HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1

Haryana State Board HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Exercise 7.1

Question 1.
Which of the following numbers are not perfect cubes:
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656.
Solution:
(i) 216 = \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 1
= 23 × 32
= (2 × 3)3
= (6)3
Which is a perfect cube.

(ii) 128 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 2
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 2
= 23 × 23 × 2
∴ 2 does not appear in a group of three.
Hence, 128 is not a percent cube.

(iii) 1000 = 2 × 2 × 2 × 5 × 5 × 5
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 3
= 2 × 5
= 10
Which is a perfect cube.

(iv) 100 = 2 × 2 × 5 × 5
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 4
Prime factor of 100 is
2 × 2 × 5 × 5
So, 100 is not a perfect cube.

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1

(v) 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 9 × 9 × 9
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 5
= 2 × 2 × 9
= 36
Which is a perfect cube.

Question 2.
Find the smallest number by which each of the following number must be multified to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100.
Solution:
(i) 243 = 3 × 3 × 3 × 3 × 3
The prime factor 3 × 3 = 9 does not appear in a group of three. Therefore 243 is not a perfect cube. To make it a cube, we need one more 3.
In that case,
243 × 3 = \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\)
= 729,
which is a perfect cube.
Hence the required smallest number by which 243 should be multiplied to make a perfect is 3.

(ii) 256 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 2 × 2
The prime factor 2 × 2 = 4 does not appear in a group of three. Therefore 256 is not a perfect cube. To make it cube, we need one more 2., In that case,
256 × 2 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
= 512
Hence the required smallest number by which 256 should be multiplied to make a perfect cube is 2.

(iii) 72 = \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3}\)
The prime factor 3 does not appear in a group of three. Therefore 72 is not a perfect cube. To make it a cube, we need one more 3. In that case
72 = \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)
or, 72 × 3
= 216,
which is a perfect cube.
Hence, the required smallest number by which 72 should be multiplied to make a perfect cube is 3.

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1

(iv) 675 = \(\underline{3 \times 3 \times 3}\) × 5 × 5
The prime factor 5 does not appear in a group of three. Therefore 675 is not a perfect cube. To make it a cube, we need one more 5. In that case
675 × 5 = \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)
= 3375.
which is a perfect cube.
Hence, the required smallest number by which 675 should be multiplied to make a perfect cube is 5.

(v) 100 = \(\underline{2 \times 2}\) × \(\underline{5 \times 5}\)
The prime factor 2 and 5 do not appear in a group of three. Therefore 100 is not a perfect cube. To make it a cube, we need one more 2 and 5 respectively. In that case
100 × 10 = \(\underline{2 \times 2 \times 2}\) × \(\underline{5 \times 5 \times 5}\)
= 1000,
which is a perfect cube. Hence the required smallest number by which 100 should be multiplied to make a perfect cube is (2 × 5) = 10.

Question 3.
Find the smallest whole number by which each of the following numbers must be divided to obtain a perfect cube :
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704
Solution:
(i) 81 = \(\underline{3 \times 3 \times 3}\) × 3
The prime factor 3 does not appear in a group of three. So 81 is not a perfect cube. In the factorisation 3 appears only one time. So if we divide 81 by 3, then the prime factorisation of the quotient will not contain 3.
81 ÷ 3 = 3 × 3 × 3
Further the perfect cube in that case is 81 , 3 = 27. Hence the smallest whole number by which 81 should be divided to make it perfect cube is 3.

(ii) 128 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 2
The prime factor 2 does not appear in a group of three. So 128 is not a perfect cube. In the factorisation 2 appear only one time. So if we divide 128 by 2, then the prime factorisation of the quotient will not contain 2
128 ÷ 2 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
Further the perfect cube in that case is
128 ÷ 2 = 64.
Hence the smallest whole number by which 128 should be divided to make it perfect cube is 2.

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1

(iii) 135 = \(\underline{3 \times 3 \times 3}\) × 5
The prime Factor 5 × 5 = 25 does not appear in a group of three. So, 135 is not a Perfect Cube. In the factorisation 5 appears only one times. So if we divide 135 by 5, then the prime factorisation of the quotient will not contain 5.
135 ÷ 5 = 3 × 3 × 3
Further then perfect cube in that case is
135 ÷ 5 = 27
Hence the smallest whole number by which 135 should be divided to make it perfect cube is 5.

(iv) 192 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 3
The prime factor 3 does not appear in a group of three, so, 192 is not a perfect cube. In the factorisation 3 appears only one times. So, if we divide 192 by 3, then the prime factorisation of the quotient will not contain 3.
192 ÷ 3 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
Further the perfect cube in that case is
(192 ÷ 3 = 64)
Hence, the smallest whole numbers by which 192 should be divided to make it perfect cube is 3.

(v) 704 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 11
The prime factor 11 does not appear in a group of three. So 704 is not a perfect cube. In the factorisation 11 appears only one time. So if we divide 704 by 11, then the prime factorisation of the quotient will not contain 11.
704 ÷ 11 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
Further the perfect cube in that case is
704 ÷ 11 = 64.
Hence the smallest whole number by which 704 should be divided to make it perfect cube is 11.

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1

Question 4.
Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboides will he need to form a cube?
Solution:
Parikshit’s cuboid = l × b × h
= 5 cm × 2 cm × 5cm
Now Parikshit’s volume of cube
= 5 cm × 5 cm × 25 cm
= 5 cm × 5 cm × 5 cm
The prime of factor of cuboid, 5 does not appear in a group of three. 50 cm3 is not a perfect cube. To make it cube, we need one more 5 cm. In that case,
Volume of cube = 5 cm × 5 cm × 5 cm
= 125 cm2
Hence the required smallest number by which 25 cm2 should be multiplied to make a perfect cube is 5 cm.

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