NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Cubes and Cube Roots |

Exercise |
Ex 7.1 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 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)** **Resolving 216 into prime factors, we find that 2**

16 = 2x2x2 x 3x3x3

Clearly, the prime factors of 216 can be grouped into triples of equal factors and no factor is left over.

∴ 216 is a perfect cube.

**(ii) Resolving 128 into prime factors, we find that**

128 = 2x2x2 x 2x2x2 x 2

Now, if we try to group together triples of equal factors, we are left with a single factor, 2.

∴ 128 is not a perfect cube.

**(iii) Resolving 1000 into prime factors, we find that**

1000 = 10 x 10 x 10 = 2x5x2x5x2x5 = 2x2x2x 5x5x5

Clearly, the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over.

∴ 1000 is a perfect cube.

**(iv) Resolving 100 into prime factors, we find that**

100 = 10×10=2x5x2x5=2x2x5x5

Clearly, the prime factors of 100 cannot be grouped into triples of equal

factors.

∴ 100 is not a perfect cube.

**(v) Resolving 46656 into prime factors, we find that**

46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x3 x 3x3x3

Clearly, the prime factors of 46656 can be grouped into prime factors and no factor is left over.

∴ 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

**Solution:**

**(i)** Writing 243 as a product of prime factors, we have

243 = 3 x 3 x 3 x 3 x 3

Clearly, to make it a perfect cube, it must be multiplied by 3.

**(ii)** Writing 256 as a product of prime factors, we have

256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Clearly, to make it a perfect cube, it must be multiplied by 2.

**(iii)** Writing 72 as a product of prime factors, we have

72 = 2x2x2 x 3 x 3

Clearly, to make it a perfect cube, it must be multiplied by 3.

**(iv)** Writing 675 as a product of prime factors, we have

675 = 3 x 3 x 3 x 5 x 5

Clearly, to make it a perfect cube, it must be multiplied by 5.

**(v)** Writing 100 as a product of prime factors, we have

100 = 2 x 2 x 5 x 5

Clearly, to make it a perfect cube, it must be multiplied 2 x 5, i.e., 10.

**Question 3.**

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

**(i)** 81

**(ii)** 128

**(iii)** 135

**(iv)** 192

**(v)** 704

**Solution:**

**(i)** Writing 81 as a product of prime factors, we have

81 = 3 x 3 x 3 x 3

Clearly, to make it a perfect cube, it must be divided by 3.

**(ii)** Writing 128 as a product of prime factors, we have

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Clearly, to make it a perfect cube, it must be divided by 2.

**(iii)** Writing 135 as a product of prime factors, we have

135 = 3 x 3 x 3 x 5

Clearly, to make it a perfect cube, it must be divided by 5.

**(iv)** Writing 192 as a product of prime factors, we have

192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Clearly, to make it a perfect cube, it must be divided by 3.

**(v)** Writing 704 as a product of prime factors, we have

704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Clearly, to make it a perfect cube, it must be divided by 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?

**Solution:**

Volume of the cuboid = 5 x 2 x 5 = 2 x 5 x 5

To make it a cube, we require 2 x 2 x 5, i.e., 20 such cuboids.

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