AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.1 Textbook Questions and Answers.
AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.1
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Question 1.
What will be the units digit of the square of the following numbers?
(i) 39
(ii) 297
(iii) 5125
(iv) 7286
(v) 8742
Solution:
| Number | Square of the units digIt | Units digit of a squared number |
| i) 39 | 92 = 9 x 9 = 81 | 1 |
| ii) 297 | 72 = 7 x 7 = 49 | 9 |
| Iii) 5125 | 52 = 5 x 5 = 25 | 5 |
| iv) 7286 | 62 = 6 x 6 = 36 | 6 |
| v) 8742 | 22 = 2 x 2 = 4 | 4 |
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Question 2.
Which of the following numbers are perfect squares?
(i) 121
(ii) 136
(iii) 256
(iv) 321
(v) 600
Solution:
| Number | Prime factorizatlon | Perfect square numbers Yes/No |
| i) 121 | 121 = 11 x 11 = 112 | yes |
| ii) 136 | 136 = 8 x 17 = 2 x 2 x 2 x 17 | No |
| iii) 256 | 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 28 = (24)2 | yes |
| iv) 321 | 321 = 3 x 107 | No |
| v) 600 | 600 = 120 x 5 = 12 x 10 x 5 = 2 x 2 x 2 x 3 x 5 x 5 | No |
Question 3.
The following numbers are not perfect squares. Give reasons?
(i) 257
(ii) 4592
(iii) 2433
(iv) 5050
(v) 6098
Solution:
i) 257 → The units digit of the number is 7. So it is not a perfect square number.
ii) 4592 → The units digit of the number is 2. So it is not a perfect square number.
iii) 2433 → The units digit of the number is 3. So, it is not a perfect square number.
Iv) 5050 → The last two digits of the number are not two zero’s. So, it is not a perfect square number.
v) 6098 → The units digit of the number is 8. So It is not a perfect square number
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Question 4.
Find whether the square of the following numbers are even or odd?
(i) 431
(ii) 2826
(iii) 8204
(iv) 17779
(v) 99998
Solution:
| Number | Units digit of square of a number | Even / Odd |
| (i) 431 | 12 = 1 | 1 , odd |
| (ii) 2826 | 62 = 36 | 6, odd |
| (iii) 8204 | 42 = 16 | 6 , even |
| (iv)17779 | 92 = 81 | 1 , odd |
| (v) 99998 | 82 = 64 | 4 , even |
Question 5.
How many numbers lie between the square of the following numbers
(i) 25; 26
(ii) 56; 57
(iii) 107;108
Solution:
The numbers lie between the square of the numbers are:
1) 25,26 → 2 x 25=50
ii) 56, 57 → 2 x 56 = 112
iii) 107, 108 → 2 x 107 = 214
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Question 6.
Without adding, find the sum of the following numbers
(i) 1 + 3 + 5 + 7 + 9 =
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 =
Solution:
(i) 1 + 3 + 5 + 7 + 9 = (5)2 = 5 x 5 = 25
[∵ Sum of ‘n’ consecutive odd number = n2]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 92 = 9 x 9 = 81
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 132 = 13 x 13 = 169