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Exercise 7.1 - Chapter 7 Cube & Cube Roots class 8 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Which of the following numbers are notperfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Answer:

(i) The prime factorisation of 216 is as follows.

2

216

2

108

2

54

3

27

3

9

3

3

 

1

216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 216 is a perfect cube.

(ii)The prime factorisation of 128 is as follows.

2

128

2

64

2

32

2

16

2

8

2

4

2

2

 

1

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, each prime factor is not appearing as many times as a perfect multiple of 3. One 2 is remaining after grouping the triplets of 2. Therefore, 128 is not a perfect cube.

(iii) The prime factorisation of 1000 is as follows.

2

1000

2

500

2

250

5

125

5

25

5

5

 

1

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.

(iv)The prime factorisation of 100 is as follows.

2

100

2

50

5

25

5

5

 

1

100 = 2 × 2 × 5 × 5

Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.

(v)The prime factorisation of 46656 is as follows.

2

46656

2

23328

2

11664

2

5832

2

2916

2

1458

3

729

3

243

3

81

3

27

3

9

3

3

 

1

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 46656 is a perfect cube.

Question 2:

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Answer:

(i) 243 = 3 × 3 × 3 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.

In that case, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube.

Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.

Then, we obtain

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) 72 = 2 × 2 × 2 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.

Then, we obtain

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) 675 = 3 × 3 × 3 × 5 × 5

Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required.

Then, we obtain

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) 100 = 2 × 2 × 5 × 5

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.

Then, we obtain

100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

Question 3:

Find the smallest 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

Answer:

(i) 81 = 3 × 3 × 3 × 3

Here, one 3 is left which is not in a triplet.

If we divide 81 by 3, then it will become a perfect cube.

Thus, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, one 2 is left which is not in a triplet.

If we divide 128 by 2, then it will become a perfect cube.

Thus, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

(iii) 135 = 3 × 3 × 3 × 5

Here, one 5 is left which is not in a triplet.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

(iv) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, one 3 is left which is not in a triplet.

If we divide 192 by 3, then it will become a perfect cube.

Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

(v) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Here, one 11 is left which is not in a triplet.

If we divide 704 by 11, then it will become a perfect cube.

Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

Question 4:

Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Answer:

Here, some cuboids of size 5 × 2 × 5 are given.

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/70/3137/Chapter%207_html_7cece766.jpg

When these cuboids are arranged to form a cube, the side of this cube so formed will be a common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.

LCM of 5, 2, and 5 = 10

Let us try to make a cube of 10 cm side.

For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width, and 2 along with its height.

Total cuboids required according to this arrangement = 2 × 5 × 2 = 20

With the help of 20 cuboids of such measures, a cube is formed as follows.

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/70/3137/Chapter%207_html_m224562d1.jpg

Alternatively

Volume of the cube of sides 5 cm, 2 cm, 5 cm

= 5 cm × 2 cm × 5 cm = (5 × 5 × 2) cm3

Here, two 5s and one 2 are left which are not in a triplet.

If we multiply this expression by 2 × 2 × 5 = 20, then it will become a perfect cube.

Thus, (5 × 5 × 2 × 2 × 2 × 5) = (5 × 5 × 5 × 2 × 2 × 2) = 1000 is a perfect cube. Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.

Also Read : Exercise-7.2-Chapter-7-Cube-&-Cube-Roots-class-8-ncert-solutions-Maths

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