Students can refer to RD Sharma Solutions Class 8 Maths which are designed by our expert team. Students can get easy and quick access to chapter-wise and exercise-wise solutions where they can practise a large number of and different types of questions from the textbook. This would ensure they come up with flying colours in their exams. In Exercise 4.5 of RD Sharma Class 8 Solutions for Chapter 4 Cubes and Cube Roots, we will discuss problems based on finding cube roots using cube root tables. Download the free PDF of RD Sharma Solutions for Class 8 Maths from the links below.
RD Sharma Solutions for Class 8 Maths Exercise 4.5 Chapter 4 Cubes and Cube Roots
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 4.5 Chapter 4 Cubes and Cube Roots
Find the cube root of the following (correct to three decimal places), using the cube root table:
1. 7
Solution:
As we know that 7 lies between 1 and 100, by using the cube root table, we get,
∛7 = 1.913
∴ the answer is 1.913.
2. 70
Solution:
As we know that 70 lies between 1 and 100, by using the cube root table from column x,
We get,
∛70 = 4.121
∴ the answer is 4.121.
3. 700
Solution:
700 = 70×10
By using the cube root table 700 will be in column ∛10x against 70.
So, we get,
∛700 = 8.879
∴ the answer is 8.879.
4. 7000
Solution:
7000 = 70×100
∛7000 = ∛(7×1000) = ∛7 × ∛1000
By using the cube root table,
We get,
∛7 = 1.913
∛1000 = 10
∛7000 = ∛7 × ∛1000
= 1.913 × 10
= 19.13
∴ the answer is 19.13.
5. 1100
Solution:
1100 = 11×100
∛1100 = ∛(11×100) = ∛11 × ∛100
By using the cube root table,
We get,
∛11 = 2.224
∛100 = 4.6642
∛1100 = ∛11 × ∛100
= 2.224 × 4.642
= 10.323
∴ the answer is 10.323.
6.780
Solution:
780 = 78×10
By using the cube root table 780 would be in column ∛10x against 78.
We get,
∛780 = 9.205
7. 7800
Solution:
7800 = 78×100
∛7800 = ∛(78×100) = ∛78 × ∛100
By using the cube root table,
We get,
∛78 = 4.273
∛100 = 4.6642
∛7800 = ∛78 × ∛100
= 4.273 × 4.642
= 19.835
∴ the answer is 19.835.
8. 1346
Solution:
Let us find the factors by using the factorisation method,
We get,
1346 = 2×673
∛1346 = ∛(2×676) = ∛2 × ∛673
Since, 670<673<680 = ∛670 < ∛673 < ∛680
By using the cube root table,
∛670 = 8.750
∛680 = 8.794
For the difference (680-670), which is 10.
So the difference in the values = 8.794 – 8.750 = 0.044
For the difference (673-670), which is 3.
So the difference in the values = (0.044/10) × 3 = 0.0132
∛673 = 8.750 + 0.013 = 8.763
∛1346 = ∛2 × ∛673
= 1.260 × 8.763
= 11.041
∴ the answer is 11.041.
9. 250
Solution:
250 = 25×100
By using the cube root table, 250 would be in column ∛10x against 25.
We get,
∛250 = 6.3
∴ the answer is 6.3.
10. 5112
Solution:
Let us find the factors by using the factorisation method,
∛5112 = ∛2×2×2×3×3×71
= ∛23×32×71
= 2 × ∛32 × ∛71
= 2 × ∛9 × ∛71
From the cube root table, we get,
∛9 = 2.080
∛71 = 4.141
∛5112 = 2 × ∛9 × ∛71
= 2 × 2.080 × 4.141
= 17.227
∴ the answer is 17.227.
11. 9800
Solution:
∛9800 = ∛98 × ∛100
From the cube root table, we get,
∛98 = 4.610
∛100 = 4.642
∛9800 = ∛98 × ∛100
= 4.610 × 4.642
= 21.40
∴ the answer is 21.40.
12. 732
Solution:
∛732
We know that the value of ∛732 will lie between ∛730 and ∛740
From the cube root table, we get,
∛730 = 9.004
∛740 = 9.045
By using the unitary method,
Difference between the values (740 – 730 = 10)
So, the difference in cube root values will be = 9.045 – 9.004 = 0.041
Difference between the values (732 – 730 = 2)
So, the difference in cube root values will be = (0.041/10) ×2 = 0.008
∛732 = 9.004+0.008 = 9.012
∴ the answer is 9.012.
13. 7342
Solution:
∛7342
We know that the value of ∛7342 will lie between ∛7300 and ∛7400
From the cube root table, we get,
∛7300 = 19.39
∛7400 = 19.48
By using the unitary method,
Difference between the values (7400 – 7300 = 100)
So, the difference in cube root values will be = 19.48 – 19.39 = 0.09
Difference between the values (7342 – 7300 = 42)
So, the difference in cube root values will be = (0.09/100) × 42 = 0.037
∛7342 = 19.39+0.037 = 19.427
∴ the answer is 19.427.
14. 133100
Solution:
∛133100 = ∛ (1331×100)
= ∛1331 × ∛100
= ∛ 113 × ∛100
= 11 × ∛100
From the cube root table, we get,
∛100 = 4.462
∛133100 = 11 × ∛100
= 11 × 4.462
= 51.062
∴ the answer is 51.062.
15. 37800
Solution:
∛37800
Firstly let us find the factors for 37800
∛37800 = ∛(2×2×2×3×3×3×175)
= ∛(23×33×175)
= 6 × ∛175
We know that the value of ∛175 will lie between ∛170 and ∛180
From the cube root table, we get,
∛170 = 5.540
∛180 = 5.646
By using the unitary method,
Difference between the values (180 – 170 = 10)
So, the difference in cube root values will be = 5.646 – 5.540 = 0.106
Difference between the values (175 – 170 = 5)
So, the difference in cube root values will be = (0.106/10) × 5 = 0.053
∛175 = 5.540 + 0.053 = 5.593
∛37800 = 6 × ∛175
= 6 × 5.593
= 33.558
∴ the answer is 33.558.
16. 0.27
Solution:
∛0.27 = ∛(27/100) = ∛27/∛100
From the cube root table, we get,
∛27 = 3
∛100 = 4.642
∛0.27 = ∛27/∛100
= 3/4.642
= 0.646
∴ the answer is 0.646.
17. 8.6
Solution:
∛8.6 = ∛(86/10) = ∛86/∛10
From the cube root table, we get,
∛86 = 4.414
∛10 = 2.154
∛8.6 = ∛86/∛10
= 4.414/2.154
= 2.049
∴ the answer is 2.049.
18. 0.86
Solution:
∛0.86 = ∛(86/100) = ∛86/∛100
From the cube root table, we get,
∛86 = 4.414
∛100 = 4.642
∛8.6 = ∛86/∛100
= 4.414/4.642
= 0.9508
∴ the answer is 0.951.
19. 8.65
Solution:
∛8.65 = ∛(865/100) = ∛865/∛100
We know that the value of ∛865 will lie between ∛860 and ∛870
From the cube root table, we get,
∛860 = 9.510
∛870 = 9.546
∛100 = 4.642
By using the unitary method,
Difference between the values (870 – 860 = 10)
So, the difference in cube root values will be = 9.546 – 9.510 = 0.036
Difference between the values (865 – 860 = 5)
So, the difference in cube root values will be = (0.036/10) × 5 = 0.018
∛865 = 9.510 + 0.018 = 9.528
∛8.65 = ∛865/∛100
= 9.528/4.642
= 2.0525
∴ the answer is 2.053.
20. 7532
Solution:
∛7532
We know that the value of ∛7532 will lie between ∛7500 and ∛7600
From the cube root table, we get,
∛7500 = 19.57
∛7600 = 19.66
By using the unitary method,
Difference between the values (7600 – 7500 = 100)
So, the difference in cube root values will be = 19.66 – 19.57 = 0.09
Difference between the values (7532 – 7500 = 32)
So, the difference in cube root values will be = (0.09/100) × 32 = 0.029
∛7532 = 19.57 + 0.029 = 19.599
∴ the answer is 19.599.
21. 833
Solution:
∛833
We know that the value of ∛833 will lie between ∛830 and ∛840
From the cube root table, we get,
∛830 = 9.398
∛840 = 9.435
By using the unitary method,
Difference between the values (840 – 830 = 10)
So, the difference in cube root values will be = 9.435 – 9.398 = 0.037
Difference between the values (833 – 830 = 3)
So, the difference in cube root values will be = (0.037/10) ×3 = 0.011
∛833 = 9.398+0.011 = 9.409
∴ the answer is 9.409.
22. 34.2
Solution:
∛34.2 = ∛(342/10) = ∛342/∛10
We know that the value of ∛342 will lie between ∛340 and ∛350
From the cube root table, we get,
∛340 = 6.980
∛350 = 7.047
∛10 = 2.154
By using the unitary method,
Difference between the values (350 – 340 = 10)
So, the difference in cube root values will be = 7.047 – 6.980 = 0.067
Difference between the values (342 – 340 = 2)
So, the difference in cube root values will be = (0.067/10) × 2 = 0.013
∛342 = 6.980 + 0.013 = 6.993
∛34.2 = ∛342/∛10
= 6.993/2.154
= 3.246
∴ the answer is 3.246.
23. What is the length of the side of a cube whose volume is 275 cm3? Make use of the table for the cube root.
Solution:
The given volume of the cube = 275cm3
Let us consider the side of the cube as ‘a’cm
a3 = 275
a = ∛275
We know that the value of ∛275 will lie between ∛270 and ∛280
From the cube root table, we get,
∛270 = 6.463
∛280 = 6.542
By using the unitary method,
Difference between the values (280 – 270 = 10)
So, the difference in cube root values will be = 6.542 – 6.463 = 0.079
Difference between the values (275 – 270 = 5)
So, the difference in cube root values will be = (0.079/10) × 5 = 0.0395
∛275 = 6.463 + 0.0395 = 6.5025
∴ the answer is 6.503cm.
Comments