Karnataka Class 8 Math Solution Chapter Cubes and Cube Roots by Experience Math Teacher. Karnataka Board Class 8 Math Part 2 PDF Chapter 1 Exercise 1.1, Exercise 1.2 Solution.
Karnataka Class 8 Math Solution Chapter Cubes and Cube Roots
Part 2 Chapter 1 Exercise 1.1
(1) Which of the following number are not perfect cubs?
(i) 216
=> Solution:- By prime factorization
216 = 2 x 2 x 2 x 3 x 3 x 3
Each factor appears 3 time
216 = 23 x 33
= (2 x 3)3
= 63 which is perfect cube.
So 216 is a perfect cube of 6.
(ii) 128
=> Solution:- by prime factorization
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
In above factorization one triplet group of 2 is not form.
So, 128 is not a perfect cube.
(iii) 1000
=> Solution:- By prime factorization
100 = 5 x 5 x 5 x 2x 2 x 2
Here each factor appear three time
1000 = 53 x 23
= (5 x 2)3
= 103 which is a perfect cube
So, 1000 is a perfect cube of 10.
(iv) 100
=> Solution:- By prime factorization
100 = 5 x 5 x 2 x 2
Here each factor repeated only twice.
So, 100 is not a perfect cube.
(v) 46656
=> Solution: – By prime factorization
46656 = 3 x 3 x 3 x 3 x 3 x 3 x 4 x 4 x 4
Here each factor appear 3 time
46656 = 33 x 33 x 43
= (3 x 3 x 4)4
= (36)3 which is a perfect cube.
So, 46656 is a perfect cube of 36.
(2) Find the smallest number by which each of the following number must be multiplied to obtain a perfect cube.
(i) Solution:- By smallest multiple method
243 = 3 x 3 x 3 x 3 x 3
Here, the prime factor 3 does not appear in a group of three, to obtain a perfect cube, we need one more 3.
In that case
243 x 3 = 3 x 3 x 3 x 3 x 3 x 3
= 729 which is a perfect cube.
Hence the smallest natural number by which 243 should be multiplied to make a perfect cube is 3.
(ii) 256
=> Solution:
Here,
The prime factor 4 does not appear in a group of there. To obtain a perfect cube we need only 2 to make a prime factor 4 i.e. (2 x 2) = 4
In that case
256 x 2 = 2 x 2 x 2 x 4 x 4 x 2 x 2
= 2 x 2 x 2 x 4 x 4x 4 x 2 x 2
= 2 x 2 x 2 x 4 x 4 x 4
= 512 which is a perfect cube.
Hence the smallest natural number by which 256 should be multiplied to make 9 perfect cube is 2.
(iii) 72
=> Solution:-72 = 2 x 2 x 2x 3 x 3
Here,
The prime factor 3 does not appear in a group of three. To obtain a perfect cube we need 3
So, 72 x 3 = 2 x 2 x 2 x 3 x 3 x 3
= 216 which is a perfect cube.
Hence the smallest natural number by which 72 should be multiplied to make a perfect cube is 3.
(iv) 675
=> Solution: 675 = 3 x 3 x 3 x 5 x 5
Here,
The prime factor 5 does not appear in a group of three. To obtain a perfect cube we need one more 5.
In that case
675 x 5 = 3 x 3 x 3 x 5 x 5 x 5
= 3375 which is a perfect cube.
Hence the smallest natural number by which 675 should be multiplied to make a perfect cube is 5.
(v) 100
=> Solution:- 100 = 5 x 5 x 2 x 2
Here both the prime factor does not form a group of three. To obtain a perfect cube we need 5 and 2 i.e. 5 x 2 = 10
Then, 100 x 5 x 2 = 5 x 5 x 5 x 2 x 2 x 2
= 1000 which is a perfect cube.
Hence the smallest natural number by which 100 should be multiplied to make a perfect cube is 5 and 2 or 10.
(3) Find the smallest number by which each of the following number must be divided to obtain a perfect cube.
(i) 81
=> Solution:- 81 = 3 x 3 x 3 x 3
In the factorization of 81 remaining prime factor 3 does not form a group of three. So, if we divide 81 by 3, then the prime factorization of the quotient will not contain 3.
So, 81 ÷ 3 = 3 x 3 x 3
= 27
Hence the smallest number by which 81 should be divided to make it perfect cube is 3.
∴ The perfect cube in that case is = 27
(ii) 128
=> Solution:-
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
In the factorization of 128 prime factor 2 does not form a group of three, so, if we divide 128 by 2, then the prime factorization of te quotients will not contain 2.
So, 128 ÷ 2 = 2 x 2 x 2 x 2 x 2 x 2
= 64
Hence the smallest number by which 128 should be divided to make it perfect cube is 2.
The perfect cube in this case is 64.
(iii) 135
=> Solution:- 135 = 3 x 3 x 3 x 5
In the factorization of 135 the prime factor 5 does not form a group of three so, if we divide 135 by 5, then the prime factorization of the quotient will not contain 5.
So, 135 ÷ 5 = 3 x 3 x 3
= 27
Hence the smallest number by which 135 should be divided to make it perfect cube is 5.
(iv) 192
=> Solution:- 192 = 4 x 4 x 4 x 3
In the factorization of 192 the prime factor 3 does not form a group of there so, if we divide 192 by 3 then the prime factorization of the quotients will not contain 3.
So, 192 ÷ 3 = 4 x 4 x 4
= 64
Hence the smallest number by which 192 should be divided to make it perfect cube is 3.
(v) 704
Solution:- 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
In the factorization of 704 the prime factor 11 does not form a group of three so, if we divide 704 by 11 then the prime factorization of the quotient will not contain 11.
704 ÷ 11 = 2 x 2 x 2 x 2 x 2 x 2
= 64
Hence the smallest number by which 704 should be divided to make it perfect cube is 11.
(4) Parikshit makes a cuboid of plasticine of sides 5cm, 2cm, 5cm, how many such cuboids will he need to form a cube?
=> Solution: Parikshit have plasticine of sides 5cm, 2cm and 5cm
Volume of cuboid is 5 x 2 x 5
= 5 x 2 x 5
= 5 x 5 x 2
Since there is only one 2 in the prime factorization so, we need 2 x 2 i.e. 4 also we need one more 5 to make a group of 5 so we can make a perfect cube
Therefore, we need 20 i.e. (5 x 4) such cuboid to make a cube.
Part 2 Chapter 1 Exercise 1.2
(1) Find the cube root of each of the following numbers by prime factories method.
(i) 64
=> Solution:- By prime factor method 64 = 2 x 2 x 2 x 2 x 2 x 2
= 23 x 23
= (2 x 2)3
3√64 = 2 x 2 = 4
Therefore, the cube root of 64 is 4
(ii) 512
=> Solution:- By prime factor method
512 = 2 x 2 x 2 x 4 x 4 x 4
= 23 x 43
= (2 x 4)3
3√512 = 8
Therefore, the cube root of 512 is 8.
(iii) 10648
=> Solution:-
By prime factor method
10648
=> Solution:- By prime factor method
10648 = 2 x 2 x 2 x 11 x 11 x 11
= 23 x 113
= (2 x 11)3
3√10648 = 2 x 11 = 22
The cube root of 10648 is 22
(iv) 27000
=> Solution:- By prime factor method
27000 = 3 x 3 x 3 x 10 x 10 x 10
= 33 x 103
= (3 x 10)3
3v27000 = 3 x 10 = 30
The cube root of 27000 is 30.
(v) 15625
=> Solution:- By prime factor method
15625 = 5 x 5 x 5 x 5 x 5 x 5
= 53 x 53
= (5 x 5)3
3√15625 = 25
The cube root of 15625 is 25.
(vi) 13824
=> Solution:- By prime factor method
13824 = 2 x 2 x 2 x 3 x 3 x 3x 4 x 4 x 4
= 23 x 33 x 43
= (2 x 3 x 4)3
3√13824 = 2 x 3 x 4 = 24
The cube root of 13824 is 24.
(vii) 110592
=> Solution:- By prime factorization method
110592 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 6 x 6 x 6
= 23 x 23 x 23 x 63
= ( 2 x 2 x 2 6)3
3√110592 = 48
The cube root of 110592 is 48.
(viii) 46656
=> Solution:- By prime factorization method
46656 = 3 x 3 x 3 x 3 x 3 x 3 x 4 x 4 x 4
= 33 x 33 x 43
= (3 x 3 x 4)3
3√46656 = 36
The cube root of 46656 is 36.
(2) State true or false
(i) Cube of any odd number is even
Solution: False.
(ii) A perfect cube does not end with two zero.
Solution: True.
(iii) If square of a number end with 5, then its cube end with 25.
Solution: False.
(iv) There is no perfect cube which ends with 8.
Solution: False.
(v) The cube of a two digit number may be three digit numbers.
Answer: False.
(vi) The cube of a two digit number may have seven or more digits.
Solution: False.
(vii) The cube of a single digit number may be a single digit number.
Solution: True.
(3) You are told that 1,332 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768
=> Solution:- 1331 is a perfect cube of 11.
Similarly,
4913 is a perfect cube of 17
12161 is a perfect cube of 23
32768 is a perfect cube of 32
