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CUBE AND CUBE ROOTS (Chapter-7) SOLUTION
EXERCISE 7.1
1. Which of the following numbers are not perfect cubes?
(i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656
Solution - In the prime factorisation of any number, if each factor appears three times, then, the number is a perfect cube.
(i) 216By prime factorisation, 216 = 2 × 2 × 2 × 3 × 3 × 3
Each factor appears 3 times. 216 = 2³ × 3³ = (2 × 3)³ = 6³
6³ = 216 is a perfect cube.
(ii) 128
By prime factorisation, 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Each factor not appears 3 times.
So, 100 is Not a perfect cube.
(iii) 1000By prime factorisation, 1000 = 2 × 2 × 2 × 5 × 5 × 5
Each factor appears 3 times. 1000 = 2³ × 5³ = (2 × 5)³ = 10³
10³ = 1000 is a perfect cube.
(iv) 100
By prime factorisation, 100 = 2 × 2 × 5 × 5
Each factor not appears 3 times.
So, 100 is Not a perfect cube.
(v) 46656By prime factorisation, 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Each factor appears 3 times. 46656 = 2³ × 2³ × 3³ × 3³ = (2 × 2³ × 3³ × 3)³ = 36³
36³ = 46656 is a perfect cube.
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
By prime factorisation, 243 = 3 × 3 × 3 × 3 × 3
The prime factor 3 does not appear in a group of three.
Therefore,
243 is not a perfect cube.
To make its a cube, we need one more 3.
In that case 243 × 3 = 3 × 3 × 3 × 3 × 3 × 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
By prime factorisation, 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
The prime factor 2 does not appear in a group of three.
Therefore,
256 is not a perfect cube.
To make its a cube, we need one more 2.
In that case 256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 which is a perfect cube.
Hence the smallest natural number by which 256 should be multiplied to make a perfect cube is 2.
(iii) 72
By prime factorisation, 72 = 2 × 2 × 2 × 3 × 3
The prime factor 3 does not appear in a group of three.
Therefore,
72 is not a perfect cube.
To make its a cube, we need one more 3.
In that case 72 × 3 = 2 × 2 × 2 × 3 × 3 × 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
By prime factorisation, 675 = 3 × 3 × 3 × 5 × 5
The prime factor 5 does not appear in a group of three.
Therefore,
675 is not a perfect cube.
To make its a cube, we need one more 5.
In that case 675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 2744 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
By prime factorisation, 100 = 2 × 2 × 5 × 5
The prime factor 2 and 5 does not appear in a group of three.
Therefore,
100 is not a perfect cube.
To make its a cube, we need one more 2 and 5 i.e. 10.
In that case 100 × 10 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 which is a perfect cube.
Hence the smallest natural number by which 100 should be multiplied to make a perfect cube is 2 and 5.
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
By prime factorisation, 81 = 3 × 3 × 3 × 3
The prime factor 3 does not appear in a group of three.
So, 81 is not a perfect cube.
In the factorisation 3 appears only one time.
If we divide the number by 3, then the prime factorisation of the quotient will not contain 3.
So, 81 ÷ 3 = 3 × 3 × 3
Hence the smallest number by which 81 should be divided to make it a perfect cube is 3.
The perfect cube in that case is = 27.
(ii) 128
By prime factorisation, 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
The prime factor 2 does not appear in a group of three.
So, 128 is not a perfect cube.
In the factorisation 2 appears only one time.
If we divide the number by 2, then the prime factorisation of the quotient will not contain 2.
So, 128 ÷ 2 = 2 × 2 × 2 × 2 × 2 × 2
Hence the smallest number by which 128 should be divided to make it a perfect cube is 2.
The perfect cube in that case is = 64.
(iii) 135
By prime factorisation, 135 = 3 × 3 × 3 × 5
The prime factor 5 does not appear in a group of three.
So, 135 is not a perfect cube.
In the factorisation 5 appears only one time.
If we divide the number by 5, then the prime factorisation of the quotient will not contain 5.
So, 135 ÷ 5 = 3 × 3 × 3
Hence the smallest number by which 135 should be divided to make it a perfect cube is 5.
The perfect cube in that case is = 27.
(iv) 192
By prime factorisation, 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
The prime factor 3 does not appear in a group of three.
So, 192 is not a perfect cube.
In the factorisation 3 appears only one time.
If we divide the number by 3, then the prime factorisation of the quotient will not contain 3.
So, 192 ÷ 3 = 2 × 2 × 2 × 2 × 2 × 2
Hence the smallest number by which 192 should be divided to make it a perfect cube is 3.
The perfect cube in that case is = 64.
(v) 704
By prime factorisation, 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
The prime factor 11 does not appear in a group of three.
So, 704 is not a perfect cube.
In the factorisation 11 appears only one time.
If we divide the number by 11, then the prime factorisation of the quotient will not contain 11.
So, 704 ÷ 11 = 2 × 2 × 2 × 2 × 2 × 2
Hence the smallest number by which 704 should be divided to make it a perfect cube is 11.
The perfect cube in that case is = 64.
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?
Solution- According to the question,
The sides of the Cuboid = 5 cm, 2 cm and 5 cm. (given)
Volume of the Cuboid = 5 cm × 2 cm × 5 cm = 50 cm³
For the prime factorisation of 50,
we have
50 = 2 × 5 × 5
To make it a Perfect Cube,
we must have, 2 × 2 × 2 × 5 × 5 × 5
= 20 × (2 × 5 × 5)
= 20 × volume of the given Cuboid
Thus, the required number of Cuboids = 20.
EXERCISE 7.2
1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64 (ii) 512 (iii) 10648 (iv) 27000 (v) 15625 (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125
Solution-(i) 64Prime factorisation of 64 is 2 × 2 × 2 × 2 × 2 × 2Here 2 formed 2 group of three.So, ³√64 = 2 × 2 = 4(ii) 512Prime factorisation of 512 is 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2Here 2 formed 3 group of three.So, ³√512 = 2 × 2 × 2 = 8(iii) 10648Prime factorisation of 10648 is 2 × 2 × 2 × 11 × 11 × 11Here 2 and 11 formed group of three.So, ³√10648 = 2 × 11 = 22(iv) 27000Prime factorisation of 27000 is 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5Here 2, 3 and 5 formed groups of three.So, ³√27000 = 2 × 3 × 5 = 30(v) 15625Prime factorisation of 15625 is 3 × 3 × 3 × 5 × 5 × 5Here 3 and 5 formed group of three.So, ³√15625 = 3 × 5 = 15(vi) 13824Prime factorisation of 13824 is 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3Here 2 formed 3 group of three and 3 formed one group of three.So, ³√13824 = 2 × 2 × 2 × 3 = 24(vii) 110592Prime factorisation of 110592 is 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3Here 2 formed 4 group of three and 3 formed one group of three.So, ³√110592 = 2 × 2 × 2 × 2 × 3 = 48(viii) 46656Prime factorisation of 46656 is 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3Here 2 and 3 formed 2 group of three.So, ³√8000 = 2 × 2 × 3 × 3 = 36(ix) 175616Prime factorisation of 175616 is 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7Here 2 formed 3 group of three and 7 formed one group of three.So, ³√175616 = 2 × 2 × 2 × 7 = 56(x) 91125Prime factorisation of 91125 is 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5Here 3 formed 2 group of three and 5 formed one group of three.So, ³√91125 = 3 × 3 × 5 = 45
2. State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(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) Cube of any odd number is even.False – Cube of any odd number is always odd, e.g., (7³) = 343(ii) A perfect cube does not end with two zeros.True – A perfect cube does not end with two zeros.(iii) If square of a number ends with 5, then its cube ends with 25.True – If a square of a number ends with 5, then its cube ends with 25, e.g., (5)² = 25 and (5³)= 625(iv) There is no perfect cube which ends with 8.False – (12)³ = 1728 (ends with 8)(v) The cube of a two digit number may be a three digit number.False – (10)³ = 1000 (4-digit number)(vi) The cube of a two digit number may have seven or more digits.False – (99)³ = 970299 (6-digit number)(vii) The cube of a single digit number may be a single digit number.True – (2)³ = 8 (1-digit number)
3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution-(i) 4913
The given number is 4913.
Step 1 - Form groups of three starting from the rightmost digit of 4913.4 913.In this case one group i.e., 913 has three digits whereas 4 has only one digit.Step 2 - Take 913. The digit 3 is at its one’s place. We take the one’s place of the required cube root as 7.Step 3 - Take the other group, i.e., 4. Cube of 1 is 1 and cube of 2 is 8. 4 lies between 1 and 8. The smaller number among 1 and 2 is 1. The one’s place of 1 is 1 itself.Take 1 as ten’s place of the cube root of 4913.Thus, ³√4913 = 17
(ii) 12167The given number is 12167.Step 1 - Form groups of three starting from the rightmost digit of 12167.12 167. In this case one group i.e., 167 has three digits whereas 12 has only two digits.Step 2 - Take 167. The digit 7 is at its one’s place. We take the one’s place of the required cube root as 3.Step 3 - Take the other group, i.e., 12 . Cube of 2 is 8 and cube of 3 is 27. 12 lies between 8 and 27. The smaller number among 2 and 3 is 2. The one’s place of 2 is 2 itself.Take 2 as ten’s place of the cube root of 12167.Thus, ³√12167 = 23
(iii) 32768The given number is 32768.Step 1 - Form groups of three starting from the rightmost digit of 32768.32 768. In this case one group i.e., 768 has three digits whereas 32 has only two digits.Step 2 - Take 768. The digit 8 is at its one’s place. We take the one’s place of the required cube root as 2.Step 3 - Take the other group, i.e., 32. Cube of 3 is 27 and cube of 4 is 64. 32 lies between 27 and 64. The smaller number among 3 and 4 is 3. The one’s place of 3 is 3 itself.Take 3 as ten’s place of the cube root of 32768.Thus, ³√32768 = 32
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