Gujarat BoardĀ 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 Ć 2 Ć 2 Ć 3 Ć 3 Ć 3

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

ā“ 216 is a perfect cube.

2. We have

128 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2

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

ā“ 128 is not a perfect cube.

3. We have 211000

1000 = 2 Ć 2 Ć 2 Ć 5 Ć 5 Ć 5

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

ā“ 1000 is a perfect cube.

4. We have

100 = 2 Ć 2 Ć 5 Ć 5

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

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

ā“ 100 is not a perfect cube.

5. We have

46656 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 3 Ć 3 Ć 3 Ć 3 Ć 3 Ć 3

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

ā“ 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 Ć 3 Ć 3 Ć 3 Ć 3

The prime factor 3 is not a group of three.

ā“ 243 is not a perfect cube.

Now, [243] Ć 3 = [3 Ć 3 Ć 3 Ć 3 Ć 3] Ć 3

or 729 = 3 Ć 3 Ć 3 Ć 3 Ć 3 Ć 3 Ć 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 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2

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

ā“ 256 is not a perfect cube.

Now, [256] Ć 2

= [2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2] Ć 2

or 512 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2

i.e; 512 is a perfect cube.

Thus, the required smallest number is 2.

3. We have 72 = 2 Ć 2 Ć 2 Ć 3 Ć 3

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

ā“ 72 is not a perfect cube.

Now, [72] Ć 3

= [2 Ć 2 Ć 2 Ć 3 Ć 3] Ć 3

or 216 = 2 Ć 2 Ć 2 Ć 3 Ć 3 Ć 3

i.e; 216 is a perfect cube.

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

4. We have 675 = 3 Ć 3 Ć 3 Ć 5 Ć 5

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

ā“ 675 is not a perfect cube.

Now, [675] Ć 5

= [3 Ć 3 Ć 3 Ć 5 Ć 5] Ć 5

or 3375 = 3 Ć 3 Ć 3 Ć 5 Ć 5 Ć 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 Ć 2 Ć 5 Ć 5

The prime factor are not in the groups of triples.

ā“ 100 is not a perfect cube.

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

or [100] Ć 10 = 2 Ć 2 Ć 2 Ć 5 Ć 5 Ć 5

1000 = 2 Ć 2 Ć 2 Ć 5 Ć 5 Ć 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 Ć 3 Ć 3 Ć 3

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

ā“ 81 is not a perfect cube.

Now, [81] + 3 [3 Ć 3 Ć 3 Ć 3] Ć· 3

or 27 – 3 Ć 3 Ć 3

i.e., 27 is a prefect cube.

Thus, the required smallest number is 3.

2. We have 2 128

128 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2

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

ā“ 128 is not a perfect cube

Now, [128] Ć· 2

= [2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2] + 2

or 64 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2

i.e.. 64 is a perfect cube.

Thus, the required smallest number is 2.

3. We have 135 = 3 Ć 3 Ć 3 Ć 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 Ć 3 Ć 3 Ć 5] + 5

or 27 = 3 Ć 3 Ć 3

i.e., 27 is a perfect cube.

Thus, the required smallest number is 5.

4. We have

192 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 3

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

ā“ 192 is not a perfect cube.

Now, [192] + 3

= [2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 3] + 3

or 64 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2

i.e., 64 is a perfect cube.

Thus, the required smallest number is 3.

5. We have

704 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 11

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

ā“ [704] + 11

= [2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 11] Ć· 11

or 64 = 2 Ć 2 Ć 2 Ć 2 Ć 2 Ć 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

ā“ Volume of the cuboid = 5 cm Ć 2cm Ć 5 cm

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

ā“ Volume of the required cube

= [5 cm Ć 5 cm Ć 2 cm] Ć 5 cm Ć 2 cm Ć 2 cm

= [5 Ć 5 Ć 2cm^{3}] = 20cm^{3}

Thus, the required number of cuboids = 20