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.
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
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
2 × 2 × 2 × 2 × 5 × 5
So, 400 is not a perfect cube.
2. 3375 = \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)
= 33 × 53
= (3 × 5)3
= (15)3
which is a perfect cube.
3. 8000 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
= 23 × 23 × 53
= (2 × 2 × 5)3
= (20)3
which is a perfect cube.
4. 15625 = \(\underline{5 \times 5 \times 5}\) × \(\underline{5 \times 5 \times 5}\)
= 53 × 33
= (5 × 5)3
= (25)3
which is a perfect cube.
5. 9000 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 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
Prime factorisation of 6859
19 × 19 × 19
So, 6859 is a perfect cube.
7. 2025 = 3 × 3 × 3 × 3 × 5 × 5
Prime factorisation of 2025 is
3 × 3 × 3 × 3 × 5 × 5
So, 2025 is not a perfect cube.
8. 10648 = \(\underline{2 \times 2 \times 2}\) × \(\underline{11 \times 11 \times 11}\)
= 23 × 113
= (2 × 11)3
= (22)3
which is a perfect Cube.