---
title: "NCERT Solutions for Class 8 Maths Chapter 1 A Square and a Cube"
url: https://www.swavid.com/maths/class/8/chapter/a-square-and-a-cube/ncert-solutions
dateModified: 2026-10-07T15:19:46+00:00
---

# NCERT Solutions for Class 8 Maths Chapter 1 A Square and a Cube

This chapter's questions cover concepts related to squares, square roots, cubes, cube roots, and patterns involving numbers.

Free PDF (20 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-1-a-square-and-a-cube-3bce2bd70e.pdf

## In-text Questions

### Question 1

*2 marks · Very short answer*

Does every number have an even number of factors?

**Solution**

1. The number of factors of a number depends on its prime factorisation and can be even or odd.
2. For example, prime numbers have exactly 2 factors (which is even), but square numbers have an odd number of factors.

**Answer:** No, not every number has an even number of factors.

> Common mistake: Assuming all numbers have an even number of factors because factors usually occur in pairs.

### Question 2

*2 marks · Very short answer*

Can you use this insight to find more numbers with an odd number of factors?

**Solution**

1. Each factor of a number has a partner factor such that their product equals the number.
2. For square numbers like $2 \times 2$ or $6 \times 6$, the two factors in the middle pair are the same, leaving one factor without a distinct partner and resulting in an odd total number of factors.

**Answer:** Yes, all square numbers have an odd number of factors.

> Common mistake: Forgetting that square numbers have identical partner factors.

### Question 3

*2 marks · Very short answer*

Write the locker numbers that remain open.

**Solution**

1. A locker remains open if it is toggled an odd number of times, which happens only when it has an odd number of factors.
2. The numbers from 1 to 100 that have an odd number of factors are the perfect squares.

**Answer:** 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

> Common mistake: Listing non-square numbers by miscounting factors.

### Question 4

*2 marks · Very short answer*

Which are these five lockers?

**Solution**

1. A locker is toggled twice if it has exactly two factors.
2. The numbers between 1 and 100 that have exactly two factors (1 and the number itself) are the prime numbers.

**Answer:** 2, 3, 5, 7, 11

> Common mistake: Including 1 as a prime number or missing 2.

### Question 5

*3 marks · Short answer*

What patterns do you notice? Share your observations and make conjectures.

**Solution**

1. Observe the units digits of the squares of the first 20 natural numbers from the table.
2. All perfect squares end with the digits 0, 1, 4, 5, 6, or 9.
3. None of the perfect squares end with the digits 2, 3, 7, or 8.

**Answer:** All perfect squares end with 0, 1, 4, 5, 6, or 9, and never with 2, 3, 7, or 8.

> Common mistake: Conjecturing that any number ending in 0, 1, 4, 5, 6, or 9 is always a square.

### Question 6

*2 marks · Very short answer*

If a number ends in $0, 1, 4, 5, 6$ or $9$, is it always a square?

**Solution**

1. Although all squares end in 0, 1, 4, 5, 6, or 9, not every number ending in these digits is a square.
2. For example, 16 is a square ending in 6, but 26 also ends in 6 and is not a square.

**Answer:** No, it is not always a square.

> Common mistake: Assuming the converse statement is true for units digits.

### Question 7

*3 marks · Short answer*

Write 5 numbers such that you can determine by looking at their units digit that they are not squares.

**Solution**

1. A number is not a perfect square if it ends with the digits 2, 3, 7, or 8.
2. We can select any 5 numbers ending with these digits, such as 22, 43, 67, 88, and 102.
3. Therefore, by looking at their units digits (2, 3, 7, 8, 2), we can determine that none of them are squares.

**Answer:** 22, 43, 67, 88, and 102

> Common mistake: Thinking numbers ending in 0, 1, 4, 5, 6, or 9 are always squares.

### Question 8

*3 marks · Short answer*

Write the next two squares. Notice that if a number has $1$ or $9$ in the units place, then its square ends in $1$.

**Solution**

1. The given squares ending in 1 are $1^2 = 1$, $9^2 = 81$, $11^2 = 121$, $19^2 = 361$, $21^2 = 441$, and $29^2 = 841$.
2. Following the pattern of numbers ending in 1 or 9, the next two squares are for $31$ and $39$.
3. Calculating them, $31^2 = 961$ and $39^2 = 1521$, both of which end in 1.

**Answer:** $31^2 = 961$ and $39^2 = 1521$

> Common mistake: Listing wrong numbers that do not have 1 or 9 in their units place.

### Question 9

*1 mark · MCQ*

Which of the following numbers have the digit $6$ in the units place?
(i) $38^2$ (ii) $34^2$ (iii) $46^2$ (iv) $56^2$ (v) $74^2$ (vi) $82^2$

- $38^2$
- $34^2$
- $46^2$
- $56^2$
- $74^2$
- $82^2$

**Solution**

1. The units digit of a square is determined by squaring the units digit of the number.
2. For $34^2$, $4^2 = 16$ (ends in 6); for $46^2$, $6^2 = 36$ (ends in 6); for $56^2$, $6^2 = 36$ (ends in 6); for $74^2$, $4^2 = 16$ (ends in 6).
3. Hence, options (ii), (iii), (iv), and (v) have 6 in their units place.

**Answer:** (ii) $34^2$, (iii) $46^2$, (iv) $56^2$, (v) $74^2$

> Common mistake: Checking the whole number instead of just its units digit.

### Question 10

*3 marks · Short answer*

Find more such patterns by observing the numbers and their squares from the table you filled earlier.

**Solution**

1. Observe the units digits of the squares of numbers ending in specific digits from the table.
2. Numbers ending in 4 or 6 have squares that always end in 6.
3. Numbers ending in 0 have squares that end with an even number of zeros.

**Answer:** Squares of numbers ending in 4 or 6 always end in 6.

> Common mistake: Failing to generalize the pattern correctly.

### Question 11

*2 marks · Very short answer*

If a number contains $3$ zeros at the end, how many zeros will its square have at the end?

**Solution**

1. When a number is squared, the number of zeros at its end gets doubled.
2. Therefore, if a number contains 3 zeros at the end, its square will have 6 zeros at the end.

**Answer:** Six zeros

> Common mistake: Adding the number of zeros instead of doubling them.

### Question 12

*3 marks · Short answer*

What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?

**Solution**

1. The number of zeros at the end of a square is always double the number of zeros at the end of the original number.
2. Yes, this will always happen because squaring a number with $k$ zeros means multiplying $10^k$ by itself, resulting in $2k$ zeros.
3. Yes, squares can only have an even number of zeros at the end.

**Answer:** Number of zeros in the square is double the number of zeros in the number. Yes. Yes.

> Common mistake: Stating that squares can have an odd number of zeros.

### Question 13

*2 marks · Very short answer*

What can you say about the parity of a number and its square?

**Solution**

1. The square of an even number is always even, and the square of an odd number is always odd.

**Answer:** Square of an even number is even and the square of an odd number is odd.

> Common mistake: Confusing the parity of numbers and thinking squares of even numbers can be odd.

### Question 14

*3 marks · Short answer*

Using the pattern above, find $36^2$, given that $35^2 = 1225$.

**Solution**

1. We know that every square is the sum of successive odd numbers starting from 1, and $36^2$ can be obtained by adding the 36th odd number to $35^2$.
2. The $n^{\text{th}}$ odd number is given by $2n - 1$, so the 36th odd number is $2(36) - 1 = 71$.
3. Adding 71 to 1225 gives $1225 + 71 = 1296$.

**Answer:** 1296

> Common mistake: Adding the wrong odd number or starting from a number other than 1.

### Question 15

*3 marks · Short answer*

Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?

**Solution**

1. Let the two consecutive perfect squares be $n^2$ and $(n + 1)^2$.
2. The numbers between $n^2$ and $(n + 1)^2$ are $(n^2 + 1), (n^2 + 2), \dots, ((n + 1)^2 - 1)$.
3. The total number of non-square numbers between $n^2$ and $(n + 1)^2$ is given by $(n + 1)^2 - n^2 - 1 = 2n$.

**Answer:** There are $2n$ numbers between the consecutive squares $n^2$ and $(n + 1)^2$.

> Common mistake: Subtracting the two squares directly without subtracting an extra 1, forgetting that both square numbers themselves are excluded.

### Question 16

*3 marks · Short answer*

How many square numbers are there between $1$ and $100$? How many are between $101$ and $200$? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of $100$. What is the largest square less than $1000$?

**Solution**

1. The square numbers between $1$ and $100$ are $1^2, 2^2, 3^2, \dots, 10^2$, which gives a total of $10$ squares.
2. The square numbers between $101$ and $200$ are $11^2 = 121, 12^2 = 144, 13^2 = 169, 14^2 = 196$, which gives a total of $4$ squares.
3. The largest square less than $1000$ is $31^2 = 961$ since $32^2 = 1024$.

**Answer:** There are 10 squares between 1 and 100, 4 squares between 101 and 200, and the largest square less than 1000 is 961.

> Common mistake: Including the boundaries improperly or forgetting that 1 is included in the first interval.

### Question 17

*3 marks · Short answer*

Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.

**Solution**

1. The sum of two consecutive triangular numbers gives a square number, for example, $1 + 3 = 2^2$, $3 + 6 = 3^2$, and $6 + 10 = 4^2$.
2. The next term in the triangular number sequence after 15 is $15 + 6 = 21$.
3. Adding the consecutive triangular numbers 10 and 15 gives $10 + 15 = 25 = 5^2$.

**Answer:** The sum of two consecutive triangular numbers is always a square number; the next term sum is $10 + 15 = 25 = 5^2$.

> Common mistake: Confusing triangular numbers with square numbers.

### Question 18

*2 marks · Very short answer*

What is the square root of $64$?

**Solution**

1. We know that $8 \times 8 = 64$ and $(-8) \times (-8) = 64$, so the square roots of 64 are $+8$ and $-8$.
2. Considering only the positive square root as per the chapter convention, the square root of 64 is 8.

**Answer:** 8 (or $\pm 8$)

> Common mistake: Writing only the negative root or forgetting that every positive number has two square roots.

### Question 19

*3 marks · Short answer*

Given a number, such as $576$ or $327$, how do we find out if it is a perfect square? If it is a perfect square, how can we find its square root?

**Solution**

1. We check if a number is a perfect square by listing sequences of squares, using successive subtraction of consecutive odd numbers starting from 1, or by prime factorisation.
2. To find its square root, we group the prime factors into two identical groups and take the product of factors from one group.

**Answer:** We can determine if a number is a perfect square and find its square root using prime factorisation or successive subtraction of odd numbers.

> Common mistake: Confusing prime factorisation grouping with cube roots.

### Question 20

*2 marks · Very short answer*

Is $324$ a perfect square?

**Solution**

1. The prime factorisation of $324$ is $2 \times 2 \times 3 \times 3 \times 3 \times 3$.
2. Since the prime factors can be grouped into two identical groups of $(2 \times 3 \times 3)$, $324$ is a perfect square.

**Answer:** Yes, $324$ is a perfect square.

> Common mistake: Incorrectly grouping prime factors.

### Question 21

*2 marks · Very short answer*

Is $156$ a perfect square?

**Solution**

1. The prime factorisation of $156$ is $2 \times 2 \times 3 \times 13$.
2. Since we cannot pair up all these prime factors into two identical groups, $156$ is not a perfect square.

**Answer:** No, $156$ is not a perfect square.

> Common mistake: Assuming numbers ending in even digits are squares.

### Question 22

*3 marks · Short answer*

Find whether $1156$ and $2800$ are perfect squares using prime factorisation.

**Solution**

1. The prime factorisation of $1156$ is $2 \times 2 \times 17 \times 17$, which can be paired into two identical groups $(2 \times 17) \times (2 \times 17)$, so it is a perfect square.
2. The prime factorisation of $2800$ is $2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7$, where the prime factor $7$ cannot be paired, so it is not a perfect square.

**Answer:** $1156$ is a perfect square, whereas $2800$ is not a perfect square.

> Common mistake: Leaving out prime factors during factorisation.

### Question 23

*2 marks · Very short answer*

How many cubes of side $1 \text{ cm}$ will make a cube of side $3 \text{ cm}$?

**Solution**

1. A cube of side $3 \text{ cm}$ has a volume of $3 \text{ cm} \times 3 \text{ cm} \times 3 \text{ cm} = 27 \text{ cu. cm}$.
2. Therefore, $27$ cubes of side $1 \text{ cm}$ are required to make a cube of side $3 \text{ cm}$.

**Answer:** 27 cubes

> Common mistake: Multiplying the side length by 3 instead of cubing it.

### Question 24

*2 marks · Very short answer*

Is $9$ a cube?

**Solution**

1. A number is a perfect cube if it is obtained by multiplying a number by itself three times.
2. Since $2 \times 2 \times 2 = 8$ and $3 \times 3 \times 3 = 27$, $9$ cannot be expressed as the product of three identical integers.

**Answer:** No, $9$ is not a cube.

> Common mistake: Confusing squares with cubes ($9 = 3^2$, not a cube).

### Question 25

*3 marks · Short answer*

Can you estimate the number of unit cubes in a cube with an edge length of $4$ units?

**Solution**

1. Each layer of the cube has $4 \times 4 = 16$ unit cubes.
2. Since there are $4$ such layers, the total number of unit cubes is $4 \times 4 \times 4 = 64$.
3. Therefore, the number of unit cubes is $64$.

**Answer:** 64 unit cubes

> Common mistake: Multiplying by 3 instead of cubing the edge length.

### Question 26

*3 marks · Short answer*

What patterns do you notice in the table above?

**Solution**

1. Observing the table of cubes, we notice that cubes of even numbers are even and cubes of odd numbers are odd.
2. The last digits of cubes follow a repeating pattern, and unlike squares where only specific digits appear, cubes can end in any digit from $0$ to $9$.
3. The number of zeros at the end of a cube of a multiple of $10$ is a multiple of $3$.

**Answer:** Cubes of even numbers are even, cubes of odd numbers are odd, and every digit from $0$ to $9$ can appear as the last digit of a cube.

> Common mistake: Confining the last digits of cubes to only positive square-like endings.

### Question 27

*3 marks · Short answer*

Similar to squares, can you find the number of cubes with $1$ digit, $2$ digits, and $3$ digits? What do you observe?

**Solution**

1. Cubes with $1$ digit are from numbers $1$ to $2$ ($1^3 = 1$ to $2^3 = 8$), total $2$ numbers.
2. Cubes with $2$ digits are from numbers $3$ to $4$ ($3^3 = 27$ to $4^3 = 64$), total $2$ numbers.
3. Cubes with $3$ digits are from numbers $5$ to $9$ ($5^3 = 125$ to $9^3 = 729$), total $5$ numbers.

**Answer:** There are 2 one-digit cubes, 2 two-digit cubes, and 5 three-digit cubes.

> Common mistake: Miscounting the range of numbers for each digit count.

### Question 28

*3 marks · Short answer*

Can a cube end with exactly two zeroes ($00$)? Explain.

**Solution**

1. When a number is cubed, its prime factors are multiplied three times.
2. Therefore, any prime factor in the prime factorization of a cube must appear in a multiple of $3$ times.
3. This means the number of trailing zeros in a cube must always be a multiple of $3$, so a cube can never end with exactly two zeros.

**Answer:** No, because the number of zeros at the end of a cube must always be a multiple of 3.

> Common mistake: Confusing the properties of squares with cubes regarding trailing zeros.

### Question 29

*3 marks · Short answer*

The next two taxicab numbers after $1729$ are $4104$ and $13832$. Find the two ways in which each of these can be expressed as the sum of two positive cubes.

**Solution**

1. For $4104$, the two ways are $2^3 + 16^3$ and $9^3 + 15^3$.
2. For $13832$, the two ways are $2^3 + 24^3$ and $18^3 + 20^3$.
3. Both numbers can be expressed as the sum of two positive cubes in two different ways.

**Answer:** $4104 = 2^3 + 16^3 = 9^3 + 15^3$ and $13832 = 2^3 + 24^3 = 18^3 + 20^3$

> Common mistake: Using negative numbers or incorrect cube pairs.

### Question 30

*3 marks · Short answer*

Can you tell what this sum is without doing the calculation?

**Solution**

1. We know that the sum of consecutive odd numbers starting from the pattern corresponds to cubes, where the sum of $n$ consecutive odd numbers equals $n^3$.
2. The given series of odd numbers from $91$ to $109$ is the $10^{\text{th}}$ set in the pattern of consecutive odd numbers summing to cubes.
3. Therefore, the sum is equal to $10^3 = 1000$.

**Answer:** 1000

> Common mistake: Attempting to add all the terms individually instead of using the cube pattern.

### Question 31

*2 marks · Very short answer*

Let us check if $3375$ is a perfect cube.

**Solution**

1. Find the prime factorisation of $3375$ as $3 \times 3 \times 3 \times 5 \times 5 \times 5$.
2. Group the prime factors into three identical groups to get $(3 \times 5) \times (3 \times 5) \times (3 \times 5) = 15^3$.
3. Therefore, $3375$ is a perfect cube.

**Answer:** Yes, 3375 is a perfect cube.

> Common mistake: Grouping into two groups instead of three groups.

### Question 32

*2 marks · Very short answer*

Is $500$ a perfect cube?

**Solution**

1. Find the prime factorisation of $500$ as $2 \times 2 \times 5 \times 5 \times 5$.
2. Observe that the factors cannot be split into three identical groups since the prime factor $2$ appears only twice.
3. Therefore, $500$ is not a perfect cube.

**Answer:** No, 500 is not a perfect cube.

> Common mistake: Assuming a number ending in zeros is always a cube.

### Question 33

*3 marks · Short answer*

Find the cube roots of these numbers:
(i) $\sqrt[3]{64} = $ (ii) $\sqrt[3]{512} = $ (iii) $\sqrt[3]{729} = $

**Part (i)**

1. Since $4 \times 4 \times 4 = 64$, the cube root of $64$ is $4$.

Answer (i): 4

**Part (ii)**

1. Since $8 \times 8 \times 8 = 512$, the cube root of $512$ is $8$.

Answer (ii): 8

**Part (iii)**

1. Since $9 \times 9 \times 9 = 729$, the cube root of $729$ is $9$.

Answer (iii): 9

**Answer:** (i) 4, (ii) 8, (iii) 9

> Common mistake: Confusing cube roots with square roots.

### Question 34

*3 marks · Short answer*

Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?

**Solution**

1. Write the sequence of perfect cubes: $1, 8, 27, 64, 125, 216, \dots$
2. Compute Level 1 differences between consecutive cubes: $8-1=7$, $27-8=19$, $64-27=37$, $125-64=61$, $216-125=91$.
3. Compute Level 2 differences: $19-7=12$, $37-19=18$, $61-37=24$, $91-61=30$.
4. Compute Level 3 differences: $18-12=6$, $24-18=6$, $30-24=6$.
5. We notice that after three levels of successive differences, all the differences become the same constant value ($6$).

**Answer:** After three levels, all the differences at a level are the same (constant value 6).

> Common mistake: Stopping at Level 2, which applies to squares instead of cubes.

## Figure it Out

### Question 1

*1 mark · MCQ*

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

- $2032$
- $2048$
- $1027$
- $1089$

**Solution**

1. We know that perfect squares can only end in $0, 1, 4, 5, 6,$ or $9$ and cannot end in $2, 3, 7,$ or $8$.
2. Among the given numbers, $2032$ ends in $2$, $2048$ ends in $8$, and $1027$ ends in $7$, so none of them can be perfect squares.

**Answer:** (i), (ii) and (iii)

> Common mistake: Thinking that numbers ending in $2, 3, 7,$ or $8$ can be perfect squares.

### Question 2

*1 mark · MCQ*

Which one among $64^2, 108^2, 292^2, 36^2$ has last digit $4$?

- $64^2$
- $108^2$
- $292^2$
- $36^2$

**Solution**

1. The units digit of the square of a number depends only on the units digit of the number itself.
2. For $108^2$, the units digit is $8 \times 8 = 64$, which ends in $4$. For $292^2$, the units digit is $2 \times 2 = 4$, which also ends in $4$.

**Answer:** $108^2$ and $292^2$

> Common mistake: Checking only one of the correct options when multiple options are valid.

### Question 3

*1 mark · MCQ*

Given $125^2 = 15625$, what is the value of $126^2$?
(i) $15625 + 126$ (ii) $15625 + 26^2$ (iii) $15625 + 253$
(iv) $15625 + 251$ (v) $15625 + 51^2$

- $15625 + 126$
- $15625 + 26^2$
- $15625 + 253$
- $15625 + 251$
- $15625 + 51^2$

**Solution**

1. We know that the difference between consecutive squares $(n+1)^2 - n^2$ is equal to $(n+1) + n = 2n + 1$.
2. For $n = 125$, $126^2 - 125^2 = 126 + 125 = 251$, which means $126^2 = 15625 + 251$.

**Answer:** (iv) $15625 + 251$

> Common mistake: Adding only 126 instead of adding both 125 and 126.

### Question 4

*3 marks · Short answer*

Find the length of the side of a square whose area is $441 \text{ m}^2$.

**Solution**

1. Area of the square = $441 \text{ m}^2$.
2. Let the side of the square be $s$. Then $s^2 = 441$.
3. Using prime factorisation, $441 = 3 \times 3 \times 7 \times 7 = (3 \times 7)^2 = 21^2$.
4. Therefore, the length of the side is $21 \text{ m}$.

**Answer:** $21 \text{ m}$

> Common mistake: Forgetting to write the unit metres with the final answer.

### Question 5

*3 marks · Short answer*

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

**Solution**

1. First, find the LCM of the numbers $4, 9,$ and $10$.
2. Prime factorisations are $4 = 2^2$, $9 = 3^2$, and $10 = 2 \times 5$.
3. LCM = $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$.
4. To make the LCM a square number, we must multiply by the missing factor $5$, so the smallest square number is $180 \times 5 = 900$.

**Answer:** $900$

> Common mistake: Finding only the LCM without multiplying by the necessary factors to make it a perfect square.

### Question 6

*3 marks · Short answer*

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**

1. Find the prime factorisation of $9408$: $9408 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 3 = 2^6 \times 7^2 \times 3$.
2. Grouping the prime factors in pairs, we find that $3$ does not have a pair.
3. Thus, $9408$ must be multiplied by $3$ to make the product a perfect square.
4. The new product is $9408 \times 3 = 28224$, and its square root is $2^3 \times 7 \times 3 = 8 \times 7 \times 3 = 168$.

**Answer:** Multiplier is $3$, and the square root of the product is $168$.

> Common mistake: Multiplying incorrectly or failing to take the square root of the resulting product.

### Question 7

*3 marks · Short answer*

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

**Part (i)**

1. Let $n = 16$ and $n+1 = 17$.
2. The number of non-square numbers lying between the squares of numbers $n$ and $(n+1)$ is given by $2n$.
3. Substitute $n = 16$ to get $2 \times 16 = 32$.

Answer (i): $32$

**Part (ii)**

1. Let $n = 99$ and $n+1 = 100$.
2. The number of non-square numbers lying between the squares of numbers $n$ and $(n+1)$ is given by $2n$.
3. Substitute $n = 99$ to get $2 \times 99 = 198$.

Answer (ii): $198$

**Answer:** (i) $32$ numbers (ii) $198$ numbers

> Common mistake: Subtracting the two numbers directly without multiplying by 2.

### Question 8

*3 marks · Short answer*

In the following pattern, fill in the missing numbers:
$1^2 + 2^2 + 2^2 = 3^2$
$2^2 + 3^2 + 6^2 = 7^2$
$3^2 + 4^2 + 12^2 = 13^2$
$4^2 + 5^2 + 20^2 = (\text{___})^2$
$9^2 + 10^2 + (\text{___})^2 = (\text{___})^2$

**Part (i)**

1. Observe the relation in the given pattern where $n^2 + (n+1)^2 + [n(n+1)]^2 = [n(n+1)+1]^2$.
2. For $4^2 + 5^2 + 20^2 = (\text{___})^2$, take $n = 4$, so the third term is $4 \times 5 = 20$ and the RHS is $20 + 1 = 21$.

Answer (i): $21$

**Part (ii)**

1. Observe the pattern for $9^2 + 10^2 + (\text{___})^2 = (\text{___})^2$.
2. Take $n = 9$, so the missing base in the LHS is $9 \times 10 = 90$ and the RHS is $90 + 1 = 91$.

Answer (ii): $90$ and $91$

**Answer:** $4^2 + 5^2 + 20^2 = (21)^2$ and $9^2 + 10^2 + (90)^2 = (91)^2$

> Common mistake: Misidentifying the relationship between the first two bases and the third base.

### Question 9

*3 marks · Short answer*

How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.

**Solution**

1. By observing the grid of tiny squares in the figure on page 11, we count the total number of tiny squares.
2. The total number of tiny squares is 144.
3. The prime factorisation of 144 is $2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$.

**Answer:** 144 tiny squares; prime factorisation is $2^4 \times 3^2$

> Common mistake: Counting errors in the grid or incorrect prime factorisation.

## Figure it Out

### Question 1

*3 marks · Short answer*

Find the cube roots of $27000$ and $10648$.

**Part (i)**

1. $27000 = 27 \times 1000 = 3^3 \times 10^3 = (3 \times 10)^3 = 30^3$.
2. Therefore, $\sqrt[3]{27000} = 30$.

Answer (i): 30

**Part (ii)**

1. Find the prime factorisation: $10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 = 2^3 \times 11^3 = (2 \times 11)^3 = 22^3$.
2. Therefore, $\sqrt[3]{10648} = 22$.

Answer (ii): 22

**Answer:** Cube root of 27000 is 30 and cube root of 10648 is 22.

> Common mistake: Grouping prime factors incorrectly into triplets.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Find the prime factorisation of $1323$: $1323 = 3 \times 3 \times 3 \times 7 \times 7 = 3^3 \times 7^2$.
2. To make it a cube number, each prime factor must appear in groups of three.
3. Since $7$ appears only twice, we need to multiply by $1$ more $7$ so that its power becomes $3$.

**Answer:** 7

> Common mistake: Dividing instead of multiplying, or finding factors of squares instead of cubes.

### Question 3

*1 mark · True or false*

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.

**Part (i)**

1. The cube of an odd number is always odd.
2. Therefore, the statement is false.

Answer (i): False

**Part (ii)**

1. The cube of $2$ is $8$, and the cube of $12$ is $1728$, which end in $8$.
2. Therefore, perfect cubes can end with $8$, making the statement false.

Answer (ii): False

**Part (iii)**

1. The smallest $2$-digit number is $10$, and its cube is $10^3 = 1000$, which has $4$ digits.
2. Therefore, the cube of a $2$-digit number cannot be a $3$-digit number, making the statement false.

Answer (iii): False

**Part (iv)**

1. The largest $2$-digit number is $99$, and its cube is $99^3 = 970299$, which has $6$ digits.
2. Therefore, the cube of a $2$-digit number cannot have seven or more digits, making the statement false.

Answer (iv): False

**Part (v)**

1. Cube numbers can have an even or odd number of factors depending on whether their prime factor exponents are even or odd.
2. Therefore, the statement is false.

Answer (v): False

**Answer:** All statements are false.

> Common mistake: Confusing the properties of squares and cubes.

### Question 4

*3 marks · Short answer*

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 $32768$.

**Part (i)**

1. $\sqrt[3]{1331} = 11$.

Answer (i): 11

**Part (ii)**

1. $\sqrt[3]{4913} = 17$.

Answer (ii): 17

**Part (iii)**

1. $\sqrt[3]{12167} = 23$.

Answer (iii): 23

**Part (iv)**

1. $\sqrt[3]{32768} = 32$.

Answer (iv): 32

**Answer:** Cube roots are 11, 17, 23, and 32 respectively.

> Common mistake: Guessing without checking the units digit or bounds.

### Question 5

*1 mark · MCQ*

Which of the following is the greatest? Explain your reasoning.
(i) $67^3 - 66^3$ (ii) $43^3 - 42^3$ (iii) $67^2 - 66^2$ (iv) $43^2 - 42^2$

- $67^3 - 66^3$
- $43^3 - 42^3$
- $67^2 - 66^2$
- $43^2 - 42^2$

**Solution**

1. We know that the difference between consecutive cubes $n^3 - (n-1)^3 = 1 + n(n-1) \times 3$.
2. For $67^3 - 66^3$, the difference is $1 + 67 \times 66 \times 3 = 13267$.
3. For $43^3 - 42^3$, the difference is $1 + 43 \times 42 \times 3 = 5419$.
4. For differences of squares, $67^2 - 66^2 = 67 + 66 = 133$ and $43^2 - 42^2 = 43 + 42 = 85$.
5. Comparing all values, $67^3 - 66^3$ is the greatest.

**Answer:** (i) $67^3 - 66^3$

> Common mistake: Choosing $67^2 - 66^2$ thinking squares grow faster than cubes for the same base numbers.

## Puzzle Time!

### Question 1

*Activity*

Try arranging the numbers $1$ to $17$ (without repetition) in a row in a similar way — the sum of every adjacent pair of numbers should be a square.

**Solution**

1. Test arrangements of numbers from 1 to 17 in a row such that the sum of every adjacent pair is a square number.
2. One such valid arrangement is: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16, 8 (or other valid chains).

**Answer:** The numbers 1 to 17 can be arranged in a row such that each adjacent pair adds up to a square.

### Question 2

*3 marks · Short answer*

Can you arrange them in more than one way? If not, can you explain why?

**Solution**

1. Certain numbers like 1 and 17 have very few possible square sums with the available numbers from 1 to 17.
2. Due to these strict boundary restrictions, the choices for the ends of the row are severely limited.
3. Therefore, the sequence can be arranged in a few specific ways depending on the choice of endpoints and branch reversals.

**Answer:** Yes, they can be arranged in more than one way because certain valid sequences can be reversed or have alternative valid branches at specific nodes.

> Common mistake: Assuming there is only a unique solution without testing potential reversals or branch swaps.

### Question 3

*Activity*

Can you do the same with numbers from $1$ to $32$ (again, without repetition), but this time arranging all the numbers in a circle?

**Solution**

1. Arrange numbers from 1 to 32 in a circular chain such that the sum of any two adjacent numbers, including the first and the last, is a square.
2. Check that each number from 1 to 32 is used exactly once without repetition.

**Answer:** The numbers 1 to 32 can be arranged in a circle such that every adjacent pair adds up to a square.

## Frequently asked questions

### How many total questions are there in NCERT Solutions for Class 8 Maths Chapter 1: A Square and a Cube?

This chapter for the 2026-27 session based on the new NCERT book contains a total of 51 questions across different sections. You can find SwaVid's free PDF and step-by-step solutions for all of them on this page only.

### Which topics do the questions cover in this Class 8 Maths chapter?

The questions cover concepts like checking perfect cubes using prime factorisation, counting square numbers, number of zeros in squares, properties of cubes, and finding square roots and cube roots. SwaVid provides detailed step-by-step solutions for every single topic right here on this page.

### What are the hardest question types in this chapter and how should we approach them?

The puzzle time and complex prime factorisation questions are often considered the trickiest by students. To approach them, you should carefully break down the numbers into their prime factors and use the step-by-step methods available in SwaVid's free PDF on this page.

### How can I write answers to get full marks in Class 8 exams for this chapter?

To secure full marks, you must write every intermediate step clearly, especially when applying prime factorisation or property rules for squares and cubes. You can refer to SwaVid's step-by-step solutions on this page to learn the correct presentation format.

### Is the free PDF for this Class 8 Maths chapter available for download?

Yes, the complete resource including the free PDF and detailed step-by-step solutions is available on this page only. It is fully updated for the new NCERT book according to the current academic session.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student