Chapter 1 — A Square and A Cube — is the opening chapter of the new Ganita Prakash Part 1 (Grade 8) textbook. It introduces square numbers and cube numbers, digit and zero patterns, the sum-of-consecutive-odd-numbers pattern, and how to find square roots and cube roots using prime factorisation. The chapter has two Figure It Out blocks (9 questions on squares, 5 on cubes — 14 in all) plus several in-text ‘Try These’ and ‘Math Talk’ questions. Below are complete, verified answers to all of them. These Class 8 Mathematics Chapter 1 solutions are also useful as quick revision notes before exams.
NCERT Solutions for Class 8 Maths Chapter 1: A Square and A Cube
Figure It Out — Square Numbers (Section 1.1)
1. Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
(i), (ii) and (iii) are not perfect squares — each ends in 2, 8 or 7, digits that a perfect square can never end in. (iv) 1089 = 33², so it is a perfect square.
2. Which one among 64², 108², 292², 36² has last digit 4?
108² = 11664 and 292² = 85264 both end in 4. (64²=4096 and 36²=1296 end in 6.)
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
(iv) 15625+251, since 126² = (125+1)² = 125² + 2×125 + 1 = 15625 + 250 + 1 = 15625 + 251.
4. Find the length of the side of a square whose area is 441 m².
441 = 3²×7², so √441 = 21. The side is 21 m.
5. Find the smallest square number that is divisible by each of 4, 9 and 10.
LCM(4,9,10) = 180 = 2²×3²×5. The lone 5 needs a partner, so multiply by 5: 180×5 = 900 (=30²).
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.
9408 = 2²×3×7² needs one more 3, so multiply by 3. The product is 28224, and √28224 = 2³×3×7 = 168.
7. How many numbers lie between the squares of (i) 16 and 17 (ii) 99 and 100?
Between n² and (n+1)² there are always 2n numbers. (i) 2×16 = 32 numbers. (ii) 2×99 = 198 numbers.
8. Fill in the missing numbers in the pattern: 1²+2²+2²=3²; 2²+3²+6²=7²; 3²+4²+12²=13²; 4²+5²+20²=( )²; 9²+10²+( )²=( )²
Pattern: for a row starting at n, the third term is n(n+1) and the result is n(n+1)+1. So 4²+5²+20² = 21², and 9²+10²+90² = 91² (9×10=90, 90+1=91).
9. How many tiny squares are in the picture, and what is the prime factorisation of that number?
81 big squares (9×9), each split into 25 tiny squares, gives 81×25 = 2025 tiny squares. Prime factorisation: 2025 = 3⁴×5² (=45²).
Figure It Out — Cubic Numbers (Section 1.2)
1. Find the cube roots of 27000 and 10648.
27000 = 2³×3³×5³ = 30³, so ∛27000 = 30. 10648 = 2³×11³ = 22³, so ∛10648 = 22.
2. What number will you multiply by 1323 to make it a cube number?
1323 = 3³×7², needing one more 7, so multiply by 7 (gives 3³×7³ = 9261 = 21³).
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.
All five statements are False. (i) odd×odd×odd = odd (e.g. 3³=27). (ii) numbers ending in 2 cube to numbers ending in 8 (e.g. 12³=1728). (iii) the smallest 2-digit number, 10, already gives 10³=1000 — 4 digits, never 3. (iv) the largest 2-digit number, 99, gives 99³=970299 — only 6 digits, never 7 or more. (v) e.g. 8=2³ has factors {1,2,4,8} — 4 factors, an even count; only perfect squares among cubes (6th powers) have an odd factor count, so the general claim is false.
4. You are told 1331 is a perfect cube. Guess its cube root without factorisation. Do the same for 4913, 12167 and 32768.
Using the last-digit rule and bracketing between known cubes: ∛1331 = 11, ∛4913 = 17, ∛12167 = 23, ∛32768 = 32.
5. Which of the following is the greatest? (i) 67³−66³ (ii) 43³−42³ (iii) 67²−66² (iv) 43²−42²
(i) 67³−66³ = 13267 is the greatest (43³−42³=5419; 67²−66²=133; 43²−42²=85). Cube differences of consecutive numbers grow much faster than square differences.
Try These & Math Talk (In-text Questions)
Which locker numbers stay open in the 100-lockers puzzle?
Exactly the perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 — because only a square number has an odd number of factors (one factor pairs with itself).
Which of 38², 34², 46², 56², 74², 82² end in digit 6?
34²=1156, 46²=2116, 56²=3136, 74²=5476 all end in 6 (38²=1444 and 82²=6724 end in 4).
If a number has 3 zeros at the end, how many zeros will its square have?
6 zeros — squaring doubles the count of trailing zeros, so squares can only have an even number of trailing zeros.
Can a cube end with exactly two zeroes?
No — cubing multiplies the trailing-zero count by 3, so a cube’s trailing zeros are always a multiple of 3 (0, 3, 6, 9…), never 2.
91+93+95+…+109 = ?
This is the 10th run of consecutive odd numbers, which always sums to n³: the answer is 10³ = 1000.
Find the cube roots of (i) 64 (ii) 512 (iii) 729.
4, 8, 9 respectively.
Taxicab numbers: express 4104 and 13832 as a sum of two positive cubes, two ways each.
4104 = 2³+16³ = 9³+15³. 13832 = 2³+24³ = 18³+20³. (These follow 1729 = 1³+12³ = 9³+10³, the famous Hardy–Ramanujan number.)
Why This Chapter Matters
Squares, cubes and their roots via prime factorisation are used constantly in later chapters — the Baudhayana–Pythagoras Theorem (Ganita Prakash Part 2) depends directly on squares, and exponent rules, mensuration (area, volume) and algebra all build on the number-pattern thinking introduced here.
Extra Questions (HOTS) | Revision Notes | Formulas Handbook | Class 8 Maths Book
Class 8 Mathematics Chapter 1 – Notes and Extra Questions
Along with these NCERT Solutions, students can also use the Class 8 Mathematics Chapter 1 Extra Questions and Class 8 Mathematics Chapter 1 Revision Notes for quick revision and extra practice.
- Chapter 2: Power Play – Free PDF Download
- Chapter 3: A Story of Numbers – Free PDF Download
- Chapter 4: Quadrilaterals – Free PDF Download
- Chapter 5: Number Play – Free PDF Download
- Chapter 6: We Distribute, Yet Things Multiply – Free PDF Download
- Chapter 7: Proportional Reasoning-1 – Free PDF Download
- Chapter 8: Fractions in Disguise (Percentages) - Ganita Prakash
- Chapter 9: The Baudhayana-Pythagoras Theorem - Ganita Prakash
- Chapter 10: Proportional Reasoning 2 - Ganita Prakash
- Chapter 11: Exploring Some Geometric Themes - Ganita Prakash
- Chapter 12: Tales by Dots and Lines - Ganita Prakash
- Chapter 13: Algebra Play - Ganita Prakash
- Chapter 14: Area - Ganita Prakash
Frequently Asked Questions
Is this the same as the old Class 8 Maths Chapter 1, ‘Rational Numbers’?
No. Under the new NEP-aligned Ganita Prakash textbook, Chapter 1 is now ‘A Square and A Cube’. Rational numbers are no longer this chapter’s topic in the current syllabus — if you’re looking for the older book’s content, note that NCERT has replaced it with this new edition.
Are the mathematical concepts (squares, cubes, roots) different from the old book?
No — the core mathematics is the same, but this edition introduces new patterns, puzzles (like the lockers problem and taxicab numbers) and a fresh question set.

