NCERT Solutions Class 8 Maths Chapter 7

NCERT Solutions for Class 8 Mathematics Chapter 7: Cubes and Cubes Roots

Mathematics teaches us a lot of exciting concepts and theorems out of which some can be extremely complex  to comprehend. Therefore, students need to upgrade their preparation strategy and focus more on important concepts. Also, class 8 is an important stage for the students as it prepares them for upcoming opportunities. Thus, students need regular practice and a deeper understanding of the topics.

NCERT Class 8 Mathematics Chapter 7: Cubes and Cubes Roots explains the pattern of cubes. It tells that a perfect cube or cube numbers are obtained when a number is multiplied by taking it three times. . Students learn about the relations made concerning natural numbers with original numbers. In addition, this chapter helps students understand the calculation of cube and cube roots, interesting facts and patterns related to them.

Our Extramarks NCERT Solution Class 8 Mathematics Chapter 7 are available for the students on our website or app. It covers essential topics and helps in revising the chapter properly. The Extramarks expert team has solved the questions from chapter 7 of the NCERT textbook in a step-by-step manner. It helps students strengthen their concepts. There are various types of questions given in the solutions  includingMCQs, long-form questions and short-form questions as well as miscellaneous questions at the end of the solutions.

Extramarks is an online learning platform for the best study materials for lakhs of students. Further, they can access other NCERT solutions, study materials, revision notes and their doubts.

Students can also visit our Extramarks’ website for exam-related updates and news. Further, students can also refer to the other solutions such as NCERT solutions class 10, NCERT solutions class 11, and NCERT solutions class 12.

Key Topics Covered in NCERT Solutions for Class 8 Mathematics Chapter 7:

Our NCERT Solutions for Class 8 Mathematics Chapter 7 is available for the students on our website and app. It covers all essential topics from chapter 7- Cubes and Cube Roots. This chapter covers all the chapter end questions along with their comprehensive answers explained with proper illustrations.

The answers deal with essential topics to build the students’ foundation while dealing with complex numbers and cube roots. Step-wise explanations of answers in simple language make this chapter exciting and fun to learn.

Some of the key topics featured in NCERT Solutions for Class 8 Mathematics Chapter 7 are:

Introduction:

Students will be introduced to various types of cubes and their cube roots. There are many other interesting patterns of cubes as well.

Cubes:

The cube of a number is  the result obtained from multiplying  the same number thrice. Let’s say if x is a number, then x3 = (X) x (X) x (X). A natural number “n” is said to be a perfect cube if n = m3 for some natural number m.

Properties of Cube Numbers

There are a few key features of cubes explained in NCERT Solutions for Class 8 Mathematics Chapter 7

• A cube of an even natural number is even.
• A cube of an odd natural number is odd.
• A cube of a negative number is always negative.
• Sum of the first n natural numbers’ cubes is always equal to the square of their sum.
• Numbers of the cubes ending with the digits 0, 1, 2, 3, 4, and 5 always end with digits 0, 1, 8, 1, 7, and 5.
• Cubes of the numbers which ends with unit digit 2 will have a unit digit  8. Whereas cubes of the numbers which ends with unit digit 8 will have a unit digit 2 Cube of the numbers which end with digit 3 or 7 their cube will end with 3 or 7.
 Number Square Number Square 1 1 7 49 2 4 8 64 3 9 9 81 4 16 10 100 5 25 11 121 6 36 12 144

Some exciting patterns elaborated in NCERT Solutions for Class 8 Mathematics Chapter 7:

• Cubes are exponents which are raised to the power of 3. It is a mathematical expression or number expressed where the base and n are the power or exponent to which a is presented.
• Each  prime factor appears three times in its cubes.
• Factors can be grouped in triples.

Cube Root

Cube root of a number is the value which when multiplied by itself three times produces the original value. Therefore, the cube for the number x becomes x3 or x-cubed. For example, when we take the number 5, we have 5 x 5  x 5= 125. Based on the above-mentioned, 125 is called the cube of 5. On the other hand, the cube root of a number can be the reverse process. It is denoted by ∛. For example, five is called the cube root of the number 125.

Cube root is also denoted as ⅓ as the component of a number. Cube roots have many properties, and we have elaborated them in our Extramarks, NCERT Solutions for Class 8 Mathematics Chapter 7. Therefore, students can refer to our solution to get a hold of the formation of cube roots.

Perfect Cubes

A perfect cube is an integer which can be expressed as the product of three same or equal integers. More precisely, 125 is a perfect cube because the cube root of 5 = 5 x 5 x 5 = 125. On the other hand, 121 is not a perfect cube because no number can give the product 121 if multiplied three times. A perfect cube is always a number whose cube root is an integer.

To understand the application of perfect cubes, students can refer to NCERT Solutions for Class 8 Mathematics Chapter 7.

Prime Factorization method for the cube root

Students can calculate the cube root of a number with the help of the prime factorization method. First, start with the prime factors of the given number. The prime factors are usually denoted as a, b, c, d and so on. Next, find the most significant prime factor from these prime factors.

Properties of Perfect cube root:

Here are some properties of perfect cube roots which students can refer to NCERT Solutions for Class 8 Mathematics Chapter 7:

• The cube root of an integer is always a natural number.
• If n is an integer and the 1, 2, 3 and n are natural numbers then (1 + 2 + 3 + 4)3 = n3.
• Each positive number has two cube roots, one positive and another negative or zero, i.e. 1, 2, 3 and 4 and 5.
• The cube roots of natural numbers are divided in half equally. For example, the cube root of 8 is 2 (2 x 8 = 16). Note: The cube root of a natural number does not necessarily equal the corresponding positive rational number.
• Equivalent to the method, we can find the cube root of a natural number by multiplying any power at least twice to it, which can be expressed as follows: This method is known as the Algebraic factorization method.

NCERT Solutions for Class 8 Mathematics Chapter 7: Exercise & Solutions

NCERT Solutions for Class 8 Mathematics Chapter 7 are available for the students on our website. Along with this, students can refer to exercises and answer solutions for better performance in the exam. The solution covers all essential concepts and topics, such as cubes, and the smallest multiple is a perfect cube. Our subject matter experts prepare the exercise and solutions.

Class 8 Mathematics chapter 7 talks contain two exercises which deal with Cubes and Cube roots. The key topics covered in this chapter include cube root through the prime factorization method and some interesting patterns. In addition, the chapter helps students understand the different numbers and the relationship between cubes and square roots.

Click the links below for specific exercises questions and solutions.

• Exercise 7.1 Solutions 4 Questions (4 Short Answer Questions)
• Exercise 7.2 Solutions 3 Questions (2 Long Answer Questions,1 Short Answer Questions)

Along with this, students can also refer to other solutions for primary and secondary classes:

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

NCERT Exemplar for Class 8 Mathematics:

Mathematics can be challenging to handle as it involves various concepts and theories. To understand the concepts and formulas, students need to solve different equations of different  difficulty levels.

Further, it helps to boost confidence and build a solid foundation for the upcoming exams. Extramarks’ NCERT Solutions is a solved key to textual questions. By referring to this, a student can accomplish any desired learning goal like – studying a concept/chapter from scratch, revising before examination and interpretation/drawing/labelling of diagrams. As NCERT Solutions by Extramarks is available for every chapter of every subject, students can understand all the topics in an enhanced manner and ace their preparation. By accessing other NCERT study material like important questions from the chapter, frequently asked questions, MCQs, past years’ papers, expert tips & tricks to grasp answers on Extramarks app or website, students can leverage higher marks in their examination.

Through regular practice, students will be able to increase their understanding of the chapter and build a strong foundation on the concepts.

Key Features of NCERT Solutions for Class 8 Mathematics Chapter 7:

Below are the key features of our our NCERT Solutions for Class 8 Mathematics Chapter 7:

• It covers all the chapter end questions along with their comprehensive answers explained with solved examples.
• Through these, students will get to learn about the properties and applications of cube roots.
• The solution will help students gain confidence in the chapter and build a strong foundation.
• We have divided the solutions into small parts so that students can grasp them quickly.
• These solutions are designed as per the current CBSE syllabus, and hence, they are considered a reliable source of learning.
• Our expert team tests and checks all the solutions to ensure that they are error-free.

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

(i) 216 (ii) 128 (iii) 1000 (iv)100 (v) 46656

Ans

$\begin{array}{l}\text{(i)The prime factorisation of 216 is as follows:}\\ \begin{array}{cc}2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ \text{216}=\underset{¯}{\text{2}×\text{2}×\text{2}}×\underset{¯}{\text{3}×\text{3}×\text{3}}\\ \text{Since all the factors appears in a group of three,so 216}\\ \text{is a perfect cube}.\end{array}$

$\begin{array}{l}\text{(ii)The prime factorisation of 128 is as follows:}\\ \begin{array}{cc}2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\\ \\ 128=\underset{¯}{\text{2}×\text{2}×\text{2}}×\underset{¯}{\text{2}×\text{2}×\text{2}}×2\\ \text{Since 2 is not appearing in a group of three,so 128}\\ \text{is not a perfect cube}.\\ \\ \text{(iii)The prime factorisation of 1000 is as follows:}\\ \begin{array}{cc}2& 1000\\ 2& 500\\ 2& 250\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 1000=\underset{¯}{\text{2}×\text{2}×\text{2}}×\underset{¯}{\text{5}×\text{5}×\text{5}}\\ \text{Since 2 and 5 appears in a group of three,so 1000}\\ \text{is a perfect cube}.\end{array}$

$\begin{array}{l}\text{(iv)The prime factorisation of 100 is as follows:}\\ \begin{array}{cc}2& 100\\ 2& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 100=2×2×5×5\\ \text{Since 2 and 5 are not appearing in a group of three,so}\\ \text{100 is not a perfect cube}.\\ \\ \text{(v)The prime factorisation of 46656 is as follows:}\\ \begin{array}{cc}2& 46656\\ 2& 23328\\ 2& 11664\\ 2& 5832\\ 2& 2916\\ 2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}46656=\underset{¯}{\text{2}×\text{2}×\text{2}}×\underset{¯}{\text{2}×\text{2}×\text{2}}×\underset{¯}{3×3×3}×\underset{¯}{3×3×3}\\ \text{Since all the factors appears in a group of three,so}\\ \text{46656 is a perfect cube}.\end{array}$

Q.2 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100

Ans

(i)

$\begin{array}{l}\text{The prime factorisation of 243 is as follows:}\\ \begin{array}{cc}3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 243=\underset{¯}{3×3×3}×3×3\end{array}$

Here, two 3s are left which does not appear in a group of three. To make 243 a cube, one more 3 is required.
In that case, we have to multiply 243 by 3
i.e. 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729, which is a perfect cube.
Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii)

$\begin{array}{l}\text{The prime factorisation of 256 is as follows:}\\ \begin{array}{cc}2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\\ \\ 256=\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×2×2\end{array}$

Here, two 2s are left which does not appear in a group of three. To make 256 a cube, one more 2 is required.

If we multiply 256 by 2, we get

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512, which is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii)

$\begin{array}{l}\text{The prime factorisation of 72 is as follows:}\\ \begin{array}{cc}2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 72=\underset{¯}{2×2×2}×3×3\end{array}$

Here, two 3s are left which are not in a group of three. To make 72 a perfect cube, one more 3 is required.

Then, we obtain

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216, which is a perfect cube.

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

(iv)

$\begin{array}{l}\text{The prime factorisation of 675 is as follows:}\\ \begin{array}{cc}3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\text{675}=\underset{¯}{\text{3}×\text{3}×\text{3}}×\text{5}×\text{5}$

Here, two 5s are left which are not in a group of three. To make 675 a cube, one more 5 is required.

If we multiply 675 by 5, we get

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375, which is a perfect cube.

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

(v)

$\begin{array}{l}\text{The prime factorisation of 100 is as follows:}\\ \begin{array}{cc}2& 100\\ 2& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ \text{1}00=\text{2}×\text{2}×\text{5}×\text{5}\end{array}$

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 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000, which is a perfect cube

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

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

(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704

Ans

(i)

$\begin{array}{l}\text{The prime factorisation of 81 is as follows:}\\ \begin{array}{cc}3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

81 = 3 × 3 × 3 × 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.

Therefore, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.

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

(ii)

$\begin{array}{l}\text{The prime factorisation of 128 is as follows:}\\ \begin{array}{cc}2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, one 2 is left which is not in a group of three.

If we divide 128 by 2, then it will become a perfect cube.

Therefore, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

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

(iii)

$\text{The prime factorisation of 135 is as follows:}$

$\begin{array}{cc}3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$

135 = 3 × 3 × 3 × 5

Here, one 5 is left which is not in a group of three.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.

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

(iv)

$\text{The prime factorisation of 192 is as follows:}$

$\begin{array}{cc}2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}$

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, one 3 is left which is not in a group of three.

If we divide 192 by 3, then it will become a perfect cube.

Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

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

(v)

$\text{The prime factorisation of 704 is as follows:}$

$\begin{array}{cc}2& 704\\ 2& 352\\ 2& 176\\ 2& 88\\ 2& 44\\ 2& 22\\ 11& 11\\ & 1\end{array}$

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Here, one 11 is left which is not in a group of three.

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

Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

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

Ans

Volume of the cube of sides 5 cm,2 cm ,5 cm

=5 cm ×5 cm ×2 cm =5 ×5 ×2 cm3

Here, two 5s and one 2 are left which are not in a triplet. If we multiply this expression by 5 × 2 × 2 = 20, then it will become a perfect cube.

Thus, 5 × 5 × 2 × 5 × 2 × 2 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, 20 cuboids of 5 cm, 2 cm, and 5 cm are required to form a cube.

Q.5 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

$\begin{array}{l}\text{(i) The prime factorisation of 64 is as follows:}\\ \begin{array}{cc}2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$

$\begin{array}{l}64=\underset{¯}{2×2×2}×\underset{¯}{2×2×2}\\ \\ \therefore \sqrt[3]{64}=2×2=4\end{array}$

$\begin{array}{l}\text{(ii) The prime factorisation of 512 is as follows:}\\ \begin{array}{cc}2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\end{array}$

$\begin{array}{l}512=\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{2×2×2}\\ \\ \therefore \sqrt[3]{512}=2×2×2=8\end{array}$

$\begin{array}{l}\text{(iii) The prime factorisation of 10648 is as follows:}\\ \begin{array}{cc}2& 10648\\ 2& 5324\\ 2& 2662\\ 11& 1331\\ 11& 121\\ 11& 11\\ & 1\end{array}\end{array}$

$\begin{array}{l}10648=\underset{¯}{2×2×2}×\underset{¯}{11×11×11}\\ \\ \therefore \sqrt[3]{10648}=2×11=22\end{array}$

$\begin{array}{l}\text{(iv)The prime factorisation of 27000 is as follows:}\\ \begin{array}{cc}2& 27000\\ 2& 13500\\ 2& 6750\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\begin{array}{l}27000=\underset{¯}{2×2×2}×\underset{¯}{3×3×3}×\underset{¯}{5×5×5}\\ \\ \therefore \sqrt[3]{27000}=2×3×5=30\end{array}$

$\begin{array}{l}\text{(v) The prime factorisation of 15625 is as follows:}\\ \begin{array}{cc}5& 15625\\ 5& 3125\\ 5& 625\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\begin{array}{l}15625=\underset{¯}{5×5×5}×\underset{¯}{5×5×5}\\ \\ \therefore \sqrt[3]{15625}=5×5=25\end{array}$

$\begin{array}{l}\text{(vi) The prime factorisation of 13824 is as follows:}\\ \begin{array}{cc}2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}13824=\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{3×3×3}\\ \\ \therefore \sqrt[3]{13824}=2×2×2×3=24\end{array}$

$\begin{array}{l}\text{(vii) The prime factorisation of 110592 is as follows:}\\ \\ \begin{array}{cc}2& 110592\\ 2& 55296\\ 2& 27648\\ 2& 13824\\ 2& 6912\\ 2& 3456\\ 2& 1728\\ 2& 864\\ 2& 432\\ 2& 216\\ 2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}110592=\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{3×3×3}\\ \\ \therefore \sqrt[3]{110592}=2×2×2×2×3=48\end{array}$

$\begin{array}{l}\text{(viii) The prime factorisation of 46656 is as follows:}\\ \\ \begin{array}{cc}2& 46656\\ 2& 23328\\ 2& 11664\\ 2& 5832\\ 2& 2916\\ 2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$

$\begin{array}{l}46656=\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{3×3×3}×\underset{¯}{3×3×3}\\ \\ \therefore \sqrt[3]{46656}=2×2×3×3=36\end{array}$

$\begin{array}{l}\text{(ix) The prime factorisation of 175616 is as follows:}\\ \\ \begin{array}{cc}2& 175616\\ 2& 87808\\ 2& 43904\\ 2& 21952\\ 2& 10976\\ 2& 5488\\ 2& 2744\\ 2& 1372\\ 2& 686\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\end{array}$

$\begin{array}{l}175616=\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{2×2×2}×\underset{¯}{7×7×7}\\ \\ \therefore \sqrt[3]{175616}=2×2×2×7=56\end{array}$

$\begin{array}{l}\text{(x) The prime factorisation of 91125 is as follows:}\\ \\ \begin{array}{cc}3& 91125\\ 3& 30375\\ 3& 10125\\ 3& 3375\\ 3& 1125\\ 3& 375\\ 5& 125\\ 5& 25\\ 5& 5\\ & 1\end{array}\end{array}$

$\begin{array}{l}91125=\underset{¯}{3×3×3}×\underset{¯}{3×3×3}×\underset{¯}{5×5×5}\\ \\ \therefore \sqrt[3]{91125}=3×3×5=45\end{array}$

Q.6 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

1. 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.

1. 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.

1. 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.

1. 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.

1. False.

Reason: The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.

1. False.

Reason: The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit 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.7 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, 13 = 1 and 23 = 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, 23 = 8 and 33 = 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, 33 = 27 and 43 = 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.

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