# Class 8 RD Sharma Solution – Chapter 4 Cubes and Cube Roots – Exercise 4.5

• Last Updated : 03 Jan, 2021

### Question 1.  7

Solution:

We know that 7 lies between 1-100 so using cube root table we will get ∛7 = 1.913

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

### Question 2.  70

Solution:

We know that 70 lies between 1-100 so using cube root table we will get ∛70 = 4.121

### Question 3.  700

Solution:

We can write 700 as 70×10

Now we can write ∛700 as ∛70 x ∛10 and we will get ∛700 = 8.879

### Question 4.  7000

Solution:

We can write 7000 as 70×100

Now we can write ∛7000 as ∛7 × ∛1000

Using cube root table and we get ∛7 = 1.913 and ∛1000 = 10

= 1.913 × 10 = 19.13

### Question 5.  1100

Solution:

We can write 1100 as 11 x 100

Now we can write ∛1100 as ∛11 × ∛100

Using cube root table we get ∛11 = 2.224 and ∛100 = 4.6642

= 2.224 × 4.642 = 10.323

### Question 6.  780

Solution:

We can write 780 as 78×10

We can write ∛780 as ∛78 and ∛10

Using cube root table we get ∛78 = 4.272 and ∛10 = 2.154

= 4.272 x 2.154 = 9.205

### Question 7.  7800

Solution:

We can write 7800 as 78×100

Now we can write ∛7800 as ∛78 × ∛100

Using cube root table we get ∛78 = 4.273 and ∛100 = 4.6642

= 4.273 × 4.642 = 19.835

### Question 8. 1346

Solution:

Let’s find the factors of 1346 and we can write it as 2×673

Now we can write ∛1346 as ∛2 × ∛673

Since 673 lies between 670 & 680 so ∛670 < ∛673 < ∛680

Now using cube root table we get

∛670 = ∛67 x ∛10 = 8.750

∛680 = ∛68 x ∛10 = 8.794

Difference between 680 & 670 is 10.

So the difference in their cube roots are : 8.794 – 8.750 = 0.044

Difference between 673 & 670  is 3.

So the difference in their cube will : (0.044/10) × 3 = 0.0132

∛673 = 8.750 + 0.013 = 8.763

We can write ∛1346 as ∛2 × ∛673

= 1.260 × 8.763 = 11.041

### Question 9.  250

Solution:

We can write 250 as 25×100 and further ∛25 x ∛10

We get 6.3

### Question 10.  5112

Solution:

Let’s find the factor of 5112,

We get ∛(2×2×2×3×3×71)

= 2 × ∛9 × ∛71

Using cube root table we get ∛9 = 2.080 & ∛71 = 4.141

= 2 × 2.080 × 4.141 = 17.227

### Question 11.  9800

Solution:

We can write ∛9800 as ∛98 × ∛100

Using cube root table we get ∛98 = 4.610 & ∛100 = 4.642

= 4.610 × 4.642 = 21.40

### Question 12.  732

Solution:

As we know that the value of ∛732 will lie between ∛730 & ∛740

Using cube root table we get ∛730 = 9.004 & ∛740 = 9.045

Now we will use unitary method,

Difference between 740 & 730: (740 – 730) = 10

So the difference between their cube roots will be : 9.045 – 9.004 = 0.041

Difference between the values: 732 – 730 = 2

Now we will calculate difference in cube root values: (0.041/10) ×2 = 0.008

∛732 = 9.004+0.008 = 9.012

### Question 13.  7342

Solution:

As we know that the value of ∛7342 will lie between ∛7300 & ∛7400

Using cube root table we will get ∛7300 = 19.39 & ∛7400 = 19.48

Now we will use unitary method,

Difference between 7400 & 7300 = 100

So the difference between their cube roots will be : 19.48 – 19.39 = 0.09

Difference between the values : 7342 – 7300 = 42

Now we will calculate difference in cube root values : (0.09/100) × 42 = 0.037

∛7342 = 19.39+0.037 = 19.427

### Question 14.  133100

Solution:

We can write ∛133100 as ∛ 1331 × ∛ 100,

= 11 × ∛100

Using cube root table we get ∛100 = 4.462

= 11 × 4.462 = 51.062

### Question 15. 37800

Solution:

Let us find the factors for 37800

We get ∛(2×2×2×3×3×3×175)

= 2 x 3 x ∛(175)

= 6 × ∛175

As we know that the value of ∛175 will lie between ∛170 & ∛180

Using cube root table we get ∛170 = 5.540 & ∛180 = 5.646

Now we will use Unitary method,

Difference between 180 & 170 = 10

So the difference between their cube roots will be : 5.646 – 5.540 = 0.106

Difference between the values 175 & 170 = 5

So the difference in their cube roots will be = (0.106/10) × 5 = 0.053

∛175 = 5.540 + 0.053 = 5.593

∛37800 = 6 × ∛175 = 6 × 5.593 = 33.558

### Question 16.  0.27

Solution:

We can write ∛0.27 as ∛27/∛100

By using cube root table we get ∛27 = 3 & ∛100 = 4.642

∛0.27 = ∛27/∛100

= 3/4.642 = 0.646

### Question 17.  8.6

Solution:

∛8.6 = ∛86/∛10

By using cube root table we get ∛86 = 4.414 & ∛10 = 2.154

∛8.6 = ∛86/∛10

= 4.414/2.154 = 2.049

### Question 18.  0.86

Solution:

∛0.86 = ∛86/∛100

By using cube root table we get ∛86 = 4.414 & ∛100 = 4.642

∛8.6 = ∛86/∛100

= 4.414/4.642 = 0.9508

### Question 19.  8.65

Solution:

∛8.65 = ∛865/∛100

As we know that value of ∛865 will lie between ∛860 & ∛870

Using cube root table we get ∛860 = 9.510 & ∛870 = 9.546 & ∛100 = 4.642

We will use Unitary method,

Difference between the values 870 & 860 = 10

So, the difference in their cube roots will be : 9.546 – 9.510 = 0.036

Difference between the values 865 – 860 = 5

So, the difference between their cube roots 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

### Question 20.  7532

Solution:

As we know that value of ∛7532 will lie between ∛7500 & ∛7600

Using cube root table we get ∛7500 = 19.57 & ∛7600 = 19.66

Now we will use Unitary method,

Difference between the values 7600 & 7500 : 7600 – 7500 = 100

So the difference between their cube roots will :19.66 – 19.57 = 0.09

Difference between the values 7532 – 7500 = 32

So the difference between their cube root will : (0.09/100) × 32 = 0.029

∛7532 = 19.57 + 0.029 = 19.599

### Question 21.  833

Solution:

As we know that value of ∛833 will lie between ∛830 and ∛840

Using cube root table we get ∛830 = 9.398 & ∛840 = 9.435

We will use Unitary method,

Difference between the values 840 & 830 = 840 – 830 = 10

So, the difference in their cube root values will be : 9.435 – 9.398 = 0.037

Difference between the values 833 & 830 = 3

So, the difference in their cube root values will be = (0.037/10) ×3 = 0.011

∛833 = 9.398+0.011 = 9.409

### Question 22.  34.2

Solution:

∛34.2 = ∛342/∛10

As we know that value of ∛342 will lie between ∛340 & ∛350

Using cube root table we get ∛340 = 6.980 & ∛350 = 7.047 & ∛10 = 2.154

We will use 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 their 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

### Question 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:

Given that,

Volume of the cube = 275cm3

Let us assume that the side of the cube as ‘a’ cm

As we know that Volume of Cube : a^3 = 275

a = ∛275

Now we know that the value of ∛275 will lie between ∛270 & ∛280

Using cube root table we get ∛270 = 6.463 & ∛280 = 6.542

We will use Unitary method,

Difference between the values 280 & 270 = 10

So the difference in their cube root will = 6.542 – 6.463 = 0.079

Difference between the values 275 & 270 = 5

So the difference in their cube roots will be = (0.079/10) × 5 = 0.0395

∛275 = 6.463 + 0.0395 = 6.5025

My Personal Notes arrow_drop_up