--- title: "NCERT Solutions for Class 8 Maths Chapter 2 Power Play (2026-27)" url: https://www.swavid.com/maths/class/8/chapter/power-play/ncert-solutions dateModified: 2026-10-07T15:32:13+00:00 --- # NCERT Solutions for Class 8 Maths Chapter 2 Power Play (2026-27) This chapter's questions cover exponential notation, exponent laws, and scientific notation applied to real-world growth and large numbers. Free PDF (36 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-2-power-play-922222abe9.pdf ## An Impossible Venture! ### Question 1 *Activity* Take a sheet of paper, as large a sheet as you can find. Fold it once. Fold it again, and again. How many times can you fold it over and over? ... Try it with different types of paper and see what happens. **Solution** 1. This activity demonstrates the rapid growth of thickness when a sheet of paper is repeatedly folded in half. 2. Observation: It is physically difficult to fold a standard sheet of paper more than 7 or 8 times due to the thickness and stiffness increasing exponentially. **Answer:** Standard paper cannot be folded more than 7 or 8 times. ### Question 2 *2 marks · Very short answer* Say you can fold a sheet of paper as many times as you wish. What would its thickness be after 30 folds? Make a guess. **Solution** 1. The thickness of the paper doubles with each fold, growing exponentially from an initial thickness of $0.001\text{ cm}$. 2. After 30 folds, the thickness reaches approximately $10.7\text{ km}$, which is roughly the typical height at which commercial airplanes fly. **Answer:** About 10.7 km > Common mistake: Assuming linear addition instead of multiplicative doubling. ## Experiencing the Power Play ... ### Question 1 *2 marks · Very short answer* Now, what do you think the thickness would be after 30 folds? 45 folds? Make a guess. **Solution** 1. The thickness of a sheet of paper doubles with each fold. 2. After 30 folds, the thickness is about $10.7$ km, and after 45 folds, it becomes extremely large. **Answer:** After 30 folds, the thickness is about $10.7$ km, and after 45 folds, it is more than $7,00,000$ km. > Common mistake: Thinking the growth is linear instead of exponential/multiplicative. ### Question 2 *3 marks · Short answer* Fill the table below. **Solution** 1. The initial thickness is $0.001$ cm and doubles after each fold. 2. For 18 folds, the thickness is $\approx 262$ cm, for 19 folds it is $\approx 524$ cm, and for 20 folds it is $\approx 10.4$ m. 3. For 21 to 26 folds, the thickness continues to double successively. **Answer:** The completed table values are approximately $262$ cm, $524$ cm, and $10.4$ m for 18, 19, and 20 folds respectively. > Common mistake: Errors in doubling the decimal values correctly for higher folds. ## Exponential Notation and Operations ### Question 1 *1 mark · MCQ* Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number $v$. - $10v$ - $10 + v$ - $2 \times 10 \times v$ - $2^{10}$ - $2^{10}v$ - $10^2v$ **Solution** 1. The initial thickness is represented by $v$. 2. Since the thickness doubles after each fold, after 10 folds it is multiplied by $2$ ten times, which is $2^{10}$. 3. Thus, the thickness after 10 folds is given by $2^{10}v$. **Answer:** (v) $2^{10}v$ > Common mistake: Choosing $10v$ assuming linear growth instead of exponential growth. ### Question 2 *2 marks · Very short answer* What is $(-1)^5$? Is it positive or negative? What about $(-1)^{56}$? **Solution** 1. When $-1$ is multiplied an odd number of times, the product is negative, so $(-1)^5 = -1$ (negative). 2. When $-1$ is multiplied an even number of times, the product is positive, so $(-1)^{56} = 1$ (positive). **Answer:** $(-1)^5$ is negative and $(-1)^{56}$ is positive. > Common mistake: Confusing the sign of even and odd powers of negative numbers. ### Question 3 *2 marks · Very short answer* Is $(-2)^4 = 16$? Verify. **Solution** 1. Expand $(-2)^4$ as $(-2) \times (-2) \times (-2) \times (-2)$. 2. Multiplying gives $16$, so $(-2)^4 = 16$ is true. **Answer:** Yes, $(-2)^4 = 16$. > Common mistake: Writing the result as $-16$ by forgetting that an even power of a negative number is positive. ### Question 1 *3 marks · Short answer* Express the following in exponential form: (i) $6 \times 6 \times 6 \times 6$ (ii) $y \times y$ (iii) $b \times b \times b \times b$ (iv) $5 \times 5 \times 7 \times 7 \times 7$ (v) $2 \times 2 \times a \times a$ (vi) $a \times a \times a \times c \times c \times c \times c \times d$ **Part (i)** 1. Count the number of times 6 is multiplied, which is 4. 2. Write in exponential form as $6^4$. Answer (i): $6^4$ **Part (ii)** 1. Count the number of times $y$ is multiplied, which is 2. 2. Write in exponential form as $y^2$. Answer (ii): $y^2$ **Part (iii)** 1. Count the number of times $b$ is multiplied, which is 4. 2. Write in exponential form as $b^4$. Answer (iii): $b^4$ **Part (iv)** 1. Count 5 multiplied twice and 7 multiplied three times. 2. Write in exponential form as $5^2 \times 7^3$. Answer (iv): $5^2 \times 7^3$ **Part (v)** 1. Count 2 multiplied twice and $a$ multiplied twice. 2. Write in exponential form as $2^2 \times a^2$. Answer (v): $2^2 \times a^2$ **Part (vi)** 1. Count $a$ three times, $c$ four times, and $d$ once. 2. Write in exponential form as $a^3 \times c^4 \times d$. Answer (vi): $a^3 \times c^4 \times d$ **Answer:** Expressed in exponential form. > Common mistake: Writing multiplication instead of exponents. ### Question 2 *3 marks · Short answer* 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 **Part (i)** 1. Find the prime factorization of 648. 2. $648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 = 2^3 \times 3^4$. Answer (i): $2^3 \times 3^4$ **Part (ii)** 1. Find the prime factorization of 405. 2. $405 = 3 \times 3 \times 3 \times 3 \times 5 = 3^4 \times 5$. Answer (ii): $3^4 \times 5$ **Part (iii)** 1. Find the prime factorization of 540. 2. $540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^3 \times 5$. Answer (iii): $2^2 \times 3^3 \times 5$ **Part (iv)** 1. Find the prime factorization of 3600. 2. $3600 = 2^4 \times 3^2 \times 5^2$. Answer (iv): $2^4 \times 3^2 \times 5^2$ **Answer:** Expressed as product of powers of prime factors. > Common mistake: Arithmetic errors during prime factorization division steps. ### Question 3 *3 marks · Short answer* 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) $(-2)^5 \times (-10)^6$ **Part (i)** 1. Evaluate $2 \times 10^3 = 2 \times 1000$. Answer (i): $2000$ **Part (ii)** 1. Evaluate $7^2 \times 2^3 = 49 \times 8$. Answer (ii): $392$ **Part (iii)** 1. Evaluate $3 \times 4^4 = 3 \times 256$. Answer (iii): $768$ **Part (iv)** 1. Evaluate $(-3)^2 \times (-5)^2 = 9 \times 25$. Answer (iv): $225$ **Part (v)** 1. Evaluate $3^2 \times 10^4 = 9 \times 10000$. Answer (v): $90000$ **Part (vi)** 1. Evaluate $(-2)^5 \times (-10)^6 = -32 \times 1000000$. Answer (vi): $-32000000$ **Answer:** Calculated numerical values. > Common mistake: Forgetting to apply the negative sign for odd powers like $(-2)^5$. ### Question 7 *2 marks · Very short answer* How many rooms were there altogether? **Solution** 1. From the diagram, the number of rooms can be found by multiplying the number of choices at each stage: 3 daughters $\times$ 3 baskets $\times$ 3 keys $\times$ 3 rooms. 2. This gives $3^4 = 243$ rooms altogether. **Answer:** $243$ > Common mistake: Multiplying the base and exponent incorrectly or adding instead of multiplying. ### Question 8 *2 marks · Very short answer* How many diamonds were there in total? Can we find out by just one multiplication using the products above? **Solution** 1. The total number of diamonds is obtained by multiplying 3 seven times. 2. In exponential form, the number of diamonds is $3^7 = 2187$. **Answer:** $3^7$ or 2187 diamonds > Common mistake: Adding the numbers instead of multiplying. ### Question 9 *2 marks · Very short answer* $3^7$ can also be written as $3^2 \times 3^5$. Can you reason out why? **Solution** 1. Write $3^7$ as the product of 3 multiplied by itself 7 times: $3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$. 2. Group them into $(3 \times 3) \times (3 \times 3 \times 3 \times 3 \times 3)$, which gives $3^2 \times 3^5$. **Answer:** $3^7 = 3^2 \times 3^5$ by grouping factors. > Common mistake: Adding the bases instead of keeping the base same and adding exponents. ### Question 10 *2 marks · Very short answer* Write the product $p^4 \times p^6$ in exponential form. **Solution** 1. Expand $p^4 \times p^6$ as $(p \times p \times p \times p) \times (p \times p \times p \times p \times p \times p)$. 2. Combining all factors gives $p^{10}$. **Answer:** $p^{10}$ > Common mistake: Multiplying the exponents $4 \times 6$ instead of adding them. ### Question 11 *3 marks · Short answer* Use this observation to compute the following. (i) $2^9$ (ii) $5^7$ (iii) $4^6$ **Part (i)** 1. Express $2^9$ as $2^4 \times 2^5$. 2. Calculate $16 \times 32 = 512$. Answer (i): 512 **Part (ii)** 1. Express $5^7$ as $5^3 \times 5^4$. 2. Calculate $125 \times 625 = 78125$. Answer (ii): 78125 **Part (iii)** 1. Express $4^6$ as $4^3 \times 4^3$. 2. Calculate $64 \times 64 = 4096$. Answer (iii): 4096 **Answer:** (i) 512, (ii) 78125, (iii) 4096 > Common mistake: Applying incorrect exponent rules. ### Question 12 *2 marks · Very short answer* Is $2^{10}$ also equal to $(2^5)^2$? Write it as a product. **Solution** 1. Expand $2^{10}$ as $(2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2 \times 2)$. 2. This can be written as $(2^5) \times (2^5) = (2^5)^2$. **Answer:** Yes, $2^{10} = (2^5)^2$ > Common mistake: Confusing power of a power rule with addition of exponents. ### Question 13 *5 marks · Long answer* Write the following expressions as a power of a power in at least two different ways: (i) $8^6$ (ii) $7^{15}$ (iii) $9^{14}$ (iv) $5^8$ **Part (i)** 1. Express the base as a power or change the factors of the exponent. 2. $8^6 = (8^2)^3$ 3. $8^6 = (2^3)^6 = 2^{18}$ Answer (i): $(8^2)^3$ and $(2^3)^6$ **Part (ii)** 1. Express the exponent as a product of two numbers. 2. $7^{15} = (7^3)^5$ 3. $7^{15} = (7^5)^3$ Answer (ii): $(7^3)^5$ and $(7^5)^3$ **Part (iii)** 1. Express the base as a square or factor the exponent. 2. $9^{14} = (9^2)^7$ 3. $9^{14} = (3^2)^{14} = 3^{28}$ Answer (iii): $(9^2)^7$ and $(3^2)^{14}$ **Part (iv)** 1. Express the exponent as a product of two numbers. 2. $5^8 = (5^2)^4$ 3. $5^8 = (5^4)^2$ Answer (iv): $(5^2)^4$ and $(5^4)^2$ **Answer:** Expressed as a power of a power in two different ways each. > Common mistake: Multiplying the base instead of the exponents when applying the power of a power rule. ### Question 14 *2 marks · Very short answer* If the pond is completely covered by lotuses on the 30th day, how much of it is covered by lotuses on the 29th day? **Solution** 1. Since the number of lotuses doubles every day, the pond should be half covered on the day before the 30th day. 2. Therefore, the pond was half covered on the 29th day. **Answer:** The pond was half covered on the 29th day. > Common mistake: Dividing 30 by 2 to get the 15th day. ### Question 15 *2 marks · Very short answer* Write the number of lotuses (in exponential form) when the pond was — (i) fully covered (ii) half covered **Part (i)** 1. The number of lotuses doubles every day for 30 days starting with 1 lotus ($2^0$). 2. Thus, when fully covered on the 30th day, the number of lotuses is $2^{30}$. Answer (i): $2^{30}$ **Part (ii)** 1. Since the pond is half covered on the 29th day, the number of lotuses is half of the fully covered amount. 2. Half of $2^{30}$ is $2^{30} \div 2^1 = 2^{29}$. Answer (ii): $2^{29}$ **Answer:** (i) $2^{30}$, (ii) $2^{29}$ > Common mistake: Writing $30^2$ instead of $2^{30}$. ### Question 16 *2 marks · Very short answer* What if Damayanti had changed the order in which she placed the flowers in the lakes? How many lotuses would be there? **Solution** 1. Damayanti placed flowers in two different lakes sequentially, first doubling and then tripling. 2. Changing the order of multiplication of powers still yields the same total number of lotuses due to the commutative property of multiplication. **Answer:** The total number of lotuses would remain the same. > Common mistake: Assuming the total would change because the order of operations changed. ### Question 17 *2 marks · Very short answer* Can this product be expressed as an exponent $m^n$, where $m$ and $n$ are some counting numbers? **Solution** 1. By regrouping the bases and exponents, products of the form $a^n \times b^n$ can be combined. 2. Yes, it can be expressed as $(3 \times 2)^4 = 6^4$ using the rule $m^a \times n^a = (mn)^a$. **Answer:** Yes, it can be expressed as $6^4$. > Common mistake: Adding the bases instead of multiplying them inside the bracket. ### Question 18 *2 marks · Very short answer* Use this observation to compute the value of $2^5 \times 5^5$. **Solution** 1. Use the rule $m^a \times n^a = (mn)^a$. 2. $2^5 \times 5^5 = (2 \times 5)^5 = 10^5 = 1,00,000$. **Answer:** $1,00,000$ > Common mistake: Adding the bases or exponents incorrectly instead of combining them into $(mn)^a$. ### Question 19 *2 marks · Very short answer* Simplify $\frac{10^4}{5^4}$ and write it in exponential form. **Solution** 1. Using the rule $\frac{n^a}{m^a} = \left(\frac{n}{m}\right)^a$, write $\frac{10^4}{5^4} = \left(\frac{10}{5}\right)^4$. 2. Simplify the division inside the bracket to get $2^4$, which equals $16$. **Answer:** $2^4 = 16$ > Common mistake: Subtracting exponents instead of dividing the bases when the exponents are the same. ### Question 20 *2 marks · Very short answer* Estu has 4 dresses and 3 caps. How many different ways can Estu combine the dresses and caps? **Solution** 1. For each cap, Estu can choose any of the 4 dresses. 2. Multiply the number of dresses and caps to find the total combinations: $4 \times 3 = 12$ ways. **Answer:** $12$ ways > Common mistake: Adding the number of dresses and caps instead of multiplying. ### Question 21 *2 marks · Very short answer* Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up? **Solution** 1. Multiply the number of choices for each item: dresses, hats, and shoes. 2. Total ways = $7 \times 2 \times 3 = 42$ ways. **Answer:** $42$ ways > Common mistake: Adding the choices instead of multiplying them according to the fundamental counting principle. ### Question 22 *2 marks · Very short answer* Estu and Roxie came across a safe containing old stamps and coins that their great-grandfather had collected. It was secured with a 5-digit password. Since nobody knew the password, they had no option except to try every password until it opened. They were unlucky and the lock only opened with the last password, after they had tried all possible combinations. How many passwords did they end up checking? **Solution** 1. Each of the 5 digits in the password has 10 possible choices from 0 to 9. 2. Total passwords = $10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 1,00,000$ passwords. **Answer:** $1,00,000$ passwords > Common mistake: Multiplying by 5 instead of raising 10 to the power of 5. ### Question 23 *2 marks · Very short answer* How many passwords are possible with such a lock? **Solution** 1. Each of the 6 slots has 26 possible choices for the letters A to Z. 2. Total possible codes = $26 \times 26 \times 26 \times 26 \times 26 \times 26 = (26)^6$. **Answer:** $(26)^6$ > Common mistake: Using 10 digits instead of 26 letters for the slots. ### Question 24 *3 marks · Short answer* Think about how many combinations are possible in different contexts. Some examples are— (i) Pincodes of places in India (ii) Mobile numbers. (iii) Vehicle registration numbers. Try to find out how these numbers or codes are allotted/generated. **Solution** 1. Pincodes in India have 6 digits, where each digit has 10 options ($0$ to $9$), resulting in $10^6$ combinations. 2. Mobile numbers generally start with specific prefixes and consist of 10 digits, giving a large number of possible combinations. 3. Vehicle registration numbers consist of state codes, district codes, unique serial numbers, and series letters, generated systematically to ensure uniqueness. **Answer:** Codes and numbers are generated using fixed lengths and choices per position to ensure uniqueness. > Common mistake: Assuming all positions in pincodes or mobile numbers allow unrestricted digits. ## The Other Side of Powers ### Question 1 *2 marks · Very short answer* What is $2^{100} \div 2^{25}$ in powers of 2? **Solution** 1. Using the general form $n^a \div n^b = n^{a - b}$, we subtract the exponent of the divisor from the exponent of the dividend. 2. So, $2^{100} \div 2^{25} = 2^{100 - 25} = 2^{75}$. **Answer:** $2^{75}$ > Common mistake: Dividing the exponents instead of subtracting them. ### Question 2 *2 marks · Very short answer* Why can’t $n$ be 0? **Solution** 1. In the expression $n^a \div n^b = n^{a-b}$, the base $n$ appears in the denominator when written as a fraction. 2. If $n = 0$, division by zero occurs, which is not defined. **Answer:** If $n = 0$, we get division by zero which is not defined. > Common mistake: Forgetting that a denominator cannot be zero. ### Question 3 *2 marks · Very short answer* We have not covered the case when the exponent is 0; for example, what is $2^0$? **Solution** 1. Using the rule $n^a \div n^a = n^{a-a} = n^0$. 2. Also, any non-zero number divided by itself equals 1, so $2^0 = 2^{4-4} = 2^4 \div 2^4 = 1$. **Answer:** $2^0 = 1$ > Common mistake: Writing the answer as 0 instead of 1. ### Question 4 *2 marks · Very short answer* Can we write $10^3 = \frac{1}{10^{-3}}$? **Solution** 1. Evaluating the expression using the definition of negative exponents, $\frac{1}{10^{-3}} = 1 \div \frac{1}{10^3}$. 2. This simplifies to $1 \times 10^3 = 10^3$, so the statement is true. **Answer:** Yes, we can write $10^3 = \frac{1}{10^{-3}}$ > Common mistake: Confusing negative exponents with negative numbers. ### Question 5 *2 marks · Very short answer* We had required $a$ and $b$ to be counting numbers. Can $a$ and $b$ be any integers? Will the generalised forms still hold true? **Solution** 1. The generalised laws of exponents derived for counting numbers continue to hold true when the exponents are extended to any integers. 2. Thus, negative integers and zero satisfy the same exponent rules. **Answer:** Yes, the generalised forms still hold true for any integers. > Common mistake: Assuming exponent rules only apply to positive integers. ### Question 6 *3 marks · Short answer* Write equivalent forms of the following. (i) $2^{-4}$ (ii) $10^{-5}$ (iii) $(-7)^{-2}$ (iv) $(-5)^{-3}$ (v) $10^{-100}$ **Part (i)** 1. Using the property $n^{-a} = \frac{1}{n^a}$, we write $2^{-4}$ with a positive exponent. Answer (i): $\frac{1}{2^4}$ **Part (ii)** 1. Using the property $n^{-a} = \frac{1}{n^a}$, we write $10^{-5}$ with a positive exponent. Answer (ii): $\frac{1}{10^5}$ **Part (iii)** 1. Using the property $n^{-a} = \frac{1}{n^a}$, we write $(-7)^{-2}$ with a positive exponent. Answer (iii): $\frac{1}{(-7)^2}$ **Part (iv)** 1. Using the property $n^{-a} = \frac{1}{n^a}$, we write $(-5)^{-3}$ with a positive exponent. Answer (iv): $\frac{1}{(-5)^3}$ **Part (v)** 1. Using the property $n^{-a} = \frac{1}{n^a}$, we write $10^{-100}$ with a positive exponent. Answer (v): $\frac{1}{10^{100}}$ **Answer:** Equivalent forms written using positive exponents. > Common mistake: Dropping the negative sign on the base along with the exponent. ### Question 7 *3 marks · Short answer* Simplify and write the answers in exponential form. (i) $2^{-4} \times 2^7$ (ii) $3^2 \times 3^{-5} \times 3^6$ (iii) $p^3 \times p^{-10}$ (iv) $2^4 \times (-4)^{-2}$ (v) $8^p \times 8^q$ **Part (i)** 1. $2^{-4} \times 2^7 = 2^{-4 + 7}$ 2. = $2^3$ Answer (i): $2^3$ **Part (ii)** 1. $3^2 \times 3^{-5} \times 3^6 = 3^{2 + (-5) + 6}$ 2. = $3^3$ Answer (ii): $3^3$ **Part (iii)** 1. $p^3 \times p^{-10} = p^{3 + (-10)}$ 2. = $p^{-7}$ Answer (iii): $p^{-7}$ **Part (iv)** 1. $2^4 \times (-4)^{-2} = 2^4 \times (-(2^2))^{-2}$ 2. $= 2^4 \times (-1)^{-2} \times 2^{-4} = 1$ Answer (iv): $1$ **Part (v)** 1. $8^p \times 8^q = 8^{p + q}$ Answer (v): $8^{p+q}$ **Answer:** Simplified expressions in exponential form for all five parts. > Common mistake: Adding or subtracting the exponents incorrectly when they have opposite signs. ### Question 8 *2 marks · Very short answer* Can we say that 16384 ($4^7$) is 16 ($4^2$) times larger than 1,024 ($4^5$)? **Solution** 1. $4^7 \div 4^5 = 4^{7-5} = 4^2 = 16$ 2. Yes, it is 16 times larger. **Answer:** Yes, since $4^7 \div 4^5 = 4^2$. > Common mistake: Multiplying the numbers instead of subtracting exponents. ### Question 9 *2 marks · Very short answer* How many times larger than $4^{-2}$ is $4^2$? **Solution** 1. $4^2 \div 4^{-2} = 4^{2 - (-2)}$ 2. $= 4^{2+2} = 4^4$ **Answer:** $4^4$ times > Common mistake: Subtracting exponents incorrectly when the divisor has a negative power. ### Question 10 *3 marks · Short answer* Use the power line for 7 to answer the following questions. **Part (i)** 1. $2,401 \times 49 = 7^4 \times 7^2$ 2. $= 7^{4+2} = 7^6$ Answer (i): $7^6$ **Part (ii)** 1. $49^3 = (7^2)^3$ 2. $= 7^{2 \times 3} = 7^6$ Answer (ii): $7^6$ **Part (iii)** 1. $343 \times 2,401 = 7^3 \times 7^4$ 2. $= 7^{3+4} = 7^7$ Answer (iii): $7^7$ **Part (iv)** 1. $\frac{16,807}{49} = \frac{7^5}{7^2}$ 2. $= 7^{5-2} = 7^3$ Answer (iv): $7^3$ **Part (v)** 1. $\frac{7}{343} = \frac{7^1}{7^3}$ 2. $= 7^{1-3} = 7^{-2}$ Answer (v): $7^{-2}$ **Part (vi)** 1. $\frac{16,807}{8,23,543} = \frac{7^5}{7^7}$ 2. $= 7^{5-7} = 7^{-2}$ Answer (vi): $7^{-2}$ **Part (vii)** 1. $1,17,649 \times \frac{1}{343} = 7^6 \times 7^{-3}$ 2. $= 7^{6-3} = 7^3$ Answer (vii): $7^3$ **Part (viii)** 1. $\frac{1}{343} \times \frac{1}{343} = 7^{-3} \times 7^{-3}$ 2. $= 7^{-3+(-3)} = 7^{-6}$ Answer (viii): $7^{-6}$ **Answer:** Solved using the power line for 7. > Common mistake: Confusing multiplication of powers with division of powers. ## Powers of 10 ### Question 1 *3 marks · Short answer* Write these numbers in the same way: (i) 172, (ii) 5642, (iii) 6374. **Part (i)** 1. $172 = 100 + 70 + 2$ 2. $172 = (1 \times 10^2) + (7 \times 10^1) + (2 \times 10^0)$ Answer (i): $(1 \times 10^2) + (7 \times 10^1) + (2 \times 10^0)$ **Part (ii)** 1. $5642 = 5000 + 600 + 40 + 2$ 2. $5642 = (5 \times 10^3) + (6 \times 10^2) + (4 \times 10^1) + (2 \times 10^0)$ Answer (ii): $(5 \times 10^3) + (6 \times 10^2) + (4 \times 10^1) + (2 \times 10^0)$ **Part (iii)** 1. $6374 = 6000 + 300 + 70 + 4$ 2. $6374 = (6 \times 10^3) + (3 \times 10^2) + (7 \times 10^1) + (4 \times 10^0)$ Answer (iii): $(6 \times 10^3) + (3 \times 10^2) + (7 \times 10^1) + (4 \times 10^0)$ **Answer:** Expanded form using powers of 10 for the given numbers. > Common mistake: Writing powers of 10 incorrectly by mixing up the exponent with the digit value. ### Question 2 *2 marks · Very short answer* How can we write 561.903? **Solution** 1. Write the whole number part using powers of 10. 2. Write the decimal part using negative powers of 10. 3. $561.903 = (5 \times 10^2) + (6 \times 10^1) + (1 \times 10^0) + (9 \times 10^{-1}) + (0 \times 10^{-2}) + (3 \times 10^{-3})$ **Answer:** $(5 \times 10^2) + (6 \times 10^1) + (1 \times 10^0) + (9 \times 10^{-1}) + (0 \times 10^{-2}) + (3 \times 10^{-3})$ > Common mistake: Using positive exponents for digits after the decimal point. ## Scientific Notation ### Question 1 *3 marks · Short answer* Write the large-number facts we read just before in this form. **Solution** 1. The distance from the centre of our Milky Way galaxy to the Sun is $30,00,00,00,00,00,00,00,00,000\text{ m} = 3 \times 10^{21}\text{ m}$. 2. The number of stars in our galaxy is $1,00,00,00,00,000 = 10^{11}$. 3. The mass of the Earth is $59,76,00,00,00,00,00,00,00,00,00,000\text{ kg} = 5.976 \times 10^{24}\text{ kg}$. **Answer:** $3 \times 10^{21}\text{ m}$, $10^{11}$, and $5.976 \times 10^{24}\text{ kg}$ > Common mistake: Counting the number of zeroes incorrectly. ### Question 2 *2 marks · Very short answer* Can you say which of the three distances is the smallest? **Solution** 1. Compare the given distances: Sun to Earth is $1.496 \times 10^{11}\text{ m}$, Sun to Saturn is $1.4335 \times 10^{12}\text{ m}$, and Saturn to Uranus is $1.439 \times 10^{12}\text{ m}$. 2. Since $10^{11}$ is smaller than $10^{12}$, the distance between the Sun and the Earth is the smallest. **Answer:** Distance between the Sun and the Earth is the smallest. > Common mistake: Comparing only the coefficient without looking at the powers of 10. ### Question 3 *3 marks · Short answer* The number line below shows the distance between the Sun and Saturn ($1.4335 \times 10^{12}\text{ m}$). On the number line below, mark the relative position of the Earth. The distance between the Sun and the Earth is $1.496 \times 10^{11}\text{ m}$. **Solution** 1. Observe the distance between the Sun and Saturn as $1.4335 \times 10^{12}\text{ m}$ (or $14.335 \times 10^{11}\text{ m}$). 2. The distance between the Sun and the Earth is $1.496 \times 10^{11}\text{ m}$, which is approximately $1.5 \times 10^{11}\text{ m}$. 3. Mark the relative position of the Earth close to the Sun on the number line since $1.496 \times 10^{11}$ is much smaller than $1.4335 \times 10^{12}$. **Answer:** The Earth is marked close to the Sun on the number line. > Common mistake: Placing the Earth incorrectly by ignoring the exponent difference. ### Question 4 *3 marks · Short answer* Express the following numbers in standard form. (i) 59,853 (ii) 65,950 (iii) 34,30,000 (iv) 70,04,00,00,000 **Part (i)** 1. $59,853 = 5.9853 \times 10,000$ Answer (i): $5.9853 \times 10^4$ **Part (ii)** 1. $65,950 = 6.595 \times 10,000$ Answer (ii): $6.595 \times 10^4$ **Part (iii)** 1. $34,30,000 = 3.43 \times 10,00,000$ Answer (iii): $3.43 \times 10^6$ **Part (iv)** 1. $70,04,00,00,000 = 7.004 \times 10^{10}$ Answer (iv): $7.004 \times 10^{10}$ **Answer:** (i) $5.9853 \times 10^4$, (ii) $6.595 \times 10^4$, (iii) $3.43 \times 10^6$, (iv) $7.004 \times 10^{10}$ > Common mistake: Writing the coefficient greater than 10 or less than 1. ## Did You Ever Wonder? ### Question 1 *3 marks · Short answer* What would be the worth (in rupees) of the donated jaggery? What would be the worth (in rupees) of the donated wheat? **Solution** 1. The worth of donated jaggery is given by Roxie's weight in kg multiplied by the cost of 1 kg of jaggery. 2. The worth of donated wheat is given by Estu's weight in kg multiplied by the cost of 1 kg of wheat. 3. Assuming Roxie's weight to be $45\text{ kg}$ and the cost of $1\text{ kg}$ of jaggery to be $\text{₹}70$, the worth of jaggery is $45 \times 70 = \text{₹}3150$. 4. Assuming Estu's weight to be $50\text{ kg}$ and the cost of $1\text{ kg}$ of wheat to be $\text{₹}50$, the worth of wheat is $50 \times 50 = \text{₹}2500$. **Answer:** Worth of jaggery is $\text{₹}3150$ and worth of wheat is $\text{₹}2500$. > Common mistake: Forgetting to state assumptions for weight and cost clearly. ### Question 2 *3 marks · Short answer* Make necessary and reasonable assumptions for the unknowns and find the answers. Remember, Roxie is 13 years old and Estu is 11 years old. **Solution** 1. Identify the knowns: Roxie is $13$ years old and Estu is $11$ years old. 2. Make reasonable assumptions for their weights based on their age: assume Roxie weighs $45\text{ kg}$ and Estu weighs $40\text{ kg}$. 3. State the relationships and calculate the values using assumed prices or weights as required by the context. **Answer:** Weights are assumed as $45\text{ kg}$ for Roxie and $40\text{ kg}$ for Estu based on their ages. > Common mistake: Making unrealistic weight assumptions for the given age. ### Question 3 *3 marks · Short answer* Roxie wonders, “Instead of jaggery if we use 1-rupee coins, how many coins are needed to equal my weight?”. How can we find out? **Solution** 1. First, determine Roxie's weight in grams by multiplying her weight in kg by $1000$. 2. Find the weight of a single $1$-rupee coin by measuring or approximation. 3. Divide Roxie's total weight in grams by the weight of a single $1$-rupee coin to find the total number of coins required. **Answer:** Total coins = $\frac{\text{Roxie's weight in grams}}{\text{weight of one } 1\text{-rupee coin in grams}}$ > Common mistake: Mixing up units of grams and kilograms during division. ### Question 4 *2 marks · Very short answer* Would the number of coins be in hundreds, thousands, lakhs, crores, or even more? Make an instinctive guess. **Solution** 1. Make an instinctive guess by considering that a person's weight in grams is tens of thousands of grams, and a coin weighs a few grams. 2. The number of coins will be in the tens of thousands or lakhs. **Answer:** In lakhs > Common mistake: Guessing a number that is too small like hundreds. ### Question 5 *3 marks · Short answer* Find the answer by making necessary and reasonable assumptions and approximations for the unknowns. Remember, we are not looking for an exact answer but a reasonably close estimate. **Solution** 1. Assume Roxie's weight is $45\text{ kg}$, which is $45,000\text{ g}$. 2. Assume the weight of one $1$-rupee coin is approximately $3\text{ g}$. 3. Compute the total number of coins: $45,000 \div 3 = 15,000$ coins. **Answer:** Approximately $15,000$ coins > Common mistake: Incorrectly approximating the weight of a coin. ### Question 6 *2 marks · Very short answer* Estu asks, “What if we use 5-rupee coins or 10-rupee notes instead? How much money could it be?” **Solution** 1. If using $5$-rupee coins instead of $1$-rupee coins, the number of coins needed will be one-fifth of the $1$-rupee coins. 2. If using $10$-rupee notes, the number of pieces of money will be one-tenth of the $1$-rupee coins, reducing the total count. **Answer:** The number of pieces of money will decrease proportionally to the denomination value. > Common mistake: Multiplying instead of dividing when changing to higher denominations. ### Question 7 *3 marks · Short answer* Make an instinctive guess first. Then find out (make necessary and reasonable assumptions about the unknown details and find the answers). **Solution** 1. Guess: The number of coins would be in lakhs. 2. Assumption: Assume Roxie's weight is $45\text{ kg}$ and the thickness of a 1-rupee coin is about $1.5\text{ mm}$ with a mass of $3.5\text{ g}$. 3. Calculation: Number of coins = $\frac{45 \times 1000\text{ g}}{3.5\text{ g}} \approx 12,857$ coins. **Answer:** Approximately $13,000$ coins > Common mistake: Not stating assumptions clearly before performing the estimation calculation. ### Question 8 *3 marks · Short answer* How many people might benefit from each of these offerings in a year? Again, guess first before finding out. **Solution** 1. Guess: A few hundred people might benefit from the annual donations. 2. Assumption: Assume a student needs $10$ notebooks worth $\text{₹}500$ a year, and the total value donated equals Roxie's weight in jaggery worth $\text{₹}3150$. 3. Calculation: Number of people who can benefit from notebooks = $\frac{3150}{500} \approx 6$ people. **Answer:** Approximately $6$ people > Common mistake: Forgetting to state the assumed cost of items per person. ### Question 9 *2 marks · Very short answer* How long ago would they have started their journey? **Solution** 1. Assumption: Assume a person walks at an average speed of $4\text{ km/h}$ for $8$ hours a day, covering about $32\text{ km}$ per day. 2. Calculation: Time taken to walk $400\text{ km} = \frac{400}{32} \approx 12.5$ days ago. **Answer:** About 12 to 13 days ago > Common mistake: Assuming non-stop walking without resting or sleeping hours. ### Question 10 *3 marks · Short answer* Find answers by making necessary assumptions and approximations. Do guess first before calculating to check how close your guess was! **Solution** 1. Let us assume a person's weight to be $50\text{ kg}$ and the cost of $1\text{ kg}$ of rice to be $40$ rupees. 2. The worth of rice equal to the person's weight is given by the product of the weight in kilograms and the cost per kilogram. 3. Worth = $50 \times 40 = 2000$ rupees. **Answer:** The worth of donated rice is ₹2000. > Common mistake: Forgetting to multiply the weight by the cost per unit mass. ### Question 11 *3 marks · Short answer* How many times can a person circumnavigate (go around the world) the Earth in their lifetime if they walk non-stop? Consider the distance around the Earth as 40,000 km. **Solution** 1. Given: Distance around the Earth is $40,000\text{ km}$. Assume a human lifetime is $70$ years and a person walks $20\text{ km}$ per day. 2. Calculation: Distance walked in a lifetime = $70 \times 365 \times 20 = 5,11,000\text{ km}$. 3. Number of circumnavigations = $\frac{5,11,000}{40,000} \approx 12.77$ times. **Answer:** Approximately $13$ times > Common mistake: Taking total lifetime days instead of active walking days. ## Linear Growth vs. Exponential Growth ### Question 1 *2 marks · Very short answer* What do you think? Make an instinctive guess first. **Solution** 1. An instinctive guess for the number of steps in a ladder reaching the Moon is a very large number, usually in crores or even more. 2. This is because the distance to the Moon is extremely large compared to the distance between consecutive steps of a ladder. **Answer:** The instinctive guess would be crores of steps. > Common mistake: Underestimating the vast distance between the Earth and the Moon. ### Question 2 *2 marks · Very short answer* Would the number of steps be in thousands, lakhs, crores, or even more? **Solution** 1. We consider the distance to the Moon and the small gap between ladder steps. 2. The number of steps will be in the range of billions, which is much more than thousands, lakhs, or crores. **Answer:** The number of steps would be far more than crores. > Common mistake: Thinking the number fits within lakhs or crores. ### Question 3 *2 marks · Very short answer* We have to find out how many 20 cm make 3,84,400 km. **Solution** 1. First convert the distance to the Moon from kilometres to centimetres: $3,84,400 \text{ km} = 3,84,400 \times 10^5 \text{ cm} = 3,84,40,00,000 \text{ cm}$. 2. Divide by the gap between steps ($20 \text{ cm}$) to get the total number of steps: $3,84,40,00,000 \div 20 = 1,92,20,00,000 \text{ steps}$. **Answer:** $1,92,20,00,000$ steps (or $192$ crore and $20$ lakh steps) > Common mistake: Forgetting to convert kilometres to centimetres. ### Question 4 *3 marks · Short answer* Can you come up with some examples of linear growth and of exponential growth? **Solution** 1. Linear growth is additive, where a fixed amount is added at each step. 2. Example of linear growth: Walking a fixed distance of $5 \text{ km}$ every day. 3. Exponential growth is multiplicative, where the quantity multiplies by a constant factor at each step, such as the folding of paper or the doubling of lotuses in a pond. **Answer:** Linear growth example: walking $5 \text{ km}$ daily. Exponential growth example: paper folding thickness. > Common mistake: Confusing multiplicative growth with additive growth. ### Question 5 *2 marks · Very short answer* With a global human population of about $8 \times 10^9$ and about $4 \times 10^5$ African elephants, can we say that there are nearly 20,000 people for every African elephant? **Solution** 1. Given global human population = $8 \times 10^9$ and African elephant population = $4 \times 10^5$. 2. Calculate the ratio: $\frac{8 \times 10^9}{4 \times 10^5} = 2 \times 10^4 = 20,000$. **Answer:** Yes, there are nearly 20,000 people for every African elephant. > Common mistake: Errors in simplifying powers of 10 during division. ### Question 6 *3 marks · Short answer* Calculate and write the answer using scientific notation: (i) How many ants are there for every human in the world? (ii) If a flock of starlings contains 10,000 birds, how many flocks could there be in the world? (iii) If each tree had about $10^4$ leaves, find the total number of leaves on all the trees in the world. (iv) If you stacked sheets of paper on top of each other, how many would you need to reach the Moon? **Part (i)** 1. Global ant population is $2 \times 10^{16}$ and human population is $8 \times 10^9$. 2. Divide ant population by human population: $\frac{2 \times 10^{16}}{8 \times 10^9} = 0.25 \times 10^7 = 2.5 \times 10^6$. Answer (i): $2.5 \times 10^6$ ants **Part (ii)** 1. Global starling population is $1.3 \times 10^9$ and each flock has $10,000$ ($10^4$) birds. 2. Divide total birds by birds per flock: $\frac{1.3 \times 10^9}{10^4} = 1.3 \times 10^5$. Answer (ii): $1.3 \times 10^5$ flocks **Part (iii)** 1. Total trees in the world = $3 \times 10^{12}$ and leaves per tree = $10^4$. 2. Multiply: $3 \times 10^{12} \times 10^4 = 3 \times 10^{16}$. Answer (iii): $3 \times 10^{16}$ leaves **Part (iv)** 1. Distance to the Moon is $3,84,400 \text{ km} = 3.844 \times 10^8 \text{ m} = 3.844 \times 10^{11} \text{ cm}$. 2. Thickness of a sheet of paper is $0.001 \text{ cm} = 10^{-3} \text{ cm}$. 3. Divide distance by thickness: $\frac{3.844 \times 10^{11}}{10^{-3}} = 3.844 \times 10^{14}$. Answer (iv): $3.844 \times 10^{14}$ sheets **Answer:** Calculated values in scientific notation for all parts. > Common mistake: Incorrectly applying laws of exponents when dividing or multiplying powers of 10. ### Question 7 *2 marks · Very short answer* If you have lived for a million seconds, how old would you be? **Solution** 1. One million seconds is equal to $10^6$ seconds. 2. Since $10^5$ seconds is approximately $1.16$ days, $10^6$ seconds is approximately $11.57$ days. 3. Therefore, you would be approximately $12$ days old or about $11.57$ days old. **Answer:** Approximately $11.57$ days (or about $12$ days) old. > Common mistake: Confusing millions with billions or dividing by 24 hours incorrectly. ### Question 8 *3 marks · Short answer* $10^5\text{ seconds} \approx 1.16\text{ days}$ and $10^6\text{ seconds} \approx 11.57\text{ days}$. Think of some events or phenomena whose time is of the order of (i) $10^5$ seconds and (ii) $10^6$ seconds. Write them in scientific notation. **Part (i)** 1. An event whose time is of the order of $10^5$ seconds (about $1.16$ days) is the lifespan of an adult butterfly or a short journey. 2. Written in scientific notation as $1 \times 10^5$ seconds. Answer (i): $1 \times 10^5$ seconds **Part (ii)** 1. An event whose time is of the order of $10^6$ seconds (about $11.57$ days) is a short vacation or a two-week camp. 2. Written in scientific notation as $1 \times 10^6$ seconds. Answer (ii): $1 \times 10^6$ seconds **Answer:** Events of the order of $10^5$ seconds and $10^6$ seconds are given in scientific notation. > Common mistake: Failing to express the time in standard scientific notation format. ### Question 9 *3 marks · Short answer* Calculate and write the answer using scientific notation: (i) If one star is counted every second, how long would it take to count all the stars in the universe? Answer in terms of the number of seconds using scientific notation. (ii) If one could drink a glass of water (200 ml) every 10 seconds, how long would it take to finish the entire volume of water on Earth? **Part (i)** 1. The estimated number of stars in the observable universe is $2 \times 10^{23}$. 2. At the rate of one star per second, the time taken in seconds is $2 \times 10^{23}$ seconds. Answer (i): $2.0 \times 10^{23}$ seconds **Part (ii)** 1. The total volume of water on Earth is estimated to be $2 \times 10^{25}$ drops. 2. Each glass contains $200\text{ ml}$, and assuming $16$ drops per millilitre, one glass has $200 \times 16 = 3200$ drops. 3. Dividing the total drops by the number of drops finished every 10 seconds gives $\frac{2 \times 10^{25}}{3200} \div 10$, which is approximately $6.25 \times 10^{22}$ seconds. Answer (ii): $6.25 \times 10^{22}$ seconds **Answer:** Calculations using scientific notation. > Common mistake: Incorrectly handling powers of 10 during division. ## Figure it Out ### Question 1 *2 marks · Very short answer* Find out the units digit in the value of $2^{224} \div 4^{32}$? [Hint: $4 = 2^2$] **Solution** 1. $2^{224} \div 4^{32} = 2^{224} \div (2^2)^{32} = 2^{224} \div 2^{64} = 2^{224-64} = 2^{160}$ 2. The units digit of powers of 2 follows the cycle 2, 4, 8, 6. Since 160 is divisible by 4, the units digit is 6. **Answer:** The units digit is 6. > Common mistake: Confusing the cycle of units digits for powers of 2. ### Question 2 *2 marks · Very short answer* There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days? **Solution** 1. Initial number of bottles = 5 2. Each day, 1 new container with 5 bottles is brought in, meaning 5 bottles are added per day. 3. Total number of bottles after 40 days = $5 + 40 \times 5 = 205$ bottles or $2 \times 10^2$ bottles approximately. **Answer:** 200 = $2 \times 10^2$ bottles > Common mistake: Treating it as a geometric progression instead of a linear addition. ### Question 3 *3 marks · Short answer* 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) $32^{-5}$ **Part (i)** 1. $64^3 = (2^6)^3 = 2^{18}$ 2. $64^3 = 2^{10} \times 2^8 = 4^5 \times 4^4 = 8^3 \times 8^3$ Answer (i): $2^{10} \times 2^8, 4^5 \times 4^4, 8^3 \times 8^3$ **Part (ii)** 1. $192^8 = (2^6 \times 3)^8 = 2^{48} \times 3^8$ 2. $192^8 = 2^{40} \times 2^8 \times 3^8 = 2^{40} \times 6^8 = 2^{48} \times 3^8$ Answer (ii): $2^{40} \times 2^8 \times 3^8, 2^{40} \times 6^8, 2^{48} \times 3^8$ **Part (iii)** 1. $32^{-5} = (2^5)^{-5} = 2^{-25}$ 2. $32^{-5} = 2^{-10} \times 2^{-15} = 2^{-5} \times 2^{-20} = 4^{-12} \times 2^{-1}$ Answer (iii): $2^{-10} \times 2^{-15}, 2^{-5} \times 2^{-20}, 4^{-12} \times 2^{-1}$ **Answer:** Various representations are possible using laws of exponents. > Common mistake: Incorrectly applying the power of a power rule. ### Question 4 *3 marks · Short answer* 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) $q^{46}$ is both a 4th power and a 6th power ($q$ is a prime number). **Part (i)** 1. Only Sometimes True. 2. This is true only for numbers that have powers as 6, such as $n^6 = (n^2)^3 = (n^3)^2$. Answer (i): Only Sometimes True. **Part (ii)** 1. Always True. 2. We can write $n^4 = (n^2)^2$. Answer (ii): Always True. **Part (iii)** 1. Always True. 2. As $\frac{n^5}{n^3} = n^2$, the fifth power is divisible by the cube. Answer (iii): Always True. **Part (iv)** 1. Always True. 2. $(n_1^3) \times (n_2^3) = (n_1 \times n_2)^3$. Answer (iv): Always True. **Part (v)** 1. Never True. 2. For $q^{46}$, 46 is not divisible by 4 or 6. Answer (v): Never True. **Answer:** Statements evaluated based on exponent properties. > Common mistake: Assuming divisibility rules for exponents apply to the base. ### Question 5 *3 marks · Short answer* Simplify and write these in the exponential form. (i) $10^{-2} \times 10^{-5}$ (ii) $5^7 \div 5^4$ (iii) $9^{-7} \div 9^4$ (iv) $(13^{-2})^{-3}$ (v) $m^5n^{12}(mn)^9$ **Part (i)** 1. $10^{-2} \times 10^{-5} = 10^{-2-5} = 10^{-7}$ Answer (i): $10^{-7}$ **Part (ii)** 1. $5^7 \div 5^4 = 5^{7-4} = 5^3$ Answer (ii): $5^3$ **Part (iii)** 1. $9^{-7} \div 9^4 = 9^{-7-4} = 9^{-11}$ Answer (iii): $9^{-11}$ **Part (iv)** 1. $(13^{-2})^{-3} = 13^{(-2 \times -3)} = 13^6$ Answer (iv): $13^6$ **Part (v)** 1. $m^5 n^{12} (mn)^9 = m^5 n^{12} m^9 n^9 = m^{5+9} n^{12+9} = m^{14} n^{21}$ Answer (v): $m^{14} n^{21}$ **Answer:** Simplified exponential forms. > Common mistake: Adding exponents when dividing or multiplying incorrectly. ### Question 6 *3 marks · Short answer* If $12^2 = 144$ what is (i) $(1.2)^2$ (ii) $(0.12)^2$ (iii) $(0.012)^2$ (iv) $120^2$ **Part (i)** 1. $(1.2)^2 = 1.44$ Answer (i): 1.44 **Part (ii)** 1. $(0.12)^2 = 0.0144$ Answer (ii): 0.0144 **Part (iii)** 1. $(0.012)^2 = 0.000144$ Answer (iii): 0.000144 **Part (iv)** 1. $(120)^2 = 14400$ Answer (iv): 14400 **Answer:** Calculated squares. > Common mistake: Misplacing the decimal point. ### Question 7 *2 marks · Very short answer* Circle the numbers that are the same— $2^4 \times 3^6 \quad 6^4 \times 3^2 \quad 6^{10} \quad 18^2 \times 6^2 \quad 6^{24}$ **Solution** 1. Express each given number in terms of its prime factors 2 and 3. 2. $2^4 \times 3^6 = 2^4 \times 3^6$, $6^4 \times 3^2 = (2 \times 3)^4 \times 3^2 = 2^4 \times 3^4 \times 3^2 = 2^4 \times 3^6$, and $18^2 \times 6^2 = (2 \times 3^2)^2 \times (2 \times 3)^2 = 2^2 \times 3^4 \times 2^2 \times 3^2 = 2^4 \times 3^6$. 3. Thus, $2^4 \times 3^6$, $6^4 \times 3^2$, and $18^2 \times 6^2$ are the same. **Answer:** $2^4 \times 3^6$, $6^4 \times 3^2$, and $18^2 \times 6^2$ > Common mistake: Failing to convert base numbers into their prime factors before comparing exponents. ### Question 8 *3 marks · Short answer* 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 $2^{100}$ **Part (i)** 1. Evaluate $4^3 = 4 \times 4 \times 4 = 64$ and $3^4 = 3 \times 3 \times 3 \times 3 = 81$. 2. Since $81 > 64$, $3^4$ is greater than $4^3$. Answer (i): $3^4$ **Part (ii)** 1. Evaluate $2^8 = 256$ and $8^2 = 64$. 2. Since $256 > 64$, $2^8$ is greater than $8^2$. Answer (ii): $2^8$ **Part (iii)** 1. Express $100^2$ as $(10^2)^2 = 10^4 = 10,000$. 2. Compare $10^4$ with $2^{100}$; since $2^{100}$ is vastly larger than $10,000$, $2^{100}$ is greater. Answer (iii): $2^{100}$ **Answer:** (i) $3^4$, (ii) $2^8$, (iii) $2^{100}$ > Common mistake: Confusing base and exponent during evaluation. ### Question 9 *3 marks · Short answer* 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? **Solution** 1. Total packets of milk to be produced in a year = $8,500,000,000$. 2. Using digits 0-9, there are 10 choices for each digit in the code. For an $n$-digit code, the total number of unique codes possible is $10^n$. 3. We need $10^n \geq 8,500,000,000$. Since $10^9 < 8,500,000,000$ and $10^{10} > 8,500,000,000$, we must have $n = 10$. 4. Hence, the code should consist of at least 10 digits. **Answer:** 10 digits > Common mistake: Choosing 9 digits because $8.5$ is close to 9, ignoring the actual place value of billion. ### Question 10 *3 marks · Short answer* 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? **Solution** 1. Check if there are other numbers that can be expressed as both $n^2$ and $m^3$. 2. Any number that is a sixth power ($x^6$) can be written as $(x^3)^2$ (a square) and $(x^2)^3$ (a cube). 3. Therefore, there are infinitely many such numbers, described in general as $n^6$ for any integer $n$. **Answer:** Yes, infinitely many numbers of the form $n^6$. > Common mistake: Stating that only 64 is both a square and a cube. ### Question 11 *3 marks · Short answer* 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? **Solution** 1. Determine the number of possible characters for each position: 26 letters and 10 digits give $26 + 10 = 36$ choices. 2. Since the passcode has a length of 5 and each position is independent, multiply the choices for each position. 3. Total possible codes = $36 \times 36 \times 36 \times 36 \times 36 = 36^5$. **Answer:** $36^5$ > Common mistake: Multiplying 26 and 10 instead of adding them to find the total character choices per slot. ### Question 12 *1 mark · MCQ* 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? - $20^9$ - $10^{11}$ - $10^{10}$ - $10^{18}$ - $2 \times 10^9$ - $10^9 + 10^9$ **Solution** 1. Population of sheep is $10^9$ and population of goats is about the same, which is $10^9$. 2. Total population is $10^9 + 10^9 = 2 \times 10^9$. 3. Option (iv) $10^{18}$ is incorrect because exponents are added only during multiplication, not addition. **Answer:** (v) $2 \times 10^9$ > Common mistake: Adding the exponents to get $10^{18}$ instead of adding the quantities to get $2 \times 10^9$. ### Question 13 *3 marks · Short answer* 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. **Part (i)** 1. Global population is approximately $8.2 \times 10^9$ and each person has 30 pieces of clothing. 2. Total pieces of clothing = $8.2 \times 10^9 \times 30 = 2.46 \times 10^{11}$. Answer (i): $2.46 \times 10^{11}$ **Part (ii)** 1. Number of bee colonies = $100 \text{ million} = 10^8$ and bees per colony = $50,000 = 5 \times 10^4$. 2. Total honeybees = $5 \times 10^4 \times 10^8 = 5.0 \times 10^{12}$. Answer (ii): $5.0 \times 10^{12}$ **Part (iii)** 1. Bacterial cells per human = $38 \text{ trillion} = 38 \times 10^{12}$ and world population = $8.2 \times 10^9$. 2. Total bacterial population = $(38 \times 10^{12}) \times (8.2 \times 10^9) = 311.6 \times 10^{21} = 3.116 \times 10^{23}$. Answer (iii): $3.116 \times 10^{23}$ **Part (iv)** 1. Assuming an average lifetime of 70 years and 1 hour of eating per day, total seconds spent eating = $3600 \times 365 \times 70 = 91980000$. 2. Expressing in scientific notation gives $9.198 \times 10^7 \text{ seconds}$. Answer (iv): $9.198 \times 10^7 \text{ seconds}$ **Answer:** Calculated total values for clothing, bees, bacteria, and lifetime eating time in scientific notation. > Common mistake: Incorrectly adjusting the decimal point when converting the coefficient into standard form between 1 and 10. ### Question 14 *2 marks · Very short answer* What was the date 1 arab/1 billion seconds ago? **Solution** 1. 1 arab or 1 billion seconds is equal to $1 \times 10^9 \text{ seconds}$. 2. Dividing by seconds in a day ($86400$), this represents approximately $11574 \text{ days}$ or about $31.7 \text{ years}$ ago. **Answer:** Approximately 31.7 years ago > Common mistake: Forgetting to convert seconds into years correctly. ## Frequently asked questions ### How many total questions are there in NCERT Solutions for Class 8 Maths Chapter 2 Power Play? This chapter contains a total of 78 questions across all sections based on the new NCERT book for the 2026-27 session. You can access SwaVid's free PDF and step-by-step solutions for all these questions on this page only. ### Which important topics and exercises are covered in this chapter? The chapter covers various topics such as exponential notation and operations, scientific notation, linear versus exponential growth, and powers of 10. These concepts are spread across different sections like Exponential Notation and Operations and Scientific Notation, featuring multiple question types. ### What are the hardest question types in Class 8 Maths Chapter 2 and how should I approach them? Long answer questions involving laws of exponents, combinatorics, and scientific notation calculations are often considered challenging by students. To approach them, you should carefully break down the given problem, apply the relevant exponent laws step by step, and verify your calculations. ### How can I write answers for full marks in the Class 8 Maths Power Play chapter? To secure full marks, make sure to write every intermediate step clearly while applying properties of exponents or scientific notation. Referring to SwaVid's free PDF and step-by-step solutions on this page only will help you understand the proper structuring of answers. ### Is the free PDF for Class 8 Maths Chapter 2 solutions available for the 2026-27 session? Yes, the complete chapter solutions aligned with the new NCERT book under the NCF 2023 guidelines are provided here. You can easily download SwaVid's free PDF and step-by-step solutions from this page only to aid your exam preparation. ## Related pages - [Class 8 Maths chapters](https://www.swavid.com/maths/class/8) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student