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Chapter 7 · Finding the Unknown

Finding an unknown by reasoning about what must balance

यह वीडियो हिंदी में भी · Watch in Hindi

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10 min.

Also recorded in Hindi.Englishहिन्दी

Two balanced pictures, both hiding a weight. Only one of them yields to arithmetic you already have.

The idea

Every one of these hanging figures can be cracked before a single letter is written down — but not all of them by the same means, and the gap between the two kinds is what §7.1 is built on. Figs. 7.1 to 7.5 hang under a ringed total, so halving it and working inward settles them, and that is ordinary arithmetic. Figs. 7.6 to 7.8 take the ring away and the level bar alone still carries the answer; in Fig. 7.7 you can instead strike a matching star off each string, which is the new move showing up early. Then the two-pan scales arrive, and from Fig. 7.10 onward the sacks stand on both plates at once, where working inside one side gets you nowhere. What gets you somewhere is a permission the page supplies as a hint: take equal weights off each plate and the balance holds. §7.2 comes straight back to these pictures on Part II's p.169 and divides them along exactly that line — some gave up their answer directly, the rest needed the equal-removal fact — and it is the second group the symbols are written for.

What you should be able to do

  • Work out an unknown weight from a hanging figure and say which step did the work
  • Use the fact that equal removals from both sides leave the agreement intact
  • Handle a figure in which the unknown appears on both sides by removing a matched copy from each
  • Chain results: use an unknown found early in a figure to unlock a later one
  • Read a nested hanger, where one branch is itself a bar with two things on it
  • Explain why the same reasoning works whether the numbers are 5 and 7 or 90 and 500
  • Distinguish reasoning that forces an answer from guessing that happens to land on one
  • State what this picture-based method still cannot do, and why symbols are needed next

Words to know

TermDefinition in one lineFirst introduced
weighing scalethe hanging device, and later the two-pan balance, the chapter reasons withprinted in Part II, §7.1, p.164
balancedsaid of a scale whose two sides agree, so it does not tipprinted in Part II, §7.1, p.165
plateone of the two pans of the two-pan scaleprinted in Part II, §7.1, pp.165–166
unknown weightthe weight the figure does not tell youprinted in Part II, §7.1, pp.164–165
sackthe repeated unknown object on the two-pan scalesprinted in Part II, §7.1, p.165
units of weightthe undeclared unit the ringed numbers count inprinted in Part II, §7.1, p.164
equal weightsthe matched amounts that may be taken off both sides at onceprinted in Part II, §7.1, p.165
letter-numbera letter used in place of a number whose value is not yet knownprinted in Part II, §7.1, p.167, and again on p.168 — p.167 is inside this brief's own span, so citing p.168 was a page late
Math Talkthe chapter's marker for a prompt meant to be argued out loudprinted in Part II, §7.1, pp.165, 168
hangerthe explanation's word for the bar-and-string figure of Figs. 7.1 to 7.8an added word; the book names no device here beyond weighing scale
strippingthe explanation's shorthand for removing matched objects from both sidesan added shorthand; the book's verb is remove
two-pan scalethe balance with a pan hanging on each side, as drawn in Figs. 7.9 to 7.12the explanation's compound; the book prints weighing scale and plate, and gives the removal move no name of its own anywhere in this chapter — all 28 printed pages, 164 to 191, were read

Where people slip up

  • "You find the unknown by trying numbers until one fits." Nothing in §7.1 works that way, and the chapter says so out loud at the top of §7.2 when it contrasts these figures with trial and error (Part II, §7.2, p.169). Each figure forces its answer.
  • "A level bar means each side weighs half the ring number." True of the hangers in Figs. 7.1–7.5, where the bar is drawn level and the ring gives the total — and it is worth stating in that form. But it is not what "balanced" means in general: the opening picture on p.164 hangs 4 and 3 under a ring marked 7, and it tips. The halving is a consequence of the bar being level, not the definition of the device.
  • "Fig. 7.10 needs a different idea because the sacks are on both pans." It needs the same idea used once more. The book's hint says as much.
  • "Big numbers need a different method." Fig. 7.12 is 90 against 60 with a 500 kg block, and the move is identical to Fig. 7.9. If the explanation makes it look harder, the point of printing it is lost.
  • "An answer is right because it comes out to a whole number." Whole answers here are a design choice by the authors, not evidence. The instruction on p.165 asks for a reason, not for a tidy value.
  • "Once you can do the pictures you can do the algebra." Not yet. The pictures hand you a physical excuse for each move; §7.2 has to argue the same moves from the meaning of the equals sign, without a scale to lean on.
Transcript1,449 words

Two pictures. Both balanced. Both hiding a weight from you. One of them you can crack with arithmetic you already have. The other cannot be cracked that way, and the reason is the whole video. The first: a bar hanging from a ring, with things on two strings below it. The second: a scale with a pan on each side, and the same unknown sack on both pans. That difference is the whole difference.

Everything before it is counting, and everything after it needs a permission nobody has given you yet. Start with the hanging bar. It hands you two facts at once. The ring above it names the whole weight hanging below. And the bar hanging level says the two sides agree. Put those together. If the ring says twenty four and the bar is level, each side carries twelve. But be careful about why: it is not because there is a ring, it is because the bar is level.

Here is a ring marked seven, with four on one string and three on the other. The ring is telling the truth: four and three really is seven. And this bar tips, so there are no halves to take. Take a real one. The ring says twenty four. On the left string: a starfish, a striped fish, a starfish. On the right: a striped fish and a grey fish. One number is given to you. A starfish weighs two.

Halve the total first. Each side carries twelve. Now work inward on the left. Two and two is four, so the striped fish makes up the rest. It is eight. The right string also carries twelve, and eight of that is the striped fish. So the grey fish is four. Notice the shape: one unknown had to be found before the other could be. Here is one with a twist. The ring says eight.

On the left, a brick. On the right, a shorter bar hanging from the first, carrying two identical cards. Halve the eight. Each side carries four. So the brick weighs four. And now look at the little bar. It is not decoration. It is a hanger in its own right, carrying four, and hanging level. Which means its two sides agree as well, and each card is two. A branch of a picture can be a whole claim by itself, solved the same way one level down.

Two more, the same reasoning entered from opposite ends. Ring eighteen. On the left, a sun and four clouds. On the right, three lightning bolts. The sun weighs five. Each side carries nine. Five of the left is the sun, so four clouds share the other four. Each cloud is one. The right side then falls out: nine shared by three bolts, so each bolt is three. Now the reversal. Ring forty, and nothing is given to you at all.

On the left, a crown and five gems. On the right, four crowns. Each side carries twenty. Nothing to start with on the left: both things on it are unknown. But the right side is four crowns making twenty, so a crown is five. Carry it back. A crown and five gems make twenty, so each gem is three. Now take the ring away entirely. A bar held level in somebody's hand. No ring. No total anywhere.

On one side, a slice of watermelon. On the other, a banana, an orange and a second banana. You are told two things: the watermelon is ten, the orange is four. The bar is level, so the two sides agree. Ten on the left means ten on the right. Four of that ten is the orange, and the two bananas share what is left. Each banana is three. No ring, no halving, and still forced.

One more of those, and something new happens. Four stars hang on the left. On the right, a ring, a star, and a second ring. A star weighs four. Read it straight off and it works. Four stars is sixteen, so the right side is sixteen. The star there takes four, and twelve is shared by two rings. Each ring is six. But there is a shorter way, and it is the one to watch.

There is a star on each side. Take one star off the left and one off the right, together. The bar does not move, because you took the same weight off both sides. And what is left is smaller. Three stars against two rings. Twelve against two rings, so each ring is six. Same answer, different move. Hold on to that move. Now the picture changes: a scale with a pan on each side.

Same idea, standing on the ground instead of hanging. Left pan: one sack and a two kilogram weight. Right pan: a ten and a two. Nobody says what the sack weighs. That is what you want. There is a two kilogram weight on each pan. So lift them both off at the same instant. Watch the beam. It does not move. Now the left pan holds only the sack, and the right holds ten kilograms.

The sack is ten. Here is the one that breaks that method. Left pan: two sacks. Right pan: a ten, a four, and one more sack. Try halving. There is no total to halve. Try working inside the left pan. It holds nothing but sacks, and you do not know what a sack weighs. The trouble is that the thing you are chasing stands on both pans at once. So use the move from the stars. There is a sack on each pan.

Take one sack off the left and one off the right, together. The beam does not move. You took the same weight off both sides, even though you do not know what that weight was. And now one sack sits against fourteen kilograms. The sack is fourteen. Do it again, harder. Left pan: five sacks. Right pan: two ten kilogram weights, a one, and two more sacks. Twenty one kilograms and two sacks, against five sacks.

This time aim at something specific: get all the sacks onto one pan. There are two sacks on the right, and certainly two among the five on the left. So take two sacks off each pan. Three sacks left on one side, twenty one kilograms on the other. Three sacks make twenty one, so each sack is seven. And the one that looks frightening and is not. Left pan: ninety sacks and a fifty kilogram block. Right pan: sixty sacks and a five hundred kilogram block.

Nothing new is being asked; only the numbers changed. There are sixty sacks on each pan. Take sixty off each. There is fifty kilograms on each pan. Take fifty off each. What is left is thirty sacks against four hundred and fifty kilograms. Thirty sacks make four hundred and fifty, so a sack is fifteen. Two moves — exactly the two you used on the small one. The size of the numbers changed nothing about the method.

One habit to kill before we finish. Somebody looks at the two sack puzzle and says: I think it is fourteen. Two fourteens is twenty eight. Ten and four and fourteen is twenty eight too. It works. That check is true, and it is not the same thing as an argument. It tells you fourteen works. It tells you nothing about fifteen. To rule fifteen out you have to try it. Then sixteen. Then keep going.

Taking a sack off each pan does something different. It does not test one answer. It rules out every other one, in a single move. Guessing that happens to land is not the same as reasoning that forces. So: where this leaves you, and where it stops. Everything in this video was decided by two moves and only two. Take equal weights off both sides, and the balance holds. Then split what is left into equal parts.

That is a real method. Not guessing, and it worked on numbers of every size. But look at what did the work: a picture of a beam that does not move. The move is right because the scale would tip if it were not, and you can see that it does not. Next you make the same move with no scale in front of you. No beam, no pans, nothing to look at.

Just the claim that two things are the same size, and an argument that you may do the same to both. Because a picture can only show what you can draw, and a few hundred of anything on one arm stops being an argument.

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

Comes up again in

The book

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