Proportional Reasoning – 2

Proportional reasoning comes down to one question asked again and again: when one quantity changes, what happens to the other? Before reaching for any formula, decide which of three situations you are looking at. Getting that decision right is most of the work, because the arithmetic that follows is usually short.

Direct proportion. Two quantities are in direct proportion when multiplying one of them by a number multiplies the other by the same number. A scooter that steadily gives 45 km on one litre of petrol is the standard example. Two litres carry you 90 km, five litres carry you 225 km and eight litres carry you 360 km. Divide the distance by the petrol in each case and the same answer keeps appearing: 90 ÷ 2 = 45, 225 ÷ 5 = 45 and 360 ÷ 8 = 45. That repeated answer is the signature of direct proportion.

Inverse proportion. Two quantities are in inverse proportion when multiplying one of them by a number divides the other by the same number. Suppose 60 sweets are shared equally in a class. Four children get 15 each, five children get 12 each, six children get 10 each and ten children get 6 each. Dividing tells you nothing useful here, but multiplying does: 4 × 15 = 60, 5 × 12 = 60, 6 × 10 = 60 and 10 × 6 = 60. The product stays fixed because the pile of sweets never changes.

Neither of the two. Plenty of everyday pairs are connected without being proportional at all, and a maths question will happily test whether you noticed. A taxi charges ₹50 as a fixed booking fee and ₹18 for every kilometre. Five kilometres cost 50 + 90 = ₹140, while ten kilometres cost 50 + 180 = ₹230. Doubling the distance did not double the fare, because the fixed ₹50 rides along unchanged. Your age and your height are connected as well, but nobody is twice as tall at sixteen as at eight.

The quick test. Take any pair of values from the question and ask what happens if the first quantity doubles. If the second doubles too, the proportion is direct. If the second halves, it is inverse. If neither happens cleanly, look for a fixed amount hiding inside the problem, like that booking fee.

Worked example. A typist types 24 pages in 2 hours, 36 pages in 3 hours and 60 pages in 5 hours. Dividing gives 24 ÷ 2 = 12, 36 ÷ 3 = 12 and 60 ÷ 5 = 12, so pages and hours are in direct proportion, at 12 pages an hour. Now take a bus that covers one fixed route in 8 hours at 30 km/h, in 6 hours at 40 km/h and in 4 hours at 60 km/h. Multiplying gives 30 × 8 = 240, 40 × 6 = 240 and 60 × 4 = 240, so speed and time are in inverse proportion, and the route is 240 km long. The constant you find is never just a number; it always means something, here the length of the route.

Once you have decided that a relationship is direct, the whole question collapses into finding a single number: the constant of proportionality, usually written as k. It is the second quantity divided by the first, and unless the two quantities are measured in the same unit it carries a unit of its own that tells you exactly what it means.

If y is directly proportional to x, then y ÷ x = k for every matching pair, which can be rewritten as y = k × x. In the scooter example k = 45 and its unit is kilometres per litre. For a shopkeeper selling identical pens, k is the price of one pen in rupees. Naming the unit out loud is a good habit, because it stops you from dividing the wrong way round.

Worked example. Twelve identical pens cost ₹216. What do seven pens cost, and how many pens can be bought for ₹450? Start with k: 216 ÷ 12 = 18, so one pen costs ₹18. Seven pens then cost 18 × 7 = ₹126. For the second part, use the constant the other way round: 450 ÷ 18 = 25 pens. One constant answered both questions, which is exactly why finding k first is worth the trouble.

The same example without finding k. You can also compare the two situations directly. Write 12 pens for ₹216 against 7 pens for ₹y, and set the ratios equal: 12 ÷ 216 = 7 ÷ y. Cross-multiplying gives 12 × y = 216 × 7 = 1,512, so y = 1,512 ÷ 12 = ₹126, the same answer as before. Use whichever route you find safer, but check that the two sides of your proportion are arranged the same way, pens above rupees on both sides.

Worked example in rupees per kilogram. Fifteen kilograms of rice cost ₹1,080. Then k = 1,080 ÷ 15 = 72, so rice costs ₹72 per kilogram. Four kilograms cost 4 × 72 = ₹288. Going backwards, ₹1,800 buys 1,800 ÷ 72 = 25 kg. Check the last step by multiplying back: 72 × 25 = 1,800, which matches.

Worked example with a decimal answer. A machine fills 840 bottles in 6 hours. Its rate is 840 ÷ 6 = 140 bottles an hour. In 5 hours it fills 140 × 5 = 700 bottles. To fill 1,050 bottles it needs 1,050 ÷ 140 = 7.5 hours, which is 7 hours and 30 minutes. Keep the units matched throughout: a rate written per hour must meet a time written in hours, never in minutes.

A shortcut worth knowing. If 5 kg of dal costs ₹350, then 20 kg is four times as much, so it costs 350 × 4 = ₹1,400. You never had to work out the price of one kilogram. Whenever the new amount is a neat multiple of the old one, scaling straight across is faster and safer than dividing.

Inverse proportion has its own constant, but this time it is a product rather than a quotient. If x and y are in inverse proportion then x × y = k for every matching pair, which can be rewritten as y = k ÷ x. Comparing two situations gives the working rule x₁ × y₁ = x₂ × y₂, and almost every inverse question is that one line plus a division.

Worked example with workers and days. Fifteen workers build a wall in 8 days. The constant is 15 × 8 = 120, and its unit is worker-days, meaning the wall needs 120 days of one person's work however you arrange it. With 20 workers the job takes 120 ÷ 20 = 6 days. If the wall must be ready in 5 days, you need 120 ÷ 5 = 24 workers. Notice that both answers came from the same number 120.

Worked example with speed and time. A car covers a certain trip in 5 hours at 60 km/h. The constant is 60 × 5 = 300, which is the distance in kilometres. At 75 km/h the same trip takes 300 ÷ 75 = 4 hours. Before writing the answer, check the direction: the car went faster, so the time had to come down, and 4 hours is indeed less than 5 hours.

Worked example with pipes. Six identical pipes fill a tank in 1 hour 20 minutes. Convert to one unit first: 1 hour 20 minutes is 80 minutes, so the constant is 6 × 80 = 480 pipe-minutes. Eight pipes would take 480 ÷ 8 = 60 minutes, that is exactly 1 hour. Ten pipes would take 480 ÷ 10 = 48 minutes.

Worked example with supplies. A hostel has food enough for 24 students for 15 days. The constant is 24 × 15 = 360 student-days. If 6 more students join on the very first day there are 30 students, so the food lasts 360 ÷ 30 = 12 days. More mouths, fewer days, exactly as expected.

Two cautions. First, inverse proportion needs a genuinely fixed total, such as one wall, one tank or one stock of food; if the job itself grows, the simple product rule no longer applies on its own. Second, these questions always assume everyone works at the same steady rate, which is a mathematical assumption rather than a fact about real building sites. State it if a question asks you to explain your reasoning.

Time and work questions are inverse proportion wearing a costume, and they all give way to one idea: stop asking how long the whole job takes, and ask instead how much of the job gets done in one day.

If A can finish a job alone in 10 days, then in one day A finishes 1/10 of it. If B can finish the same job alone in 15 days, then B finishes 1/15 of it in a day. Working together they finish 1/10 + 1/15 in a day. The LCM of 10 and 15 is 30, so this is 3/30 + 2/30 = 5/30 = 1/6. If they complete one sixth of the job every day, the whole job takes 6 days.

The LCM method, which avoids fractions. Pretend the job is built from a convenient number of units, namely the LCM of the given days. Here take 30 units of work. A does 30 ÷ 10 = 3 units a day and B does 30 ÷ 15 = 2 units a day, so together they do 5 units a day and finish 30 units in 30 ÷ 5 = 6 days. Same answer, and no fraction ever appeared. Because it stays in whole numbers, this version is usually quicker to write out under time pressure.

Worked example with an untidy answer. A takes 12 days and B takes 18 days. The LCM of 12 and 18 is 36, so call the job 36 units. A does 36 ÷ 12 = 3 units a day, B does 36 ÷ 18 = 2 units a day, and together 5 units a day. The job takes 36 ÷ 5 = 7.2 days, that is 7 and one fifth days. Checking with fractions: 1/12 + 1/18 = 3/36 + 2/36 = 5/36, and turning 5/36 upside down gives 36/5 = 7.2. The two methods agree.

When part of the work is already done. Suppose A works for 4 days on a job he alone would finish in 10 days, then leaves. He has completed 4 × 1/10 = 4/10 = 2/5 of it, so 3/5 is left. B alone would need 15 days for the whole job, so B does 1/15 in a day. The days B needs are (3/5) ÷ (1/15) = 3/5 × 15 = 9 days.

Pipes and leaks. A leak is simply negative work. If a pipe fills a tank in 6 hours it adds 1/6 of the tank an hour, and if a leak empties the full tank in 9 hours it removes 1/9 an hour. Together the net filling is 1/6 − 1/9 = 3/18 − 2/18 = 1/18 of the tank an hour, so the tank fills in 18 hours instead of 6.

Sharing the wages. A and B, who alone need 10 and 15 days, finish a job together in 6 days and are paid ₹3,000 in all. Wages follow the work done, and their daily rates were 3 units and 2 units, a ratio of 3 : 2. Over 6 days A does 18 units and B does 12 units, which adds to the full 30 units. So A receives 3/5 of 3,000 = ₹1,800 and B receives 2/5 of 3,000 = ₹1,200, and the two shares add back to ₹3,000.

Speed is a constant of proportionality that you already use every day. It is distance divided by time, and it tells you how many kilometres belong to one hour. From that single sentence come all three forms: speed = distance ÷ time, distance = speed × time and time = distance ÷ speed.

Worked example. A bus covers 240 km in 4 hours, so its speed is 240 ÷ 4 = 60 km/h. At that speed it would cover 60 × 3.5 = 210 km in three and a half hours, and it would need 350 ÷ 70 = 5 hours for a 350 km journey at 70 km/h.

Which proportion is hiding here. If the speed is fixed, distance and time are in direct proportion, so doubling the time doubles the distance. If the distance is fixed, speed and time are in inverse proportion, so their product stays equal to that distance. A 300 km trip takes 300 ÷ 50 = 6 hours at 50 km/h and 300 ÷ 60 = 5 hours at 60 km/h, and 50 × 6 = 60 × 5 = 300 confirms it. Deciding which quantity is fixed is the whole trick.

Changing units. To turn km/h into m/s, multiply by 5/18, because 1 km/h is 1000 m in 3600 s, and 1000/3600 simplifies to 5/18. So 72 km/h is 72 × 5 ÷ 18 = 360 ÷ 18 = 20 m/s, and 54 km/h is 54 × 5 ÷ 18 = 270 ÷ 18 = 15 m/s. Going the other way, multiply by 18/5: a runner at 15 m/s is moving at 15 × 18 ÷ 5 = 270 ÷ 5 = 54 km/h. The two conversions undo each other, which is a handy way to check yourself.

Average speed, and the usual mistake. Average speed is total distance divided by total time. Whenever the two legs of a journey take different times, it is not the average of the two speeds. A cyclist rides 30 km to a town at 15 km/h, taking 30 ÷ 15 = 2 hours, and returns the same 30 km at 10 km/h, taking 30 ÷ 10 = 3 hours. The whole trip is 60 km in 5 hours, so the average speed is 60 ÷ 5 = 12 km/h. Averaging 15 and 10 would have given 12.5 km/h, which is wrong because the cyclist spent more hours at the slower speed.

Working with clock times. A family leaves home at 7:30 am for a place 45 km away and drives at 30 km/h. The journey takes 45 ÷ 30 = 1.5 hours, that is 1 hour 30 minutes, so they arrive at 9:00 am. Convert the decimal part of an hour into minutes before you write a clock time, since 1.5 hours is 1 hour 30 minutes and not 1 hour 5 minutes.

Scaling is direct proportion applied to a whole set of numbers at once. The rule is short: find the multiplier once, then apply the very same multiplier to every quantity. The usual slip is scaling some of the numbers and forgetting the rest.

Scaling a recipe. A kheer recipe for 4 people uses 1 litre of milk, 100 g of rice and 150 g of sugar. To cook for 10 people the multiplier is 10 ÷ 4 = 2.5, so you need 2.5 litres of milk, 100 × 2.5 = 250 g of rice and 150 × 2.5 = 375 g of sugar. For 6 people the multiplier is 6 ÷ 4 = 1.5, giving 1.5 litres of milk, 150 g of rice and 225 g of sugar. If milk costs ₹60 a litre, the milk for the ten-person version costs 2.5 × 60 = ₹150.

Reading a map. A map often states its scale in words, such as 1 cm represents 25 km. Two towns drawn 6.4 cm apart are then 6.4 × 25 = 160 km apart on the ground. Going the other way, a real distance of 90 km appears as 90 ÷ 25 = 3.6 cm on the map.

Reading a ratio scale. A scale written as 1 : 50,000 means one unit on the map stands for 50,000 of the same units on the ground. So 1 cm on the map is 50,000 cm on the ground, which is 500 m or 0.5 km. A gap of 7 cm on that map is 7 × 0.5 = 3.5 km. To go backwards, 12 km would be drawn as 12 ÷ 0.5 = 24 cm. Always convert the ground measurement into sensible units before answering, since nobody reports a distance in centimetres.

Models and enlargements. A model car is built to the scale 1 : 24. A real car 4.2 m long is 420 cm long, so the model is 420 ÷ 24 = 17.5 cm long. A photograph 12 cm wide and 8 cm tall is enlarged so that its width becomes 30 cm. The multiplier is 30 ÷ 12 = 2.5, so the height must become 8 × 2.5 = 20 cm. Using a different multiplier for the height would stretch faces out of shape, which is why the same factor must be used for both sides.

One caution about area. Lengths and areas do not scale by the same number. On a floor plan drawn at 1 cm to 2 m, a room drawn 4.5 cm by 3 cm is really 9 m by 6 m. Its drawn area is 4.5 × 3 = 13.5 square centimetres, while its real area is 9 × 6 = 54 square metres. Each square centimetre stands for 2 m by 2 m, which is 4 square metres, and 13.5 × 4 = 54 confirms the answer. Lengths were multiplied by 2 metres per centimetre, but areas were multiplied by 4.

Harder questions change two things at once: more workers and a bigger job, or more machines and a longer shift. There are two reliable ways through, and it is worth knowing both so that each can check the other.

Method one: find the amount of work in one unit. Combine the people and the time into a single quantity such as worker-days or machine-hours, then work out how much gets produced per worker-day. Six tailors stitch 45 shirts in 5 days. That is 6 × 5 = 30 tailor-days for 45 shirts, so one tailor-day produces 45 ÷ 30 = 1.5 shirts. For 108 shirts you need 108 ÷ 1.5 = 72 tailor-days, and with 9 tailors that is 72 ÷ 9 = 8 days.

Method two: change one thing at a time. Keep the tailors at six and increase only the shirts. The order grew from 45 to 108 shirts, a multiplier of 108 ÷ 45 = 2.4, and shirts and days are in direct proportion, so the days grow to 5 × 2.4 = 12 days. Now change only the number of tailors, from 6 to 9. Tailors and days are in inverse proportion, so 6 × 12 = 72 and the days become 72 ÷ 9 = 8. Both methods land on 8 days, so the answer is safe.

Worked example with machines. Five machines pack 300 boxes in 2 hours. That is 5 × 2 = 10 machine-hours for 300 boxes, so one machine packs 300 ÷ 10 = 30 boxes an hour. Eight machines running for 3 hours give 8 × 3 = 24 machine-hours, so they pack 24 × 30 = 720 boxes. If instead 900 boxes must be packed in 5 hours, you need 900 ÷ 30 = 30 machine-hours, and 30 ÷ 5 = 6 machines.

Worked example where the situation changes midway. A camp of 500 people has food for 30 days, which is 500 × 30 = 15,000 person-days of food. After 10 days they have used 500 × 10 = 5,000 person-days, leaving 10,000. At that moment 100 people leave, so 400 remain, and the food that is left will last 10,000 ÷ 400 = 25 more days. Do not divide the original 30 days by anything; only the food still in store can be shared out.

Checking your direction. After every step, ask whether the answer moved the way common sense expects. More workers should mean fewer days. A bigger order should mean more days. A faster speed should mean less time. If a step pushes the number the wrong way, you have multiplied where you should have divided, and catching it right there costs you ten seconds instead of the whole question.