NCERT Solutions for Class 8 Maths Chapter 2 - Power Play

NCERT Solutions for Class 8 Maths Chapter 2: Power Play provides students with a clear, concise understanding of powers and exponents. The goal of this chapter is to provide students with a convenient method for representing repetitive multiplication and large values easily.

This will develop students' understanding of power, build their numerical skills and logical reasoning abilities, and prepare students for the study of topics in advanced mathematics such as algebraic expressions, scientific notation, and the laws of exponents in subsequent years.

Each NCERT Solution contains a detailed description and a procedure for solution development to aid students in preparing for their board examinations.

1.0NCERT Solutions for Class 8 Maths Chapter 2 PDF Download

Students can download the Power Play Class 8 PDF with answers from the table below. The NCERT Solutions for Class 8 Maths Chapter 2 – Power Play PDF are meticulously prepared by ALLEN’s expert faculty members, who possess in-depth subject knowledge and a strong understanding of the CBSE curriculum.

These solutions aim to simplify complex ideas, improve conceptual clarity, and help students perform well in school assessments and exams.

2.0Quick Insights About the Chapter - PowerPlay

This chapter introduces students to the concept of powers, which is an essential mathematical tool for simplifying repeated multiplication and understanding large numbers.

• Definition of powers and exponents and their notation
• Understanding the relationship between base and exponent
• Representation of repeated multiplication using powers
• Learning basic laws of exponents
• Simplifying numerical expressions involving powers
• Applying power concepts to solve mathematical problems
• By mastering this chapter, students develop strong analytical skills and a solid foundation for future mathematics learning.

3.0NCERT Solutions for Class 8 Maths Chapter 2 Power Play : All Exercises

4.0NCERT Questions with Solutions Class 8 Maths Chapter 2 - Power Play - Detailed Solutions

FIGURE IT OUT-01

1. Express the following in exponential form: (i) $6\times 6\times 6\times 6$ (ii) $y\times y$ (iii) $\mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}$ (iv) $5\times 5\times 7\times 7\times 7$ (v) $2\times 2\times a\times a$ (vi) $a\times a\times a\times c\times c\times c\times c\times d$

Sol. (i) $6\times 6\times 6\times 6=6^4$ (ii) $y\times y=y^2$ (iii) $\mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}={\mathrm{b}}^4$ (iv) $5\times 5\times 7\times 7\times 7=5^2\times 7^3$ (v) $2\times 2\times a\times a=2^2\times a^2$ (vi) $a\times a\times a\times c\times c\times c\times c\times d$ $=a^3\times c^4\times d^1$

2. Express each of the following as a product of powers of their prime factors in exponential form. (i) 648 (ii) 405 (iii) 540 (iv) 3600

Sol. (i) Prime factors of 648 $=2\times 2\times 2\times 3\times 3\times 3\times 3$ Exponential form of $648=2^3\times 3^4$

(ii) Prime factors of 405 $=3\times 3\times 3\times 3\times 5$ Exponential form of 405=34 × 5

(iii) Prime factors of 540 $=2\times 2\times 3\times 3\times 3\times 5$ Exponential form of 540 $=2^2\times 3^3\times 5$

(iv) Prime factors of 3600 $=2\times 2\times 2\times 2\times 3\times 3\times 5\times 5$

Exponential form of 3600 $=2^4\times 3^2\times 5^2$

3. Write the numerical value of each of the following: (i) $2\times 10^3$ (ii) $7^2\times 2^3$ (iii) $3\times 4^4$ (iv) $(-3{)}^2\times (-5{)}^2$ (v) $3^2\times 10^4$ (vi) $(-2{)}^5\times (-10{)}^6$

Sol. (i) $2\times 10^3=2\times 10\times 10\times 10$ $\begin{array}{rl}& =2\times 1000\\ & =2000\end{array}$ (ii) $7^2\times 2^3=7\times 7\times 2\times 2\times 2$ $\begin{array}{rl}& =49\times 8\\ & =392\end{array}$ (iii) $3\times 4^4=3\times 4\times 4\times 4\times 4$ $\begin{array}{rl}& =3\times 256\\ & =768\end{array}$ (iv) $(-3{)}^2\times (-5{)}^2=-3\times -3\times -5\times -5$ $\begin{array}{rl}& =9\times 25\\ & =225\end{array}$ Negative numbers raised to even powers become positive. (v) $3^2\times 10^4=3\times 3\times 10\times 10\times 10\times 10$ $\begin{array}{rl}& =9\times 10,000\\ & =90,000\end{array}$ (vi) $(-2{)}^5\times (-10{)}^6$ $\begin{array}{rl}=& (-2)\times (-2)\times (-2)\times (-2)\times (-2)\times \\ & (-10)\times (-10)\times (-10)\times (-10)\times \\ & (-10)\times (-10)\\ =& (-32)\times (10,00,000)\\ =& -3,20,00,000\end{array}$ Odd power of a negative number remains negative, even power becomes positive.

FIGURE IT OUT-02

1. Find out the units digit in the value of $2^{224}÷4^{32}$ ? [Hint: $4=2^2$ ]

Sol. The expression is $2^{224}÷4^{32}$. We can rewrite the base 4 as $2^2$. $4^{32}={\left(2^2\right)}^{32}$ Using the exponent rule ${\left(a^m\right)}^n=a^{m\times n}$ : $4^{32}=2^{2\times 32}=2^{64}$ Now substitute this back into the division: $2^{224}÷4^{32}=\frac{2^{224}}{2^{64}}$ Using the exponent rule $\frac{a^m}{a^n}=a^{m-n}$ : $\frac{2^{224}}{2^{64}}=2^{224-64}=2^{160}$ The units digits of powers of 2 follow a repeating cycle of length 4 as $2,4,8,6$ $\left(\because 2^1=2,2^2=4,2^3=8,2^4=16,2^5=32\right)$ To find the units digit of $2^{160}$, we need to find the remainder of the exponent (160) when divided by the cycle length (4). Remainder $=160÷4$ We can calculate this: $160÷4=40$ with a remainder of 0 When the remainder is 0 , the units digit is the last digit in the cycle, which corresponds to the units digit of 24 . The last digit in the cycle $(2,4,8,6)$ is 6 . Therefore, the units digit of $2^{160}$ is 6 . The units digit in the value of $2^{224}÷4^{32}$ is 6 .

2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days? Sol. Start with 1 container having 5 bottles, 1 new container is added every day, so after 40 days. Number of containers after 40 days $=40$ containers × 5 bottles each $=200$ bottles

3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers. (i) $64^3$ (ii) $192^8$ (iii) $32^{-5}$ Sol.

• 4. Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True', or ‘Never True’. Explain your reasoning. (i) Cube numbers are also square numbers. (ii) Fourth powers are also square numbers. (iii) The fifth power of a number is divisible by the cube of that number. (iv) The product of two cube numbers is a cube number. (v) ${\mathrm{q}}^{46}$ is both a $4^{\text{th }}$ power and a $6^{\text{th }}$ power ( $q$ is a prime number).
  
  Sol. (i) Cube numbers are also square numbers only sometimes true reason: (a) $64=4^3=8^2\to$ both cube and square (b) $8=2^3\to$ not a square. A number must be both a square and a cube, i.e., a sixth power, to satisfy both. Not all cubes are sixth powers.
  (ii) Fourth powers are also square numbers - Always True Reason: Any fourth power is the form of $a^4={\left(a^2\right)}^2$, which is clearly a square. Fourth powers are squares because squaring a square gives a fourth power.
  (iii) The fifth power of a number is divisible by the cube of that number. Always True Reason: $a^5÷a^3=a^{5-3}=a^2$, which is valid for any $\mathrm{a}\ne 0$. The fifth power contains at least three powers of the base, so it's divisible by its cube.
  (iv) The product of two cube numbers is a cube number. - Always True Reason: The product of two cubes is also a cube, just raise the product of their bases to the third power.
  (v) ${\mathrm{q}}^{46}$ is both a $4^{\text{th }}$ power and a $6^{\text{th }}$ power ( $q$ is a prime number)- Never True Reason: 46 is not divisible by 4 or 6 → can't be a $4^{\text{th }}$ or $6^{\text{th }}$ power.
• 5. Simplify and write these in the exponential form. (i) $10^{-2}\times 10^{-5}$ (ii) $5^7÷5^4$ (iii) $9^{-7}÷9^4$ (iv) ${\left(13^{-2}\right)}^{-3}$ (v) ${\mathrm{m}}^5{\mathrm{n}}^{12}(\mathrm{mn}{)}^9$
• Sol. (i) $10^{-2}\times 10^{-5}=10^{-2-5}=10^{-7}$ $\left({\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right)$ (ii) $5^7÷5^4=5^{7-4}=5^3$ $\left(a^m÷a^n=a^{m-n}\right)$ (iii) $9^{-7}÷9^4=9^{-7-4}=9^{-11}$ $\left({\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right)$ (iv) ${\left(13^{-2}\right)}^{-3}=(13{)}^{-2\times (-3)}=(13{)}^6$ $\left[{\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right]$ (v) $m^5{n}^{12}(mn{)}^9=m^5{n}^{12}{m}^9{n}^9$ $\begin{array}{rl}& =m^{5+9}\cdot n^{12+9}\\ & \\ & =m^{14}{n}^{21}\\ & \left[{\left(a^m\right)}^n=a^{mn}\&a^m\times a^n=a^{m+n}\right]\end{array}$
  
  6. If $12^2=144$, what is (i) $(1.2{)}^2$ (ii) $(0.12{)}^2$ (iii) $(0.012{)}^2$ (iv) $120^2$
  Sol. (i) $(1.2{)}^2={\left(\frac{12}{10}\right)}^2=\frac{144}{100}=1.44$ (ii) $(0.12{)}^2={\left(\frac{12}{100}\right)}^2=\frac{144}{10000}=0.0144$ (iii) $(0.012{)}^2={\left(\frac{12}{1000}\right)}^2=\frac{144}{1000000}$ $\text{ = }0.000144$ (iv) $120^2=(12\times 10{)}^2$ $=144\times 100=14400$
  
  7. Circle the numbers that are the same

• $2^4\times 3^6$
• $6^4\times 3^2$
• $6^{10}$
• $18^2\times 6^2$
• $6^{24}$
  Sol. $\begin{array}{rl}& 6^4\times 3^2=(2\times 3{)}^4\times 3^2\\ & =2^4\times 3^4\times 3^2=2^4\times 3^6\\ & 6^{10}=(2\times 3{)}^{10}\\ & 18^2\times 6^2={\left(3^2\times 2\right)}^2\times (2\times 3{)}^2\\ & =3^4\times 2^2\times 2^2\times 3^2=24\times 3^6\\ & 6^{24}=(2\times 3{)}^{24}=2^{24}\times 3^{24}\end{array}$
  
  8. Identify the greater number in each of the following: (i) $4^3$ or $3^4$ (ii) $2^8$ or $8^2$ (iii) $100^2$ or $2^{100}$ Sol.
  (i) $4^3=64;3^4=81$ $81>64$ $\therefore 3^4>4^3$ (ii) $2^8=256;8^2={\left(2^3\right)}^2=2^6=64$ $256>64$ $\therefore 2^8>8^2$ (iii) $100^2=10,000$ $2^{100}\sim 1.27\times 10^{30}$ A number with 30 zeros is much larger than 10,000. $\therefore 2^{100}>100^2$
  
  9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0 9, how many digits should the code consist of? Sol.
  Total no. of packets produced in a year $=8.5$ billion ∴ Number of codes required $=8.5\times 10^9$ codes For ID digits to be taken from 0 to 9 . So, each digit has 10 choices. To make code with n digits, the total number of possible codes $=10^{\mathrm{n}}$ $\begin{array}{ll}\therefore & 10^n\ge 8.5\times 10^9\\ & 10^9=1,00,00,00,000\end{array}$ The smallest value of $n$ that satisfies $10^{\mathrm{n}}\ge 8.5\times 10^9$ is 10 . Hence number of digits the code should consist of $10^{10}$.
  
  10. 64 is a square number ( $8^2$ ) and a cube number ( $4^3$ ). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
  
  Sol.

Yes, other numbers are both squares and cubes. General Rule: A number is both a perfect square and a perfect cube if and only if it is a perfect sixth power, i.e., it can be written as ${\mathrm{x}}^6$ for some integer x.

11. A digital locker has an alphanumeric (it can have both digits and letters) pass code of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?
    Sol: Number of letters (A-Z) = 26 Number of digits (0-9) = 10 So, each character in the passcode can be any of the 36 alphanumeric characters. Also, each of the 5 positions has 36 options. Total codes $=36^5=6,04,66,176$ Hence number of possible 5-character alphanumeric passcodes is 6,04,66,176.
12. The worldwide population of sheep (2024) is about $10^9$, and that of goats is also about the same. What is the total population of sheep and goats? (i) $20^9$ (ii) $10^{11}$ (iii) $10^{10}$ (iv) $10^{18}$ (v) $2\times 10^9$ (vi) $10^9+10^9$ Sol. Sheep population $=10^9$ Goat population $=10^9$ Total population of sheep and goats = $10^9+10^9=2\times 10^9$
13. Calculate and write the answer in scientific notation: (i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing. (ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees. (iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world. (iv) Total time spent eating in a lifetime in seconds.
    Sol.
    (i) World population $=8.2$ billion $=8.2\times 10^9$ Pieces of clothing $=30$ pieces per person Total pieces of clothing $=8.2\times 10^9\times 30$ $=246\times 10^9$ $=2.46\times 10^{11}$ pieces of clothing
    (ii) No. of bee colonies = 100 million $=1\times 10^8$ No. of bee per colony $=50,000$ $=5\times 10^4$ Total no. of bees $=1\times 10^8\times 5\times 10^4$ $=5\times 10^{8+4}$ $=5\times 10^{12}$ honeybees
    (iii) No. of bacterial cells per human body $=38$ trillion $=3.8\times 10^{13}$ World population (approx.) $=8.2$ billion $=8.2\times 10^9$ Total bacterial population $=3.8\times 10^{13}\times 8.2\times 10^9$ $=31.16\times 10^{13+9}$ $=31.16\times 10^{22}$
    (iv) Let average eating time per day = 1.5 hours In seconds $=1.5\times 60\times 60=5400$ $=5.4\times 10^3$ and an average person's lifetime (approx.) = 70 years In seconds $=70\times 365\times 24\times 60\times 60$ = 161,148,960,000 $=1.61\times 10^{11}$ Total time spent in eating $=5.4\times 10^3\times 1.61\times 10^{11}$ $=8.694\times 10^{3+11}$ $=8.7\times 10^{14}$ seconds
14. What was the date 1 arab/ 1 billion seconds ago? Sol.
    1 arab / 1 billion = $10^9$ minutes $=1,000,000,000/60$ hours $=1,000,000,000/(60\times 60)$ days $=1,000,000,000/(60\times 60\times 24)$ years $=1,000,000,000/(365\times 60\times 60\times 24)$ Now, we go back to the calendar. So, it is approx. 31 years, 8 months, and 15 days. Let today's date = 1 Jan, 2026 After 31 years and 8 months and 15 days before the date was $24{}^{\text{th }}$ April, 1994 (approx.)

5.0
Benefits of NCERT Solutions Class 8 Maths Chapter 2

Providing students with an understanding of powers and exponents through NCERT Solutions for Class 8 Maths Chapter 2 – Power Play is a vital part of their middle school math education.

1. Improving the Understanding of Powers and Exponents - The explanations of powers are easy to follow and provide students with a structured approach that allows for the understanding of repeated multiplication.
2. Aligning Solutions with NCERT and CBSE Guidelines - All the solutions will adhere to the most recent NCERT syllabus and the CBSE exam format, allowing for exam relevant preparation.
3. Enhancing Accurate Calculations - By providing step-by-step explanations, students are able to apply exponent rules accurately and avoid many typical mistakes.
4. Laying the Groundwork for Advanced Mathematics - A strong understanding of powers will provide a solid foundation for students to learn about algebra, scientific notation, and other higher-level mathematical topics.
5. Improved Mathematical Problem-Solving Skills - By working through comprehensive NCERT questions, students will develop strong reasoning skills and numerical efficiency.
6. Effective Self-Study and Revision - The comprehensive explanations of all NCERT Solutions provide a means for independent studying and revising material before examinations by providing detailed solutions to the most frequently asked questions.