Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 7 Cube and Cube Roots Ex 7.1

Question 1.

Which of the following numbers are not perfect cubes?

- 216
- 128
- 1000
- 100
- 46656

Solution:

1. We have

216 = 2 × 2 × 2 × 3 × 3 × 3

Grouping the prime factors of 216 into triples, no factor is left over.

∴ 216 is a perfect cube.

2. We have

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor.

∴ 128 is not a perfect cube.

3. We have 211000

1000 = 2 × 2 × 2 × 5 × 5 × 5

Grouping the prime factors of 1000 into triples, we are not left over with any factor.

∴ 1000 is a perfect cube.

4. We have

100 = 2 × 2 × 5 × 5

Grouping the prime factors into triples, we do not get any triples.

Factors 2 × 2 and 5 × 5 are not in triples.

∴ 100 is not a perfect cube.

5. We have

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

Grouping the prime factors of 46656 in triples we are not left over with any prime factor.

∴ 46656 is a perfect cube.

Question 2.

Find the smallest number by which each of the following numbers must obtain a perfect cube?

- 243
- 256
- 72
- 675
- 100

Solution:

1. We have

243 = 3 × 3 × 3 × 3 × 3

The prime factor 3 is not a group of three.

∴ 243 is not a perfect cube.

Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3

or 729 = 3 × 3 × 3 × 3 × 3 × 3 × 3

Now, 729 becomes a perfect cube.

Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.

2. We have

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 256 in triples, we are left over with 2 × 2.

∴ 256 is not a perfect cube.

Now, [256] × 2

= [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2

or 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

i.e; 512 is a perfect cube.

Thus, the required smallest number is 2.

3. We have 72 = 2 × 2 × 2 × 3 × 3

Grouping the prime factors of 72 in triples, we are left over with 3 × 3.

∴ 72 is not a perfect cube.

Now, [72] × 3

= [2 × 2 × 2 × 3 × 3] × 3

or 216 = 2 × 2 × 2 × 3 × 3 × 3

i.e; 216 is a perfect cube.

∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

4. We have 675 = 3 × 3 × 3 × 5 × 5

Grouping the prime factors of 675 to triples, we are left over with 5 × 5.

∴ 675 is not a perfect cube.

Now, [675] × 5

= [3 × 3 × 3 × 5 × 5] × 5

or 3375 = 3 × 3 × 3 × 5 × 5 × 5

Now, 3375 is a perfect cube.

Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

5. We have

100 = 2 × 2 × 5 × 5

The prime factor are not in the groups of triples.

∴ 100 is not a perfect cube.

Now, [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5

or [100] × 10 = 2 × 2 × 2 × 5 × 5 × 5

1000 = 2 × 2 × 2 × 5 × 5 × 5

Now, 1000 is a perfect cube.

Thus, the required smallest number is 10.

Question 3.

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

- 81
- 128
- 135
- 192
- 704

Solution:

1. We have 81 = 3 × 3 × 3 × 3

Grouping the prime factors of 81 into triples, we are left with 3

∴ 81 is not a perfect cube.

Now, [81] + 3 [3 × 3 × 3 × 3] ÷ 3

or 27 – 3 × 3 × 3

i.e., 27 is a prefect cube.

Thus, the required smallest number is 3.

2. We have 2 128

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 128 into triples, we are left with 2.

∴ 128 is not a perfect cube

Now, [128] ÷ 2

= [2 × 2 × 2 × 2 × 2 × 2 × 2] + 2

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e.. 64 is a perfect cube.

Thus, the required smallest number is 2.

3. We have 135 = 3 × 3 × 3 × 5

Grouping the prime factors of 135 into triples, we are left over with 5.

135 is not a perfect cube.

Now, [135] + 5 = [3 × 3 × 3 × 5] + 5

or 27 = 3 × 3 × 3

i.e., 27 is a perfect cube.

Thus, the required smallest number is 5.

4. We have

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Grouping the prime factors of 192 into triples, 3 is left over.

∴ 192 is not a perfect cube.

Now, [192] + 3

= [2 × 2 × 2 × 2 × 2 × 2 × 2 × 3] + 3

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e., 64 is a perfect cube.

Thus, the required smallest number is 3.

5. We have

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Grouping the prime factors of 704 into triples, 11 is left over.

∴ [704] + 11

= [2 × 2 × 2 × 2 × 2 × 2 × 11] ÷ 11

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e; 64 is a perfect cube.

Thus, the required smallest number is 11.

Question 4.

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

Solution:

Sides of the cuboid are: 5 cm, 2 cm, 5 cm

∴ Volume of the cuboid = 5 cm × 2cm × 5 cm

To form it as a cube its dimension should be in the group of triples.

∴ Volume of the required cube

= [5 cm × 5 cm × 2 cm] × 5 cm × 2 cm × 2 cm

= [5 × 5 × 2cm^{3}] = 20cm^{3}

Thus, the required number of cuboids = 20