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

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Class 8 Maths Chapter 1 A Square and A Cube Solutions

Ganita Prakash Class 8 Chapter 1 Solutions

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

Intext (NCERT Page 2)

Question 1.
Does every number have an even number of factors?
Solution:
No, not every number has an even number of factors. Perfect squares have an odd number of factors.

Question 2.
For instance, 36 has a factor pair 6 x 6 where both numbers are 6. Does this number have an odd number of factors? If every factor of 36 other than 6 has a different factor as its partner then we can be sure that 36 has an odd number of factors. Check if this is true.
Solution:
The factors of 36 are 1,2, 3,4, 6, 9,12,18 and 36.
∴ Number of factors of 36 = 9, which is odd.
So, 36 has an odd number of factors.
Now, pair of partner factors of 36 are (1, 36), (2,18),
(3, 12), (4, 9) and (6, 6).
All pairs of factors except 6 have a distinct partner factor. The factor 6 is paired with itself, forming a double factor in the list.
This phenomenon of a factor being paired with itself occurs when the number is a perfect square and it leads to an odd number of factors.
So, the statement is true.

Intext (NCERT Page 4)

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

Answer:
The following table as shown below.

(ii) What patterns do you notice? Share your observations and make conjectures.
Solution:
The unit digits of perfect squares follow a repeating pattern : 0, 1, 4, 5, 6, 9.
Conjecture : A natural number is a perfect square only if its unit digits is 0, 1, 4, 5, 6 or 9.

Intext (NCERT Page 5)

Question 1.
Which of the following numbers have the digit 6 in the units place?
(i) 38²
(ii) 34²
(iii) 46²
(iv) 56²
(v) 74²
(vi) 82²
Solution:
We know that if the unit digit of a number is 4 or 6, the unit digit of the square of the number is 6.
Here, Unit digit of 34 = 4
Unit digit of 46 = 6
Unit digit of 56 = 6
Unit digit of 74 = 4
Hence, unit digits of 34², 46², 56² and 74² is 6.

Question 2.
If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Solution:
Let the numbers be 1000 and 2000.
∴ (1000)² = 1000000 and (2000)² = 4000000
Hence, the number of zeros in square of number contains 3 zeros at the end is 6.

Figure it Out (NCERT Page 10 and 11)

Question 1.
Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
Solution:
We know that a number having 2, 3, 7 or 8 at unit place is never a perfect square.
So, 2032, 2048 and 1027 are not a perfect square.

Now, by prime factorisation
1089 = 3 × 3 × 11 × 11
So, 1089 is a perfect square number.

Question 2.
Which one among 642, 1082, 2922, 362 has last digit 4?
Solution:
We know that if the unit digit of a number is 2 or 8 then the unit digit of the square of the number is 4.
So, the numbers with the last digit 4 are 108 2 and 2922.

Question 3.
Given,125² = 15625, what is the value ofl262?
(i) 15625 + 126
(ii) 15625 + 26²
(iii) 15625 + 253
(iv) 15625 + 251
(v) 15625 + 51²
Solution:
(iv) We have, 125² = 15625
∴ 126² = 15625 + 126th odd number
= 15625 + (2 × 126 – 1) [∵ nth odd number = 2n – 1]
= 15625 + 25² – 1
= 15625 + 251

Question 4.
Find the length of the side of a square whose area is 441m2.
Solution:
Given, Area of square = 441

Let side of the square be a.
a × a = 441 [∵ area of square = side × side]
⇒ a² = 441
⇒ a = \(\sqrt{441}\)
⇒ a = \(\sqrt{3 \times 3 \times 7 \times 7}\)
⇒ a = \(\sqrt{(3 \times 7)^2}\)
⇒ a = 21m

Question 5.
Find the smallest square number that is divisible by each of the following numbers : 4, 9 and 10.
Solution:
Given numbers are 4, 9 and 10. 2 4, 9, 10

LCM of 4, 9 and 10 = 2 × 2 × 3 × 3 × 5 =180
Here, the prime factor 5 does not have any pair.
So, 180 is multiplied by 5 to make pair.
∴ 180 × 5 = 900
Hence, 900 is the smallest square number that is 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:
By prime factorisation

∴ 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7
Here, the prime factor 3 does not have any pair.
So, we multiply 9408 by 3 to get the required number
i.e. 9408×3 = 28224
Now, square root of 28224
= \(\sqrt{28224}=\sqrt{(2 \times 2 \times 2 \times 3 \times 7)^2}\)
= 2 × 2 × 2 × 3 × 7
= 168

Question 7.
How many numbers lie between the squares of the following numbers?
(i) 16 and 17
Solution:
Given, two numbers are 16 and 17.
Here, n= 16 and n + 1 = 17
We know that total numbers lie between
n² and(n + 1)² = 2n
Total numbers lie between squares of 16 and 17
= 2 × 16
= 32

(ii) 99 and 100
Solution:
Given, two numbers are 99 and 100.
Here, n = 99 and n + 1 = 100
∴ Total numbers lie between squares of 99 and 100.
= 2 × 99
= 198

Question 8.
In the following pattern, fill in the missing number.

Solution:
Here, the third number is the product of the first two numbers and the fourth number is obtained by adding 1 to the third number.
∴ 4² + 5² + 20² = 21²
and 9² + 10² + 90² = 91²

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

Solution:
Here, total number of large squares = 9 × 9 = 81,
number of rows in one large square = 5
and number of columns in one large square = 5
∴ Total number of tiny squares in one large square
A = 5 ×5 = 25
Now, total number of tiny squares in 81 large squares = 25 × 81 = 2025
Prime factorisation of 2025
= 3 × 3 × 3 × 3 × 5 × 5

Intext (NCERT Page 11-13)

Question 1.
How many cube of side 1 cm make a cube of side 2 cm?
Solution:
We know that
A 1 unit cube is the smallest cube and 2-unit cube has 8 small cubes.
∴ Required number of cubes = \(\frac{8}{1}\) = 8

Question 2.
How many cubes of side 1 cm will make a cube of side 3 cm?
Solution:
We know that
A 1-unit cube is the smallest cube and a 3-unit cube has 27 small cubes.
∴ Required number of cubes = \(\frac{27}{1}\) = 27

Question 3.
(i) Complete the table below.

Answer:

(ii) What patterns do you notice in the table above?
Answer:
Here, we observe the following patterns.
(a) Cubes of the numbers ending in digit 0, 1, 4, 5, 6 and 9 are the numbers ending in the same digit.
(b) Cubes of numbers ending in digit 2 ends in digit 8 and vice versa.
(c) Cubes of the numbers ending in digits 3 and 7 end in digits 7 and 3, respectively.

Question 4.
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:
Here, 0³ = 0
∴ Last digit is 0.
1³ = 1
∴ Last digit is 1.
2³ = 8
∴ Last digit is 8.
3³ = 27
∴ Last digit is 7.
4³ = 64
∴ Last digit is 4.
5³ = 125
∴ Last digit is 5.
6³ = 216
∴ Last digit is 6.
7³ = 343
∴ Last digit is 3.
8³ = 512
∴ Last digit is 2.
9³ = 729
∴ Last digit is 9.
So, the possible last digits of cubes are 0, 1,2, 3, 4, 5, 6, 7, 8 and 9.

Question 5.
Similarly to squares, can you find the number of cubes with 1 digit, 2 digits and 3 digits? What do you observe?
Solution:
We known that
1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125,
6³ =216, 7³ = 343, 8³ = 512, 9³ = 729
∴ There are two 1-digit cube numbers, two 2-digit cube numbers and five 3-digit cube numbers.

Question 6.
Can a cube end with exactly two zeroes (00)? Explain
Solution:
No, because a cube always and with multiple of 3 zeroes.

Question 7.
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:
Given number are 4104 and 13832.
∴ 4104 = 2³ + 16³
= 9³ + 15³
and 13832 = 2³ + 24³
= 18³ + 20³

Intext (NCERT Page 15)

Question 1.
Find the cube roots of these numbers.
(i) \(\sqrt[3]{64}\)
Solution:
By prime factorisation,
64 = 2 × 2 × 2 × 2 × 2 × 2
= (2 × 2 × 2) × (2 × 2 × 2)
∴ \(\sqrt[3]{64}\) = 2 × 2 = 4

(ii) \(\sqrt[3]{512}\)
Solution:
By prime factorisation,
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)
∴ \(\sqrt[3]{512}\) = 2 × 2 × 2 = 8

(iii) \(\sqrt[3]{729}\)
Solution:
By prime factorisatian
729 = 3 × 3 × 3 × 3 × 3 × 3
= (3 × 3 × 3) × (3 × 3 × 3)
∴ \(\sqrt[3]{729}\) = 3 × 3 = 9

Question 2.
Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?

Solution:

Here, the differences of third level are constant and equal to 6.
∴ For a cubic sequence (n³), the differences of third level are constant. Hence, for a sequence of degree n, the differences of nth level are constant.

Figure it Out (NCERT Page 16 and 17)

Question 1.
Find the cube roots of 27000 and 10648.
Solution:
By prime factorisation,

27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
= (2 × 2 × 2) × (3 × 3 × 3) × (5 × 5 × 5)
∴ \(\sqrt[3]{27000}\) = 2 × 3 × 5 = 30
Now, by prime factorisation

10648 = 2 × 2 × 2 × 11 × 11 × 11
= (2 × 2 × 2) × (11 × 11 × 11)
∴ \(\sqrt[3]{10648}\) = 2 × 11 = 22

Question 2.
What number will you multiply by 1323 to make it a cube number?
Solution:
By prime factorisation, 1323 = 3 × 3 × 3 × 7 × 7

Here, 7 is without group of three.
So, 1323 is not a perfect cube.
To make perfect cube, we multiply 1323 by 7.
Thus, 1323 × 7 = 3 × 3 × 3 × 7 × 7 × 7 = 9261

Question 3.
State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
Solution:
False
The cube of any odd number is odd. e.g. (3)³ = 27, which is odd.

(ii) There is no perfect cube that ends with 8.
Solution:
False
The cubes of all the numbers having their unit place digit 2 will end with 8.
e.g. (12)³ = 1728, which is even.

(iii) The cube of a 2-digit number may be a 3-digit number.
Solution:
False
The cube of a 2-digit number may have 4-digits to 6-digits.
e.g. (10)³ = 1000 and (99)³ =970299

(iv) The cube of a 2-digit number may have seven or more digits.
Solution:
False
The cube of a 2-digit number may have 4-digits to 6-digits.

(v) Cube numbers have an odd number of factors.
Solution:
False
We know that 2³ = 8.
∴ Factors of 8 are 1, 2, 4, 8
∴ Number of factors of 8 = 4, which is even.
So, cube numbers do not necessarily have an odd number of factors.

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:
Last digit of 1331 = 1
The cube of number ending in digit 1 also ends in 1 …
(i) Since, 1331 is a 4-digit number.
So, its cube root will be a 2-digit number.
We know that 10³ = 1000 and 20³ = 8000.
Since, 1331 is between 1000 and 8000.
So, its cube root is between 10 and 20. …(ii)
∴ Cube root of 1331 = 11 [from Eqs. (i) and (ii)]

Last digit of 4913 = 3
The cube of number ending in digit 3 ends in digit 7. …(iii)
Since, 4913 is a 4-digit number.
So, its cube root will be a 2-digit number.
We know that 10³ = 1000 and 20³ = 8000
Since, 4913 is between 1000 and 8000.
So, its cube root is between 10 and 20. … (iv)
∴ Cube root of 4913 = 17 [from Eqs. (iii) and (iv)]

Last digit of 12167 = 7
The cube of number ending in digit 7 ends in digit 3. … (v)
Since, 12167 is a 5-digit number.
So, its cube root will be a 2-digit number.
We know that 20³ = 8000 and 30³ = 27000
Since, 12167 is between 8000 and 27000.
So, its cube root is between 20 and 30. … (vi)
∴ Cube root of 12167 = 23 [from Eqs. (v) and (vi)]

Last digit of 32768 = 8
The cube of number ending in digit 8 ends in digit 2. … (vii)
Since, 32768 is a 5-digit number.
So, its cube root will be a 2-digit number.
We know that 30³ = 27000 and 40³ = 64000
Since, 32768 is between 27000 and 64000.
So, its cube root is between 30 and 40. … (viii)
∴ Cube root of32768 = 32 [from Eqs. (vii) and (viii)]

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:
(i) We have, 67³ – 66³
We know that b³ – a³ = 1 + 3ab, b > a …(i)
where, a and b are consecutive numbers.
On putting a = 66 and b = 67, in Eq. (i),
we get 67³ – 66³ = 1 + 3 × 67 × 66
= 1 + 13266
= 13267

(ii) We have, 43³ – 42³
On putting a = 42 and b = 43 in Eq. (i),
we get 43³ – 42³ = 1 + 3 × 43 × 42
= 1 + 5418
= 5419

(iii) We have, 67² – 66²
We know that the difference of the squares of two consecutive numbers is equal to their sum.
∴ 67² -66²
= 67 + 66
= 133

(iv) We have, 43² – 42² = 43 + 42 = 85
Since, 13267 > 5419 > 133 > 85
i. e. 67³ – 66³ > 43³ – 42³ > 67² – 66² > 43² – 42²
Hence, 67³ – 66³ is the greatest.