by Tiwari Academy
NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 1 A Square and A Cube for Session 2025–26. It explains the concepts of squares and cubes clearly. It covers perfect squares, perfect cubes, square roots and cube roots. NCERT Maths Ganita Prakash chapter 1 includes puzzles, patterns and examples to help students understand better. Questions are solved in easy steps. Important properties and tricks are also included. These solutions help in practice and revision. They are useful for exams and concept clarity.
Class 8 Maths Ganita Prakash Chapter 1 MCQs
Class 8 Maths NCERT Solutions
Class 8 Ganita Prakash Chapter 1 Solutions
Page 10
Figure it Out
1. Which of the following numbers are not perfect squares?
(i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
See Solutions(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, could be a perfect square.
Let’s check: 33² = 1089
To find which numbers are not perfect squares, we know that:
Perfect squares end with digits: 0, 1, 4, 5, 6 or 9.
If a number ends with 2, 3, 7 or 8, it cannot be a perfect square.
Check the square root of the number mentally or via estimation for confirmation.
2. Which one among 64², 108², 292², 36² has last digit 4?
See Solutions64² = 4096 → ends in 6
108² = 11664 → ends in 4
292² = 85264 → ends in 4
36² = 1296 → ends in 6
So, 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
See SolutionsWe know that to go from n² to (n + 1)², we use the identity:
(n + 1)² = n² + 2n + 1
This is based on expansion: (n + 1) × (n + 1) = n² + 2n + 1
Now, n = 125
So, 126² = 125² + 2 × 125 + 1
= 15625 + 250 + 1
= 15625 + 251
Therefore, the Correct Answer: (iv) 15625 + 251
Explanation of Class 8 Ganita Prakash Chapter 1
4. Find the length of the side of a square whose area is 441 m².
See SolutionsTo find the length of the side of a square whose area is 441 m², we need to find the square root of 441.
This is because:
The square root of the area gives the length of the side.
Now, we check perfect squares we know:
20 × 20 = 400
21 × 21 = 441
So, √441 = 21
The length of the side of the square is 21 metres, because 21 × 21 = 441.
5. Find the smallest square number that is divisible by each of the following numbers: 4, 9 and 10.
See SolutionsPrime Factorisation of the numbers:
4 = 2 × 2
9 = 3 × 3
10 = 2 × 5
The LCM (Least Common Multiple)
To find a number divisible by all three, we have to take their LCM:
So, the LCM = 2 × 2 × 3 × 3 × 5 = 180
But 180 is not a perfect square, because the prime factor 5 appears only once.
To make 180 a perfect square, every prime factor must appear in pairs. Only 5 is unpaired.
So, multiply by another 5 to complete the pair: 180 × 5 = 900
Therefore, the smallest square number divisible by 4, 9 and 10 is 900.
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.
See SolutionsPrime factorise 9408
9408 ÷ 2 = 4704
4704 ÷ 2 = 2352
2352 ÷ 2 = 1176
1176 ÷ 2 = 588
588 ÷ 2 = 294
294 ÷ 2 = 147
147 ÷ 3 = 49
49 ÷ 7 = 7
7 ÷ 7 = 1
So, 9408 = 2⁶ × 3 × 7²
For a perfect square, all primes must appear in even powers.
In 9408:
2⁶ is in even powers
3¹ is not in even powers – needs one more 3.
7² is in even powers
So, to make it a perfect square, we must multiply by one more 3.
Multiply to get the perfect square 9408 × 3 = 28224
Now, 28224 = 2⁶ × 3² × 7²
So, √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168
Therefore, the Smallest number to multiply 9408 is 3.
Square root of the product (28224): 168
8th Ganita Prakash Chapter 1 Question Answers
7. How many numbers lie between the squares of the following numbers?
(i) 16 and 17 (ii) 99 and 100
See SolutionsWe know that, if we have given two consecutive numbers, n and n+1:
n² is the smaller square
(n+1)² is the bigger square
The count of numbers between them is (n+1)² – n² – 1
(i) Between 16² and 17²
16² = 256
17² = 289
Numbers between them = 289 – 256 – 1 = 32
(ii) Between 99² and 100²
99² = 9801
100² = 10000
Numbers between them = 10000 – 9801 – 1 = 198
8. In the following pattern, fill in the missing numbers:
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = (___)²
9² + 10² + (___)² = (___)²
See Solutions4² + 5² + 20² = (21)²
9² + 10² + (90)² = (91)²
9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
See SolutionsThere are total of 81 squares.
For prime factorisation:
81 ÷ 3 = 27
27 ÷ 3 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1
So, 81 = 3 × 3 × 3 × 3
Therefore, the prime factorisation of 81 = 3⁴
8th Ganita Prakash Chapter 1 Exercises
Page 16
Figure it Out
1. Find the cube roots of 27000 and 10648.
See SolutionsCube root of 27000:
27000 = 27 × 1000
Now, 27 = 3 × 3 × 3 = 3³ and 1000 = 10 × 10 × 10 = 10³
So, 27000 = (3 × 10)³ = 30³
Hence, ∛27000 = 30
Cube root of 10648:
10648 = (2 × 2 × 2)× (11 × 11 × 11)
So, 10648 = (2 × 11)³ = 22³
Thus, 10648 = 22³
Hence, ∛10648 = 22
2. What number will you multiply by 1323 to make it a cube number?
See SolutionsFor Prime factorisation of 1323:
1323 ÷ 3 = 441
441 ÷ 3 = 147
147 ÷ 3 = 49
49 ÷ 7 = 7
7 ÷ 7 = 1
So, 1323 = 3³ × 7²
We have to make all exponents multiples of 3
We already have:
3³ (complete triplet)
7² (needs one more 7 to make 7³)
So, we need to multiply by 7.
Therefore, Multiply 1323 by 7 to make it a perfect cube.
The cube number will be 1323 × 7 = 9261
Thus, ∛9261 = 21
3. State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
See Solutions(i) The cube of any odd number is even. False
Because: The cube of an odd number is always odd.
Example:
3³ = 27 (odd)
5³ = 125 (odd)
So, odd × odd × odd = odd, not even.
(ii) There is no perfect cube that ends with 8.
See Solutions(ii) There is no perfect cube that ends with 8. False
Because: Some perfect cubes do end in 8.
Example:
2³ = 8
12³ = 1728
So, a perfect cube can end with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
See Solutions(iii) The cube of a 2-digit number may be a 3-digit number. True
Because: The cube of 10 = 1000 → 4-digit
But the cube of 4 = 64 (2-digit)
Now try 5 to 9:
5³ = 125
6³ = 216
9³ = 729
So yes, some 2-digit numbers (like 5 to 9) give 3-digit cubes.
(iv) The cube of a 2-digit number may have seven or more digits.
See Solutions(iv) The cube of a 2-digit number may have seven or more digits. True
Example:
99³ = 970299 → 6-digit
100³ = 1000000 → 7-digit
So cube of a high 2-digit number (like 99) is close to 7 digits and cube of 100 (a 3-digit number) gives exactly 7 digits.
So, cube of a 2-digit number can have 7 digits, especially when close to 100.
(v) Cube numbers have an odd number of factors.
See Solutions(v) Cube numbers have an odd number of factors. False
Only perfect squares have an odd number of factors.
Cube numbers usually have an even number of factors, unless they are also perfect squares.
Example:
8 = 2³ → factors = 1, 2, 4, 8 → total = 4 (even)
27 = 3³ → factors = 1, 3, 9, 27 → total = 4 (even)
Exercises Question Answers of Class 8 Ganita Prakash
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.
See SolutionsCube Root of 1331:
Ends in 1 → Cube of a number ending in 1 also ends in 1
Try: 11³ = 1331
So, Cube root of 1331 = 11
Cube Root of 4913:
Ends in 3 → Cube of a number ending in 7 ends in 3 (because 7³ = 343)
Try: 17³ = 4913
So, Cube root of 4913 = 17
Cube Root of 12167:
Ends in 7 → Cube of a number ending in 3 ends in 7 (because 3³ = 27)
Try: 23³ = 12167
So, Cube root of 12167 = 23
Cube Root of 32768:
Ends in 8 → Cube of a number ending in 2 ends in 8 (because 2³ = 8)
Try: 32³ = 32768
So, Cube root of 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
See SolutionsCube Difference: a³ – b³ = (a – b)(a² + ab + b²)
(i) 67³ – 66³
= (67 – 66)(67² + 67×66 + 66²)
= 1 × (4489 + 4422 + 4356)
= 1 × 13267
(ii) 43³ – 42³
= (43 – 42)(43² + 43×42 + 42²)
= 1 × (1849 + 1806 + 1764)
= 1 × 5419
Square Difference: a² – b² = (a – b)(a + b)
(iii) 67² – 66²
= (67 – 66)(67 + 66) = 1 × 133 = 133
(iv) 43² – 42²
= (43 – 42)(43 + 42) = 1 × 85 = 85
So, (i) 67³ – 66³ is the greatest, because it has the highest value as 13267.
Is Class 8 Maths Ganita Prakash Chapter 1 A Square and A Cube difficult to understand for average students?
No, it’s not very difficult—especially if you enjoy number patterns. The chapter uses fun stories, puzzles like the “locker problem” and real-life patterns to explain square and cube numbers. Most students find it easier than other chapters because it starts with basic ideas and slowly builds up. If you understand how multiplication works and remember a few squares and cubes, you will find it quite interesting. With regular practice and a few visual tricks, even average students can master this chapter easily.
How can I make Class 8 Maths Ganita Prakash Chapter 1 easy and interesting to learn?
Start by learning the squares and cubes of numbers up to 20. Then explore the patterns shown in the chapter—like how square numbers come from adding odd numbers or how cube numbers grow. Try to understand the puzzles, like the locker activity or the taxicab number story, as they make the concepts more enjoyable. Use colors to highlight square vs. cube numbers in your notebook. Watch short animated videos if needed. Also, solve at least 5 practice problems daily to stay confident. Once you start noticing the number patterns, the chapter becomes really fun.
How much time does it usually take to complete Class 8 Maths Ganita Prakash Chapter 1 with proper understanding?
On average, it takes 4 to 6 days to complete this chapter thoroughly if you study regularly for 45–60 minutes a day. First two days can be used for learning perfect squares and square roots. Then spend two days on cubes and cube roots. The remaining time can be used for practicing puzzles, revising and solving exercise questions. If you already know basic squares and multiplication, you might finish it faster. The key is to focus on understanding patterns rather than just memorizing facts.
What are the most important topics to focus on in Class 8 Maths Ganita Prakash Chapter 1 for scoring well in exams?
Some key topics that often come in exams from this chapter include:
► Recognizing perfect squares and perfect cubes.
► Prime factorisation method to check for square or cube numbers.
► Finding the smallest number to multiply or divide to make a number a perfect square/cube.
► Estimating square roots and cube roots.
► Application-based puzzles like the 100 lockers and the taxicab number.
If you understand these concepts clearly and practice questions based on them, you can easily score full marks from this chapter.
Are puzzles like the locker problem in Class 8 Maths Ganita Prakash Chapter 1 helpful in understanding square numbers better?
Absolutely yes! The locker puzzle is a very smart way to show how square numbers behave. It helps you understand that square numbers have an odd number of factors, which is why their lockers stay open while others close. Instead of just memorizing definitions, puzzles like these help you “see” how square numbers work. They also make math feel like a game, not just a subject. So, yes—solving such puzzles not only makes your concept strong but also builds logical thinking and interest in math.
