HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

Haryana State Board HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

Try These (Page 111)

Question 1.
Find the one’s 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:
We know, that the cube of a number ending in,
(a) 0, 1, 4, 5, 6 and 9 ends in 0, 1, 4, 5, 6 and 9 respectively.
(b) 2 ends in 8 and vice-versa.
(c) 3 or 7 ends in 7 or 3 respectively.
(i) ∵ 3331 ends in 1, therefore, its cube ends in 1.
Hence, the required unit’s digit of the cube is 1.

(ii) ∵ 8888 ends in 8, therefore, its cube ends in 2.
Hence, the required unit’s digit of the cube is 2.

(iii) ∵ 149 ends in 9, therefore, its cube ends is 9.
Hence, the required unit’s digit of the cube is 9.

(iv) ∵ 1005 ends in 5, therefore, its cube ends is 5.
Hence, the required unit’s digits of the cube is 5.

(v) ∵ 1024 ends is 4, therefore, its cube ends is 4.
Hence, the required unit’s digit of the cube is 4

(vi) ∵ 77 ends is 7, therefore, its cube ends is 3.
Hence, the required unit’s digits of the cube is 3.

(vii) ∵ 5022 ends is 2, therefore, its cube . ends is 8.
Hence, the required unit’s digits of the cube is 8.

(viii) ∵ 53 ends is 3, therefore, its cube ends is 7.
Hence, the required unit’s digits of the cube is 7.

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

Question 2.
Express the following numbers as the sum of odd numbers using the above pattern ?
(a) 63
(b) 83
(c) 73
Solution:
Observe the following pattern of sums of odd numbers :
1 = 1 = 13
3 + 5 = 8 = 23
7 + 9 + 11 = 27 = 33
13 + 15 + 17 + 19 = 64 = 43
21 + 23 + 25 + 27 + 29 = 125 = 53
31 + 33 + 35 + 37 + 39 + 41 = 216 = 63
— — — — — — = 343 = 73
— — — — — — = 512 = 83
— — — — — — = 729 = 93
— — — — — — = 1000 = 103
According to above pattern, we have
(a) 63 = 31 + 33 +35 + 37 + 39 + 41
= 216
(b) 83 = 57 + 59 + 61 + 63 + 65 + 67 + 69+71
= 512
(c) 73 = 43 + 45 + 47 + 49 + 51 + 53 + 55
= 343

Question 3.
Consider the following pattern.
23 – 13 = 1 + 2 × 1 × 3
33 – 23 = 1 + 3 × 2 × 3
43 – 33 = 1 + 4 × 3 × 3
Using the above pattern, find the value of the following:
(i) 73 – 62
(ii) 123 – 113
(iii) 203 – 193
(iv) 513 – 503
Solution:
(i) 73 – 62 = 1 + 7 × 6 × 3
= 1 + 126 = 127

(ii) 123 – 113 = 1 + 12 × 11 × 3
= 1 + 396 = 397

(iii) 203 – 193 = 1 + 20 × 19 × 3
= 1 + 1140= 1141

(iv) 513 – 503 = 1 + 51 × 50 × 3
= 1 + 7650 = 7651

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

Try These (Page 112)

Question 1.
Which of the following are perfect cubes ?
1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648
Solution:
1. 400 = 2 × 2 × 2 × 2 × 5 × 5
Prime factorisation of 400 is
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 1
2 × 2 × 2 × 2 × 5 × 5
So, 400 is not a perfect cube.

2. 3375 = \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 2
= 33 × 53
= (3 × 5)3
= (15)3
which is a perfect cube.

3. 8000 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 3
= 23 × 23 × 53
= (2 × 2 × 5)3
= (20)3
which is a perfect cube.

4. 15625 = \(\underline{5 \times 5 \times 5}\) × \(\underline{5 \times 5 \times 5}\)
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 4
= 53 × 33
= (5 × 5)3
= (25)3
which is a perfect cube.

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

5. 9000 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 5
Prime factorisation of 9000 is
2 × 2 × 2 × 3 × 3 × 5 × 5 × 5
So, 9000 is not a perfect cube.

6. 6859 = 19 × 19 × 19
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 6
Prime factorisation of 6859
19 × 19 × 19
So, 6859 is a perfect cube.

7. 2025 = 3 × 3 × 3 × 3 × 5 × 5
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 7
Prime factorisation of 2025 is
3 × 3 × 3 × 3 × 5 × 5
So, 2025 is not a perfect cube.

HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

8. 10648 = \(\underline{2 \times 2 \times 2}\) × \(\underline{11 \times 11 \times 11}\)
HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions 8
= 23 × 113
= (2 × 11)3
= (22)3
which is a perfect Cube.

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