Class 8 Maths Ganita Prakash Chapter 3 Solutions (NCERT 2026–27) – A Story of Numbers

These Class 8 Maths Ganita Prakash Chapter 3 solutions cover A Story of Numbers from the new NCF-2023 textbook (2026–27). Every “Figure it Out” exercise, Math Talk and Try This box is solved step by step, with each numeral conversion verified, so you can understand how number systems evolved and revise the whole chapter quickly.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 3 Exercises: Figure it Out (3.1–3.4) Session: 2026–27
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Chapter 3 Overview

Chapter 3 of Ganita Prakash, A Story of Numbers, traces how humans learned to count and write numbers, from the Stone Age to the modern day. Through Reema’s curiosity about a strange piece of writing, the chapter walks through the mechanism of counting (sticks, sounds and symbols, one-to-one mapping), early systems built on body parts, tally marks and counting in twos, the Roman numerals with their landmark numbers, the Egyptian base-10 system and the general idea of a base-n number system, and finally place value systems — Mesopotamian (Babylonian), Mayan and Chinese — leading to the Hindu (Indian) number system and the revolutionary idea of zero. The Class 8 Maths Ganita Prakash Chapter 3 solutions below work through every exercise in the chapter.

Key Concepts & Definitions

Number system: a standard sequence of objects, names or written symbols with a fixed order, used to count by one-to-one mapping.

Numerals: the symbols used to represent numbers in a written number system (e.g. 0, 1, 5, 36, 193).

One-to-one mapping: matching each object to be counted with exactly one item of the standard sequence (e.g. one cow → one stick), with no two objects mapped to the same item.

Landmark numbers: easily recognised reference numbers that get their own symbol and are used as anchors to build other numbers (e.g. Roman I, V, X, L, C, D, M).

Base-n number system: a system whose landmark numbers are the powers of n — the first landmark number is 1, and every next landmark number is n times the current one.

Decimal (base-10) system: a base-n system with n = 10, e.g. the Egyptian and Hindu systems.

Place value / positional system: a system with a base that uses the position of a symbol to decide which landmark number (power of the base) it stands for. Used by the Mesopotamians, Mayans, Chinese and Indians.

Zero (0): the placeholder showing “nothingness” that makes a place value system unambiguous; in the Hindu system 0 is also a number in its own right.

Important Facts & Landmark Numbers

Roman landmark numbers: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1,000.

Egyptian (base-10) landmark numbers: 1, 10, 102 = 100, 103 = 1000, …, up to 107 (a crore), each with its own symbol.

Base-5 landmark numbers: 50 = 1, 51 = 5, 52 = 25, 53 = 125, 54 = 625, 55 = 3125.

Base-n landmark numbers (general): n0 = 1, n, n2, n3, … (the powers of n).

Mesopotamian (sexagesimal, base-60): 1, 60, 602 = 3600, 603 = 216000, … (1 hour = 60 min, 1 min = 60 s come from this).

Mayan (almost base-20): 1, 20, 20×18 = 360, 202×18 = 7200, 203×18 = 144000.

Product of landmark numbers in a base-n system: always another landmark number (e.g. 10a × 10b = 10a+b) — which is why multiplication is easy in a base system.

Figure it Out (Page 54)

1. Suppose you are using the number system that uses sticks to represent numbers, as in Method 1. Without using either the number names or the numerals of the Hindu number system, give a method for adding, subtracting, multiplying and dividing two numbers or two collections of sticks.

SOLUTION Represent each number by a collection of sticks (one stick per object), then: Addition: push both collections of sticks together into one pile — the combined pile is the sum. Subtraction: from the larger collection, remove (pair off) as many sticks as there are in the smaller collection; the sticks left over are the difference. Multiplication: make as many equal groups as the first number, each group having as many sticks as the second number, then combine all the groups into one pile. Division: from the given pile, repeatedly take out groups each of the divisor’s size; the number of complete groups is the quotient and the leftover sticks are the remainder.

2. One way of extending the number system in Method 2 is by using strings with more than one letter—for example, we could use ‘aa’ for 27. How can you extend this system to represent all the numbers? There are many ways of doing it!

SOLUTION The alphabet gives only 26 single letters (a–z = 1–26). To go further, keep adding more letters to each string in order. One way: a, b, c, …, z (1–26), then aa, bb, cc, …, zz (27–52), then aaa, bbb, ccc, …, zzz (53–78), and so on — each round adds one more repeated letter, so every number gets a name. (Any consistent rule that never runs out of new strings is acceptable, e.g. a, b, …, z, aa, ab, ac, … like spreadsheet column names.)

3. Try making your own number system.

SOLUTION This is an open activity. A simple example using shapes as a base-3 system: Pick three symbols, say △ = 1, □ = 2, ◯ = 0, and group in threes (landmark numbers 1, 3, 9, 27, …). Then read positions like place value: for example, △◯△ would mean (1)×9 + (0)×3 + (1)×1 = 10. Any clear, consistent rule is correct.

Figure it Out (Page 59)

1. Represent the following numbers in the Roman system. (i) 1222    (ii) 2999    (iii) 302    (iv) 715

SOLUTION (i) 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1 = MCCXXII. (ii) 2999 = 1000 + 1000 + (1000 − 100) + (100 − 10) + (10 − 1) = MM + CM + XC + IX = MMCMXCIX. (iii) 302 = 100 + 100 + 100 + 1 + 1 = CCCII. (iv) 715 = 500 + 100 + 100 + 10 + 5 = DCCXV.

Figure it Out (Page 60–61)

2. Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s. Come up with ways of performing the different arithmetic operations (+, –, ×, ÷) for numbers occurring in this system, without using Hindu numerals. Use this to evaluate the following: (i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon) (ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar-ukasar) (iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar) (iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)

SOLUTION In the Gumulgal system, urapon = 1 and ukasar = 2, and numbers are written as a sum of 2s and 1s (count by 2s). Method: add/remove the “ukasar” (2) and “urapon” (1) words like sticks, then regroup (two extra “urapon” become one “ukasar”); multiply by repeated addition and divide by repeated grouping. (i) Value = (2+2+2+2+1) + (2+2+2+1) = 9 + 7 = 16 = eight 2s ⇒ ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar. (ii) Value = 9 − (2+2+2) = 9 − 6 = 3 = 2 + 1 ⇒ ukasar-urapon. (iii) Value = 9 × 4 = 36 = eighteen 2s ⇒ ukasar repeated 18 times (ukasar-ukasar-…-ukasar, 18 times). (iv) Value = 16 ÷ 4 = 4 = 2 + 2 ⇒ ukasar-ukasar.

3. Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.

SOLUTION The Hindu number system (0, 1, 2, …, 9) is far more efficient than the Roman system (I, V, X, L, C, D, M):
Hindu Number SystemRoman Numerals
Place value system — a symbol’s value depends on its positionFixed value — each symbol always means the same amount
Has a symbol and number for zero (0)No symbol for zero
Easy to add, subtract, multiply and divideCalculation (especially × and ÷) is extremely difficult
Only 10 digits represent all numbers, however largeNeeds new symbols for larger and larger numbers

Figure it Out (Page 62)

1. Represent the following numbers in the Egyptian system: 10458, 1023, 2660, 784, 1111, 70707.

SOLUTION Group each number into powers of 10 (each power has its own Egyptian symbol; you write a symbol as many times as that power occurs): 10458 = 10000 + 400 + 50 + 8 = one 104, four 100s, five 10s, eight 1s. 1023 = 1000 + 20 + 3 = one 1000, two 10s, three 1s. 2660 = 2000 + 600 + 60 = two 1000s, six 100s, six 10s. 784 = 700 + 80 + 4 = seven 100s, eight 10s, four 1s. 1111 = 1000 + 100 + 10 + 1 = one 1000, one 100, one 10, one 1. 70707 = 70000 + 700 + 7 = seven 104, seven 100s, seven 1s.

2. What numbers do these numerals stand for? (i) two “100” symbols, six “10” symbols and sixteen “1” symbols (ii) four “1000” symbols, three “100” symbols and two “10” symbols

SOLUTION (i) = 2×100 + 6×10 + 16×1 = 200 + 60 + 16 = 276. (ii) = 4×1000 + 3×100 + 2×10 + 2×1 = 4000 + 300 + 20 + 2 = 4322. (The two numerals shown in the book are made of the Egyptian symbols listed above.)

Figure it Out (Page 63)

1. Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50, 137, 293, 651.

SOLUTION Base-5 landmark numbers are 1, 5, 25, 125, 625. Group each number using as many of the largest landmark numbers as possible: (i) 15 = 5 + 5 + 5 = three 5s (i.e. three “5” symbols). (ii) 50 = 25 + 25 = two 25s. (iii) 137 = 125 + 5 + 5 + 1 + 1 = one 125, two 5s, two 1s. (iv) 293 = 125 + 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1 = two 125s, one 25, three 5s, three 1s. (v) 651 = 625 + 25 + 1 = one 625, one 25, one 1.

2. Is there a number that cannot be represented in our base-5 system above? Why or why not?

SOLUTION Yes — zero (0) cannot be represented, because this system has no symbol for 0; it only has symbols for the landmark numbers 1, 5, 25, …. Every other positive whole number can be represented.

3. Compute the landmark numbers of a base-7 system. In general, what are the landmark numbers of a base-n system?

SOLUTION Base-7: 70 = 1, 71 = 7, 72 = 49, 73 = 343, 74 = 2401, … So the landmark numbers are 1, 7, 49, 343, 2401, … In general, the landmark numbers of a base-n system are the powers of n: n0 = 1, n1, n2, n3, …

Figure it Out (Page 65)

1. Add the following Egyptian numerals. (i) (three 103, six 102, eight 10s) + (two 102, seven 10s) (ii) (one 103, eight 10s) + (four 10s, six 1s)

SOLUTION Add like symbols, then regroup: every 10 symbols of one power make one symbol of the next power. (i) Totals: three 103, (6 + 2) = eight 102, (8 + 7) = fifteen 10s. Regroup the 10s: 15 tens = ten 10s + five 10s = one 102 + five 10s. Now 102 count = 8 + 1 = nine. Result = three 103, nine 102, five 10s (the number 3950). Check: 3680 + 270 = 3950. ✓ (ii) Totals: one 103, (8 + 4) = twelve 10s, six 1s. Regroup the 10s: 12 tens = ten 10s + two 10s = one 102 + two 10s. Result = one 103, one 102, two 10s, six 1s (the number 1126). Check: 1080 + 46 = 1126. ✓

2. Add the following numerals that are in the base-5 system that we created (remember, in this system 5 times a landmark number gives the next one): (one 125, one 25, one 25, one 5, two 1s) + (one 125, one 25, one 5, one 5, one 1).

SOLUTION The two base-5 numerals shown are 161 and 162. First numeral = 125 + 25 + 5 + 1 + 1 = 157; second = 125 + 25 + 25 + 5 + 5 + 1 = 186. (Reading the book’s symbols.) Method: add the matching landmark symbols, then regroup any five of a kind into one of the next landmark number. Sum = 157 + 186 = 343 = 125 + 125 + 25 + 25 + 25 + 5 + 5 + 5 + 1 + 1 + 1, and after regrouping five 25s → one 125 and so on, this is 2 × 125 + 3 × 25 + 3 × 5 + 3 × 1. The key idea: regrouping in base 5 is exactly like “carrying” in our decimal addition, but five of one symbol make the next.

Figure it Out (Page 69–70)

1. Can there be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times? Why not?

SOLUTION No. In the Egyptian (base-10) system, ten copies of any landmark symbol equal exactly one symbol of the next landmark number (e.g. ten 1s = one 10, ten 10s = one 100). So whenever a symbol would appear 10 or more times, we regroup it into the next higher symbol. In a correctly written numeral, each symbol therefore occurs at most 9 times.

2. Create your own number system of base 4, and represent numbers from 1 to 16.

SOLUTION Base-4 landmark numbers are 40 = 1, 41 = 4, 42 = 16. Let 1 = ∟, 4 = △, 16 = □ (any symbols work).
NumberGroupingNumberGrouping
1∟9△ △ ∟
2∟ ∟10△ △ ∟ ∟
3∟ ∟ ∟11△ △ ∟ ∟ ∟
4△12△ △ △
5△ ∟13△ △ △ ∟
6△ ∟ ∟14△ △ △ ∟ ∟
7△ ∟ ∟ ∟15△ △ △ ∟ ∟ ∟
8△ △16□
Rule: four 1s (∟) regroup into one 4 (△); four 4s regroup into one 16 (□). Any consistent base-4 scheme is correct.

3. Give a simple rule to multiply a given number by 5 in the base-5 system that we created.

SOLUTION In a base-5 system, the landmark numbers are powers of 5, so multiplying any landmark number by 5 gives the next landmark number (1→5→25→125→…). Rule: to multiply a number by 5, simply replace every symbol by the next-higher landmark symbol (each 1 becomes a 5, each 5 becomes a 25, each 25 becomes a 125, and so on). No regrouping is needed. This is just like “adding a zero” (a left shift) when we multiply by 10 in our decimal system.

Figure it Out (Page 73)

1. Represent the following numbers in the Mesopotamian system. (i) 63    (ii) 132    (iii) 200    (iv) 60    (v) 3605

SOLUTION The Mesopotamian system is base-60 (landmark numbers 1, 60, 3600, …), using one symbol for 1 and one for 10. Group each number into powers of 60, then write each group with the 1- and 10-symbols: (i) 63 = (1)×60 + 3 ⇒ one 60, then three 1s. (ii) 132 = (2)×60 + 12 = (2)×60 + 10 + 2 ⇒ two 60s, then one 10 and two 1s. (iii) 200 = (3)×60 + 20 ⇒ three 60s, then two 10s. (iv) 60 = (1)×60 ⇒ one 60 (and a blank in the 1s place). (v) 3605 = (1)×3600 + 5 ⇒ one 3600, a blank in the 60s place, then five 1s. Check: (i) 60+3 = 63; (ii) 120+12 = 132; (iii) 180+20 = 200; (v) 3600+5 = 3605. ✓

Figure it Out (Page 76)

Represent the following numbers using the Mayan system. (i) 77    (ii) 100    (iii) 361    (iv) 721

SOLUTION Mayan landmark numbers are 1, 20, 360, 7200, … (written one below the other, lowest = 1s). A dot = 1, a bar = 5, a shell = 0 are used for each position (1–19). (i) 77 = (3)×20 + (17)×1 ⇒ position-20: 3; position-1: 17. (ii) 100 = (5)×20 + (0)×1 ⇒ position-20: 5; position-1: 0 (shell). (iii) 361 = (1)×360 + (0)×20 + (1)×1 ⇒ position-360: 1; position-20: 0 (shell); position-1: 1. (iv) 721 = (2)×360 + (0)×20 + (1)×1 ⇒ position-360: 2; position-20: 0 (shell); position-1: 1. Check: (i) 60+17 = 77; (ii) 100; (iii) 360+0+1 = 361; (iv) 720+0+1 = 721. ✓

Figure it Out (Page 80)

1. Why do you think the Chinese alternated between the Zong and Heng symbols? If only the Zong symbols were to be used, how would 41 be represented? Could this numeral be interpreted in any other way if there is no significant space between two successive positions?

SOLUTION The Chinese alternated between Zong (vertical) and Heng (horizontal) rod numerals so that neighbouring place values look different, making it easy to tell where one position ends and the next begins. If only Zong symbols were used, 41 = (4)×10 + 1 would be written as four vertical rods then one vertical rod — i.e. five identical-looking vertical rods in a row. With no clear space between positions, this could be misread as 23, 32 or even 122 — which is exactly why the Zong/Heng alternation (or a clear gap) is needed.

2. Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the digits. Compare this system with that of the Gumulgal’s.

SOLUTION A base-2 place value system needs two digits, one for 0 and one for 1. Let urapon (ur) = 0 and ukasar (uk) = 1; place values are 20, 21, 22, …
NumberBase-2 (ur = 0, uk = 1)Gumulgal system
1ukur
2uk uruk
3uk ukuk-ur
4uk ur uruk-uk
5uk ur ukuk-uk-ur
6uk uk uruk-uk-uk
7uk uk ukuk-uk-uk-ur
8uk ur ur uruk-uk-uk-uk
Comparison: both use only two basic words, but the base-2 system is a place value system (position matters and it has a 0), whereas the Gumulgal system just counts in 2s by repetition (no place value, no zero). The place value system stays short for large numbers; the Gumulgal one grows longer and longer.

3. Where in your daily lives, and in which professions, do the Hindu numerals, and 0, play an important role? How might our lives have been different if our number system and 0 hadn’t been invented or conceived of?

SOLUTION Hindu numerals and 0 are used everywhere — money and shopping, telling time, phone numbers, measuring, banking, and in professions such as engineering, accounting, banking, science, medicine, computer programming (binary uses 0 and 1) and architecture. Without an efficient number system and zero, calculations would be slow and clumsy (like the Roman system), and modern science, technology and computers — which depend on 0 and place value — would have been extremely hard to develop.

4. The ancient Indians likely used base 10 for the Hindu number system because humans have 10 fingers, and so we can use our fingers to count. But what if we had only 8 fingers? How would we be writing numbers then? What would the Hindu numerals look like if we were using base 8 instead? Base 5? Try writing the base-10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write it in base-2?

SOLUTION With 8 fingers we would probably use a base-8 system, with digits 0–7 and place values 80, 81, 82, … A base-5 system would use digits 0–4 and place values 50, 51, 52, … Base-8: 25 = (3)×8 + (1)×1, so 25 = 31 in base 8. Base-5: 25 = (1)×52 + (0)×5 + (0)×1, so 25 = 100 in base 5. Base-2: 25 = 16 + 8 + 1 = (1)×24 + (1)×23 + (0)×22 + (0)×21 + (1)×20, so 25 = 11001 in base 2. Check: 11001 = 16 + 8 + 0 + 0 + 1 = 25. ✓

Math Talk & Try This Boxes

Math Talk — Counting with sticks (Page 51) How will you use such sticks to answer Q2 (do we have fewer cows than our neighbour?) and Q3 (how many more cows do we need?) Answer. Q2: Pair off your sticks one-to-one with your neighbour’s sticks. If your sticks run out first, you have fewer cows. Q3: The number of your neighbour’s sticks left unpaired is exactly how many more cows you need.
Math Talk — Letters as numbers (Page 53) How many numbers can you represent using the sounds of the letters of your language? Answer. Only as many as the number of letters — e.g. English has 26 letters, so single letters represent just 1 to 26. (Hindi/Devanagari has more letters, so a few more, but it is still limited.) This is the limitation that pushes us towards systems that can go on forever.
Math Talk — Extending Method 3 (Page 53) Do you see a way of extending this method (the Roman-style symbols I, II, III, …) to represent bigger numbers? Answer. Yes — by introducing new symbols for larger landmark numbers (V = 5, X = 10, L = 50, C = 100, …) and combining them, just as the Roman system does. This avoids writing huge strings of I’s.
Math Talk — Grouping at a glance (Page 57) Up to what group size could you immediately see the number of objects without counting? Answer. Most people can instantly recognise groups of up to about 4 objects; from 5 or more we usually need to count. This human limit is one reason number systems group in small sizes (like 5s) and give them new symbols.
Math Talk — Counting only by 5s (Page 58) What could be the difficulties with a system that counts only in a single group size? How would you represent 1345 in a system that counts only by 5s? Answer. Using only one group size, large numbers need very many symbols and become hard to read and compute with. To write 1345 by 5s you would need 269 fives (269×5 = 1345) — far too long. This shows why we need a sequence of landmark numbers (a base), not just one group size.
Try This — Roman addition (Page 60) Do it yourself: (b) LXXXVII + LXXVIII. Answer. LXXXVII = 87 and LXXVIII = 78, so the sum = 165 = 100 + 50 + 10 + 5 = CLXV. (Working with the symbols: collect the Ls, Xs, Vs and Is, regroup — two Ls = C, the Vs and Is regroup — to get CLXV.)
Try This / Math Talk — Roman products (Page 60) Find the products of the landmark numbers: V × L, L × D, V × D, VII × IX. Answer. V × L = 5 × 50 = 250 = CCL. L × D = 50 × 500 = 25000 = twenty-five M’s (MMM…M, 25 times). V × D = 5 × 500 = 2500 = MMD. VII × IX = 7 × 9 = 63 = LXIII. (These show why Roman multiplication is awkward.)
Math Talk — Products in base-n (Pages 66–68) What is any landmark number × 10 (and × 102) in the Egyptian system? Does the “product of two landmark numbers is a landmark number” rule hold in the base-5 system, and in any base? Answer. Multiplying a landmark number (a power of 10) by 10 gives the next landmark number (10a × 10 = 10a+1); multiplying by 102 raises the power by 2, and so on. So the product of any two landmark numbers is again a landmark number. Yes, this holds in the base-5 system and in any base-n system, because na × nb = na+b — which is exactly why multiplication is easy in a base system.
Math Talk — Different counts for different objects (Page 60) A group of indigenous people in a Pacific island use different sequences of number names to count different objects. Why do you think they do this? Answer. Different objects (e.g. people, animals, long objects, round objects) were counted using different “counting words” that suited the kind of thing being counted — a cultural way of keeping track of what was being counted, not just how many. (Accept reasonable explanations.)
Math Talk — The Mesopotamian 3600 (Page 73) Look at the representation of 60. What will be the representation for 3,600? Answer. In the compact Mesopotamian system, 3600 = 602 was written with the same single “1” symbol as 1 and 60 — only its position (the third place from the right) told you it meant 3600. This sameness is exactly the defect that made their numerals ambiguous before the placeholder (zero) idea.

Common Mistakes to Avoid

Watch out for these

  • Letting an Egyptian symbol appear 10 or more times — always regroup ten of one power into one of the next.
  • Confusing landmark numbers with base: in a base-n system the landmark numbers must be the powers of n (1, n, n2, …).
  • Forgetting that the Mayan system is not a true base-20 (its third landmark is 360, not 400), so its multiplication is not as easy as a real base.
  • Writing Roman numerals carelessly — e.g. 2999 is MMCMXCIX, not MM999 or IMMM; use the subtractive forms CM, XC, IX.
  • Reading a Mesopotamian or Chinese numeral wrongly when a blank place (zero) is missing — without a placeholder, 122, 23 and 32 can look alike.
  • Thinking 0 is “just nothing” — in the Hindu system 0 is both a placeholder and a number you can compute with.
  • Mixing up which Chinese symbols are Zong (vertical, for 1s, 100s, …) and Heng (horizontal, for 10s, 1000s, …).

Practice MCQs & Assertion–Reason

1. The symbols used to write numbers in a written number system are called:

(a) digits only    (b) numerals    (c) tallies    (d) bases

2. The Roman numeral for 2999 is:

(a) MMCMXCIX    (b) MMIM    (c) MMCMIC    (d) MMMCMXC

3. The landmark numbers of a base-n number system are:

(a) multiples of n    (b) the powers of n    (c) only 1 and n    (d) n and 10

4. The Egyptian number system is a:

(a) base-5 system    (b) base-20 system    (c) base-10 (decimal) system    (d) base-60 system

5. The Mesopotamian (Babylonian) system was a:

(a) base-10 system    (b) base-20 system    (c) base-60 (sexagesimal) system    (d) base-2 system

6. In base 8, the number 25 (base 10) is written as:

(a) 25    (b) 31    (c) 100    (d) 41

7. The third landmark number of the Mayan system is:

(a) 400    (b) 360    (c) 100    (d) 8000

8. A number system that uses the position of a symbol to decide its value is called a:

(a) tally system    (b) place value (positional) system    (c) body-part system    (d) Roman system

9. The base-5 representation of 25 (base 10) is:

(a) 25    (b) 50    (c) 100    (d) 31

10. The use of 0 as a number in its own right was codified in 628 CE by:

(a) Aryabhata    (b) Al-Khwārizmī    (c) Brahmagupta    (d) Fibonacci

Answer key: 1-(b), 2-(a), 3-(b), 4-(c), 5-(c), 6-(b), 7-(b), 8-(b), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: In the Egyptian system, no landmark symbol can appear 10 or more times.

Reason: Ten copies of any power of 10 equal one copy of the next power of 10.

A-R 2. Assertion: The Mayan number system is a true base-20 system.

Reason: Its landmark numbers are 1, 20, 360, 7200, …, where the third is 360 instead of 400.

A-R 3. Assertion: A place value system needs a placeholder symbol like 0 to be unambiguous.

Reason: A blank space alone can be misread, making different numbers look the same.

A-R 4. Assertion: Multiplication is easier in a base-n system than in the Roman system.

Reason: In a base system the product of two landmark numbers is again a landmark number.

A-R 5. Assertion: The Hindu number system originated in India around 2000 years ago.

Reason: The Roman numerals were invented in India and later carried to Europe.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(C).

Quick Revision Summary

  • A number system is a fixed-order sequence of objects, names or symbols; we count by one-to-one mapping. Its written symbols are numerals.
  • Early ideas: counting with sticks/tally marks, with body parts (Papua New Guinea), and counting in 2s (Gumulgal).
  • Landmark numbers are reference numbers with their own symbols (Roman I, V, X, L, C, D, M).
  • A base-n system has landmark numbers that are the powers of n; the Egyptian and Hindu systems are base-10 (decimal).
  • In a base system the product of two landmark numbers is a landmark number, which makes multiplication easy.
  • Place value (positional) systems use a symbol’s position to fix its value — Mesopotamian (base-60), Mayan (almost base-20), Chinese (base-10), Indian (base-10).
  • The Hindu number system with the digit 0 (placeholder and number) lets us write all numbers unambiguously with just 10 symbols — one of history’s greatest inventions.

How to score full marks in this chapter

For every numeral question, first write the number as a sum of the system’s landmark numbers (e.g. 2999 = MM + CM + XC + IX), then convert — and always check by converting back to base 10. State the base before you start, regroup carefully (ten 1s → one 10 in base 10; five in base 5; sixty in base 60), and remember to write a 0/placeholder for any empty position in a place value system.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 3 about?

Chapter 3, A Story of Numbers, tells how counting and number-writing evolved — from sticks, tally marks and body parts to the Roman, Egyptian, Mesopotamian, Mayan, Chinese and finally the Hindu (Indian) number system — and explains landmark numbers, the idea of a base, place value and the role of zero.

What is a base-n number system?

A base-n system is one whose landmark numbers are the powers of n: the first landmark number is 1, and every next landmark number is n times the current one (1, n, n2, n3, …). The Egyptian and Hindu systems are base-10.

How many exercises does Chapter 3 have?

The chapter has several “Figure it Out” sets (on pages 54, 59, 60–61, 62, 63, 65, 69–70, 73, 76 and 80) plus many Math Talk and Try This boxes — all solved on this page.

Are these Class 8 Maths Ganita Prakash Chapter 3 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for 2026–27.

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