NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.2
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NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots (EX 7.2) Exercise 7.2
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Access NCERT Solutions for Class 8 Mathematics Chapter 7 – Cubes and Cube Roots
NCERT exercises are designed to test students’ understanding of the ideas presented in a particular chapter. For board exams, NCERT solutions are essential. If you have any trouble understanding Cubes And Cube Roots, consult the study materials for Mathematics Chapter 7 Class 8 like the NCERT Solutions For Class 8 Maths Chapter 7 Exercise 7.2 on the Extramarks website and app. Since this chapter is challenging for many students, they consult the NCERT Solutions For Class 8 Maths Chapter 7 Exercise 7.2, which provides answers.
It makes sense that without assistance, students would struggle to comprehend the ideas in the brandnew, challenging chapter on linear equations in one variable. To study a new chapter, which is already quite challenging, students would put in more work than is necessary. In the test preparation, the NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.2, covers a wide range of important subjects. This will benefit students since they will score higher on the NCERT Class 8 Maths Chapter 7 Exercise 7.2 problems on the board exams. The goal of the NCERT Solutions For Class 8 Maths Chapter 7 Exercise 7.2, is to help students go past a challenge they are now experiencing, so they may continue learning and applying other topics from various textbooks. They could complete Class 8 Maths Exercise 7.2 with the help of the NCERT Solutions For Class 8 Maths Chapter 7 Exercise 7.2, allowing them more time to prepare for tests and take practice exams. If students do this, they might perform better in the Class 8 Maths Chapter 7 Exercise 7.2 final board exams.
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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
Cube And Cube Roots is the title of Chapter 7 in the NCERT Mathematics textbook for Class 8. There will be an overview of the chapter in the NCERT Solutions For Class 8 Maths Chapter 7 Exercise 7.2. This chapter begins with a brief introduction before covering a number of concepts, including how to solve equations with linear expressions on one side and numbers on the other, how to apply the aforementioned exercise, and how to solve equations with variables on both sides. Equations that can be transformed into a simpler form or a linear form. The second exercise helps students put what they learned in the first exercise into practice. It is the basis for the NCERT Solutions For Class 8 Maths Chapter 7 Exercise 7.2. Therefore, it’s critical for students to comprehend the NCERT Solutions For Class 8 Maths Chapter 7 Exercise 7.2 in order to perform well in tests.
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Q.1 Find the cube root of each of the following numbers by prime factorization method.
(i) 64 (ii) 512 (iii) 10648 (iv)27000 (v) 15625
(vi)13824 (vii) 110592 (viii) 46656 (ix) 175616
(x) 91125
Ans
$\begin{array}{l}\text{(i) The prime factorisation of 64 is as follows:}\\ \begin{array}{cc}2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$
$\begin{array}{l}64=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\\ \\ \therefore \sqrt[3]{64}=2\times 2=4\end{array}$
$\begin{array}{l}\text{(ii) The prime factorisation of 512 is as follows:}\\ \begin{array}{cc}2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$
$\begin{array}{l}512=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\\ \\ \therefore \sqrt[3]{512}=2\times 2\times 2=8\end{array}$
$\begin{array}{l}\text{(iii) The prime factorisation of 10648 is as follows:}\\ \begin{array}{cc}2& 10648\\ 2& 5324\\ 2& 2662\\ 11& 1331\\ 11& 121\\ 11& 11\\ & 1\end{array}\end{array}$
$\begin{array}{l}10648=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{11\times 11\times 11}\\ \\ \therefore \sqrt[3]{10648}=2\times 11=22\end{array}$
$\begin{array}{l}\text{(iv)The prime factorisation of 27000 is as follows:}\\ \begin{array}{cc}2& 27000\\ 2& 13500\\ 2& 6750\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$
$\begin{array}{l}27000=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{3\times 3\times 3}\times \underset{\xaf}{5\times 5\times 5}\\ \\ \therefore \sqrt[3]{27000}=2\times 3\times 5=30\end{array}$
$\begin{array}{l}\text{(v) The prime factorisation of 15625 is as follows:}\\ \begin{array}{cc}5& 15625\\ 5& 3125\\ 5& 625\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$
$\begin{array}{l}15625=\underset{\xaf}{5\times 5\times 5}\times \underset{\xaf}{5\times 5\times 5}\\ \\ \therefore \sqrt[3]{15625}=5\times 5=25\end{array}$
$\begin{array}{l}\text{(vi) The prime factorisation of 13824 is as follows:}\\ \begin{array}{cc}2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$
$\begin{array}{l}13824=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{3\times 3\times 3}\\ \\ \therefore \sqrt[3]{13824}=2\times 2\times 2\times 3=24\end{array}$
$\begin{array}{l}\text{(vii) The prime factorisation of 110592 is as follows:}\\ \\ \begin{array}{cc}2& 110592\\ 2& 55296\\ 2& 27648\\ 2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$
$\begin{array}{l}110592=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{3\times 3\times 3}\\ \\ \therefore \sqrt[3]{110592}=2\times 2\times 2\times 2\times 3=48\end{array}$
$\begin{array}{l}\text{(viii) The prime factorisation of 46656 is as follows:}\\ \\ \begin{array}{cc}2& 46656\\ 2& 23328\\ 2& 11664\\ 2& 5832\\ 2& 2916\\ 2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$
$\begin{array}{l}46656=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{3\times 3\times 3}\times \underset{\xaf}{3\times 3\times 3}\\ \\ \therefore \sqrt[3]{46656}=2\times 2\times 3\times 3=36\end{array}$
$\begin{array}{l}\text{(ix) The prime factorisation of 175616 is as follows:}\\ \\ \begin{array}{cc}2& 175616\\ 2& 87808\\ 2& 43904\\ 2& 21952\\ 2& 10976\\ 2& 5488\\ 2& 2744\\ 2& 1372\\ 2& 686\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\end{array}$
$\begin{array}{l}175616=\underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{2\times 2\times 2}\times \underset{\xaf}{7\times 7\times 7}\\ \\ \therefore \sqrt[3]{175616}=2\times 2\times 2\times 7=56\end{array}$
$\begin{array}{l}\text{(x) The prime factorisation of 91125 is as follows:}\\ \\ \begin{array}{cc}3& 91125\\ 3& 30375\\ 3& 10125\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$
$\begin{array}{l}91125=\underset{\xaf}{3\times 3\times 3}\times \underset{\xaf}{3\times 3\times 3}\times \underset{\xaf}{5\times 5\times 5}\\ \\ \therefore \sqrt[3]{91125}=3\times 3\times 5=45\end{array}$
Q.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.
Ans
 False.
Reason: The cube of any odd number is an odd number because when we find the cube of any odd number then we are multiplying its unit’s digit three times and the unit’s digit of any odd number is also an odd number. Therefore, the product will again be an odd number.
For example: The cube of 9 (i.e., an odd number) is 729, which is again an odd number.
 True.
Reason: A perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3.
For example: The cube of 10 is 1000 and there are 3 zeroes at the end of it.
The cube of 100 is 1000000 and there are 6 zeroes at the end of it.
 False.
Reason: It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.
For example: the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.
 False.
Reason: The cubes of all the numbers having their unit’s digit as 2 will end with 8.
For example: The cube of 22 is 10648 and the cube of 32 is 32768.
 False.
Reason: The smallest twodigit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.
 False.
Reason: The largest twodigit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any twodigit number cannot have 7 or more digits in it.
(vii)True
Reason: As the cube of 1 and 2 are 1 and 8 respectively.
Q.3 You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Ans
Firstly, we will make groups of three digits starting from the rightmost digit of the number 1331.
There are 2 groups, 1 and 331, in it.
Consider the first group 331.
The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1.
Now we will take the second group i.e., 1.
The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the one’s place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.
$\text{Hence,}\sqrt[3]{1331}=11$ $\sqrt[3]{4913}$
First make groups of three digits starting from the rightmost digit of 4913.
The groups are 4 and 913.
Consider the first group 913.
The number 913 ends with 3. We know that if a perfect cube number ends with 3, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7.
Now, take the second group i.e., 4.
We know that, 1^{3} = 1 and 2^{3} = 8
Also, 1 < 4 < 8
Therefore, we take the one’s place, of the smaller number 1 as the ten’s place of the required cube root.
$\text{Hence,}\sqrt[3]{4913}=17$
Now,
$\sqrt[3]{12167}$
First make groups of three digits starting from the rightmost digit of 12167.
The groups are 12 and 167.
Consider the first group 167.
167 end with 7. We know that if a perfect cube number ends with 7, then its cube root will have its unit’s digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3.
Take the second group i.e., 12.
We know that, 2^{3} = 8 and 3^{3} = 27
Also, 8 < 12 < 27
2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root.
$\text{Hence,}\sqrt[3]{12167}=23$ $\text{}\sqrt[3]{32768}$
We will make groups of three digits starting from the rightmost digit of the number 32768.
The groups are 32 and 768.
Consider the first group 768.
768 end with 8. We know that if a perfect cube number ends with 8, then its cube root will have its unit’s digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2.
Taking the other group i.e., 32,
We know that, 3^{3} = 27 and 4^{3} = 64
Also, 27 < 32 < 64
3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.
$\text{Hence,}\sqrt[3]{32768}=32$
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