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NCERT Solutions for Class 8 Maths Chapter 7: Cubes and Cube Roots - Exercise 7.2

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NCERT Solutions for Class 8 Maths Chapter 7 (EX 7.2)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.2 (EX 7.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Vedantu is a platform that provides free NCERT Solution and other study materials for students.Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.


Class:

NCERT Solutions for Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 7 - Cubes and Cube Roots

Exercise:

Exercise - 7.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

Access NCERT Solutions for Class 8 Mathematics Chapter 7 - Cubes and Cube Roots

Exercise 7.2

1. Find the cube root of each of the following numbers by prime factorisation method

i. $64$

Ans: Expand $64$ in factors of prime numbers.

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 $

$= {2^3} \times {2^3} $

Take cube root on both sides of equation.

$ \because 64 = {2^3} \times {2^3} $

  $\therefore \sqrt[3]{{64}} = 2 \times 2 = 4 $

The cube root of $64$ is $4.$

ii. $512$

Ans: Expand $512$ in factors of prime numbers.

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 $

$= {2^3} \times {2^3} \times {2^3} $

Take cube root on both sides of equation.

$\because 512 = {2^3} \times {2^3} \times {2^3} $

$\therefore \sqrt[3]{{512}} = 2 \times 2 \times 2 = 8 $

The cube root of $512$ is $8.$

iii. $10648$

Ans: Expand $10648$ in factors of prime numbers.

$10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 $

$= {2^3} \times {11^3} $

Take cube root on both sides of equation.

$\because 10648 = {2^3} \times {11^3} $

$\therefore \sqrt[3]{{10648}} = 2 \times 11 = 22 $

The cube root of $10648$ is $22.$

iv. $27000$

Ans: Expand $27000$ in factors of prime numbers.

$ 27000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 $

$= {2^3} \times {3^3} \times {5^3} $

Take cube root on both sides of equation.

$\because 27000 = {2^3} \times {3^3} \times {5^3} $

$\therefore \sqrt[3]{{27000}} = 2 \times 3 \times 5 = 30 $

The cube root of $27000$ is $30.$

v. $15625$

Ans: Expand $15625$ in factors of prime numbers.

$  15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 $

$ = {5^3} \times {5^3} $

Take cube root on both sides of equation.

$\because 15625 = {5^3} \times {5^3} $

$\therefore \sqrt[3]{{15625}} = 5 \times 5 = 25 $

The cube root of $15625$ is $25.$

vi. $13824$

Ans: Expand $13824$ in factors of prime numbers.

$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {2^3} \times {3^3} $

Take cube root on both sides of equation.

$  \because 13284 = {2^3} \times {2^3} \times {2^3} \times {3^3} $

$\therefore \sqrt[3]{{13284}} = 2 \times 2 \times 2 \times 3 = 24 $

The cube root of $13824$ is $24$

vii. $110592$

Ans: Expand $110592$ in factors of prime numbers.

$115092 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {2^3} \times {2^3} \times {3^3} $

Take cube root on both sides of equation.

$\because 110592 = {2^3} \times {2^3} \times {2^3} \times {2^3} \times {3^3} $

$\therefore \sqrt[3]{{110592}} = 2 \times 2 \times 2 \times 2 \times 3 = 48 $

The cube root of $110592$ is $48.$

viii. $46656$

Ans: Expand $46656$ in factors of prime numbers.

$46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 $

$= {2^3} \times {2^3} \times {3^3} \times {3^3} $

Take cube root on both sides of equation.

$\because 46656 = {2^3} \times {2^3} \times {3^3} \times {3^3} $

$\therefore \sqrt[3]{{46656}} = 2 \times 2 \times 3 \times 3 = 36 $

The cube root of $46656$ is $36.$

ix. $175616$

Ans: Expand $175616$ in factors of prime numbers.

$175616 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 $

$= {2^3} \times {2^3} \times {2^3} \times {7^3} $

Take cube root on both sides of equation.

$\because 175616 = {2^3} \times {2^3} \times {2^3} \times {7^3} $

$\therefore \sqrt[3]{{175616}} = 2 \times 2 \times 2 \times 7 = 56 $

The cube root of $175616$ is $56.$

x. $91125$

Ans: Expand $91125$ in factors of prime numbers.

$91125 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 $

$= {3^3} \times {3^3} \times {5^3} $

Take cube root on both sides of equation.

$\because 91125 = {3^3} \times {3^3} \times {5^3} $

$\therefore \sqrt[3]{{91125}} = 3 \times 3 \times 5 = 45 $

The cube root of $91125$  is $45.$

2. State true or false.

i) Cube of any odd number is even. 

Ans: False. As multiplying an odd number three times will yield an odd number.

For example, the cube of $5$ which is an odd number is \[25\], which is also an odd number.

ii) A perfect cube does not end with two zeroes. 

Ans: True. Perfect cube will always terminate with multiple of $3$ numbers of zeroes.

For example, the cube of \[100\] is $1000000$ and there are $6$ zeros at the end of it.

iii) If square of a number ends with \[5\], then its cube ends with \[25\]. 

Ans: False, it is not always certain that if the square of a number ends with \[5\], then its cube will end with \[25\].

For examples, square of $55$ ends with 5, $3025$ but its cube,$166375$ does not end with \[25\].

iv) There is no perfect cube which ends with \[8\]. 

Ans: False, all the numbers having \[2\] at its unit digit place will have \[8\] in end as cube.


v) The cube of a two digit number may be a three digit number. 

Ans: False, as cube of even smallest two digit number, $10$ is a four digit number,$1000$.

vi) The cube of a two digit number may have seven or more digits. 

Ans: False, as cube of even largest two digit number, $99$ is a six digit number, $970299$

vii) The cube of a single digit number may be a single digit number. 

Ans: True, as a cube of first two natural numbers, $1$ and $2$ are $1$ and $8$ respectively.

3. You are told that \[{\mathbf{1331}}\] is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of \[{\mathbf{4913}},{\text{ }}{\mathbf{12167}},{\text{ }}{\mathbf{3276}}8\]

Ans:  Form groups of three digits starting from the rightmost digit of the number.

\[\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\cdot}$}}{1} \;\underline {331} \]

\[2\] groups are formed, \[1\] and \[331\], in particular.

Take first group for consideration \[331\],

The digit at unit place of \[331\] is \[1\]. If the digit \[1\] is at the unit place of a perfect

cube number, then its cube root will also have \[1\] at its unit place. 

Take second group for consideration \[1\],

The cube of \[1\] is equal to the number of the second group. 

So, the tens digit of cube root of  \[1331\]will be \[1\].

Hence, \[\sqrt[3]{{1331}} = 11\]


4. Calculate cube root of $4913$.

Ans: Make groups having three digit numbers starting from the rightmost digit of the number.

\[\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\cdot}$}}{4} \;\underline {913} \]

\[2\] groups are formed, \[4\]and \[913\], in particular.

Take first group for consideration\[913\],

The digit at unit place of \[913\] is \[3\]. If the digit \[3\]is at the unit place of a perfect

cube number, then its cube root will also have\[7\] at its unit place. 

Take second group for consideration\[4\],

Note that 

${1^3} = 1$ and ${2^3} = 8$

Also $1 < 4 < 8$.

So, taking the smaller number into account.

So, the tens digit of cube root of  \[1331\]will be \[1\].

Hence,\[\sqrt[3]{{4913}} = 17\]


5. Calculate cube root of \[12167\].

Ans: Make groups having three digit numbers starting from the rightmost digit of the number.

\[\underline {12} \;\underline {167} \]

\[2\] groups are formed, \[12\]and \[167\], in particular.

Take first group for consideration\[167\]

The digit at unit place of \[167\] is \[7\]. If the digit \[7\]is at the unit place of a perfect

cube number, then its cube root will also have\[3\] at its unit place. 

Take second group for consideration\[12\],

Note that 

${2^3} = 8$ and ${3^3} = 27$

Also $8 < 12 < 27$.

So, taking the smaller number into account.

So, the tens digit of cube root of \[12167\] will be \[2\].

Hence,\[\sqrt[3]{{12167}} = 23\]


6. Calculate cube root of \[32768\].

Ans: Make groups having three digit numbers starting from the rightmost digit of the number.

\[\underline {32} \;\underline {768} \]

\[2\] groups are formed, \[32\] and \[768\], in particular.

Take first group for consideration \[768\].

The digit at unit place of \[768\] is \[8\]. If the digit \[8\] is at the unit place of a perfect

cube number, then its cube root will also have\[2\] at its unit place. 

Take second group for consideration \[32\],

Note that 

${3^3} = 27$ and ${4^3} = 64$

Also $27 < 32 < 64$.

So, taking the smaller number into account.

So, the tens digit of cube root of  \[32768\] will be \[3\].

Hence,\[\sqrt[3]{{32768}} = 32\]

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.2

Opting for the NCERT solutions for Ex 7.2 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.2 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 7 Exercise 7.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 7 Exercise 7.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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