. Conversely, if someone's one day work is 1/50, they need fifty days to finish.
So remember these fundamental formulas. One day's work equals one divided by the number of days required to complete the work. And, the number of days required equals one divided by one day's work.
Also, time required for a certain work equals the work to be done divided by the work done in unit time.
Let us apply this. Person A completes work in four days, and person B in six days. How long if they work together?
A's one day work is 1/4. B's one day work is 1/6. Together, they complete 1/4 + 1/6 equals 5/12 per day. So they need 12/5 days, which is 2 2/5 days.
Here is another scenario. Ajay and Vijay together paint a hall in six days. Ajay alone takes eight days. How long for Vijay alone?
Together they do 1/6 per day. Ajay does 1/8 per day. So Vijay does 1/6 − 1/8 equals 1/24 per day. Therefore, Vijay alone needs twenty-four days.
Finally, consider pipes and cisterns. One tap fills a cistern in four hours. Another empties it in six hours. If both are open, what happens?
The filling tap contributes 1/4 per hour. The emptying tap removes 1/6 per hour. Net effect: 1/4 − 1/6 equals 1/12 per hour. So the cistern fills in twelve hours.
Let us recap the essential takeaways from today's lesson.
First, the unitary method means finding the value of one unit first, then scaling to what you need. Second, direct variation: quantities increase or decrease together. Third, inverse variation: one quantity rises as the other falls. Fourth, in time and work problems, one day's work equals one divided by total days. Fifth, combined work adds up individual work rates. Sixth, for filling and emptying together, subtract the rates appropriately.
You now have a powerful toolkit for solving a wide range of practical problems. Keep practicing, stay curious, and remember — mathematics is about understanding relationships, not just memorising steps. Until next time, happy learning!