NCERT Solutions Class 8 Maths Chapter 02 Power Play

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Question 1. Express the following in exponential form: (i) 6 × 6 × 6 × 6 (ii) y × y (iii) b × b × b × b (iv) 5 × 5 × 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d
Answer:
(i) \( 6^4 \)
(ii) \( y^2 \)
(iii) \( b^4 \)
(iv) \( 5^2 \times 7^3 \)
(v) \( 2^2 \times a^2 \)
(vi) \( a^3 \times c^4 \times d \)
In simple words: When you multiply the same number many times, you can write it as a power. The number tells you how many times to multiply.

Exam Tip: Count how many times each base appears. That count becomes the exponent. Write bases in order (numbers first, then letters).

Question 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
Answer:
(i) 648: First divide by prime factors. 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3, which simplifies to \( 2^3 \times 3^4 \)
(ii) 405: Dividing by prime factors gives 405 = 3 × 3 × 3 × 3 × 5, which simplifies to \( 3^4 \times 5 \)
(iii) 540: Breaking it down yields 540 = 2 × 2 × 3 × 3 × 3 × 5, which simplifies to \( 2^2 \times 3^3 \times 5 \)
(iv) 3600: Prime factorization gives 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5, which simplifies to \( 2^4 \times 3^2 \times 5^2 \)
In simple words: Break each number into prime factors (small numbers that divide evenly). Then group the same factors together and write them as powers.

Exam Tip: Always divide by the smallest prime (2, then 3, then 5, etc.). Keep dividing until you get 1. This ensures you find all prime factors.

Question 3. Write the numerical value of each of the following: (i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴ (iv) (–3)² × (–5)² (v) 3² × 10⁴ (vi) (–2)⁵ × (–10)⁶
Answer:
(i) \( 2 \times 10^3 = 2 \times 1000 = 2000 \)
(ii) \( 7^2 \times 2^3 = 49 \times 8 = 392 \)
(iii) \( 3 \times 4^4 = 3 \times 256 = 768 \)
(iv) \( (-3)^2 \times (-5)^2 = 9 \times 25 = 225 \)
(v) \( 3^2 \times 10^4 = 9 \times 10,000 = 90,000 \)
(vi) \( (-2)^5 \times (-10)^6 = (-32) \times 1,000,000 = -32,000,000 \)
In simple words: Calculate the power first (like \( 2^3 = 8 \)). Then do the multiplication. A negative number raised to an even power gives a positive result.

Exam Tip: Work carefully with negative numbers. Remember: odd powers keep the negative sign, even powers make it positive.

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Question 1. Find out the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]
Answer: To solve this, rewrite everything using base 2. Since \( 4 = 2^2 \), we have \( 4^{32} = (2^2)^{32} = 2^{64} \). The expression becomes \( 2^{224} \div 2^{64} = 2^{224-64} = 2^{160} \). Now find the units digit of \( 2^{160} \). Powers of 2 follow a repeating pattern in their units digits: \( 2^1 = 2 \), \( 2^2 = 4 \), \( 2^3 = 8 \), \( 2^4 = 16 \) (units digit 6), then it repeats as 2, 4, 8, 6. The cycle length is 4. Divide the exponent by 4: \( 160 \div 4 = 40 \) with remainder 0. When the remainder is 0, the units digit is the 4th number in the cycle, which is 6.
In simple words: Powers of 2 repeat their last digit every 4 times: 2, 4, 8, 6, 2, 4, 8, 6... Since 160 divides evenly by 4, the units digit is 6.

Exam Tip: Always find the cycle pattern for the units digit first. Then use division to locate where your exponent falls in that cycle.

Question 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?
Answer: Each container holds 5 bottles. One new container gets added every single day. After 40 days have passed, the total count of containers is 40. Therefore, the total number of bottles is \( 40 \times 5 = 200 \).
In simple words: In 40 days, you get 40 containers. Each has 5 bottles. Multiply: 40 times 5 equals 200 bottles.

Exam Tip: Read carefully - the question asks about a specific time period. Count the containers first, then multiply by bottles per container.

Question 3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers: (i) 64³ (ii) 192⁸ (iii) 32⁻⁵
Answer:
(i) For \( 64^3 \): We know \( 64 = 2^6 \), so \( 64^3 = (2^6)^3 = 2^{18} \). Also, since \( 8 = 2^3 \), we can write \( 64^3 = 8^6 \). And since \( 4 = 2^2 \), we get \( 64^3 = 4^9 \).
(ii) For \( 192^8 \): After prime factorization, \( 192 = 2^6 \times 3 \), so \( 192^8 = (2^6 \times 3)^8 = 2^{48} \times 3^8 \). We can also write it as \( (64 \times 3)^8 = 64^8 \times 3^8 \) since \( 64 = 2^6 \). Or as \( (2^3 \times 2^3 \times 3)^8 = 2^{24} \times 2^{24} \times 3^8 = 2^{48} \times 3^8 \).
(iii) For \( 32^{-5} \): We know \( 32 = 2^5 \), so \( 32^{-5} = (2^5)^{-5} = 2^{-25} \). We can also express it as \( (2^3 \times 2^2)^{-5} \) since \( 32 = 2^5 \). Or as \( (4^3 \times 2^{-1})^{-5} \).
In simple words: Find different bases that equal the original number or its parts. Use exponent rules to rewrite in new ways. You can break numbers into smaller prime factors too.

Exam Tip: Remember that \( (a^m)^n = a^{mn} \) and \( a^m \times a^n = a^{m+n} \). Use these rules to create different forms of the same number.

Question 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) q⁴⁶ is both a 4th power and a 6th power (where q is a prime number).
Answer:
(i) Only Sometimes True - A cube number has the form \( n^3 \) while a square number has the form \( m^2 \). For a number to be both a perfect cube and a perfect square, it must have an exponent that is a multiple of 6 (the LCM of 2 and 3). The number 64 works: \( 64 = 4^3 = 8^2 \). However, 8 equals \( 2^3 \), which is not a square. So only certain cube numbers are also squares.
(ii) Always True - Any number raised to the 4th power can be written as \( n^4 = (n^2)^2 \), which means every 4th power is automatically a perfect square.
(iii) Always True - Any number \( a^5 \) can be factored as \( a^5 = a^3 \cdot a^2 \). Since \( a^3 \) is a factor of \( a^5 \), it always divides evenly (for all \( a \neq 0 \)).
(iv) Always True - If we have two cube numbers \( a^3 \) and \( b^3 \), their product is \( a^3 \times b^3 = (ab)^3 \), which is also a perfect cube.
(v) Never True - A number is both a 4th power and a 6th power only when its exponent is a multiple of LCM(4, 6) = 12. Since 46 = 2 × 23, it is not divisible by either 4 or 6. Therefore, \( q^{46} \) cannot be both a 4th and a 6th power.
In simple words: Check each claim by testing examples or using rules about exponents. See if one rule covers all cases (always), some cases (sometimes), or no cases (never).

Exam Tip: Always give a reason - cite a concrete example that works or explain why a statement must always/never work using exponent rules.

Question 5. Simplify and write these in the exponential form: (i) 10⁻² × 10⁻⁵ (ii) 5⁷ ÷ 5⁴ (iii) 9⁻⁷ ÷ 9⁴ (iv) (13⁻²)⁻³ (v) m⁵n¹²(mn)⁹
Answer:
(i) \( 10^{-2} \times 10^{-5} = 10^{-2-5} = 10^{-7} \)
(ii) \( 5^7 \div 5^4 = 5^{7-4} = 5^3 \)
(iii) \( 9^{-7} \div 9^4 = 9^{-7-4} = 9^{-11} \)
(iv) \( (13^{-2})^{-3} = 13^{(-2)(-3)} = 13^6 \)
(v) \( m^5 n^{12} (mn)^9 = m^5 n^{12} \times m^9 \times n^9 = m^{5+9} n^{12+9} = m^{14} n^{21} \)
In simple words: Use the rules: multiply same bases - add exponents. Divide same bases - subtract exponents. Power of a power - multiply exponents.

Exam Tip: Apply one rule at a time and be careful with negative exponents. When dividing, subtract the bottom exponent from the top exponent.

Question 6. If 12² = 144 what is (i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²
Answer:
(i) \( (1.2)^2 = (12/10)^2 = 12^2 / 10^2 = 144/100 = 1.44 \)
(ii) \( (0.12)^2 = (12/100)^2 = 12^2 / 100^2 = 144/10,000 = 0.0144 \)
(iii) \( (0.012)^2 = (12/1000)^2 = 12^2 / 1000^2 = 144/1,000,000 = 0.000144 \)
(iv) \( 120^2 = (12 \times 10)^2 = 12^2 \times 10^2 = 144 \times 100 = 14,400 \)
In simple words: Write the decimal number as a fraction. Square both numerator and denominator separately using the fact that 12² = 144.

Exam Tip: Look for the base number (12) in each expression. Rewrite using fractions, then apply the power rule \( (a/b)^n = a^n/b^n \).

Question 7. Circle the numbers that are the same - 2⁴ × 3⁶, 6⁴ × 3², 6¹⁰, 18² × 6², 6²⁴
Answer: Simplify each expression by breaking it into prime factors:
• \( 2^4 \times 3^6 \) - already in prime form
• \( 6^4 \times 3^2 = (2 \times 3)^4 \times 3^2 = 2^4 \times 3^4 \times 3^2 = 2^4 \times 3^6 \) - same as first
• \( 6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10} \) - different
• \( 18^2 \times 6^2 = (2 \times 3^2)^2 \times (2 \times 3)^2 = 2^2 \times 3^4 \times 2^2 \times 3^2 = 2^4 \times 3^6 \) - same as first
• \( 6^{24} = (2 \times 3)^{24} = 2^{24} \times 3^{24} \) - different
The matching numbers are: \( 2^4 \times 3^6 \), \( 6^4 \times 3^2 \), and \( 18^2 \times 6^2 \).
In simple words: Break each number into prime factors (2, 3, 5, etc.). Then compare - if the prime factors and their powers match, the numbers are the same.

Exam Tip: Always factor composite numbers completely before comparing. Use the rule \( (ab)^n = a^n b^n \) to expand expressions.

Question 8. Identify the greater number in each of the following - (i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰
Answer:
(i) \( 4^3 = 64 \) and \( 3^4 = 81 \). Since 81 is larger, \( 3^4 \) is greater.
(ii) \( 2^8 = 256 \) and \( 8^2 = 64 \). Since 256 is larger, \( 2^8 \) is greater.
(iii) \( 100^2 = 10,000 \) and \( 2^{100} = (2^{10})^{10} = 1024^{10} \). Since \( 2^{10} = 1024 \), the value \( 2^{100} \) is an enormously large number, making it far greater than 10,000.
In simple words: Calculate or simplify each power. Compare the results to find which is bigger. For very large exponents, recognize that they grow much faster than small bases.

Exam Tip: Don't assume the larger base always gives the larger result - the exponent matters greatly. For very large exponents, even a small base produces a huge number.

Question 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?
Answer: The dairy will produce 8.5 billion = \( 8.5 \times 10^9 \) packets. Each packet needs its own unique code made from digits 0-9. With n digits available, the number of unique codes that can be formed is \( (10)^n \). We need to find the smallest n where \( 10^n \geq 8.5 \times 10^9 \). Testing values: If \( n = 9 \), then \( 10^9 = 1,000,000,000 \), which is less than 8.5 billion. If \( n = 10 \), then \( 10^{10} = 10,000,000,000 \), which exceeds 8.5 billion. Therefore, the code must contain at least 10 digits, as \( 10^{10} \) is the smallest power of 10 that is greater than 8.5 billion.
In simple words: Find the smallest power of 10 that is bigger than 8.5 billion. Nine-digit codes give only 1 billion combinations, which is not enough. Ten-digit codes give 10 billion combinations, which is enough.

Exam Tip: Set up an inequality and test powers of 10 systematically. Remember that each additional digit multiplies the total combinations by 10.

Question 10. 64 is a square number (8²) and a cube number (4³). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
Answer: Many numbers are both perfect squares and perfect cubes. Some examples include: 1 = 1² = 1³, 64 = 8² = 4³, 729 = 27² = 9³, 4096 = 64² = 16³, and 15625 = 125² = 25³. To describe such numbers in general: A number is both a perfect square and a perfect cube if and only if it is a sixth power. This is because the exponent must be divisible by both 2 (to be a square) and 3 (to be a cube). The least common multiple of 2 and 3 is 6. Therefore, all numbers of the form \( n = k^6 \) (such as \( 1^6, 2^6, 3^6, \) and so on) are both perfect squares and perfect cubes.
In simple words: A number that is both a square and a cube must be a sixth power. That means its exponent in prime factorization must be divisible by 6.

Exam Tip: Look for the LCM of the exponents needed (2 for square, 3 for cube). If a number's exponent is a multiple of 6, it is both a square and a cube.

Question 11. A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW and 003AZ. How many such codes are possible?
Answer: Each passcode has exactly 5 characters. Each position can be filled with any alphanumeric character - either a letter (A-Z: 26 choices) or a digit (0-9: 10 choices). This gives a total of 26 + 10 = 36 choices per position. Since each of the 5 positions can independently be filled in 36 different ways, the total number of possible codes is \( 36 \times 36 \times 36 \times 36 \times 36 = 36^5 \). Computing this: \( 36^5 = 60,466,176 \). Therefore, 60,466,176 different codes are possible.
In simple words: Each spot in the code can be one of 36 choices (26 letters plus 10 digits). With 5 spots, multiply 36 by itself 5 times.

Exam Tip: Use the multiplication principle: if one position has m choices and another position has n choices, together they have m × n choices. Extend this to all 5 positions.

Question 12. The worldwide population of sheep (2024) is about 10⁹ and that of goats is also about the same. What is the total population of sheep and goats? (i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹
Answer: (v) and (vi) 2 × 10⁹ and 10⁹ + 10⁹
Given: Population of sheep = \( 10^9 \) and Population of goats = \( 10^9 \). The total population is \( 10^9 + 10^9 = 2 \times 10^9 \). Both expressions (v) and (vi) represent the same total and are correct.
In simple words: When you add the same power of 10 twice, you get two times that power. Both \( 2 \times 10^9 \) and \( 10^9 + 10^9 \) say the same thing.

Exam Tip: Remember that \( 10^n + 10^n = 2 \times 10^n \). Do not incorrectly combine exponents - you add or subtract the coefficients, not the powers.

Question 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.
Answer:
(i) World population \( \approx 8 \times 10^9 \). Each person has 30 clothes. Total clothes = \( 30 \times 8 \times 10^9 = 240 \times 10^9 = 2.4 \times 10^{11} \)
(ii) Number of bee colonies = 100 million = \( 1 \times 10^8 \). Bees per colony = 50,000 = \( 5 \times 10^4 \). Total bees = \( (1 \times 10^8) \times (5 \times 10^4) = 5 \times 10^{12} \)
(iii) World population \( \approx 8 \times 10^9 \). Bacterial cells per human = 38 trillion = \( 3.8 \times 10^{13} \). Total bacteria = \( (8 \times 10^9) \times (3.8 \times 10^{13}) = 30.4 \times 10^{22} = 3.04 \times 10^{23} \)
(iv) Average eating time per day = 1.5 hours = \( 1.5 \times 60 \times 60 = 5,400 \) seconds. Average lifespan = 70 years. Days in 70 years = \( 70 \times 365 = 25,550 \) days. Total eating time = \( 5,400 \times 25,550 = 137,970,000 \) seconds = \( 1.3797 \times 10^8 \) seconds
In simple words: Convert all numbers to scientific notation first. Multiply coefficients together and add exponents. Rearrange the result to standard scientific notation.

Exam Tip: Always express large numbers in scientific notation: a number between 1 and 10 multiplied by a power of 10. This makes calculations easier and answers clearer.

Question 14. What was the date 1 arab/1 billion seconds ago?
Answer: In the Indian number system, 1 arab = 1,00,00,00,000 = \( 10^9 \) = 1 billion seconds. Convert \( 10^9 \) seconds into years. Known facts: 1 minute = 60 seconds, 1 hour = 60 minutes = 3,600 seconds, 1 day = 24 hours = 86,400 seconds, 1 year \( \approx \) 365.25 days (accounting for leap years). Seconds in one year = \( 365.25 \times 24 \times 60 \times 60 = 31,557,600 \) seconds. Number of years in \( 10^9 \) seconds = \( 10^9 / 31,557,600 = 1,00,00,00,000 / 31,557,600 = 31.7 \) years. If today is July 29, 2025, go back 31.7 years - approximately 31 years and 8.5 months. Subtract 31 years to reach July 29, 1994. Then go back approximately 8.5 months to reach mid-November 1993. Therefore, 1 arab seconds ago, the date was around November 1993.
In simple words: Convert seconds into years by dividing by the number of seconds in a year. Then count backwards from today to find the date.

Exam Tip: Use 365.25 days per year to account for leap years. Break the fractional years into months for precision: 0.7 years is about 8-9 months.