A Square and A Cube Class 8 Solutions Ganita Prakash Maths Chapter 1

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NCERT Class 8 Maths Chapter 1 A Square and A Cube Solutions Question Answer

Ganita Prakash Class 8 Chapter 1 Solutions A Square and A Cube

NCERT Class 8 Maths Ganita Prakash Chapter 1 A Square and A Cube Solutions Question Answer

1.1 Square Numbers

Patterns and Properties of Perfect Squares

NCERT In-Text Questions (Pages 4-5)

Question 1.
Find the squares of the first 30 natural numbers and fill in the table below.

Solution:

Question 2.
What patterns do you notice? Share your observations and make conjectures.
Solution:
Observations:

• A perfect square number ends in 0, 1, 4, 5, 6, or 9 only.
• The square of numbers ending in 0, 1, 5, and 6 will always end in 0, 1, 5, and 6, respectively.
• The squares of numbers ending with 2, 3, 4, 7, 8, and 9 will always end in 4, 9, 6, 9, 4, and 1, respectively.
• The difference between any two consecutive squares is always an odd number.
• For example, 4 – 1 = 3, 9 – 4 = 5, 16 – 9 = 7, and so on.
• Squares of even numbers are even, and squares of odd numbers are odd.

Question 3.
If a number ends in 0, 1, 4, 5, 6, or 9, is it always a square?
Solution:
No. Some numbers end in 0, 1, 4, 5, 6, or 9 but are not squares.
For example, 10, 14, 135, 4006, 90109, etc.

Question 4.
Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
Solution:
Square numbers would never end with 2, 3, 7, or 8.
So, some numbers that are not squares are 222, 4503, 27, 68, and 142.

Question 5.
Let us consider square numbers ending in 6: 16 = 4², 36 = 6², 196 = 14², 256 = 16², 576 = 24², and 676 = 26².
Which of the following numbers has the digit 6 in the units place?
(i) 38²
(ii) 34²
(iii) 46²
(iv) 56²
(v) 74²
(vi) 82²
Solution:
From the table of the square of the first 30 natural numbers, we observe that the square of a number ending with 4 or 6 ends with the digit 6.
Therefore, numbers (ii) 34², (iii) 46², (iv) 56², and (v) 74² would have the digit 6 in the units place among the given numbers.

Question 6.
Find more such patterns by observing the numbers and their squares from the table you filled earlier.
Solution:
The table is given below:

Observations/Patterns:

• The square of a single-digit number is a two-digit number except for 1, 2, and 3, whose squares are single-digit numbers.
• The square of a two-digit number is a three-digit number.
• The square of a two-digit number with 0 at the unit’s place ends with two trailing zeros. For example, 20² = 400, 30² = 900.

Question 7.
If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Solution:
Such a number will have 6 zeros at the end.
For example, 1000² = 1000000, 2000² = 4000000

Question 8.
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:
We have observed in the table above that the squares of two-digit numbers that have zero at the units place have two trailing zeros. This will always happen.
The square of a number with ‘n’ zeros at the end will end with ‘2n’ trailing zeros.
So, we can say that a perfect square of a zero ending number will always have an even number of zeros at the end.

Question 9.
What can you say about the parity of a number and its square?
Solution:
Based on the squares of the first 30 natural numbers, we can say that:
The square of any even number is always even.
An even number can be represented algebraically as 2n, where n is any integer.
When we square this number, we get: (2n)² = 2² × n² = 4n² = 2(2n²)
Since the result is a multiple of 2, it is always an even number.
The square of any odd number is always odd.
An odd number can be represented algebraically as 2n + 1, where n is any integer.
When we square this number, we get:
(2n + 1)² = (2n + 1)(2n + 1) = 4n² + 4n + 1 = 2(2n² + 2n) + 1
The term 2(2n² + 2n) is a multiple of 2, so it is always an even number.
When we add 1 to an even number, the result is always an odd number.

Perfect Squares and Odd Numbers

NCERT In-Text Questions (Page 7)

Question 1.
Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
Solution:
Look at the table below:

From this, we conclude that between the squares of a number n and (n + 1), there will be 2n numbers.
For example, between 3² and 4², we have 2 × 3 = 6 numbers.
The pattern is that the number of integers between two consecutive squares is always twice the smaller of the two square roots.

Question 2.
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:
We find the number of perfect squares in each block of 100 as shown below in the table:

The largest square less than 1000 is 31² = 961.

Perfect Squares and Triangular Numbers

NCERT In-Text Questions (Page 7)

Question 1.
Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.

Solution:
The sum of any two consecutive triangular numbers will always be a square number.

Square Roots

NCERT In-Text Questions (Page 9)

Question 1.
Find whether 1156 and 2800 are perfect squares using prime factorisation.
Solution:
Prime factorisation of 1156:
1156 = 2 × 2 × 17 × 17
= 2² × 17²
= (2 × 17)²
Now, the prime factors can be divided into two equal groups, i.e., (2 × 17) and (2 × 17).
Therefore, 1156 is a perfect square.
Similarly, the prime factorisation of 2800:
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7 = 2² × 2² × 5² × 7
These factors cannot be divided into equal groups because 7 appears only once.
Therefore, 2800 is not a perfect square.

Figure it Out (Pages 10-11)

Question 1.
Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
Solution:
To check if a number is a perfect square, we can look at its last digit or use prime factorisation.
Perfect squares only end with 0, 1, 4, 5, 6, or 9.

• 2032 ends with 2 → Not a perfect square.
• 2048 ends with 8 → Not a perfect square.
• 1027 ends with 7 → Not a perfect square.
• 1089 ends with 9 → Could be a perfect square.

Now, prime factorisation of 1089:
3 × 3 × 11 × 11 = 3² × 11² = (3 × 11)²
The prime factors can be equally grouped into two parts. So, 1089 is a perfect square.
So, (i) 2032, (ii) 2048, and (iii) 1027 are not perfect squares.

Question 2.
Which one among 64², 108², 292², 36² has the last digit 4?
Solution:
Look at the table below:

On examining the square of the last digits of the given numbers, we find that only 108² and 292² will have 4 as the last digit.

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
Solution:
We know that the square of a number n is the sum of the first n odd numbers.
So, 125² = sum of the first 125 odd numbers = 15625.
Now, if we add 126th odd number to 15625, we get 126².
The nth odd number is given by the expression 2n – 1.
So, 126th odd number = 2 × 126 – 1 = 251
Therefore, 126² = 125² + 251 = 15625 + 251 = 15876
So, option (iv) is correct.

Question 4.
Find the length of the side of a square whose area is 441 m².
Solution:
The area of a square is given by:
Area = side × side
So, we need to find a number whose square is 441 m².
Let us find the square root of 441 using prime factorisation.
441 = 3 × 3 × 7 × 7 = 3² × 7² = (3 × 7)²
Now, we can divide the prime factors into two equal groups: (3 × 7) and (3 × 7).
Therefore, the side of the square is 3 × 7 = 21 m.

Question 5.
Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Solution:
If we know the LCM of the given numbers, we can find the smallest square divisible by them by multiplying the LCM by the smallest possible number that makes it a perfect square.
To find the LCM of 4, 9, and 10, we write the prime factorisation of these numbers:
4 = 2 × 2
9 = 3 × 3
10 = 2 × 5
Now, the LCM of 4, 9, and 10 = 2 × 2 × 3 × 3 × 5 = 180.
Now, check whether 180 is a perfect square.
Prime factors of 180: 2 × 2 × 3 × 3 × 5.
Here, all prime factors appear in pairs except 5.
To make it a perfect square, multiply by one more 5: 180 × 5 = 900.
Therefore, 900 is the smallest square number divisible by 4, 9, and 10.

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.
Solution:
To find the smallest number by which 9408 must be multiplied so that the product is a perfect square, we first need to find the prime factorization of 9408 and check whether its prime factors can be divided into two equal groups.
Prime factorisation of 9408: 2⁶ × 3 × 7²
Here, 3 appears only once, so we multiply 9408 by 3 to make all prime factors in pairs, i.e., 9408 × 3 = 28224, which is a square number.
Prime factorisation of 28224: 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Grouping the prime factors into equal groups, we get: (2 × 2 × 2 × 3 × 7) × (2 × 2 × 2 × 3 × 7)
So, the square root of 28224 is 2 × 2 × 2 × 3 × 7 = 168.

Question 7.
How many numbers lie between the squares of the following numbers?
(i) 16 and 17
(ii) 99 and 100
Solution:
The number of integers between n² and (n + 1)² is 2n.
(i) Between 16² and 17² = 2 × 16 = 32 numbers.
(ii) Between 99² and 100² = 2 × 99 = 198 numbers.

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
Solution:
We analyse the given pattern as follows:
n² + m² + (n × m)² = [(n × m) + 1]², where n and m are consecutive numbers.
Therefore, 4² + 5² + (20)² = (21)², 9² + 10² + (90)² = (91)²

Question 9.
How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.

Solution:
There are 25 tiny squares in each set, and (9 × 9) such sets are shown in the given picture.
So, total number of tiny squares = 9 × 9 × 25 = 2025
The prime factorisation of 2025 = 3 × 3 × 3 × 3 × 5 × 5 = 3² × 3² × 5²

1.2 Cubic Numbers

NCERT In-Text Questions (Pages 11-13)

Question 1.
How many cubes of side 1 cm make a cube of side 2 cm?

Solution:
By looking at the arrangement, we find that there are 2 × 2 unit cubes at the top layer, and there are 2 such layers.
So, a total of 2 × 2 × 2 = 8 cubes of side 1 cm make a cube of side 2 cm.

Question 2.
How many cubes of side 1 cm will make a cube of side 3 cm?
Solution:
In the arrangement of a cube of side 3 cm, there will be 3 × 3 unit cubes at the top layer, and there will be 3 such layers.
So, there will be 3 × 3 × 3 = 27 cubes of side 1 cm that will make a cube of side 3 cm.

Question 3.
Complete the table below:

Solution:

Question 4.
What patterns do you notice in the table above?
Solution:
Observations:

• Cube of a number having 0 at the units place, will have three trailing zeros.
• The cube of an even number is even, and the cube of an odd number is odd.
• There are only four perfect cubes up to 100 and ten perfect cubes up to 1000.

Question 5.
We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are the possible last digits of cubes?
Solution:
Cubic numbers can end in any digit from 0 to 9, unlike square numbers.
Additionally, the cube of a number ending with 0, 1, 4, 5, 6, or 9 also ends with 0, 1, 4, 5, 6, or 9, respectively.
But the cube of a number ending with 2, 3, 7, and 8 ends with 8, 7, 3, and 2, respectively.

Question 6.
Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
Solution:
Yes. By looking at the above table, we find that:

• There are 2 one-digit cubic numbers, i.e., 1, 8.
• There are 2 two-digit cubic numbers, i.e., 27, 64.
• There are 5 three-digit cubic numbers, i.e., 125, 216, 343, 512, 729.

Question 7.
Can a cube end with exactly two zeroes (00)? Explain.
Solution:
No. We know that to get a cube of a number, we multiply that number by three times.
Therefore, the cube of any number having zeros at the end will have 3 times the number of zeros at the end.
For example, 20³ (one zero) = 8000 (three zeros), 2000³ (three zeros) = 8000000000 (nine zeros).
Hence, the number of zeros at the end of a perfect cube must always be a multiple of 3 (0, 3, 6, 9, etc).
Since 2 is not a multiple of 3, a cube can never end with exactly two zeros.

Taxicab Numbers

NCERT In-Text Questions (Page 13)

Question 1.
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:
4104 and 13832 can be expressed as the sum of two positive cubes in two ways, shown below:
4104 = 4096 + 8 = 16³ + 2³
Also, 4104 = 3375 + 729 = 15³ + 9³
Similarly, 13832 = 8000 + 5832 = 20³ + 18³
Also, 13832 = 13824 + 8 = 24³ + 2³

Perfect Cubes and Consecutive Odd Numbers

NCERT In-Text Questions (Page 14)

Question 1.
Can you tell what this sum is without doing the calculation?
Solution:
It is an addition of ten consecutive odd numbers and its first term, 91 = 10 × 9 + 1.
∴ 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109 = 10³ = 1000
Alternative Method:
We see from the pattern that at each step, n consecutive odd numbers are added, the first number is n² – n + 1 and the sum is n³.
10² – 10 + 1 = 91
So here, the sum will be 10³ = 1000.

Cube Roots

NCERT In-Text Questions (Page 15)

Question 1.
Find the cube roots of these numbers:
(i) \(\sqrt[3]{64}\) = ___
(ii) \(\sqrt[3]{512}\) = ___
(iii) \(\sqrt[3]{729}\) = ___
Solution:
(i) Prime factorisation of 64 = 2 × 2 × 2 × 2 × 2 × 2
This can be divided into three equal groups as
64 = (2 × 2) × (2 × 2) × (2 × 2) = (2 × 2)³
⇒ \(\sqrt[3]{64}\) = (2 × 2) = 4

(ii) Prime factorisation of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
This can be divided into three equal groups as
512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = (2 × 2 × 2)³
⇒ \(\sqrt[3]{512}\) = (2 × 2 × 2) = 8

(iii) Prime factorisation of 729 = 3 × 3 × 3 × 3 × 3 × 3
This can be divided into three equal groups as
729 = (3 × 3) × (3 × 3) × (3 × 3) = (3 × 3)³
⇒ \(\sqrt[3]{729}\) = (3 × 3) = 9

Successive Differences

NCERT In-Text Questions (Page 15)

Question 1.
Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?

Solution:
Look at the following differences.
First-level differences are not in any pattern, but second-level differences are multiples of 6, 2nd multiple onwards.
Then the level 3 differences are the same, i.e., 6.

Thus, 6 is the constant difference found at the third level.

1.3 A Pinch of History

Figure it Out (Pages 16-17)

Question 1.
Find the cube roots of 27000 and 10648.
Solution:
To find cube roots, we use prime factorisation
27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
= (2 × 3 × 5) × (2 × 3 × 5) × (2 × 3 × 5)
= (2 × 3 × 5)³
∴ Cube root of 27000 = (2 × 3 × 5) = 30
10648 = 2 × 2 × 2 × 11 × 11 × 11
= (2 × 11) × (2 × 11) × (2 × 11)
= (2 × 11)³
Thus, cube root of 10648 = 2 × 11 = 22

Question 2.
What number will you multiply by 1323 to make it a cube number?
Solution:
By prime factorisation, we have
1323 = 3 × (3 × 7) × (3 × 7)
Here, we observe that to make 1323 a perfect cube, we need one more 7 in the factorisation of 1323.
So, we multiply 1323 by 7.
1323 × 7 = 9261, which is a cubic number having cube root as 21.

Question 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.
Because Odd number × Odd number = Odd number.
Therefore, (Odd × Odd × Odd) = (Odd × Odd) × Odd
⇒ Odd × Odd = Odd.
For example, 3³ = 27, which is odd.

(ii) False.
The cube of any number ending in 2 will end with 8.
For example, 2³ = 8, 12³ = 1728

(iii) False.
The cube of the smallest 2-digit number, i.e., 10³ = 1000, which is a 4-digit number.
So, the cube of all the 2-digit numbers will have at least 4 digits.

(iv) False.
The largest 2-digit number is 99.
The cube of 99 is 99³ = 970299, which has 6 digits.
So, the cube of a 2-digit number can have 7 or more digits.

(v) False.
A perfect cube does not necessarily have an odd number of factors.
8 is a cubic number. It has an even number of factors (4), i.e., 1, 2, 4, and 8.
64 is also a cubic number, but a square number too, i.e., 64 = 4³ = 8².
Therefore, it has an odd number of factors, i.e., 1, 2, 4, 8, 16, 32, 64, a total of 7 factors.
Therefore, the cubic numbers that are perfect squares have an odd number of factors.
So, the given statement is false.

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.
Solution:
1331

• It has 4 digits, so it should be the cube of a 2-digit number.
• It has 1 at the end, so the cube root should have 1 at the units place.
• Try the first such number, 11 × 11 × 11, we get 1331.
• So, the cube root of 1331 is 11.

4913

• It has 4 digits, so it should be the cube of a 2-digit number.
• It has 3 at the end, so the cube root should have 7 at the units place.
• Cube of 20 is 8000, so the cube root should be less than 20.
• We try 17 × 17 × 175, we get 4913.
• So, the cube root of 4913 is 17.

12167

• It has 5 digits, so it should be the cube of a 2-digit number.
• It has 7 at the end, so the cube root should have 3 at the units place.
• Cube of 25 is 15625, so the cube root should be less than 25.
• We try 23 × 23 × 23, we get 12167.
• So, the cube root of 12167 is 23.

32768

• It has 5 digits, so it should be the cube of a 2-digit number.
• It has 8 at the end, so the cube root should have 2 at the units place.
• Cube of 30 is 27000, so the cube root should be more than 30.
• We try 32 × 32 × 32, we get 32768.
• So, the cube root of 32768 is 32.

Question 5.
Which of the following is the greatest? Explain your reasoning.
(i) 67³ – 66³
(ii) 43³ – 42³
(iii) 67² – 66²
(iv) 43² – 42²
Solution:
We know that (n + 1)³ – n³ = 3 × (n + 1) × n + 1 and (n + 1)² – n² = (n + 1) + n
Therefore,
(i) 67³ – 66³ = 3 × 67 × 66 + 1 = 13267
(ii) 43³ – 42³ = 3 × 43 × 42 + 1 = 5419
(iii) 67² – 66² = 67 + 66 = 133
(iv) 43² – 42² = 43 + 42 = 85
Clearly, the value of 67³ – 66³ is the greatest.

Puzzle Time (Page 18)

Question 1.
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:
Here is an arrangement of the numbers 1 to 17 where the sum of every adjacent pair is a perfect square:
16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17

Question 2.
Can you arrange them in more than one way? If not, can you explain why?

Solution:
No, since there is only a single such path, the only other arrangement possible is to traverse the same path in the opposite direction, which gives us the reversed sequence.

Question 3.
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: