NCERT Solutions for Class 8 Maths Chapter 7: Cubes and Cube Roots - Exercise 7.1

Exercise 7.1

Refer the page 1 to 12 for the exercise 7.1 in the PDF

1. Which of the following numbers are not perfect cubes?

I. \[{\mathbf{216}}\]

And:The prime factorisation of $216$ is as follows.

\[216\]

\[ = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]

\[ = 23 \times 33\]

Here, as each prime factor is appearing as many times as a perfect multiple of $3$, therefore, $216$ is a perfect cube.

II. \[{\mathbf{128}}\]

Ans: The prime factorisation of \[128\] is as follows

\[128 = 2 \times 2 \times 2 \times {\text{2}} \times 2 \times 2 \times 2\]

Here, each prime factor is not appearing as many times as a perfect multiple of $3$. 

One\[\;2\] is remaining after grouping the triplets of \[\;2\]. 

Therefore, \[128\] is not a perfect cube.

III. \[{\mathbf{1000}}\]

Ans: The prime factorisation of \[1000\] is as follows

\[1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5\]

Here, as each prime factor is appearing as many times as a perfect multiple of $3$,

therefore, \[1000\] is a perfect cube.

IV. \[{\mathbf{100}}\]

Ans: The prime factorisation of \[100\] is as follows. 

\[100 = 2 \times 2 \times 5 \times 5\]

Here, each prime factor is not appearing as many times as a perfect multiple of$3$.

Two \[2s\] and two \[5s\] are remaining after grouping the triplets. 

Therefore, \[100\] is not a perfect cube.

V. \[{\mathbf{46656}}\]

The prime factorisation of \[46656\] is as follows.

\[46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3\]

Here, as each prime factor is appearing as many times as a perfect multiple of $3$, therefore, \[46656\] 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. \[{\mathbf{243}}\]

Ans: \[243 = 3 \times 3 \times 3 \times 3 \times 3\]

Here, two 3s are left which are not in a triplet. To make \[243\] a cube, one more $3$ is required. 

In that case, \[243 \times 3 = 3 \times 3 \times 3 \times 3 \times {\text{3}} \times 3\]

\[ = 729\]

is a perfect cube. 

Hence, the smallest natural number by which \[243\] should be multiplied to make it a perfect cube is $3$.

II. \[{\mathbf{256}}\]

Ans: \[256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Here, two 2s are left which are not in a triplet. To make \[256\] a cube, one more 2 is required. 

Then, we obtain \[256 \times 2 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

\[ = 512\]

\[512\]is a perfect cube.

III. 72

Ans: \[72 = 2 \times 2 \times 2 \times 3 \times 3\]

Here, two 3s are left which are not in a triplet. To make \[72\] a cube, one more $3$ is required. 

Then, we obtain \[72 \times 3 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]

\[ = 216\]

\[216\]is a perfect cube. 

Hence, the smallest natural number by which \[72\] should be multiplied to make it a perfect cube is $3$.

IV. \[{\mathbf{675}}\]

Ans: \[675 = 3 \times 3 \times 3 \times 5 \times 5\]

Here, two 5s are left which are not in a triplet. To make \[675\] a cube, one more $5$ is required. 

Then, we obtain \[675 \times 5 = 3 \times 3 \times 3 \times 5 \times 5 \times 5\]

\[ = 3375\]

\[3375\] is a perfect cube. 

Hence, the smallest natural number by which \[675\] should be multiplied to make it a perfect cube is $5$.

V. \[{\mathbf{100}}\]

Ans:\[100 = 2 \times 2 \times 5 \times 5\]

Here, two \[2s\] and two \[5s\] are left which are not in a triplet. To make $100$ a cube, we require one more $2$ and one more $5$. 

Then, we obtain \[100 \times 2 \times 5 = 2 \times 2 \times 2 \times 5 \times 5 \times 5\]

\[ = 1000\]

\[1000\]is a perfect cube 

Hence, the smallest natural number by which \[100\] should be multiplied to make it a perfect cube is\[2 \times 5 = 10\].

3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. 

I. 81

Ans: \[81 = 3 \times 3 \times 3 \times 3\]

Here, one $3$ is left which is not in a triplet. If we divide $81$ by $3$, then it will become a perfect cube. 

Thus, \[\dfrac{{81}}{3} = {\text{2}}7\]

\[ = 3 \times 3 \times 3\]

Whichis a perfect cube. 

Hence, the smallest number by which $81$ should be divided to make it a perfect cube is $3$.

II. \[{\mathbf{128}}\]

Ans: \[128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Here, one $2$ is left which is not in a triplet. If we divide $128$ by $2$, then it will become a perfect cube. 

Thus, \[\dfrac{{128}}{2} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Which is a perfect cube. 

Hence, the smallest number by which $128$ should be divided to make it a perfect cube is $2$.

III. \[{\mathbf{135}}\]

Ans: \[135 = 3 \times 3 \times 3 \times 5\]

Here, one $5$ is left which is not in a triplet. If we divide \[135\] by $5$, then it will become a perfect cube. 

Thus, \[\dfrac{{135}}{5} = {\text{2}}7 = 3 \times 3 \times 3\]

Which is a perfect cube. 

Hence, the smallest number by which \[135\] should be divided to make it a perfect cube is $5$.

IV. \[{\mathbf{192}}\]

Ans: \[192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3\]

Here, one $3$ is left which is not in a triplet. If we divide $192$ by $3$, then it will become a perfect cube. 

Thus, \[\dfrac{{192}}{3} = {\text{6}}4 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Which is a perfect cube. 

Hence, the smallest number by which \[192\] should be divided to make it a perfect cube is $3$.

V. \[{\mathbf{704}}\]

Ans: \[704 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11\]

Here, one \[11\] is left which is not in a triplet. 

If we divide \[704\] by\[11\], then it will become a perfect cube. 

Thus, \[\dfrac{{704}}{{11}} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Which  is a perfect cube.

4. Parikshit makes a cuboid of plasticine of 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Ans: Here, some cuboids of size \[5 \times 2 \times 5\]are given.

(Image Will Be Updated Soon)

When these cuboids are arranged to form a cube, the side of this cube so formed will be a common multiple of the sides (i.e., \[5,2,\]and $5$) of the given cuboid. 

LCM of \[5,2,\] and \[5 = 10\]

Let us try to make a cube of $10$ cm. 

For this arrangement, we have to put $2$ cuboids along with its length, $5$ along with its width, and $2$ along with its height.

Total cuboids required according to this arrangement \[ = 2 \times 5 \times 2 = 20\]

With the help of \[20\] cuboids of such measures, a cube is formed as follows.

Class 8 Maths Exercise 7.1 Cubes and Cube Roots - Exercise Overview in a Quick Glimpse

The chapter ‘Cubes and Cube Roots’ holds an important place in building the foundation of Maths as a subject for students. It is crucial for students to be acquainted with various number systems and the four commonly used mathematical operations that aid in defining the other advanced concepts in Maths - cubes being one of them.

Soon after you learn about squaring a number which is multiplying a number with itself, you will learn about cubes here. All you need to do here is multiply the same number with itself twice. It is better to have practical experience of solving numerical than merely reading the theories, hence take a look at our NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1. It will offer you with the much-required clarity before you delve into solving related numerical.

Contrarily, cube root is the other way round as it is that number, which, when multiplied to itself twice, gives the original number whose cube root is asked for. For better clarity on these technicalities, you can refer to our standard solutions for Class 8 Maths Exercise 7.1. 

We have made sure that the explanations are clear and to the point so that you can get an in-depth idea of the particular concept. It will also help you in minimising unnecessary confusion and silly errors while answering during exams.

Also, our materials are freely available on the internet, and they can be downloaded in PDF form future reference as well. All these features make our solutions stand out and most preferred by students, which is the primary motive behind providing these study materials.

Class 8 Ex 7.1 Solution – All Questions

• NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 - Question 1

The first question of Class 8 Maths Ex 7.1 has five numbers, where you are asked to find which of the given numbers are not perfect cubes. To find whether a number is a perfect cube or not, you need to go through a pre-determined process.

Finding prime factors of that particular number is one of those ways. However, students should consider that they are required to have a prior understanding of prime and composite numbers. You can get that from our NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 that will help you build a stronghold on these topics.

Here, in this question, after you find the prime numbers, arrange each of the numbers in a set of three if applicable. Likewise, you can find the cube root of a number.

• Class 8 Ex 7.1 Solution - Question 2

You are given five numbers and asked to find the smallest number to be multiplied to each of these to make them into a perfect cube. Students often opt for trial and error methods in these questions when they are not sure about the approach for such tricky questions. 

However, a trial and error method is not at all recommended considering that it takes a lot of time, and is prone to errors. Therefore, to avoid such chaotic situations, you can use a shortcut technique that is included in our NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1. It aims at easier solving over a shorter time frame.

It will not only solve the sum faster but also lessen the chances of making errors. Also, for future reference, you can download these tricks in your device and access them at any time. Hence, after using this method, you can also boost your exam scores effortlessly. Additionally, you gain the required confidence prior to taking your upcoming exam.

• Exercise 7.1 Class 8 Maths - Question 3

This question is quite similar to the previous one, with only one difference. Instead of adding, you have to find the smallest number by dividing by each number to obtain a perfect square. 

Yet again, make sure you do not use the trial and error method for such twisted questions. Take help from our NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1, where you will find step by step solutions to these questions.

We have explained the technique for each number given in the question. It will ease your understanding of the concept. Plus, you will be acquainted with every type of number, whether it is a two-digit number of a three-digit number.

• NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 - Question 4

It is the last question of this exercise. It is a subjective question, and you need to find the cubes that will make up a cuboid of a given dimension. At first glance, it might seem impossible to calculate the number of cubes. However, with closer analysis and repeated practice, these solutions become an easy affair.

Go through our NCERT study materials for Class 8 Ex 7.1 Solution and get to know the process of solving this type of sum. You will be guided thoroughly on the step of calculation in the following way -

Step 1: Find the volume of a cube from the dimensions given.

Step 2: Find the prime numbers to be multiplied to it to make it a perfect cube.

Step 3: Form sets of three numbers 

Step 4: Find the cube root.

For a higher score, make sure you carefully follow the rules in a stepwise manner. Our NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 will help you in developing a habit of using this method appropriately in the exam.

Besides, we have provided an alternative procedure to this numerical as well, for your convenience. The process is elaborately explained with proper statements to assist you in understanding the concept from a deeper perspective.

It will build your fundamentals and also help you in dealing with such trickier questions during exams. Hence, no questions will seem challenging for you, and you can improve your grasp to fetch more marks in these questions.

Exercise 7.1 Class 8 Maths - Other Exercises

There is a second exercise in this chapter which has three questions in all. They involve similar concepts of cubes and cube roots, and you can improve your understanding of the overall topic once you solve these two exercises entirely.

Here is a brief overview of what you are likely to deal with when going through the second exercise.

• Question 1: You are given ten numbers and asked to find the cube root of each using the prime factorisation method.

• Question 2: Seven statements are provided, and you have to determine whether they are true or false.

• Question 3: You are asked to guess the cube root of specific numbers without using the factorisation method.

The solutions to each of these questions are clearly explained in our NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1. We have provided explanations for both true and false statements. Also, a trick has been provided to guess the cube root of each number.

Students need to go through each reasoning and explanation to build a firm grasp on the concept. Downloading a PDF format will further simplify the understanding of students as they can access these notes from anywhere.

Our materials are framed by subject experts, and therefore, a glance at these shortcut techniques and tricks will help them memorise the same for a longer time.

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Our NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 penned by them is of the best standard and as per the latest syllabus put forth by CBSE. These solutions will act as a quick guide in helping you to sail through tough chapters like cubes and cube roots. Besides, our study notes are available for other subjects and classes as well. With a simple language for explanations and appropriate representation of data, our solutions have become the most preferred notes by students. 

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