*Exercise 7.1 *

*Q1: *

*Mention the numbers that are not perfect cubes. *

*(A) 216*

*(B) 128*

*(C) 1000*

*(D) 100*

*(E) 46656*

**Solution:**

(A) 216

Prime factors of 216: 2x2x2x3x3x3

Here all the factors are in the groups of 3’s

Therefore, 216 is said to be a perfect cube number.

02 | 0216 |

02 | 0108 |

02 | 054 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(B) 128

The prime factor of 128 = 2x2x2x2x2x2x2

Here one factor 2 does not appear in groups of 3

Hence, 128 is not a perfect cube.

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 1000

The prime factors of 1000 = 2x2x2x 5x5x5

Here all the factors are in groups of 3

Hence, 1000 is said to be a perfect cube.

02 | 01000 |

02 | 0500 |

02 | 0250 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

(D) 100

The prime factors of 100 is 2×2 x 5×5

Here all the factors do not appear in groups of 3.

Hence, 100 is not a perfect cube.

02 | 0100 |

02 | 050 |

05 | 025 |

05 | 05 |

01 |

(E) 46656

The prime factors of 46656 = 2x2x2x2x2x2x 3 x3x3x3x3x 3

Here all the factors are in groups of 3

Hence, 46656 is said to be a perfect cube.

02 | 046656 |

02 | 023328 |

02 | 011664 |

02 | 05832 |

02 | 02916 |

02 | 01458 |

03 | 0729 |

03 | 0243 |

03 | 081 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

*Q2 : *

*Find the smallest number when multiplied to obtain a perfect cube: *

* (A) 243*

*(B) 256*

*(C) 72 *

*(D) 675*

*(E) 100*

**Solution:**

(A) 243

The prime factors of 243 = 3x3x3x3x 3

Here 3 does not appear in groups of 3

Hence, For 243 to be a perfect cube it should be multiplied by 3.

03 | 0243 |

03 | 081 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(B) 256

The prime factors of 256 is 2x2x2x2x2x 2 x2 x 2

Here one factor of 2 is required for it to make groups of 3.

Hence, for 256 to be a perfect cube it should be multiplied by 2.

02 | 0256 |

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 72

The prime factors for 72 = 2 x2x2x 3x 3

Here the factor 3 does not appear in groups of 3

Hence, For 72 to be a perfect cube it should be multiplied by 3.

(D) 675

The prime factors for 675 = 3x3x3x 5×5

Here the factor 5 does not appear in groups of 3

Hence, for 675 to be a perfect cube it should be multiplied by 5.

03 | 0675 |

03 | 0225 |

03 | 075 |

05 | 025 |

05 | 05 |

01 |

(E) 100

The prime factors for 100 = 2x2x5x5

Here both the factors 2 and 5 are not in groups of 3

Hence, for 100 to be a perfect cube it should be multiplied by 2 and 5. ( i.e. 2 x 5 =10 )

02 | 0100 |

02 | 050 |

05 | 025 |

05 | 05 |

01 |

*Q3: *

*Find the smallest number by which when divided obtain a perfect cube. *

*(A) 81 *

*(B) 128*

*(C) 135*

*(D) 192 *

*(E) 704 *

**Solution:**

(A) 81

The prime factors for 81 = 3 x 3 x 3 x 3

Here, there is one factor of 3 which extra from the group of 3

Hence, for 81 to be a perfect cube it should be divided by 3.

03 | 081 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(B) 128

The prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Here there is one factor of 2 which in not in the group of 3

Hence, for 128 to be a perfect cube then it should be divided by 2.

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 135

The prime factors of 135 = 3 x 3 x 3 x 5

Here there is one factor of 5 which is not appearing with its group of 3.

Hence, for 135 to be a perfect cube it should be divided by 5.

03 | 0135 |

03 | 045 |

03 | 15 |

05 | 05 |

01 |

(D)192

The prime factors for 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Here there is one factor of 3 which does not appearing with its group of 3.

Hence for 192 to be a perfect cube then it should be divided by 3.

02 | 0192 |

02 | 096 |

02 | 048 |

02 | 024 |

02 | 012 |

02 | 06 |

03 | 03 |

01 |

(E) 704

The prime factor for 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Here there is one factor of 11 which is not appearing with its group of 3.

Hence for 704 to be a perfect cube it should be divided by 11.

02 | 0704 |

02 | 0352 |

02 | 0176 |

02 | 088 |

02 | 044 |

02 | 022 |

02 | 011 |

01 |

*Q4:*

*Reuben makes a cuboid of clay of sides 5 cm , 2 cm , 5 cm. If Reuben wants to form a cube how many such cuboids will be needed?*

**Solution:**

The numbers given: 5 x 2 x 5

Since the factors of 2 and 4 are both not in groups of 3.

Then, the number should be multiplied by 2 x 2 x 5 = 20 for it to be made a perfect cube.

Hence Reuben needs 20 cuboids.

*Exercise 7.2 *

*Q1 : *

*By the method of prime factorization find the cube root for the following. *

*(A) 64 *

*(B) 512 *

*(C) 10648*

*(D) 27000*

*(E) 15625 *

*(F) 13824 *

*(G) 110592 *

*(H) 46656*

*(I) 175616*

*(J) 91125*

**Solution:**

(A) 64

\(\sqrt[3]{64}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\\\sqrt[3]{64}=2\times 2\)= 4

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(B) 512

\(\sqrt[3]{512}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\)= 2 x 2 x 2

= 8

02 | 0512 |

02 | 0256 |

02 | 0128 |

02 | 064 |

02 | 032 |

02 | 016 |

02 | 08 |

02 | 04 |

02 | 02 |

01 |

(C) 10648

\sqrt[3]{10648}=\sqrt[3]{2\times 2\times 2\times 11\times 11\times 11}

= 2 x 11

=22

02 | 010648 |

02 | 05324 |

02 | 02662 |

011 | 01331 |

011 | 0121 |

011 | 011 |

01 |

(D) 27000

\(\sqrt[3]{27000}=\sqrt[3]{2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 5}\)=>2 x 3 x 5

=>30

02 | 027000 |

02 | 013500 |

02 | 06750 |

03 | 03375 |

03 | 01125 |

03 | 0375 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

(E) 15625

\(\sqrt[3]{15625}=\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5}\)=> 5 x 5

=> 25

05 | 015625 |

05 | 03125 |

05 | 0625 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

(F) 13824

\(\sqrt[3]{13824}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3}\)=> 2 x 2 x 2 x 3

=> 24

02 | 13824 |

02 | 06912 |

02 | 03456 |

02 | 01728 |

02 | 0864 |

02 | 0432 |

02 | 0216 |

02 | 0108 |

02 | 054 |

03 | 27 |

03 | 09 |

03 | 03 |

01 |

(G) 110592

\(\sqrt[3]{110592}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \times3\times 3\times 3}\)=> 2 x 2 x 2 x 2 x 3

=> 48

02 | 0110592 |

02 | 055296 |

02 | 027648 |

02 | 013824 |

02 | 06912 |

02 | 03456 |

02 | 01728 |

02 | 0864 |

02 | 0432 |

02 | 0216 |

02 | 0108 |

02 | 054 |

03 | 027 |

03 | 09 |

03 | 03 |

01 |

(H) 46656

\(\sqrt[3]{46656}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3}\)=> 2 x 2 x 2 x 3 x 3 x 3

=> 36

(I) 175616

\(\sqrt[3]{175616}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 7\times 7}\)=> 2 x 2 x 2 x 7

=> 56

02 | 0175616 |

02 | 087808 |

02 | 043904 |

02 | 021952 |

02 | 010976 |

02 | 05488 |

02 | 02744 |

02 | 01372 |

02 | 0686 |

07 | 0343 |

07 | 049 |

07 | 07 |

01 |

(J) 91125

\(\sqrt[3]{91125}=\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3\times 5\times 5\times 5}\)=> 3 x 3 x 5

=> 45

03 | 091125 |

03 | 010125 |

03 | 03375 |

03 | 01125 |

03 | 0375 |

05 | 0125 |

05 | 025 |

05 | 05 |

01 |

*Q2: *

*State whether the following is true of false:*

*(A) Any off number of a cube is even.*

*(B) When a number end with two zeros, it is never a perfect cube. *

*(C) If the square of a given number ends with 5 then its cube will end with 25. *

*(D) There is no number that ends with 8 which is a perfect cube.*

*(E) The cube of a given two digit number will always be a three digit number. *

*(F) The cube of a two digit number will have either seven or more digits.*

*(G) The cube of single digit number may also be a single digit number.*

**Solution:**

(A) The statement given is false.

Since, 1^{3 } = 1, 3^{3} = 27, 5^{3 }= 125, . . . . . . . . . . are all odd.

(B) The given statement is true.

Since, a perfect cube ends with three zeroes.

Eg. 10^{3} = 1000, 20^{3 }= 8000, 30^{3 } = 27,000 , . . . . . . . . . . . . so on.

(C) The given statement is false

Since, 5^{2} = 25, 5^{3} = 125 , 15^{2 }= 225, 15^{3 }= 3375 ( Did not end with 25)

(D) the given statement is false.

Since 12^{3} = 1728 [the number ends with 8]

22^{3} = 10648 [ the number ends with 8]

(E) The given statement is false

Since, 10^{3} = 1000 [Four digit number]

And 11^{3}= 1331 [four digit number]

(F) The statement is False.

Since 99^{3} = 970299 [Six digit number]

(G) the given statement is true

1^{3} = 1 [single digit]

2^{3 }= 8 [single digit]

*Q3 :*

*1331 is told to be a perfect cube. What are the factorization methods in which you can find its cube root? Similarly, find the cube roots for *

*(i)4913*

*(ii)12167*

*(iii)32768.*

**Solution:**

We know that 10^{3 }= 1000 and possible cute of 11^{3 } = 1331

Since, the cube of units digit is 1^{3 } = 1

Then, cube root of 1331 is 11

(i) 4913

We know that 7^{3} is 343

Next number that comes with 7 as the units place is 17^{3 } = 4913

Therefore the cube root of 4913 is 17

(ii) 12167

Since we know that 3^{3} = 27

Here in cube, the ones digit is 7

Now the next number with 3 In the ones digit is 13^{3} = 2197

And the next number with 3 in the ones digit is 23^{3 } = 12167

Hence the cube root of 12167 is 23

(iii) 32768

We know that 2^{3 }= 8

Here in the cube, the ones digit is 8

Now the next number with 2 in the ones digit is 12^{3 } = 1728

And the next number with 2 as the ones digit 22^{3 }= 10648

And the next number with 2 as the ones digit 32^{3} = 32768

Hence the cube root of 32768 is 32

Explanation of 1st sum of rational numbers 1st main

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