Class 8 Maths Ganita Prakash Chapter 1 Solutions (NCERT 2026–27) – A Square and A Cube

These Class 8 Maths Ganita Prakash Chapter 1 solutions cover A Square and A Cube from the new NCF-2023 textbook (Reprint 2026–27). Every Figure it Out question is solved step by step, with worked square roots, cube roots, prime factorisations and number patterns, so you can master the chapter and revise it quickly.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 1 Exercises: Figure it Out (p. 10–11), Figure it Out (p. 16–17) Session: 2026–27
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Chapter 1 Overview

Chapter 1 of Ganita Prakash, A Square and A Cube, opens with Queen Ratnamanjuri’s locker puzzle and uses it to discover why only square numbers have an odd number of factors. From there it builds the ideas of perfect squares (their units digits, the link with consecutive odd numbers, triangular numbers and parity), square roots (by estimation and prime factorisation), perfect cubes and cube roots, the famous taxicab number 1729, and a pinch of history of the Sanskrit words varga, ghana and mula. The Class 8 Maths Ganita Prakash Chapter 1 solutions below work through every Figure it Out, Math Talk and Try This question step by step.

Key Concepts & Definitions

Square number: a number that can be written as the product of a number with itself, e.g. 1, 4, 9, 16, 25, … In general n × n = n2.

Perfect square: the square of a natural number. Every perfect square ends in 0, 1, 4, 5, 6 or 9 (never 2, 3, 7 or 8) and has an even number of zeros at the end.

Square root: the inverse of squaring — if y = x2, then x is a square root of y, written √y. Every perfect square has two integer roots (+ and −); we use the positive root, e.g. √49 = 7.

Cube number / perfect cube: a number obtained by multiplying a number by itself three times, e.g. 1, 8, 27, 64, … In general n × n × n = n3.

Cube root: the inverse of cubing — if y = x3, then x is the cube root of y, written 3√y, e.g. 3√8 = 2.

Odd-factor fact: only square numbers have an odd number of factors, because one factor pairs with itself.

Taxicab number: a number expressible as a sum of two positive cubes in two different ways; the smallest is 1729 = 13 + 123 = 93 + 103 (the Hardy–Ramanujan number).

Important Formulas & Patterns (Chapter 1)

Square & cube: n2 = n × n  •  n3 = n × n × n.

Square as a sum of odd numbers: n2 = 1 + 3 + 5 + … + (2n − 1). The k-th odd number is 2k − 1.

Step from one square to the next: (n + 1)2 = n2 + (2n + 1).

Numbers between two squares: between n2 and (n + 1)2 there are exactly 2n whole numbers.

Square root by prime factorisation: split prime factors into two identical groups; the product of one group is √n.

Cube root by prime factorisation: split prime factors into three identical groups; the product of one group is 3√n.

Cubes as a sum of odd numbers: 13 = 1, 23 = 3 + 5, 33 = 7 + 9 + 11, … (consecutive odd numbers).

Figure it Out — Square Numbers & Square Roots (Page 10–11)

Questions are reproduced verbatim from the NCERT Ganita Prakash textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. Which of the following numbers are not perfect squares? (i) 2032   (ii) 2048   (iii) 1027   (iv) 1089

SOLUTION A perfect square can end only in 0, 1, 4, 5, 6 or 9. (i) 2032 ends in 2 → not a perfect square. (ii) 2048 ends in 8 → not a perfect square. (iii) 1027 ends in 7 → not a perfect square. (iv) 1089 ends in 9 (a valid units digit) and 332 = 1089, so it is a perfect square. ∴ (i), (ii) and (iii) are not perfect squares.

2. Which one among 642, 1082, 2922, 362 has last digit 4?

SOLUTION The units digit of a square depends only on the units digit of the number: 42 = 16 (ends 6), 82 = 64 (ends 4), 22 = 4 (ends 4), 62 = 36 (ends 6). 642 ends in 6; 1082 ends in 4; 2922 ends in 4; 362 ends in 6. ∴ 1082 and 2922 have last digit 4.

3. Given 1252 = 15625, what is the value of 1262? (i) 15625 + 126   (ii) 15625 + 262   (iii) 15625 + 253   (iv) 15625 + 251   (v) 15625 + 512

SOLUTION Use (n + 1)2 = n2 + (2n + 1) with n = 125: 1262 = 1252 + (2 × 125 + 1) = 15625 + 251. ∴ the correct option is (iv) 15625 + 251 (= 15876).

4. Find the length of the side of a square whose area is 441 m2.

SOLUTION Side = √(area) = √441. 441 = 3 × 3 × 7 × 7 = (3 × 7)2 = 212. ∴ side = 21 m.

5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.

SOLUTION First find the LCM of 4, 9 and 10. LCM = 22 × 32 × 5 = 180. For a perfect square every prime must appear an even number of times. In 180 = 22 × 32 × 5, the prime 5 appears once, so multiply by 5. Required number = 180 × 5 = 900, and 900 = 302. It is the smallest square divisible by 4, 9 and 10.

6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.

SOLUTION Prime factorise: 9408 = 26 × 3 × 72. Pairing the primes: 26 (three pairs) and 72 (one pair) are paired, but 3 is unpaired. So multiply by 3: product = 9408 × 3 = 28224 = 26 × 32 × 72. √28224 = 23 × 3 × 7 = 8 × 3 × 7 = 168.

7. How many numbers lie between the squares of the following numbers? (i) 16 and 17   (ii) 99 and 100

SOLUTION Between n2 and (n + 1)2 there are exactly 2n whole numbers. (i) Between 162 = 256 and 172 = 289: 2 × 16 = 32 numbers. (ii) Between 992 = 9801 and 1002 = 10000: 2 × 99 = 198 numbers.

8. In the following pattern, fill in the missing numbers: 12 + 22 + 22 = 32 22 + 32 + 62 = 72 32 + 42 + 122 = 132 42 + 52 + (___)2 = (___)2 92 + 102 + (___)2 = (___)2

SOLUTION Look at each row: the third base is the product of the first two bases, and the right-hand base is (that product + 1). For bases n and n + 1, n × (n + 1) is the third number and n(n + 1) + 1 is the answer. Row 4 (n = 4): third number = 4 × 5 = 20, answer = 20 + 1 = 21. So 42 + 52 + 202 = 212 (16 + 25 + 400 = 441 = 212). ✓ Row 5 (n = 9): third number = 9 × 10 = 90, answer = 90 + 1 = 91. So 92 + 102 + 902 = 912 (81 + 100 + 8100 = 8281 = 912). ✓

9. Devise a way to compute the number of tiny squares on the diagonal of a larger square grid, using the prime factorisation of the number of tiny squares.

SOLUTION If a large square is divided into a perfect-square number N of equal tiny squares, then the grid is √N tiny squares along each side. Find √N from the prime factorisation: split the primes of N into two identical groups; the product of one group is the number of tiny squares along a side, which is exactly the number lying on the main diagonal. Example. For N = 144 = 24 × 32, √144 = 22 × 3 = 12, so there are 12 tiny squares on the diagonal of a 12 × 12 grid. (An open exploratory question; method shown with a worked example.)

Math Talk & Try This — Answered

These are the in-text reflective and short tasks in the chapter; the determinate ones are answered, the open ones are guided.

Math Talk — Units digit of squares Which of the numbers 16, 24, 25, 36, 46, 49 have the digit 6 in the units place of their square? Answer. A square ends in 6 only when the number ends in 4 or 6 (42 = 16, 62 = 36). 26 itself ends in 6 but is not a square — so the units digit alone never proves a number is a square; it can only rule one out (e.g. anything ending in 2, 3, 7 or 8 is not a square).
Math Talk — Zeros at the end of a square If a number ends in 3 zeros, how many zeros does its square end in? What is the link between the zeros of a number and of its square? Answer. A number with 3 trailing zeros is (a number) × 103; squaring gives 106, i.e. 6 zeros. In general a number with k trailing zeros has a square with 2k trailing zeros. So a perfect square can end only in an even number of zeros — this always happens.
Math Talk — Parity What can you say about the parity of a number and of its square? Answer. The square of an even number is even and the square of an odd number is odd — squaring preserves parity, because even × even is even and odd × odd is odd.
Try This — Numbers not squares by units digit Write 5 numbers that you can tell are not squares just by looking at the units digit. Answer. Any numbers ending in 2, 3, 7 or 8 work, e.g. 32, 53, 87, 128 and 2043 — no perfect square ends in 2, 3, 7 or 8.
Math Talk — Numbers between consecutive squares How many numbers lie between two consecutive perfect squares? Do you notice a pattern? Answer. Yes — between n2 and (n + 1)2 there are exactly 2n numbers (e.g. between 32 = 9 and 42 = 16 there are 6 numbers: 10, 11, 12, 13, 14, 15).
Try This — Cube ending in two zeros Can a cube end with exactly two zeros (00)? Explain. Answer. No. Trailing zeros come from factors of 10 = 2 × 5. In a cube each prime appears a multiple of 3 times, so the number of trailing zeros is always a multiple of 3 (0, 3, 6, …). Exactly two zeros is impossible.
Math Talk — Cubes as a sum of consecutive odd numbers Later in the odd-number pattern for cubes we reach 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. Can you tell what this sum is without doing the calculation? Answer. In the pattern 13 = 1, 23 = 3 + 5, 33 = 7 + 9 + 11, …, the n3 row is made of n consecutive odd numbers and equals n3. This row has 10 consecutive odd numbers, so it is the 103 row and the sum is 103 = 1000.
Try This — Cube roots by factorisation Find the cube roots of these numbers: 64, 343, 1728. Answer. 3√64 = 3√(43) = 4;   3√343 = 3√(73) = 7;   3√1728 = 3√(26 × 33) = 22 × 3 = 12.
Try This — Taxicab numbers Express 1729 as the sum of two positive cubes in two different ways. Answer. 1729 = 13 + 123 = 1 + 1728  and  1729 = 93 + 103 = 729 + 1000. This is the smallest taxicab number, known as the Hardy–Ramanujan number.

Figure it Out — Cube Numbers & Cube Roots (Page 16–17)

1. Find the cube roots of 27000 and 10648.

SOLUTION 27000 = 27 × 1000 = 33 × 103 = (3 × 10)3 = 303, so 3√27000 = 30. 10648 = 23 × 113 = (2 × 11)3 = 223, so 3√10648 = 22.

2. What number will you multiply by 1323 to make it a cube number?

SOLUTION Prime factorise: 1323 = 33 × 72. For a perfect cube every prime must appear a multiple of 3 times. Here 33 is complete, but 7 appears only twice, so we need one more 7. Multiply by 7: 1323 × 7 = 9261 = 33 × 73 = 213, a perfect cube.

3. State true or false. Explain your reasoning. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more digits. (v) Cube numbers have an odd number of factors.

SOLUTION (i) False. odd × odd × odd is odd (e.g. 33 = 27), so the cube of an odd number is odd. (ii) False. 23 = 8 and 123 = 1728 both end in 8; cubes can end in any digit 0–9. (iii) False. The smallest 2-digit number is 10 and 103 = 1000 (4 digits), so the cube of a 2-digit number always has at least 4 digits — never just 3. (iv) False. The largest 2-digit number is 99 and 993 = 970299 (6 digits), so a 2-digit cube has at most 6 digits, never seven or more. (v) False. Having an odd number of factors is the property of square numbers, not cubes. For example 27 has the factors 1, 3, 9, 27 — an even number (4) of factors.

4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

SOLUTION Method (last-digit + grouping): the units digit of a cube fixes the units digit of its root (1→1, 8→2, 7→3, 4→4, 5→5, 6→6, 3→7, 2→8, 9→9, 0→0). The tens digit of the root comes from the part of the number to the left of the last three digits. 1331: ends in 1 → root ends in 1; left part “1” lies in 13, so tens digit 1 → 3√1331 = 11. 4913: ends in 3 → root ends in 7; left part “4” lies between 13=1 and 23=8, so tens digit 1 → 3√4913 = 17. 12167: ends in 7 → root ends in 3; left part “12” lies between 23=8 and 33=27, so tens digit 2 → 3√12167 = 23. 32768: ends in 8 → root ends in 2; left part “32” lies between 33=27 and 43=64, so tens digit 3 → 3√32768 = 32.

5. Which of the following is the greatest? Explain your reasoning. (i) 673 − 663   (ii) 433 − 423   (iii) 672 − 662   (iv) 432 − 422

SOLUTION Use the identities a3 − b3 = (a − b)(a2 + ab + b2) and a2 − b2 = (a − b)(a + b). Here a − b = 1 in every part. (i) 673 − 663 = 672 + 67 × 66 + 662 = 4489 + 4422 + 4356 = 13267. (ii) 433 − 423 = 432 + 43 × 42 + 422 = 1849 + 1806 + 1764 = 5419. (iii) 672 − 662 = 67 + 66 = 133. (iv) 432 − 422 = 43 + 42 = 85. ∴ (i) 673 − 663 is the greatest — the cube differences dominate, and the larger bases (67, 66) make (i) bigger than (ii).

Common Mistakes to Avoid

Watch out for these

  • Thinking a units digit of 0, 1, 4, 5, 6 or 9 proves a number is a square — it only tells you when a number is not a square (e.g. 26 ends in 6 but is not a square).
  • For the smallest-multiplier questions, forgetting to prime factorise first — supply only the primes needed to complete pairs (squares) or triplets (cubes).
  • Mixing up the tests: two identical groups for a square root, three identical groups for a cube root.
  • Saying cubes have an odd number of factors — that is the property of squares, not cubes.
  • Assuming a cube can end in any even count of zeros; a perfect cube’s trailing zeros are always a multiple of 3.
  • In “numbers between squares”, counting the endpoints — the count strictly between n2 and (n + 1)2 is 2n.

Practice MCQs & Assertion–Reason

1. Which of these is a perfect square?

(a) 1024    (b) 1027    (c) 2048    (d) 2032

2. A perfect square can never end in:

(a) 4    (b) 6    (c) 7    (d) 9

3. How many numbers lie between 202 and 212?

(a) 20    (b) 40    (c) 41    (d) 21

4. The square root of 1296 is:

(a) 34    (b) 36    (c) 38    (d) 46

5. Using 1252 = 15625, the value of 1242 is:

(a) 15625 − 249    (b) 15625 − 124    (c) 15625 + 249    (d) 15625 − 250

6. Which number has an odd number of factors?

(a) 12    (b) 18    (c) 36    (d) 40

7. The cube root of 1728 is:

(a) 8    (b) 12    (c) 14    (d) 18

8. The smallest number that 1323 must be multiplied by to become a perfect cube is:

(a) 3    (b) 7    (c) 9    (d) 21

9. The smallest taxicab number (Hardy–Ramanujan number) is:

(a) 1024    (b) 1331    (c) 1728    (d) 1729

10. The number of trailing zeros in a perfect cube is always:

(a) odd    (b) a multiple of 2    (c) a multiple of 3    (d) exactly 2

Answer key: 1-(a), 2-(c), 3-(b), 4-(b), 5-(a), 6-(c), 7-(b), 8-(b), 9-(d), 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: 2048 is not a perfect square.

Reason: A perfect square cannot end in the digit 8.

A-R 2. Assertion: Only square numbers have an odd number of factors.

Reason: In a square, one factor pairs with itself, leaving an odd factor count.

A-R 3. Assertion: The cube of every odd number is even.

Reason: The product of three odd numbers is odd.

A-R 4. Assertion: 1729 = 13 + 123 = 93 + 103.

Reason: 1729 is the smallest number expressible as a sum of two positive cubes in two different ways.

A-R 5. Assertion: To make 1323 a perfect cube it must be multiplied by 7.

Reason: 1323 = 33 × 72, and the factor 7 needs one more to form a triplet.

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

Quick Revision Summary

  • A square number is n × n = n2; squares of natural numbers are perfect squares.
  • Perfect squares end only in 0, 1, 4, 5, 6 or 9, and have an even number of trailing zeros.
  • Only square numbers have an odd number of factors (one factor pairs with itself).
  • Every square is the sum of consecutive odd numbers from 1; (n + 1)2 = n2 + (2n + 1).
  • Between n2 and (n + 1)2 there are exactly 2n numbers.
  • Square root: split prime factors into two identical groups; cube root: split into three identical groups.
  • A cube number is n3; its trailing-zero count is a multiple of 3; 1729 is the smallest taxicab number.

How to score full marks in this chapter

Always prime factorise before answering smallest-multiplier and root questions, and write the factors in pairs (for squares) or triplets (for cubes). For “is it a square/cube” checks, state the units-digit or trailing-zero rule you are using. Use (n + 1)2 = n2 + 2n + 1 and the identity a3 − b3 = (a − b)(a2 + ab + b2) to avoid long multiplication, and keep your working tidy so each step earns its mark.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 1 about?

Chapter 1, A Square and A Cube, covers square numbers and their properties, square roots, perfect cubes and cube roots, finding roots by estimation and prime factorisation, number patterns (odd numbers, triangular numbers) and the famous taxicab number 1729.

How many Figure it Out exercises are there in Chapter 1?

There are two main “Figure it Out” sets — one on square numbers and square roots (pages 10–11) and one on cube numbers and cube roots (pages 16–17) — plus several Math Talk and Try This tasks, all solved on this page.

Why do only square numbers have an odd number of factors?

Factors usually come in partner pairs whose product is the number. For a perfect square, one factor is the square root, which pairs with itself, so it is counted once — making the total number of factors odd.

Are these Class 8 Maths Ganita Prakash Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for the 2026–27 session, with answers verified against the book’s answer key.

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