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## KSEEB Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots InText Questions

Try These (Page 111)

Question 1.

Find the one digit of the cube of each of the following numbers.

(i) 3331

(ii) 8888

(iii) 149

(iv) 1005

(v) 1024

(vi) 77

(vii) 5022

(viii) 53

Solution:

(a) As we know that cubes of numbers ending with digit 0, 1, 4, 5, 6, and 9 ends in digit 0, 1, 4, 5, 6, and 9, respectively.

(b) Numbers ending with digit 2 end in 8 and vice-versa.

(c) Number is ending digit 3 or 7 ends in 7 or 3, respectively.

(i) 3331, end digits 1

Hence, a cube of this number ends with 1

∴ One’s digit of the cube of 3331 is 1

(ii) 8888, end digit is 8

Cube of this number ends with 2

∴ One’s digit of the cube of 8888 is 2

(iii) 149, end digit is 9

Cube of this number ends with 9

∴ One’s digit of the cube of 149 is 9.

(iv) 1005, end digit is 5

Cube of this number ends with 5

∴ One’s digit of the cube of 1005 is 5

(v) 1024, end digit is 4

Cube of this number ends with 4

∴ One’s digit of the cube of 1024 is 4

(vi) 77, end digit is 7

Cube of this number ends with 3

∴ One’s digit of the cube of 77 is 3

(vii) 5022, end digit is 2

Cube of this number ends with 2

∴ One’s digit of the cube of 5022 is 8

(viii) 53, end digit is 3

Cube of this number ends with 7

∴ One’s digit of the cube of 53 is 7

Try These (Page 111)

Question 1.

Express the following numbers as the sum of odd numbers using the pattern given below?

(a) 63

(b) 83

(c) 73

Adding consecutive odd numbers

1 = 1 = 1^{3}

3 + 5 = 8 = 2^{3}

7 + 9 + 11 = 27 = 3^{3}

13 + 15 + 17 + 19 = 64 = 4^{3}

Solution:

(a) 6^{3} = 216 = 31 + 33 + 35 + 37 + 39 + 41

(b) 8^{3} = 512 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71

(c) 7^{3} = 343 = 43 + 45 + 47 + 49 + 51 + 53 + 55

Question 2.

Consider the following pattern.

2^{3} – 1^{3} = 1 + 2 × 1 × 3

3^{3} – 2^{3} = 1 + 3 × 2 × 3

4^{3} – 3^{3} = 1 + 4 × 3 × 3

Using the above pattern, find the. value of the following:

(i) 7^{3} – 6^{3}

(ii) 12^{3} – 11^{3}

(iii) 20^{3} – 19^{3}

(iv) 51^{3} – 50^{3}

Solution:

(i) 7^{3} – 6^{3} = 1 + 7 × 6 × 3 = 1 + 126 = 127

(ii) 12^{3} – 11^{3} = 1 + 12 × 11 × 3 = 1 + 396 = 397

(iii) 20^{3} – 19^{3} = 1 + 20 × 19 × 3 = 1 + 1140 = 1141

(iv) 51^{3} – 50^{3} = 1 + 51 × 50 × 3 = 1 + 7650 = 7651

Try These (Page 112)

Question 1.

Which of the following are perfect cubes?

(i) 400

(ii) 3375

(iii) 8000

(iv) 15625

(v) 9000

(vi) 6859

(vii) 2025

(viii) 10648

Solution:

(i) 400 = 2 × 2 × 2 × 2 × 5 × 5

It is not a perfect cube.

(ii) 3375

= 3 × 3 × 3 × 5 × 5 × 5

= 3^{3} × 5^{3}

= (3 × 5)^{3}

= (15)^{3}

It is a perfect cube.

(iii) 8000

= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5

= 2^{3} × 2^{3} × 5^{3}

= (2 × 2 × 5)^{3}

= (20)^{3}

It is a perfect cube.

(iv) 15625

= 5 × 5 × 5 × 5 × 5 × 5

= 5^{3} × 5^{3}

= (25)^{3}

It is a perfect cube.

(v) 9000

= 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5

It is not a perfect cube.

(vi) 6859 = 19 × 19 × 19 = (19)^{3}

So, it is a perfect cube.

(vii) 2025 = 3 × 3 × 3 × 3 × 5 × 5

It is not a perfect cube.

(viii) 10648 = 2 × 2 × 2 × 11 × 11 × 11

= 2^{3} × 11^{3}

= (2 × 11)^{3}

= (22)^{3}

It is a perfect cube.