NCERT Solution for class 8 maths chapter 7-Cubes and Cube Roots

NCERT solutions for class 8 maths chapter 7 Cubes and Cube Roots is prepared by academic team of Physics Wallah. We have prepared solutions for all exercise of chapter 7. Given below is step by step solutions of all questions given in NCERT textbook for chapter 7. Read chapter 7 theory make sure you have gone through the theory part of chapter 7 from NCERT textbook and you have learned the formula of the given chapter. Physics Wallah prepared a detail notes and additional questions, maths doubts for class 8 maths with short notes of all maths formula of class 8 maths. Do read these contents before moving to solve the exercise of NCERT chapter 7

NCERT Solutions for Class 8 Maths Exercise 7.1

Question 1.

Which of the following numbers are not perfect cubes:
(i)216
(ii)128
(iii)1000
(iv)100
(v)46656
 

Solution :

(i)216
Prime factors of 216 = 2×2×2×3×3×3
Here all factors are in groups of 3’s (in triplets)
Cubes and Cube Roots/image002.png
Therefore, 216 is a perfect cube number.
(ii) 128
Prime factors of 128
= 2×2×2×2×2×2×2
Here one factor 2 does not appear in a 3’s group.
Cubes and Cube Roots/image004.png
Therefore, 128 is not a perfect cube.
(iii) 1000
Prime factors of 1000 = 2×2×2×3×3×3
Here all factors appear in 3’s group.
Therefore, 1000 is a perfect cube.
Cubes and Cube Roots/image005.png
 (iv) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.
Cubes and Cube Roots/image006.png
(v) 46656
Prime factors of 46656
= 2x2x2x2x2x2x3x3x3x3x3x3
Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.
Cubes and Cube Roots/image005.png

Question 2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:
(i)243
(ii)256
(iii)72
(iv)675
(v)100
 

Solution :

(i)243
Prime factors of 243 = 3x3x3x3x3
Here 3 does not appear in 3’s group.
Therefore, 243 must be multiplied by 3 to make it a perfect cube.
Cubes and Cube Roots/image010.png
(ii) 256
Prime factors of 256
= 2×2×2×2×2×2×2×2
Here one factor 2 is required to make a 3’s group.
Therefore, 256 must be multiplied by 2 to make it a perfect cube.
Cubes and Cube Roots/image012.png
(iii) 72
Prime factors of 72 = 2×2×2×3×3
Here 3 does not appear in 3’s group.
Therefore, 72 must be multiplied by 3 to make it a perfect cube.
Cubes and Cube Roots/image014.png
 
(iv) 675
Prime factors of 675 = 3 × 3 × 3 ×  5 × 5
Here factor 5 does not appear in 3’s group.
Therefore 675 must be multiplied by 3 to make it a perfect cube.
Cubes and Cube Roots/image016.png
 
(v)  100
Prime factors of 100 =2 × 2 × 5 × 5
Here factor 2 and 5 both do not appear in 3’s group.
Therefore 100 must be multiplied by Cubes and Cube Roots/image018.png = 10 to make it a perfect cube.
Cubes and Cube Roots/image019.png
 

Question 3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube:
(i)81
(ii)128
(iii)135
(iv)192
(v)704
Solution :

(i)81
Prime factors of 81 = 3 × 3 × 3 × 3
Here one factor 3 is not grouped in triplets.
Therefore 81 must be divided by 3 to make it a perfect cube.
Cubes and Cube Roots/image021.png
(ii) 128
Prime factors of 128 =Cubes and Cube Roots/image022.png
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 must be divided by 2 to make it a perfect cube.
Cubes and Cube Roots/image023.png
(iii) 135
Prime factors of 135 =Cubes and Cube Roots/image024.png
Here one factor 5 does not appear in a triplet.
Therefore, 135 must be divided by 5 to make it a perfect cube.
Cubes and Cube Roots/image025.png
 (iv) 192
Prime factors of 192 =Cubes and Cube Roots/image026.png
Here one factor 3 does not appear in a triplet.
Therefore, 192 must be divided by 3 to make it a perfect cube.
Cubes and Cube Roots/image027.png
 (v) 704
Prime factors of 704
=Cubes and Cube Roots/image028.png
Here one factor 11 does not appear in a triplet.
Therefore, 704 must be divided by 11 to make it a perfect cube.
Cubes and Cube Roots/image029.png

Question 4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Ans. Given numbers =Cubes and Cube Roots/image030.png
Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by 2 × 5 × 2 = 20 to make it a perfect cube.
Hence he needs 20 cuboids.
 

 

NCERT Solutions for Class 8 Maths Exercise 7.2

 


Question 1.

Find the cube root of each of the following numbers by prime factorization method:
(i) 64     
(ii) 512
(iii) 10648     
(iv) 27000
(v) 15625     
(vi) 13824
(vii) 110592    
(viii) 46656
(ix) 175616    
(x) 91125

Solution : (i) 64
Cubes and Cube Roots/image001.png
= 4
Cubes and Cube Roots/image004.png
 
(ii) 512
Cubes and Cube Roots/image005.png


(iii) 10648
Cubes and Cube Roots/image008.png

(iv) 27000
Cubes and Cube Roots/image011.png

(v) 15625

=5 × 5
= 25
Cubes and Cube Roots/image016.png
(vi) 13824

=Cubes and Cube Roots/image017.png
= 24
Cubes and Cube Roots/image019.png
(vii) 110592
Cubes and Cube Roots/image020.png
= 2×2×2×2×3
= 48
Cubes and Cube Roots/image022.png
 
(viii) 46656
 Cubes and Cube Roots/image023.png
= 36
Cubes and Cube Roots/image025.png
(ix) 175616
Cubes and Cube Roots/image026.png
= 56

(x) 91125
Cubes and Cube Roots/image029.png
 

 

Question 2.

State true or false:
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeroes.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.

Solution : (i) False
Since,Cubes and Cube Roots/image032.png  are all odd.
(ii) True
Since, a perfect cube ends with three zeroes. e.g. Cubes and Cube Roots/image033.png so on
 
(iii) False
Since,Cubes and Cube Roots/image034.png
(Did not end with 25)
 
(iv) False
Since   Cubes and Cube Roots/image035.png= 1728
[Ends with 8]
And  = 10648
[Ends with 8]
 
(v) False Since  Cubes and Cube Roots/image037.png= 1000
[Four digit number]
And  11³ = 1331
[Four digit number]
(vi) False Since  99³ = 970299
[Six digit number]
 
(vii) True
  = 1
[Single digit number]
 2³ = 8
[Single digit number]
 

Question 3.

You are told that 1,331 is a perfect cube. Can you guess with factorization what is its cube root? Similarly guess the cube roots of 4913, 12167, 32768.
Solution :

We know that 10³ = 1000 and Possible cube of  11³ = 1331
Since, cube of unit’s digit = 1
Therefore, cube root of 1331 is 11.
4913
We know that  7³ = 343
Next number comes with 7 as unit place  17³ = 4913
Hence, cube root of 4913 is 17.
12167
We know that = 27
Here in cube, ones digit is 7
Now next number with 3 as ones digit
 13³ = 2197
And next number with 3 as ones digit
23³ = 12167
Hence cube root of 12167 is 23.
32768
We know that  2³ = 8
Here in cube, ones digit is 8
Now next number with 2 as ones digit
 12³ = 1728
And next number with 2 as ones digit
 22³ = 10648
And next number with 2 as ones digit
32³ = 32768
Hence cube root of 32768 is 32.


 

Notes,worksheet and solved question for Maths class 8

 

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  4. class 8 maths notes on chapter Square and square roots
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  6. class 8 maths notes on chapter practical Geometry
  7. class 8 maths notes on chapter commericial maths
  8. class 8 maths notes on chapter solid shape
  9. class 8 maths notes on chapter quadrilaterals
  10. class 8 maths notes on chapter exponents
  11. class 8 maths notes on chapter factorisation
  12. class 8 maths notes on chapter inverse proporation
  13. class 8 maths notes on chapter cube and cube roots

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