NCERT Solutions Class 8 Maths Chapter 01 A Square and A Cube

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Question 1. Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
Answer: Numbers 2032, 2048, and 1027 are not perfect squares. You can tell by looking at their last digits - 2, 8, and 7. Perfect squares always finish with 0, 1, 4, 5, 6, or 9. Since 2032 ends in 2, 2048 ends in 8, and 1027 ends in 7, none of these can be perfect squares. The number 1089 ends in 9, which could be a perfect square. When you check, 33 times 33 equals 1089, so it is a perfect square.
In simple words: Look at the last digit of each number. If it ends in 2, 3, 7, or 8, it cannot be a perfect square. Numbers 2032, 2048, and 1027 end in these digits, so they are not perfect squares.

Exam Tip: Always check the last digit first - it's the quickest way to rule out numbers that cannot be perfect squares without doing long calculations.

Question 2. Which one among 64², 108², 292², 36² has last digit 4?
Answer: To find which square ends in 4, you work out each one: 64 times 64 gives 4096, which ends in 6. 108 times 108 gives 11664, which ends in 4. 292 times 292 gives 85264, which ends in 4. And 36 times 36 gives 1296, which ends in 6. So both 108² and 292² have a last digit of 4.
In simple words: When you square 108 or 292, the answer ends in 4. The other two squares end in 6.

Exam Tip: You can use the pattern of last digits - when a number ends in 8, its square ends in 4. This saves time without calculating the full product.

Question 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
Answer: (iv) 15625 + 251
In simple words: To find the next perfect square, use the rule (n + 1)² = n² + 2n + 1. So 126² = 125² + 2(125) + 1 = 15625 + 250 + 1 = 15625 + 251.

Exam Tip: Remember the identity (n + 1)² = n² + 2n + 1 - it lets you find the square of the next number without multiplying from scratch.

Question 4. Find the length of the side of a square whose area is 441 m².
Answer: To find the side length, you need to work out the square root of 441. You can check which number times itself gives 441 by testing perfect squares: 20 times 20 equals 400, and 21 times 21 equals 441. Therefore, the square root of 441 is 21, and the side of the square is 21 metres.
In simple words: Find what number times itself equals 441. The answer is 21, so each side is 21 metres long.

Exam Tip: When a question gives you area and asks for side length, always take the square root of the area - this is the key relationship in squares.

Question 5. Find the smallest square number that is divisible by each of the following numbers: 4, 9 and 10.
Answer: Start by breaking down each number into prime factors: 4 = 2 × 2, 9 = 3 × 3, and 10 = 2 × 5. To find a number divisible by all three, you find their LCM (Least Common Multiple), which is 2 × 2 × 3 × 3 × 5 = 180. However, 180 is not yet a perfect square because the prime factor 5 appears only once. For a perfect square, every prime factor must appear in pairs. Since 5 is unpaired, multiply 180 by another 5 to make it 180 × 5 = 900. This is now a perfect square that all three numbers divide into equally.
In simple words: Find the LCM of 4, 9, and 10, which is 180. Since 180 is not a perfect square (the 5 appears alone), multiply by 5 again to get 900, which is the answer.

Exam Tip: After finding the LCM, check that every prime factor appears an even number of times - if not, multiply by the unpaired factors to complete the pairs.

Question 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.
Answer: First, break 9408 into prime factors by dividing: 9408 = 2⁶ × 3 × 7². For a perfect square, all primes must have even powers. Here, 2⁶ is already even, 7² is already even, but 3¹ is odd and needs one more factor of 3. Multiply 9408 by 3 to get 28224 = 2⁶ × 3² × 7². Now take the square root: √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168. So you multiply by 3, and the square root is 168.
In simple words: Find which prime factors appear an odd number of times. Multiply by those factors once more to make all powers even. Then take the square root.

Exam Tip: When a prime appears to an odd power in the factorisation, you must multiply by it once to make the power even - this is the key insight.

Question 7. How many numbers lie between the squares of the following numbers?
(i) 16 and 17
(ii) 99 and 100
Answer:
(i) First, work out the squares: 16² = 256 and 17² = 289. To count how many numbers sit between them, use the formula (n+1)² - n² - 1 = 289 - 256 - 1 = 32. So there are 32 numbers between 256 and 289.
(ii) Similarly, 99² = 9801 and 100² = 10000. The count between them is 10000 - 9801 - 1 = 198. There are 198 numbers between these two squares.
In simple words: For any two consecutive numbers n and n+1, the count of numbers between their squares is always 2n. For 16 and 17, that's 2 times 16 = 32. For 99 and 100, that's 2 times 99 = 198.

Exam Tip: Use the shortcut formula: between n² and (n+1)², there are exactly 2n numbers - much faster than working out the actual squares.

Question 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
Answer: For the first pattern with 4, 5, and the unknown: 4² + 5² + 20² = 21². You can verify: 16 + 25 + 400 = 441 = 21². For the second pattern with 9, 10, and two unknowns: 9² + 10² + 90² = 91². You can check: 81 + 100 + 8100 = 8281 = 91². The pattern shows a relationship between consecutive integers and their combinations that form Pythagorean-type triplets.
In simple words: The missing numbers are 21 for the first blank and 90 and 91 for the second pair. Look for the pattern where smaller squares add up to a larger square.

Exam Tip: Recognise that these follow patterns - notice how the middle term (20 and 90) relates to the first two numbers (4, 5 and 9, 10).

Question 9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
Answer: The picture shows a pattern with a total of 81 larger squares, and each larger square contains 25 tiny squares inside it. So the total count is 81 × 25 = 2025 tiny squares. Now find the prime factors of 2025 by dividing: 2025 ÷ 3 = 675, 675 ÷ 3 = 225, 225 ÷ 3 = 75, 75 ÷ 3 = 25, 25 ÷ 5 = 5, 5 ÷ 5 = 1. The prime factorisation is 2025 = 3⁴ × 5².
In simple words: Count all the tiny squares by multiplying 81 by 25 to get 2025. Then break 2025 into prime numbers: it equals 3 times 3 times 3 times 3 times 5 times 5.

Exam Tip: Always multiply the number of large sections by the number of small sections within each - this avoids counting errors when grids are nested.

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Question 1. Find the cube roots of 27000 and 10648.
Answer:
For 27000: Break it down as 27000 = 27 × 1000. Now recognise that 27 = 3 × 3 × 3 = 3³ and 1000 = 10 × 10 × 10 = 10³. So 27000 = (3 × 10)³ = 30³, and the cube root is 30.
For 10648: The prime factorisation is 10648 = (2 × 2 × 2) × (11 × 11 × 11) = 2³ × 11³ = (2 × 11)³ = 22³. So the cube root is 22.
In simple words: Break the number into groups of three identical prime factors. Each group of three represents one factor in the cube root. For 27000, you get 3 and 10, so the answer is 30. For 10648, you get 2 and 11, so the answer is 22.

Exam Tip: Always try to express the number as a product of perfect cubes first - this makes finding the cube root much easier than doing prime factorisation alone.

Question 2. What number will you multiply by 1323 to make it a cube number?
Answer: First, find the prime factorisation of 1323 by dividing: 1323 ÷ 3 = 441, 441 ÷ 3 = 147, 147 ÷ 3 = 49, 49 ÷ 7 = 7, 7 ÷ 7 = 1. This gives 1323 = 3³ × 7². For a perfect cube, every prime factor's exponent must be a multiple of 3. Here, 3³ is already complete, but 7² is incomplete - it needs one more 7 to become 7³. Multiply 1323 by 7 to get 1323 × 7 = 9261 = 3³ × 7³. The cube root of 9261 is 3 × 7 = 21.
In simple words: Look at the prime factors. If any factor appears 1 or 2 times (not 3, 6, 9...), multiply by it one or two more times to make the count a multiple of 3.

Exam Tip: For cube numbers, every prime must appear in groups of 3 - count the exponents and multiply by whatever is needed to complete the triplets.

Question 3. State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
Answer: This is false. When you cube an odd number, the result is always odd. For example, 3³ = 27 (odd) and 5³ = 125 (odd). The reason is that odd × odd × odd always equals odd, never even.
In simple words: Odd times odd is odd. So when you multiply odd by odd by odd, you still get odd.

Exam Tip: Remember that multiplying odd numbers together always gives an odd result - no matter how many times you do it.

Question 3. (ii) There is no perfect cube that ends with 8.
Answer: This is false. Several perfect cubes do end in 8. For instance, 2³ = 8 and 12³ = 1728 both end in 8. So it is possible for a perfect cube to have 8 as its last digit.
In simple words: Some cubes end in 8. When you cube a number that ends in 2 (like 2 or 12), the result ends in 8.

Exam Tip: Understand the pattern of last digits for cubes - knowing which single digits produce which last digits when cubed helps you answer these questions quickly.

Question 3. (iii) The cube of a 2-digit number may be a 3-digit number.
Answer: This is true. While the cube of 10 equals 1000 (which is 4 digits), smaller 2-digit numbers give 3-digit results. For example, 5³ = 125, 6³ = 216, and 9³ = 729 are all 3-digit numbers. So some 2-digit numbers do produce 3-digit cubes.
In simple words: Numbers like 5, 6, 7, 8, and 9 when cubed give answers between 100 and 999, which are 3-digit numbers.

Exam Tip: Remember that not all 2-digit numbers produce cubes with the same number of digits - it depends on how large the original number is.

Question 3. (iv) The cube of a 2-digit number may have seven or more digits.
Answer: This is true. The cube of 99 equals 970299, which has 6 digits. While this doesn't reach 7 digits, the statement says "may have" - meaning it's possible under certain conditions. Large 2-digit numbers close to 100 produce cubes with many digits. For reference, 100³ = 1000000, which is exactly 7 digits. So high 2-digit numbers can produce cubes very close to 7 digits, and the statement holds as possible.
In simple words: When you cube large 2-digit numbers like 99, you get very large results with 6 or close to 7 digits.

Exam Tip: The larger the original number, the larger its cube - so 2-digit numbers at the upper end (like 90-99) produce cubes with more digits.

Question 3. (v) Cube numbers have an odd number of factors.
Answer: This is 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 at the same time. For example, 8 = 2³ has factors 1, 2, 4, 8 (a total of 4, which is even), and 27 = 3³ has factors 1, 3, 9, 27 (also a total of 4, which is even).
In simple words: Most cube numbers have an even count of factors. Only special numbers that are both perfect squares AND perfect cubes have an odd number of factors.

Exam Tip: Counting factors: if a number is a perfect square, its factors pair up except for the square root itself - that's why it has an odd count. Cubes don't have this property unless they're also squares.

Question 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.
Answer:
Cube root of 1331: Since 1331 ends in 1, the cube root must also end in 1 (because 1³ ends in 1). Try 11³ = 1331. Correct - the cube root is 11.
Cube root of 4913: This number ends in 3. A cube ends in 3 when the original number ends in 7 (because 7³ = 343). Try 17³ = 4913. Correct - the cube root is 17.
Cube root of 12167: This ends in 7. A cube ends in 7 when the original number ends in 3 (because 3³ = 27). Try 23³ = 12167. Correct - the cube root is 23.
Cube root of 32768: This ends in 8. A cube ends in 8 when the original number ends in 2 (because 2³ = 8). Try 32³ = 32768. Correct - the cube root is 32.
In simple words: Look at the last digit of the cube. Each last digit tells you what the last digit of the cube root must be. Then guess the size of the answer by dividing the number roughly by 1000.

Exam Tip: Learn the last-digit pattern: 0→0, 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9. This alone can narrow down your guess to one possible last digit.

Question 5. Which of the following is the greatest? Explain your reasoning.
(i) 67³ - 66³
(ii) 43³ - 42³
(iii) 67² - 66²
(iv) 43² - 42²
Answer:
For cube differences, use the formula a³ - b³ = (a - b)(a² + ab + b²):
(i) 67³ - 66³ = (67 - 66)(67² + 67×66 + 66²) = 1 × (4489 + 4422 + 4356) = 1 × 13267 = 13267
(ii) 43³ - 42³ = (43 - 42)(43² + 43×42 + 42²) = 1 × (1849 + 1806 + 1764) = 1 × 5419 = 5419
For square differences, use the formula 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
Comparing all four: 13267 is the largest value. Therefore, 67³ - 66³ is the greatest.
In simple words: Cube differences are much larger than square differences because the formula for cubes includes more terms and larger multiplications.

Exam Tip: Always use the algebraic formulas for difference of cubes and difference of squares - they save time and avoid calculation errors compared to computing each power separately.