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Chapter 7 · Proportional Reasoning-1

Sharing a whole in a given ratio

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9 min.

Also recorded in Hindi.Englishहिन्दी

A ratio counts groups, not objects. Share twelve counters as 3 : 1 and you have been told about groups, not counters.

The idea

A ratio counts groups, not objects. Told to share in the ratio 3 : 1, you are being told that one share is made of three equal groups and the other of one group of the same size — so the whole is four groups, and once you divide the whole by four the shares are no longer a matter of choice. That single step, turning the ratio into a group count, is the whole method, and it explains both why the ratio alone can never say how much anybody gets and why the answer, once the whole is fixed, is the only answer there is: two shares that must add to the whole and must stand in the given ratio leave exactly one group size available.

What you should be able to do

  • Divide a given quantity between two shares in a stated ratio, showing the group size as a separate step
  • Explain why the number of groups is the sum of the terms of the ratio
  • Verify an answer by reducing the ratio of the two shares back to the given ratio
  • Use the general form — m × x / (m + n) and n × x / (m + n) — and say what each factor is counting
  • Recover the amounts in a mixture from its total and its ratio
  • Change a mixture to a new ratio by adding to one ingredient only, identifying first which quantity is unchanged
  • Show that a ratio by itself does not determine the shares, using two problems with the same ratio and different wholes
  • Notice when a sharing problem needs an assumption that the question does not state

Words to know

TermDefinition in one lineFirst introduced
wholethe total quantity being shared outprinted in this chapter (Part I, §7.5, p.173), where it is set in quotation marks in the diagram
partsthe two shares the whole is divided intoprinted in this chapter (Part I, §7.5, p.173)
countersthe countable objects — coins, seeds or pebbles — used in the activityprinted in this chapter (Part I, §7.5, p.172)
simplest formthe reduced ratio, used here both to set the problem and to check the answerprinted in this chapter (Part I, §7.3, p.161)
group sizethe common size of the equal groups a ratio countsdescribed but not named in this chapter; Part I p.174 speaks of the size of each group without making a term of it
number of groupsthe sum of the terms of the ratioprinted in this chapter (Part I, §7.5, pp.173–174)
unchanged quantitythe ingredient that stays fixed while a mixture is adjustedan added term for the anchor of Example 12; the chapter has no label for it

Where people slip up

  • "Sharing in the ratio 3 : 1 means giving one person 3 and the other 1." That is one round of dealing, not the answer, unless the whole happens to be 4.
  • "The ratio tells me the shares." It does not until the whole is named. The chapter puts 3 : 1 on 12 counters and on ₹4,000 within three pages.
  • "Divide by the bigger term." The divisor is the sum of the terms, because that is how many groups the whole contains.
  • "Divide by 2, because there are two people." Only for 1 : 1. The two people do not get equal groups; they get equal-sized groups in unequal numbers.
  • "3 : 1 means three quarters and one quarter, so the ratio is a pair of fractions." The fractions come out of the ratio and are not the ratio. Keep the group step visible or students will guess the wrong denominator.
  • "To change a mixture from 3 : 1 to 5 : 2, change both ingredients." The problem says only cement is added, so the sand is the anchor. Choosing the wrong anchor is the commonest error on Example 12.
  • "Adding to a mixture is the same as adding to both terms of a ratio." Adding yellow paint adds to one term only. This is where this topic and the ratio-is-not-a-difference topic meet.
  • "The answer must come out in whole objects." For counters and cups of rice it does; for a length it need not — a height of 152 cm split 4 : 6 gives 60.8 cm and 91.2 cm, and the answer is still correct.
Transcript1,364 words

Put twelve counters on a table and share them between two people. Six each is the obvious one, and written as a ratio that is six to six, which reduces to one to one. But it is only one of many, and none of the others is any less a sharing. Three and nine. Eight and four. Two and ten. Each of those is a perfectly good way of dividing the pile between two people.

Every one of those is a genuine split, because every one of them still adds to twelve. Five and seven is a split too, and its ratio does not reduce at all. So the question is never how many splits there are, because there are as many as you like. The question is what a ratio is telling you to do when it is handed to you as an instruction.

A ratio can be a description of a split that already happened. Nine and three describes a pile that has already been divided, and reduced it is three to one. But the same three to one can arrive the other way round, as an instruction. Share these twelve counters in the ratio three to one. Now nothing has happened yet, and something has to be worked out. And the first thing most people do with that instruction is wrong.

They give one person three counters and the other person one, and stop. Handing out three and one is not the answer. It is one round of dealing. So keep dealing. Three to you, one to me, and eight are left on the table. Three to you, one to me, and four are left. Three to you, one to me, and the table is empty. Three rounds, so nine counters against three.

Nine and three is the answer, and notice that nobody chose it. The dealing chose it, and the dealing stopped exactly when the counters ran out, which is the only thing that ever decided the answer. Now look at what the dealing actually did, because it did something very simple. Every round moved four counters off the table. Three to one person, one to the other. Four is not a number anybody wrote down. It is three plus one.

And twelve came apart into three lots of four. Which is the same twelve counters seen the other way round, as four groups of three, and a group is three counters. One person got three of those groups, and the other got one. That is what a ratio is counting, and it is worth saying plainly. Not objects. Groups. So there are four things on the board and only one of them was ever in doubt.

The whole, which is twelve. The ratio, three to one, which is four groups. The size of one group, which is twelve divided by four, and that is three. And the parts, which are three groups and one group, so nine and three. Everything hangs on the middle step, and the middle step is a division nobody mentions when they state the problem. Add the terms of the ratio, and that is how many groups you are dividing into.

Try that on a bigger one. Forty-two, shared in the ratio four to three. You could deal it. Four and three is seven off the table each round, and forty-two takes six rounds. It works. It is just slow, and slow is not the real problem. Take ten counters and share them in the ratio three to one. Deal them and you take three, I take one, and six are left. Again, and two are left.

Two is not enough for another round, and the dealing has nowhere to go. So dealing is not the method at all. It was only ever a way of watching the method happen slowly enough to see it. Go back to the groups, and forty-two stops being difficult. Four to three is seven groups. Forty-two divided by seven is six, so each group is six. One share is four of those groups and the other is three.

Four sixes is twenty-four, and three sixes is eighteen. Twenty-four and eighteen, without dealing out a single counter, and it took one division and two multiplications. And the ten that defeated the dealing? Four groups, each of two and a half, so seven and a half against two and a half. That is now general, so write it for any whole and any ratio. Call the whole x, and the ratio m to n.

The number of groups is m plus n. The size of one group is x divided by m plus n. And the two parts are m times that, and n times that. Three symbols, one division, two multiplications, and nothing else in it at all. The only thing you have to see to use it is that m plus n is a count of groups and not a count of anything you can pick up.

There are two checks worth doing, and they are not the same check twice. The first is that the two parts add back to the whole. The second is that the two parts, reduced, give back the ratio you were asked for. Either one of them, on its own, can be passed by an answer that is wrong. Six and six adds to twelve and reduces to one to one, so it passes the first and fails the second.

Six and two reduces to three to one and adds to eight, so it passes the second and fails the first. Only nine and three does both, and that is not luck. Two conditions, one unknown group size, and exactly one value of it survives. Two people set up a food cart. One puts in seventy-five thousand and the other puts in twenty-five thousand. The first month makes four thousand in profit, to be shared the way the money went in.

Seventy-five thousand to twenty-five thousand reduces to three to one. Which is the counters again. Four groups. But the group is four thousand divided by four, and that is a thousand. So the shares are three thousand and one thousand. Same ratio as the counters, same four groups, and completely different amounts. The ratio was never going to tell you the amounts, and it was never trying to. Forty kilograms of mixture holds sand and cement in the ratio three to one.

Four groups, ten kilograms each, so thirty of sand and ten of cement. Now it has to be five to two instead, and only cement may be added. The sand is untouched, which makes thirty the number everything else has to fit around. In a five to two ratio, the second amount is two fifths of the first. Two fifths of thirty is twelve, so twelve kilograms of cement are wanted and ten are there already.

Add two, and notice what that step did. It compared the two terms inside one ratio, rather than comparing one ratio with another. Five to try, and the last one is not like the others. Four thousand five hundred in the ratio two to three is eighteen hundred and two thousand seven hundred. Two hundred and forty millilitres of acid and water at one to five is forty and two hundred.

Forty millilitres of green at three to five is fifteen of blue and twenty-five of yellow, and pouring in twenty more of yellow makes fifteen to forty-five, which is one to three. Six cups of rice and lentils at two to one is four and two. And a bucket of orange paint, three parts red to five parts yellow, with another bucket of yellow poured in. What is the new ratio?

If the second bucket is the same size as the first it is three to thirteen. If it is half the size it is one to three, and if it is twice the size it is one to seven. The question never says, so the answer is not in the arithmetic. It is in an assumption, and the useful habit is to say which one you made out loud before you start.

Where this fits

Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.

Builds on

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