Unitary Method and Chain Rule

This is the foundation idea put to work. Whatever the question gives you, bring it down to one unit of the thing that is asked about, and then build up.

A scooter runs 45 km on 3 litres of petrol. One litre gives 45 ÷ 3 = 15 km, so 7 litres give 15 × 7 = 105 km.

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Now turn the question round. If the money is given and the quantity is asked, you divide in the second step. Radhika pays ₹96 for 8 kg of potatoes, so 1 kg costs ₹12. With ₹156 she can buy 156 ÷ 12 = 13 kg.

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Many questions need two or three hops. "Distance per litre is known, petrol price is known, find the cost of a trip" means km → litres → rupees. Write each hop on its own line and keep the units with every number. You will never multiply where you should divide.

Units first. A dozen is 12, a score is 20, 1 hour is 60 minutes, 1 kg is 1,000 g. Change everything into the same unit before the first division. A machine rate given per minute and a time given in hours is the most common slip in this type.

You do not always need the value of one. If the asked quantity is a simple multiple of the given one (32 litres is 8 times 4 litres), multiply straight away. The unitary method is the safe road; the multiple is the short cut when it is visible.

When a question says directly proportional, varies directly or at the same rate, it is telling you that the ratio of the two quantities does not change.

If y ∝ x, then yx = k, a fixed number. So y₁x₁ = y₂x₂, which means y₂ = y₁ × x₂x₁. You multiply the old answer by the new-to-old ratio of the other quantity.

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Look at the table in the figure. Quantity goes from 4 to 8 (× 2), so cost goes from 60 to 120 (× 2). Quantity goes from 8 to 12 (× 1.5), so cost goes from 120 to 180. Down every column, cost ÷ quantity is 15.

This one idea covers many settings:

• Maps and scale. 2 cm stands for 5 km, so 9 cm stands for 5 × 92 = 22.5 km. For an area on the map, both length and breadth are scaled, so area is scaled by the square of the map ratio.
• Parts of a whole. If 34 of a drum holds 30 litres, the whole drum holds 30 ÷ 34 = 40 litres, and 12 of it holds 20 litres.
• Exchange. "7 pens cost the same as 4 diaries" gives a fixed ratio of prices. Change everything into one item, then use the unitary method.
• Groups in the same proportion. If 6 apples and 4 mangoes cost ₹150, then 3 apples and 2 mangoes (exactly half of each) cost ₹75. You do not need the price of either fruit.

Before you use the rule, make sure both quantities really rise together in the same proportion. A taxi fare with a fixed starting charge rises with distance, but it is not directly proportional to it, because the ratio changes.

In inverse variation one quantity goes up exactly as fast as the other comes down, so their product never changes.

If y ∝ 1x, then x × y = k. So x₁ y₁ = x₂ y₂, and y₂ = y₁ × x₁x₂. Notice the fraction is now old-to-new, the opposite of direct variation.

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The rectangles in the figure have the same area, 60. Make one side half as long and the other side must become twice as long. Every inverse question is this picture in disguise. Ask yourself what the fixed "area" is:

• Men and days for the same job. Men × days is fixed (the total work). Fewer men, more days.
• Speed and time for the same distance. Speed × time = distance, which is fixed.
• Price and quantity for the same money. Price × quantity = money spent. A higher price buys less.
• Rows and seats for the same hall. Rows × seats per row = total seats.
• Size of one piece and number of pieces. A rope of fixed length, a floor of fixed area: bigger pieces, fewer of them. For square tiles, the area of one tile is side², so the number of tiles goes with the inverse of the square of the side.
• Wheels. Revolutions × circumference = distance. A wheel of double radius turns half as many times.

The Time and Work chapter works with men and days in full detail. Here the point is to recognise the fixed product in any setting, even when the words "inversely proportional" never appear in the question.

When a question asks "by how much must it increase", find the new value first and then subtract the old one. Many students stop at the new value.

A stock of food, fodder or fuel is a fixed amount, and the people (or animals) using it decide how long it lasts. More eaters, fewer days. That is inverse variation, and the fixed product is eaters × days.

The man-days formula for food questions is taught in the Time and Work chapter (Type 11, "The MDH formula"). Here we look at the same stock with the more–less view, which is quicker when the group changes in the middle.

Step 1. Use up the days that have passed. If food was for 40 days and 10 days have gone, the food left is enough for the same group for 30 days. This step is the whole question; most wrong answers skip it.

Step 2. Change the group. Now only the people change. More people means fewer days, so multiply the remaining days by old groupnew group.

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In the figure, 90 men have food for 40 days: 3,600 man-days. After 10 days, 900 man-days are used and 2,700 are left. If 30 men leave, 60 men share 2,700 man-days, which is 45 days. With the ratio: 30 × 9060 = 45 days.

Two more settings come up:

• Reinforcements. If extra people arrive and the food now lasts a given number of days, find the new group from the product, then subtract the old group.
• Cutting the ration. If the same people must make the food last longer, each person's daily share must fall in the inverse ratio of the days. From 30 days to 40 days, the share becomes 34 of the old share, which is a cut of 14.

When the group changes twice, repeat Step 1 and Step 2 for each phase, one after the other.

The chain rule joins several proportions into one line. When three or more quantities are linked, change one column at a time and multiply the effects.

In this type the unknown is an output: bags made, fodder eaten, milk given, money earned or a bill paid. For such an unknown, almost every other column works in the same direction. More workers make more bags; more days also make more bags. So every fraction has the bigger number on top when that column increases.

10 workers make 30 bags in 3 days. How many bags will 20 workers make in 6 days?

• Workers 10 → 20: more workers, more bags. × 2010
• Days 3 → 6: more days, more bags. × 63

Bags = 30 × 2010 × 63 = 120.

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The same result comes from the unitary method in steps: 1 worker makes 3 bags in 3 days, so 1 bag a day; 20 workers in 6 days make 20 × 6 = 120 bags. The chain line does all the steps at once.

As a formula, when output Q depends directly on M and D (and H, if hours are given):

Q₁M₁ × D₁ × H₁ = Q₂M₂ × D₂ × H₂

Wages and bills are outputs too. A mess bill grows with the number of students and with the number of days. Earnings grow with persons, hours a day and days. If a rate is "per person per day", the bill is people × days × rate, which is exactly this chain.

Sometimes the unknown is not the output but one of the inputs (how many typists are needed). Then that column is the answer and the output column becomes a direction question: more pages need more typists, less time needs more typists. That mixed case is the next type.

Now the unknown is an input: days, men or hours a day. Some columns push the answer up and some push it down, and you must decide each one separately.

The full men–days–hours–work formula, M₁ D₁ H₁W₁ = M₂ D₂ H₂W₂, is taught in the Time and Work chapter (Type 11, "The MDH formula"). Both methods give the same answer. The more–less line is handy when the unknown sits in the middle of the formula, or when one column (such as the work) is given as a fraction or a multiple.

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Read the figure as a balance. On the left, 12 men working 10 days of 8 hours finish 1 job. On the right, an unknown number of men work 15 days of 4 hours on 2 jobs. Both sides must give the same men × days × hours per job: 12 × 10 × 81 = M × 15 × 42, so M = 32.

Now the more–less way. 6 men finish a job in 20 days working 8 hours a day. How long will 8 men take for double the job, working 6 hours a day?

• Men 6 → 8: more men, fewer days. × 68
• Hours 8 → 6: fewer hours a day, more days. × 86
• Work 1 → 2: more work, more days. × 21

Days = 20 × 68 × 86 × 2 = 40.

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Workers of two kinds. If 3 men or 5 women can do a job in 24 days, then 3 men = 5 women in working power. Change everyone into one kind and then use the inverse rule. 3 men + 5 women = 3 men + 3 men = 6 men, which is double the original 3 men, so they need 24 ÷ 2 = 12 days.

Work given as a fraction. "Half the work" means the work column goes 1 → 12. "Thrice the work" means 1 → 3. Write it as a column like any other.

Sometimes a quantity depends on the square (or square root) of another, or on two quantities at once. The words tell you exactly how.

• "y varies directly as x" ⇒ y = kx
• "y varies inversely as x" ⇒ y = kx
• "y varies as the square of x" ⇒ y = kx²
• "y varies as the square root of x" ⇒ y = k√x
• "y varies directly as x and inversely as z" ⇒ y = kxz

Method 1: find k. Put the given values into the equation, find k, then use the same k for the new values. If y varies as x² and y = 12 when x = 2, then 12 = k × 4, so k = 3. When x = 5, y = 3 × 25 = 75.

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Method 2: ratio. Skip k and compare the two situations directly: y₂y₁ = (x₂x₁)². Here x goes 2 → 5, so y goes × 254: 12 × 254 = 75. For two variables, multiply the effects, exactly like the chain rule.

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Breaking into pieces. If the value of a stone varies as the square of its weight, breaking it always lowers the total value, because the squares of the parts add up to less than the square of the whole. A ₹36,000 diamond broken into two pieces in the ratio 1 : 2 has pieces worth (13)² and (23)² of the original, that is 19 and 49. Together they are worth 59 × 36,000 = ₹20,000, a loss of ₹16,000.

For a pizza priced by area, area varies as the square of the diameter. Doubling the diameter makes it four times as costly, not twice.

When the product of two quantities is fixed (price × quantity = money, speed × time = distance, wage × workers = budget), a percentage change in one forces a fixed change in the other. Write the change as a fraction and flip it.

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In the figure, the price falls by 20%, so it becomes 45 of the old price. The same money then buys 54 of the old quantity, which is 25% more. The rule in fraction form:

• If one quantity rises by 1n, the other falls by 1n + 1.
• If one quantity falls by 1n, the other rises by 1n − 1.

So a rise of 3313% (13) in price means 25% (14) less sugar for the same money.

Extra quantity for the same money. "A reduction of 10% lets a buyer get 2 kg more for ₹450." The 10% saved is ₹45, and it buys the extra 2 kg at the new price. New price = 45 ÷ 2 = ₹22.5 per kg. Original price = 22.5 ÷ 0.9 = ₹25 per kg.

Speed and time. If speed rises by 15 (20%), time falls by 16. If that saves 20 minutes, 16 of the usual time is 20 minutes, so the usual time is 2 hours.

Keeping expenditure the same. If the price rises by r%, consumption must fall by r100 + r × 100%. This is the same flip rule written with percentages.

These questions also sit in Percentage. Here the idea to see is that they are inverse variation: the product is fixed, so the factors move in opposite ratios.