NCERT Solutions for Class 8 Maths Chapter 3 A Story of Numbers (PDF) – 2027

Since the Stone Age — long before writing existed. The oldest evidence is bones with tally marks:

They counted whatever they had to keep track of, and whatever they had to predict.

• Food: the quantity of grain, fruit and vegetables in store.
• Livestock: the number of animals in the herd, so a missing animal could be spotted.
• Trade: details of goods exchanged with other groups.
• Rituals: the number of offerings given.
• Time: the passing days, so that the new moon, the full moon or the onset of a season could be known and predicted in advance.

About 2000 years ago, in India.

The Mesopotamian system is base 60. It has just two marks: a wedge for 1 (we write Y) and a corner wedge for 10 (we write <). Numbers up to 59 sit in a single position; from 60 onwards a second position to the left is used.

So 20 and 50 need only the ones position. But 100 is bigger than 60, so it needs two positions: one 60, and 40 left over.

Keep one stick for every cow, and match the sticks against the cows as they come back.

• In the morning, as each cow leaves, drop one stick into a pot. Nothing is counted and nothing is named — one cow, one stick.
• In the evening, as each cow returns, take one stick out of the pot.
• If the pot is empty when the last cow is in, every cow is back. If sticks are left over, that leftover collection is the number of missing cows.

Pair off the sticks — no counting is needed.

• Make our pile of sticks — one per cow. Let the neighbour make his.
• Now pair the piles off: take one stick from ours and one from his, again and again.
• Whichever pile runs out first belongs to the person with fewer cows. If both run out together, the herds are equal.

The sticks left over in the neighbour’s pile after the pairing are the answer.

Carry that leftover collection of sticks to the market, and buy one cow for each stick. Now the two herds can be matched exactly.

By pairing one collection against the other and looking at what is left.

Exactly as many as your language has letters — no more.

Whichever language you take, the answer is a fixed, finite number. Count the letters of your own alphabet and that is your limit.

Yes — the way the Roman system itself did it. Keep repeating the symbols, and bring in a fresh symbol whenever the repetition gets too long.

So the extension works in two moves: (a) repeat the symbol you already have, and (b) when about five repetitions have piled up, replace that whole group by a brand-new symbol.

Every one of the four operations can be done by moving sticks about. Call the two collections A and B.

Use all the strings of a given length before moving to the next length, in dictionary order.

The first two-letter string is aa, and it lands on 27 — exactly as the book suggests.

(your own system may look quite different): choose three symbols, each standing for a group ten times the one before.

Counting the nine boxes on page 57, row by row:

Up to about 4. Most humans find it difficult to take in a group of 5 or more objects in a single glance.

The difficulty is that the numeral stays almost as long as the number itself.

You would have to write the “five” symbol 269 times. Grouping by 5 shortens the tally from 1345 marks to 269, but that is still far too long to write, read or compare easily.

Group each number into landmark numbers, largest first. For 4s and 9s use the “one less” form (IV, IX, XL, XC, CM).

Pool the symbols of both numerals, then regroup from the smallest upwards — remembering that the Roman exchange rates are not all the same.

Now exchange, one landmark at a time:

There is no general shortcut, because the Roman landmark numbers are not the powers of any one number. Each product has to be worked out separately by repeated addition or by the distributive law.

First read the two numerals.

Break the second numeral into its landmark parts and use the distributive law:

Product = 427812. Written in Roman numerals it needs 427 M’s followed by DCCCXII.

Because in such languages the number word is still tied to the thing being counted — the count has not yet been separated from what is counted.

• The number word carries the object with it. One series for fish, another for coconuts or canoes tells the listener both how many and of what.
• Different goods come in different natural groups — pairs, bunches, baskets — so each series suits the way that good is stacked and traded.
• Some series were kept for special things, such as people or objects used in rituals.

urapon = 1, ukasar = 2. A number is a string: one ukasar for each 2 in it, then one urapon if it is odd.

• Add: join the two strings; two urapon make one ukasar.
• Subtract: strike off ukasar against ukasar and urapon against urapon (split an ukasar into two urapon if needed).
• Multiply: write the first string as many times as the second number says; then turn each pair of urapon into an ukasar.
• Divide: cut the string into equal bunches, each matching the divisor; the number of bunches is the answer.

Five features, and each of them fixes a specific Roman weakness.

Sample answer. Take the system from page 54 (● = 1, ■ = 10, ★ = 100) and refine it with the ideas of this section.

Refinement 1 — make the landmark numbers powers of one number. Ours already are: 1, 10, 100. So extend the same way instead of inventing unrelated jumps: the next landmarks must be 1000, 10000, 100000, and so on.

Refinement 2 — test it.

One problem remains: every new power of 10 still needs a new symbol. Section 3.4 removes it by letting the position of a symbol show its landmark.

The Egyptian symbols stand for the powers of 10:

Group each number into powers of 10 and draw that many of each symbol.

Count each kind of symbol and multiply by its landmark value.

Yes. Grouping 5 collections of the previous landmark each time gives landmarks that are the powers of 5 — a base-5 system:

Yes, by any positive integer n greater than 1: the landmarks are then 1, n, n², n³, … (a base-n system; the Egyptian one has n = 10). With n = 1 every landmark is 1, so nothing new forms.

Start from the largest landmark number that is smaller than 143. That is 5³ = 125, since the next one, 625, is too big.

The symbols are △ = 1, □ = 5, ⬡ = 25, ○ = 125, ∿ = 625, ↑ = 3125. Each time, take as many of the largest possible landmark as you can, then move down.

Yes — zero. There is no symbol for it, and there is no way to build it, because a numeral here is a collection of symbols and zero would have to be an empty collection, which cannot be seen on the page.

Every counting number 1, 2, 3, … can be written: take the largest landmark that fits, subtract, and repeat until nothing is left.

Start at 1 and multiply by 7 each time.

So the landmark numbers of a base-7 system are 1, 7, 49, 343, 2401, 16807, …

In general, the landmark numbers of a base-n system are the powers of n:

(i) Read the two numerals first.

(ii)

Read the two numerals:

Now pool them, symbol by symbol:

Split each numeral into the landmark numbers it is made of, multiply every landmark of the first by every landmark of the second, then pool the results and regroup.

This works because the product of two landmark numbers is again a landmark number (10^a × 10^b = 10^a+b), so each small product is a single symbol.

Multiplying a landmark by 10 gives the next landmark, because it raises the power of 10 by one.

Multiplying by 10² moves a landmark up by two steps.

Add the powers of 10 each time.

Thus the product of any two landmark numbers is another landmark number.

Yes in the base-5 system, and yes in every base-n system.

Every symbol in the numeral is replaced by the next symbol up, and the number of each kind stays the same.

(i) The numeral in the bracket has 5 coils, 2 arches and 2 strokes: 500 + 20 + 2 = 522. Multiplying by ∩ (10), each symbol steps up to the next one:

(ii) lotus ∩ = 1000 + 10 = 1010

Replace every symbol by the next symbol in the list, keeping the count of each symbol exactly as it was.

In Hindu numerals the very same rule reads: write a 0 at the end.

Slide the counters from the right of the vertical partition across onto the matching lines on the left, then tidy up each line.

On the board, 2907 is on the left and 43 on the right:

Two exchanges keep the board readable:

• Ten counters on a line are taken off and replaced by one counter on the next line up — the next power of 10.
• Five counters on a line are taken off and replaced by one counter just above that line, since a counter above a line stands for 5.

No — not in the tidied-up form that the Egyptians actually wrote.

So ten identical symbols would always be swapped for one symbol of the next kind. Each symbol can therefore appear at most 9 times.

(The one exception is the largest symbol, the sun, 10⁷. There is no symbol for 10⁸, so numbers of ten crore or more would need the sun 10 or more times — the shortcoming described on page 69.)

Sample answer. Take three symbols for the first three landmarks of base 4:

Replace every symbol by the next symbol up, keeping the counts unchanged.

Worked example. Take ○ □□ △  =  125 + 5 + 5 + 1 = 136.

Yes. Since 640 = (10) × 60 + 40, write the numeral for ten, then the 60-symbol just once, then the numeral for forty:

It is read as “ten 60s and one 40”, exactly as the equation says.

Write Y for the 1-wedge and < for the 10-wedge. Reading right to left, the slots stand for 1s, 60s, 3600s.

Checks: 60 + 3 = 63 ✓   120 + 12 = 132 ✓   180 + 20 = 200 ✓   3600 + 0 + 5 = 3605 ✓

A single wedge Y again — this time standing in the 3600s slot, with two empty slots to its right.

All three look the same on the tablet. Only the context could tell a reader which was meant.

Landmarks: 1, 20, 360 (not 400). A dot is 1, a bar is 5, a shell is 0. The lowest set counts 1s, the set above counts 20s, the next counts 360s.

Where it figures: right at the end of the story — it is the one system that carries every idea in the chapter at once, and adds the last one of its own.

Its landmark numbers are the powers of 10:

Yes, it is a place value system. The book’s own example:

Why they alternated: because the rod numerals were actual sticks laid out on a counting board, and a blank space between two places is invisible. Turning the rods through a right angle at every place makes the boundary between places visible instead.

• Zong (upright rods) — units, hundreds, ten-thousands …
• Heng (flat rods) — tens, thousands, hundred-thousands …

41 with Zong only: 41 = 4 tens and 1 one, so you would put down Zong-4 followed by Zong-1:

Yes, it can be read in many other ways if the gaps are not clear. Five rods can be split up as:

A base-2 system needs exactly two digits, one of which must stand for zero. Let

• Gumulgal: position means nothing — each ukasar adds 2 and each urapon adds 1; there are only two landmarks (1, 2) and no zero.
• Base-2 place value: position is everything — ukasar means 1, 2, 4 or 8 depending on its place; the landmarks go on for ever (1, 2, 4, 8, 16, …) and urapon works as 0.
• So numerals are much shorter: 16 needs 8 words in the Gumulgal system but only 5 digits in base 2.

Sample answer.

• Daily life: prices and bills, dates, clock time, phone numbers and PIN codes, marks, cricket scores, page numbers.
• Professions: shopkeepers and accountants, bankers, engineers and surveyors, doctors and chemists, scientists, and computer programmers (computers work with only 0 and 1).
• Without them: every sum would need an abacus and a trained person; there would be no column addition or long division on paper, no decimals, and so no modern science, computers or digital payments.

With 8 fingers we would almost certainly have settled on base 8. Everything else about the system would be unchanged — only the base would differ.

Writing 25:

Checks: 3 × 8 + 1 = 25 ✓   1 × 25 = 25 ✓   16 + 8 + 0 + 0 + 1 = 25 ✓