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CUBE AND CUBE ROOTS (Chapter-7) SOLUTION
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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