Important Questions: A Square and A Cube | Mathematics Class 8- New NCERT (Ganita Prakash) PDF Download

Q1: Give a reason to show that the number given below is a perfect square: 5963 
Sol: The unit digit of the square numbers will be 0, 1, 4, 5, 6, or 9 if we examine the squares of numbers from 1 to 10. Thus, the unit digit for all perfect squares will be 0, 1, 4, 5, 6, or 9, and none of the square numbers will end in 2, 3, 7, or 8.
Given 5963
We have the property of a perfect square, i.e. a number ending in 3 is never a perfect square.
Therefore the given number 5963 is not a perfect square.

Q2: 2025 plants are to be planted in a garden in a way that each of the rows contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Sol:
Let the number of rows be x.
Thus, the number of plants in each row = x.
Total many contributed by all the students = x  ×  x = x²
Given, x² = Rs.2025
x² = 3 × 3 × 3 × 3 × 5 × 5
⇒ x² = (3 × 3) × (3 × 3) × (5 × 5)
⇒ x² = (3 × 3 × 5) × (3 × 3 × 5)
⇒ x² = 45 × 45
⇒ x = √(45 × 45)
⇒ x = 45
Therefore,
Number of rows = 45
Number of plants in each row = 45

Q3: Find out the cube root of 13824 by the prime factorisation method.
Sol: First, let us prime factorise 13824:
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 2 ³ × 2 ³ × 2 ³ × 3 ³
3√13824 = 2 × 2 × 2 × 3 = 24

Q4: (13/10) ³ 
Sol: The cube of a rational number is the result of multiplying a number by itself three times.
To evaluate the cube of (13/10) ³
Firstly we need to convert into proper fractions, i.e.(13/10) ³
We need to multiply the given number three times, i.e. (13/10) × (13/10) × (13/10) = (2197/1000)
∴ the cube of (1 3/10) is (2197/1000)

​Q5: By what least number should the number be divided to obtain a number with a perfect square? In this, in each case, find the number whose square is the new number 4851.
Sol: The number is a perfect square if and only if the prime factorization creates pairs; it is not exactly a perfect square if it is not paired up.
Given 4851,
Resolving 4851 into prime factors, we obtain
4851 = 3 X 3 X 7 X 7 X 11
= (32 X 72 X 11)
To obtain a perfect square, we need to divide the above equation by 11
we obtain, 9075 = 3 X 3 X 7 X 7
The new number = (9 X 49)
= (3² X 7² )
Taking squares on both sides of the above equation, we obtain
∴ The new number = (3 X 7)²
= (21)²
Therefore, the new number is a square of 21

Q6: Find the cube root of 10648 by the prime factorisation method.
Sol:10648 = 2 × 2 × 2 × 11 × 11 × 11
Grouping the factors in triplets of number equal factors,
10648 = (2 × 2 × 2) × (11 × 11 × 11)
Here, 10648 can be grouped into triplets of number equal factors,
∴ 10648 = 2 × 11 = 22
Therefore, the cube root of 10648 is 22.

Q7: Without adding, find the sum of the following:
(1+3+5+7+9+11+13+15+17+19+21+23)
Sol:   (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)
As per the given property of perfect square, for any natural number n, we have some of the first n odd natural numbers = n²
But here n = 12
By applying the above the law, we get
thus, (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)
= 12² 
= 144

Q8: By what least number should the given number be divided to get a perfect square number? In each of the following cases, find the number whose square is the new number 1575.
Sol: A method for determining the prime factors of a given number, such as a composite number, is known as prime factorisation.
Given 1575,
Resolve 1575 into prime factors, we get
1575 = 3 X 3 X 5 X 5 X 7 = (3² X 5² X 7)
To obtain a perfect square, we have to divide the above equation by 7
Then we get, 3380 = 3 X 3 X 5 X 5
New number = (9 X 25) = (3² X 5² )
Taking squares on both sides of the above equation, we get
∴ New number = (3 X 5)² = (15)²

​Q9:If m is the required square of a natural number given by n, then n is
(a) the square of m
(b) greater than m
(c) equal to m
(d) √m
Ans: (d)
Sol: n² = m
Then,
= n = √m
Q10: The cube of 100 will have _________ zeroes.
Sol: The cube of 100 will have  six zeroes.
= 1003
= 100 × 100 × 100
= 1000000

Q11: Use the following identity and find the square of 189.
(a – b)² = a² – 2ab + b²
Sol: 189 = (200 – 11) ²
= 40000 – 2 x 200 x 11 + 112
= 40000 – 4400 + 121
= 35721

Q12: What would be the square root of the number 625 using the identity
(a +b)² = a² + b² + 2ab?
Sol: (625)²
= (600 + 25)²
= 600² + 2 x 600 x 25 +25²
= 360000 + 30000 + 625
= 390625

Q13:Show that the sum of two consecutive natural numbers is 13².
Sol: Let 2n + 1 = 13
So, n = 6
So, ( 2n + 1)² = 4n² + 4n + 1
= (2n² + 2n) + (2n² + 2n + 1)
Substitute n = 6,
(13)² = ( 2 x 6² + 2 x 6) + (2 x 6² + 2 x 6 + 1)
= (72 + 12) + (72 + 12 + 1)
= 84 + 85

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Q14: (1.2) ³ = _________.
Sol: (1.2) ³  = 12/10
= (12/10) × (12/10) × (12/10)
= 1728/1000
= 1.728

Q15: Find the cube root of 91125 by the prime factorisation method.
Sol: 91125 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
By grouping the factors in triplets of equal factors, 91125 = (3 × 3 × 3) × (3 × 3 × 3) × (5 × 5 × 5)
Here, 91125 can be grouped into triplets of equal factors,
∴ 91125 = (3 × 3 × 5) = 45
Thus , 45 is the cube root of 91125.

Q16: A cuboid of plasticine made by Parikshit with sides 5 cm, 2 cm, and 5 cm. How many such cuboids will be needed to form a cube?
Sol: The given side of the cube is 5 cm, 2 cm and 5 cm.
Therefore, volume of cube = 5 × 2 × 5 = 50
The prime factorisation of 50 = 2 × 5 × 5
Here, 2, 5 and 5 cannot be grouped into triples of equal factors.
Therefore, we will multiply 50 by 2 × 2 × 5 = 20 to get the perfect square.
Hence, 20 cuboids are needed to form a cube.

Q17: State true or false.
(i) The cube of any odd number is even
(ii) A perfect cube never ends with two zeros.
(iii) If the square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two-digit number may be a three-digit number.
(vi) The cube of a two-digit number may have seven or more digits.
(vii) The cube of a single-digit number may be a single-digit number.
Sol:
(i) This statement is false.
Taking a cube of any required odd numbers
3³= 3 x 3 x 3 = 27
7³=7 x 7 x 7= 343
5³=5 x 5 x 5=125
All the required cubes of any given odd number will always be odd.
(ii) This statement is true.
10³= 10 x 10 x 10= 1000
20³ = 20 x 20 x 20 = 2000
150³ =150 x150 x150 = 3375000
Hence a perfect cube will never end with two zeros.
(iii) This statement is false.
15²= 15 x15= 225
15³= 15 x 15 x 15= 3375
Thus, the square of any given number ends with 5; then the cube ends with the number 25 is an incorrect statement.
(iv) This statement is false.
2³= 2x2x2= 8
12³ = 12 x 12 x 12= 1728
Accordingly, There are perfect cubes ending with the number 8
(v) This statement is false.
The minimum two digits number is 10
And 
10³=1000→4 Digit number.
The maximum two digits number is 99
And 
99³=970299→6 Digit number
Accordingly, the cube of two-digit numbers can never be a three-digit number.
(vi) This statement is false 
10³=1000→4 Digit number.
The maximum two digits number is 99
And 
99³=970299→6 Digit number
Accordingly, the cube of two-digit numbers can never have seven or more digits.
(vii) This statement is true
1³ = 1 x 1 x 1= 1
2³ = 2 x 2 x 2= 8
According to the cube, a single-digit can be a single-digit number.

Q18: Find the cube of 3.5.
Sol: 3.53 = 3.5 x 3.5 x 3.5
= 12.25 x 3.5
= 42.875

Q19: Find the smallest whole number from which 1008 should be multiplied to obtain a perfect square number. Also, find out the square root of the square number so obtained.
Sol:
Let us factorise the number 1008.1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 
= ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 ) × 7
Here, 7 cannot be paired.
Therefore, we will multiply 1008 by 7 to get a perfect square.
New number so obtained = 1008 ×7 = 7056
Now, let us find the square root of 70567056 = 2 × 2 × 2 × 2 × 3 × 3× 7 × 7
7056 = (2 ×  2 ) × ( 2 × 2 ) × ( 3 × 3 ) ×( 7 × 7 )
7056 = 2² × 2² × 3² × 7²
7056 = (2 × 2 × 3 × 7)²
Therefore;
√7056 = 2 × 2 × 3 × 7
= 84

Q20: √(1.96) = _________.
Sol: We have,
= √(1.96)
= √(196/100)
= √((14 × 14)/(10 × 10))
= √(14² / 10²)
= 14/10
= 1.4


Q21: There are _________ perfect cubes between 1 and 1000.
Sol: 
There are 8 perfect cubes between 1 and 1000.
2 × 2 × 2 = 8
3 × 3 × 3 = 27
4 × 4 × 4 = 64
5 × 5 × 5 = 125
6 × 6 × 6 = 216
7 × 7 × 7 = 343
8 × 8 × 8 = 512
9 × 9 × 9 = 729

Q22: Is 392 a perfect cube? If not, find the smallest natural number by which 392 should be multiplied so that the product is a perfect cube.
Sol: The prime factorisation of 392 gives:
392 = 2 x 2 x 2 x 7 x 7
As we can see, number 7 cannot be paired in a group of three. Therefore, 392 is not a perfect cube.
We must multiply the 7 by the original number to make it a perfect cube.
Thus,
2 x 2 x 2 x 7 x 7 x 7 = 2744, which is a perfect cube, such as 23 x 73 or 143.
Hence, the smallest natural number, which should be multiplied by 392 to make a perfect cube, is 7.

Q23: Which of the following numbers are in perfect cubes? In the case of a perfect cube, find the number whose cube is the given number 256 
Sol: A perfect cube can be expressed as a product of three numbers of equal factors
Resolving the given number into prime factors, we obtain
256 = 2 × 2 × 2 × 2 × 2× 2 × 2 × 2
Since the number 256 has more than three factors
∴ 256 is not a perfect cube.

Q24: Find the smallest number by which 128 must be divided to get a perfect cube.
Sol:  The prime factorisation of 128 is given by:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
By grouping the factors in triplets of equal factors,
128 = (2 × 2 × 2) × (2 × 2 × 2) × 2
Here, 2 cannot be grouped into triples of equal factors.
Therefore, to obtain a perfect cube, we will divide 128 by 2.

​Q25: There are _________ perfect squares between 1 and 100.
Sol: There are 8 perfect squares between 1 and 100.
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
6 × 6 = 36
7 × 7 = 49
8 × 8 = 64
9 × 9 = 81

Q26: Show that each of the numbers is a perfect square. In each case, find the number whose square is the given number:
7056
Sol: 7056,
A perfect square is always expressed as a product of pairs of prime factors.
Resolving 7056 into prime factors, we obtain
7056 = 11 X 539 
= 12 X 588 
= 12 X 7 X 84 
= 84 X 84 
= (84)²
Thus, 84 is the number whose square is 5929
Therefore,7056 is a perfect square.