Class 8 RD Sharma Solution - Chapter 4 Cubes and Cube Roots - Exercise 4.4 | Set 2

Exercise 4.4 | Set 2 of Chapter 4 in RD Sharma's Class 8 Mathematics textbook continues to build on the concepts of cubes and cube roots, presenting students with even more challenging problems. This set focuses on advanced applications, requiring students to synthesize their knowledge of cubes and cube roots with other mathematical concepts. The problems in this set are designed to enhance critical thinking, problem-solving skills, and the ability to apply mathematical concepts to complex, real-world scenarios.

Question 9. Fill in the blanks:

(i) ∛(125 × 27) = 3 × …

(ii) ∛(8 × …) = 8

(iii) ∛1728 = 4 × …

(iv) ∛480 = ∛3 × 2 × ∛..

(v) ∛… = ∛7 × ∛8

(vi) ∛..= ∛4 × ∛5 × ∛6

(vii) ∛(27/125) = …/5

(viii) ∛(729/1331) = 9/…

(ix) ∛(512/…) = 8/13

Solution:

(i) ∛(125 × 27) = 3 × ....

Let's solve LHS,

 ∛(125 × 27) = ∛(5 × 5 × 5 × 3 × 3 × 3) 

= 5 × 3

So missing value in RHS is 5

(ii) ∛(8 × …) = 8

Let's solve LHS,

 ∛(8 × ...) = ∛8 × 8 × 8 = 8

So missing value in LHS is 8

(iii) ∛1728 = 4 × …

Let's solve LHS,

 ∛1728 = ∛2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 

= 2 × 2 × 3 = 4 × 3

So missing value in RHS is 3

(iv) ∛480 = ∛3×2× ∛..

Let's solve LHS,

 ∛480 = ∛2 × 2 × 2 × 2 × 2 × 3 × 5 = 2 × ∛3 × ∛2× 2 × 5 

= 2 × ∛3 × ∛20

So missing value in RHS is 20

(v) ∛… = ∛7 × ∛8

Let's solve RHS,

∛7 × ∛8 = ∛(7 × 8) = ∛56

So missing value in LHS is 56

(vi) ∛..= ∛4 × ∛5 × ∛6

Let's solve RHS,

∛4 × ∛5 × ∛6 = ∛(4 × 5 × 6) = ∛120

So missing value in LHS is 120

(vii) ∛(27/125) = …/5

Let's solve LHS,

∛(27/125)= ∛(3 × 3 × 3)/(5 × 5 × 5) = 3/5 

So missing value in RHS is 3

(viii) ∛(729/1331) = 9/…

Let's solve LHS,

∛(729/1331)= ∛(9 × 9 × 9)/(11 × 11 × 11) = 9/11 

So missing value in RHS is 11

(ix) ∛(512/…) = 8/13

Let's solve RHS,

Taking cube and cube root of 8/13 at the same time

So, ∛(8/13)³ =  ∛(8 × 8 × 8)/(13 × 13 × 13) = ∛512/2197

So missing value in LHS is 2197

Question 10. The volume of a cubical box is 474. 552 cubic meters. Find the length of each side of the box.

Solution:

Let a be the length of each side of the box

The volume of cubical box = 474.552 m³

As, we know volume of cube = (side)³ 

So, a³ = 474552/1000

a = ∛474552/1000 

Solving the cube root of 474552/1000

= ∛(2 × 2 × 2 × 3 × 3 × 3 × 13 × 13 × 13)/(10 × 10 × 10)

= 2 × 3 × 13/10 = 7.8 m

Hence, the length of each side of the box is 7.8 m

Question 11. Three numbers are to one another 2:3:4. The sum of their cubes is 0.334125. Find the numbers

Solution:

Let the three numbers be 2a, 3a, and 4a

It is given that the sum of their cubes is 0.334125

So, 2a³ + 3a³ + 4a³ = 334125/1000000

99a³ = 334125/1000000 or a = ∛334125/(1000000 × 99)

= ∛3375/1000000 = ∛15 × 15 × 15/10 × 10 × 10 × 10 × 10 × 10

= 15/100 = 0.15

So, first number = 2a = 2 × 0.15 = 0.3

The second number = 3a = 3 × 0.15 = 0.45

The third number = 4a = 4 × 0.15 = 0.6

Hence, the numbers are 0.3, 0.45, and 0.6

Question 12. Find the side of a cube whose volume is 24389/216m³.

Solution:

Let a be the length of each side of the cube

The volume of cube = 24389/216 m³

As, we know volume of cube = (side)³ 

So, a³ = 24389/216

a = ∛24389/216 

Solving the cube root of 24389/216

= ∛(29 × 29 × 29)/(2 × 2 × 2 × 3 × 3 × 3)

= 29/2 × 3 = 29/6 = 4.84 m

Hence, the length of each side of the cube is 4.84 m

Question 13. Evaluate:

(i) ∛36 × ∛384

(ii) ∛96 × ∛144

(iii) ∛100 × ∛270

(iv) ∛121 × ∛297

Solution :

(i) ∛36 × ∛384

We know that, ∛a × ∛b = ∛ab

= ∛(36 × 384) = ∛(2 × 2 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 2 × 3)  

= ∛2³ × 2³ × 2³ × 3³

= 2 × 2 × 2 × 3 = 24

Hence, ∛36 × ∛384 = 24

(ii) ∛96 × ∛144

We know that, ∛a × ∛b = ∛ab

= ∛(96 × 144) = ∛(2 × 2 × 2 × 2 × 2 × 3) × (2 × 2 × 2 × 2 × 3 × 3)  

= ∛2³ × 2³ × 2³ × 3³

= 2 × 2 × 2 × 3 = 24

Hence, ∛96 × ∛144 = 24

(iii) ∛100 × ∛270

We know that, ∛a × ∛b = ∛ab

= ∛(100 × 270) = ∛27000

= ∛3 × 3 × 3 × 10 × 10 × 10 = 3 × 10 = 30

Hence, ∛100 × ∛270 = 30

(iv) ∛121 × ∛297

We know that, ∛a × ∛b = ∛ab

= ∛(121 × 297) = ∛(11 × 11) × (3 × 3 × 3 × 11)  

= ∛3³ × 11³ 

= 3 × 11 = 33

Hence, ∛121 × ∛297 = 33

Question 14. Find the cube roots of the numbers 2460375, 20346417, 210644875, 57066625 using the fact that

(i) 2460375 = 3375 × 729

(ii) 20346417 = 9261 × 2197

(iii) 210644875 = 42875 × 4913

(iv) 57066625 = 166375 × 343

Solution:

(i) 2460375 = 3375 × 729

Cube root of 2460375 will be written as,

∛2460375 = ∛3375 × 729

= ∛(3 × 3 × 3 × 5 × 5 × 5) × (9 × 9 × 9)

= ∛3³ × 5³ × 9³

= 3 × 5 × 9 = 135

Hence, cube root of 2460375 is 135

(ii) 20346417 = 9261 × 2197  

Cube root of 20346417 will be written as,

∛20346417 = ∛9261 × 2197  

= ∛(3 × 3 × 3 × 7 × 7 × 7) × (13 × 13 × 13)

= ∛3³ × 7³ × 13³

= 3 × 7 × 13 = 273

Hence, cube root of 20346417 is 273

(iii) 210644875 = 42875 × 4913 

Cube root of 210644875 will be written as,

∛210644875 = ∛42875 × 4913 

= ∛(5 × 5 × 5 × 7 × 7 × 7) × (17 × 17 × 17)

= ∛5³ × 7³ × 17³

= 5 × 7 × 17 = 595

Hence, cube root of 210644875 is 595

(iv) 57066625 = 166375 × 343

Cube root of 57066625 will be written as,

∛57066625 = ∛166375 × 343

= ∛(5 × 5 × 5 × 11 × 11 × 11) × (7 × 7 × 7)

= ∛5³ × 11³ × 7³

= 5 × 11 × 7 = 385

Hence, cube root of 57066625 is 385

Question 15. Find the unit of the cube root of the following numbers:

(i) 226981

(ii) 13824

(iii) 571787

(iv) 175616

Solution:

(i) 226981 

Since the unit digit of the given number is 1

So, the unit digit of cube root of 226981 is 1

(ii) 13824

Since the unit digit of the given number is 4

So, the unit digit of cube root of 13824 is 4

(iii) 571787

Since the unit digit of the given number is 7

So, the unit digit of cube root of 571787 is 7

(iv) 175616

Since the unit digit of the given number is 6

So, the unit digit of cube root of 175616 is 6

Question 16. Find the tens digit of the cube root of each of the numbers in Q.No.15.

(i) 226981

(ii) 13824

(iii) 571787

(iv) 175616

Solution:

(i) 226981 

Since the unit digit of the given number is 1

So, the unit digit of cube root of 226981 is 1

Let's remove the unit, tens, and hundreds digit of the given number, we get 226 

As we know the number 226 lies between cube root of 6 and 7 (6³ < 226 < 7³)

So, 6 is the largest number whose cube root will be less than or equal to 226

Hence, the tens digit of the cube root of 226981 is 6

(ii) 13824

Since the unit digit of the given number is 4

So, the unit digit of cube root of 13824 is 4

Let's remove the unit, tens, and hundreds digit of the given number, we get 13

As we know the number 13 lies between cube root of 2 and 3 (2³ < 13 < 3³)

So, 2 is the largest number whose cube root will be less than or equal to 13

Hence, the tens digit of the cube root of 13824 is 2

(iii) 571787

Since the unit digit of the given number is 7

So, the unit digit of cube root of 571787 is 3

Let's remove the unit, tens, and hundreds digit of the given number, we get 571

As we know the number 571 lies between cube root of 8 and 9 (8³ < 571 < 9³)

So, 8 is the largest number whose cube root will be less than or equal to 571

Hence, the tens digit of the cube root of 571787 is 8

(iv) 175616

Since the unit digit of the given number is 6

So, the unit digit of cube root of 175616 is 6

Let's remove the unit, tens, and hundreds digit of the given number, we get 175

As we know the number 175 lies between cube root of 5 and 6 (5³ < 571 < 6³)

So, 5 is the largest number whose cube root will be less than or equal to 175

Hence, the tens digit of the cube root of 175616 is

Summary

Exercise 4.4 | Set 2 presents a diverse range of problems that challenge students to apply their understanding of cubes and cube roots in sophisticated ways. The questions involve calculating cube roots of complex expressions, solving multi-step word problems, working with algebraic equations involving cubes, and analyzing relationships between volume, surface area, and edge lengths of cubic objects. This set also introduces problems that combine concepts from geometry and algebra, requiring students to think creatively and apply their knowledge in new contexts. By working through these problems, students will develop a deeper understanding of three-dimensional geometry, improve their algebraic manipulation skills, and enhance their ability to solve complex mathematical problems.