RBSE Solutions for Class 8 Maths Chapter 2 Cube and Cube Roots In Text Exercise is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 2 Cube and Cube Roots In Text Exercise.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 2 |
| Chapter Name | Cube and Cube Roots |
| Exercise | In Text Exercise |
| Number of Questions | 9 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 2 Cube and Cube Roots In Text Exercise
Page No: 24
Question 1.
Fill in the blanks
| No. of unit cubes in one side of large cube | No. of unit cubes in making of large cubes |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | — |
| 5 | — |
Solution.
| No. of unit cubes in one side of large cube | No. of unit cubes in making of large cubes |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
Page No: 25
Question 2.
Find the cube and fill in the blanks
| Cube Numbers | Numbers |
| 1 | 1³ = 1×1×1 = 1 |
| 2 | 2³ = 2×2×2 = 8 |
| 3 | 3³ = 3×3×3 = 27 |
| 4 | 4³ = 4×4×4 = 64 |
| 5 | 5³ = 5×5×5 = — |
| 6 | 6³ = 6×6×6 = — |
| 7 | 7³ = 7×7×7 = — |
| 8 | 8³ = 8×8×8 = — |
| 9 | 9³ = 9×9×9 = — |
| 10 | 10³ = 10×10×10 = — |
Solution.
| Cube Numbers | Numbers |
| 1 | 1³ = 1×1×1 = 1 |
| 2 | 2³ = 2×2×2 = 8 |
| 3 | 3³ = 3×3×3 = 27 |
| 4 | 4³ = 4×4×4 = 64 |
| 5 | 5³ = 5×5×5 = 125 |
| 6 | 6³ = 6×6×6 = 216 |
| 7 | 7³ = 7×7×7 = 343 |
| 8 | 8³ = 8×8×8 = 512 |
| 9 | 9³ = 9×9×9 = 729 |
| 10 | 10³ = 10×10×10 = 1000 |
Question 3.
We know that 2² = 4 where 4 = 2 x 2
or 4 = 2 + ?
Similarly, 2³ = 8 where 8 = 2 x 2 x 2
Does this(RBSESolutions.com)equal to (2 + 2 + 2)?
Solution.
∴ 8 = 2 x 2 x 2 = 2³
But, 2 + 2 + 2 = 6 ≠ 2 x 2 x 2
∴ 2³ ≠ 2 + 2 +2 is false
and 2³ = 2 x 2 x 2 is true.
Page No: 26
Question 4.
Find the unit digit of cubes given below
(i) 1331
(ii) 4444
(iii) 159
(iv) 1005
Solution.
(i) In the given number wilt digit is 1
∴ Unit digit of cube of given number will also 1
(ii) In the given number unit digit is 4
∴ unit digit of(RBSESolutions.com)cube of given number will also 4
(iii) In the given number unit digit is 9
∴ Unit digit of cube of given number will also 9
(iv) In the given number unit digit is 5
∴ Unit digit of cube of given number will also 5
Question 5.
The Cube of 46 will be even or odd.
Solution.
Given number 46 is even number
∴ Cube of given even number will also even number .
∵ Cube of even(RBSESolutions.com)number is always even number.
Question 6.
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
Think on this pattern tell : How many consecutive odd numbers will be needed to get the sum of 10³ ?
Solution.
From above pattern, we sec cube of 1 consists is 1 odd number
Cube of 2 Consists 2 odd numbers 3, 5
Cube of 3 Consists 3 odd(RBSESolutions.com)numbers 7, 9, 11
Cube of 4 Consists 4 odd numbers 13, 15, 17, 19
_____________________________
∴ Cube of 10 consists 10 consecutive odd numbers.
Page No: 26
Question 7.
According to the above pattern, find the below in form of addition of odd numbers.
(i) 7³
(ii) 8³
Solution
Addition of consecutive odd numbers as per given pattern :
(i) 7³ = 43 + 45 + 47 + 49 + 51 + 53 + 55
(ii) 8³ = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71
Question 8.
Fill in the blanks
| Number | Cubes |
| 4 = 2 x 2 | 4³ = 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2³ x 2³ |
| 6 = 2 x 3 | 6³ = 216 = 2 x 2 x 2 x 3 x 3 x 3 = 2³ x 3³ |
| 10 = 2 x 5 | 10³ = 1000 = 2 x 2 x 2 x 5 x 5 x 5 = 2³ x 5³ |
| 12 = 2 x 2 x 3 | 12³ = 1728 = ___ |
Solution.
1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
= 2³ x 2³ x 3³
Note: It is clear from above pattern that each factor in cube of a number repeats three times.
Page No:28
Question 9.
Check the perfect cube in the following numbers?
(i) 2700
(ii) 16000
(iii) 64000
(iv) 900
(v) 125000
(vi) 36000
(vii) 21600
(viii) 10000
(ix) 27000000
(x) 11000
Solution.
(i) 2700
Resolving 2700 into Prime factor as follows
The Prime factor 2 and 5 do not appear in a groups of three. Therefore 2700 is not a perfect cube number.
(ii) 16000
Resolving 16000 into Prime factor as follows
Factor 2 is not group of three
∴ 16000 is not a perfect cube
(iii) 64000
Resolving 64000 into Prime factor as follows
There are perfect cubes of each factor 2, 2, 2 and 5.
∴ 64000 is a perfect cube.
(iv) 900
Resolving 900 into Prime factors as follows
Factors 2, 3 and 5 are not groups of three.
∴ 900 is not a(RBSESolutions.com)perfect cube.
(v) 125000
Resolving 125000 into Prime factors as follows
Factors 2, 5 and 5 are in(RBSESolutions.com)groups of three
∴ 125000 is a perfect cube
(vi) 36000
Resolving 36000 into Prime factors as follows
Factors 2 and 3 are not groups of three
∴ 36000 is not a perfect cube.
(vii) 21600
Resolving 21600 into Prime factors as follows
Factors 2 and 5 are not groups of three
∴ 21600 is not a(RBSESolutions.com)perfect cube.
(viii) 10000
Resolving 10000 into Prime factors as follows
Factors 2 and 5 are not groups(RBSESolutions.com)of three.
Therefore, 10000 is not a perfect cube.
(ix) 27000000
Resolving 27000000 into Prime factors as follows
Factors 2, 3 and 5 are groups of three.
Therefore, 27000000 is a perfect cube.
(x) 11000
Resolving 11000 into Prime factors as follows
Here, 11 is not in groups(RBSESolutions.com)of three.
∴ 11000 is not a perfect cube.
We hope the given RBSE Solutions for Class 8 Maths Chapter 2 Cube and Cube Roots In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 2 Cube and Cube Roots In Text Exercise, drop a comment below and we will get back to you at the earliest.
Leave a Reply