CBSE Class 8 Mathematics Power Play Long Answer and Case Study Questions with Answers

Long Answer Questions

Teacher's Note:
1. 2¹⁰ = 1024, so 2²⁰ = 1024 × 1024.
2. Divide by 100 to change cm into m.

Question 2. The chapter builds a set of general rules for working with powers.
(i) State the five rules for \( n^{a} \times n^{b} \), \( n^{a} \div n^{b} \), \( (n^{a})^{b} \), \( n^{a} \times m^{a} \) and n⁰, naming the condition needed on n wherever one is needed.
(ii) Illustrate each rule with one numerical example of your own and work out its value.
(iii) Explain with the help of the expanded products why 2³ × 2⁴ = 2⁷, and not 2¹².
Answer:
(i) The five rules:
\( n^{a} \times n^{b} = n^{a+b} \)
\( n^{a} \div n^{b} = n^{a-b} \), where n ≠ 0
\( (n^{a})^{b} = (n^{b})^{a} = n^{a \times b} \)
\( n^{a} \times m^{a} = (n \times m)^{a} \), and similarly \( n^{a} \div m^{a} = \left(\frac{n}{m}\right)^{a} \) with m ≠ 0
\( n^{0} = 1 \), where n ≠ 0
(ii) Examples:
2³ × 2⁴ = 2⁷ = 128
7⁶ ÷ 7⁴ = 7² = 49
(3²)³ = 3⁶ = 729
2⁴ × 5⁴ = 10⁴ = 10000
9⁰ = 1
(iii) 2³ = 2 × 2 × 2 and 2⁴ = 2 × 2 × 2 × 2. Writing the product out gives seven factors of 2 in a row, so the answer is 2⁷.
Multiplying two powers joins the two lists of factors, so the number of factors is added, not multiplied. Multiplying the exponents would mean a power of a power, which is a different operation: (2³)⁴ = 2¹².
In simple words: Each law comes from counting factors, which is why exponents add when powers of the same base are multiplied.

Teacher's Note:
1. Learn each rule together with its condition.
2. Expanding the factors is the best way to check a rule.

Question 3. Powers can have exponents that are zero or negative.
(i) Starting from a line of length 2⁴ units and halving it again and again, show how the meanings of 2⁰, 2⁻¹ and 2⁻² arise.
(ii) Write the general rules \( n^{-a} = \frac{1}{n^{a}} \) and \( n^{a} = \frac{1}{n^{-a}} \), and give one example of each.
(iii) Find the value of 5⁻² + 2⁻³ as a fraction in its simplest form, and also as a decimal.
Answer:
(i) The line is 2⁴ = 16 units long. Each halving divides the length by 2.
Halving 4 times: 2⁴ ÷ 2⁴ = 2⁴⁻⁴ = 2⁰, and the length left is 16 ÷ 16 = 1 unit, so 2⁰ = 1.
Halving 5 times: 2⁴ ÷ 2⁵ = 2⁻¹, and the length left is \( \frac{16}{32} \) = \( \frac{1}{2} \) unit, so 2⁻¹ = \( \frac{1}{2} \).
Halving 6 times: 2⁴ ÷ 2⁶ = 2⁻², and the length left is \( \frac{16}{64} \) = \( \frac{1}{4} \) unit, so 2⁻² = \( \frac{1}{4} \).
(ii) \( n^{-a} = \frac{1}{n^{a}} \) and \( n^{a} = \frac{1}{n^{-a}} \), where n ≠ 0.
Examples: 10⁻³ = \( \frac{1}{10^{3}} \) = \( \frac{1}{1000} \), and 7² = \( \frac{1}{7^{-2}} \) = 49.
(iii) \( 5^{-2} + 2^{-3} = \frac{1}{25} + \frac{1}{8} = \frac{8}{200} + \frac{25}{200} = \frac{33}{200} \) = \( \frac{33}{200} \) = 0.165.
A negative exponent never makes a number negative; it only makes it a fraction less than 1 (for a base greater than 1).
In simple words: Each halving lowers the exponent by 1, so going past 2⁰ = 1 gives fractions with negative exponents.

Teacher's Note:
1. The pattern of halving explains the definitions; they are not arbitrary.
2. Use the LCM 200 to add the two fractions.

Question 4. Very large and very small numbers are written in standard form x × 10^y, with 1 ≤ x < 10 and y an integer.
(i) Write 92,00,000 and 0.00045 in standard form.
(ii) Write 7.3 × 10⁵ and 2.4 × 10⁻³ in the usual decimal form.
(iii) The population of a town is reported as 2 × 10⁷. Explain why the exponent 7 carries more information than the coefficient 2, using the effect of changing each of them.
Answer:
(i) 92,00,000 = 9.2 × 10⁶ and 0.00045 = 4.5 × 10⁻⁴.
(ii) 7.3 × 10⁵ = 7,30,000 and 2.4 × 10⁻³ = 0.0024.
(iii) In 2 × 10⁷ the coefficient is 2 and the exponent is 7.
If the coefficient changes from 2 to 3, the population changes from 2 crore to 3 crore, an increase of one-half.
If the exponent changes from 7 to 8, the population becomes 2 × 10⁸ = 20 crore, which is ten times as large.
So a change of 1 in the exponent is far bigger than a change of 1 in the coefficient. The exponent tells the number of digits (how big the quantity is), while the coefficient only gives the leading digits. That is why the exponent is read first.
In simple words: Changing the exponent by 1 multiplies the number by 10, while changing the coefficient only adjusts it a little.

Teacher's Note:
1. A positive exponent moves the decimal point right; a negative one moves it left.
2. Order of size is decided by the exponent.

Question 5. A school has 4 houses. Each house runs 4 clubs, each club has 4 teams, and each team has 4 members.
(i) Find the number of clubs, teams and members, writing each answer as a power of 4 and as a number.
(ii) If every member designs 4 posters, write the total number of posters as a power of 4 and find its value.
(iii) Explain how the chain “houses to clubs to teams to members” shows why the exponent increases by 1 at every stage.
Answer:
(i) Houses = 4 = 4¹.
Clubs = 4 × 4 = 4² = 16.
Teams = 4² × 4 = 4³ = 64.
Members = 4³ × 4 = 4⁴ = 256.
(ii) Each of the 4⁴ members makes 4 posters, so the total is 4⁴ × 4 = 4⁵ = 1024 posters.
(iii) At each stage every item of the previous stage gives rise to 4 new items, so the new count is the old count multiplied by 4. Multiplying by 4 once more adds one more factor of 4, which by the rule 4^a × 4 = 4^a+1 raises the exponent by exactly 1.
This is why a chain of 5 stages of branching by 4 gives 4⁵ and not 4 × 5: the stages multiply, they do not add.
In simple words: Every level multiplies the count by 4, so each new level adds 1 to the exponent.

Teacher's Note:
1. Draw a small tree diagram to see the branching.
2. The exponent equals the number of stages.

Question 6. In a class test, every question must be attempted.
(i) The test has 6 true/false questions. Find the number of different answer sheets that are possible, writing it as a power of 2.
(ii) The teacher changes the paper so that each of the 6 questions now has 4 options. Find the number of possible answer sheets as a power of 4 and as a number.
(iii) Using the laws of exponents, show how many times the second number is larger than the first, and write that factor as a power of 2.
Answer:
(i) Each question can be answered in 2 ways and there are 6 questions, so the number of answer sheets = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶ = 64.
(ii) Now each question can be answered in 4 ways, so the number of answer sheets = 4⁶ = 4096.
(iii) Since 4 = 2², 4⁶ = (2²)⁶ = 2¹². Therefore \( \frac{4^{6}}{2^{6}} = \frac{2^{12}}{2^{6}} = 2^{12-6} = 2^{6} = 64 \).
So the four-option paper has 2⁶ = 64 times as many possible answer sheets. Check: 4096 ÷ 64 = 64.
Doubling the choices per question does not double the total; it multiplies the total by 2⁶, because the extra choice is available in all six questions at once.
In simple words: Two choices per question give 2⁶ sheets, four choices give 2¹², which is 2⁶ times more.

Teacher's Note:
1. Choices per question form the base; the number of questions is the exponent.
2. Write 4 as 2² to compare with powers of 2.

Question 7. Consider the “power line” of 3, with the powers 3⁻², 3⁻¹, 3⁰, 3¹, 3², 3³, 3⁴ placed in order.
(i) Write the value of each of these seven powers.
(ii) Describe what happens to the value as you move one step to the right along the line, and what happens as you move one step to the left.
(iii) How many times larger is 3⁴ than 3⁻²? Answer as a power of 3 and as a number.
(iv) Write \( \frac{2187}{27} \) and \( \frac{1}{243} \) as powers of 3.
Answer:
(i) 3⁻² = \( \frac{1}{9} \), 3⁻¹ = \( \frac{1}{3} \), 3⁰ = 1, 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81.
(ii) One step to the right multiplies the value by 3 (the exponent rises by 1). One step to the left divides the value by 3 (the exponent falls by 1). Going left past 3⁰ = 1 gives fractions, not negative numbers; that is what negative exponents mean.
(iii) 3⁴ ÷ 3⁻² = 3^{4 − (−2)} = 3⁶ = 729. So 3⁴ is 729 times 3⁻². Check: 81 ÷ \( \frac{1}{9} \) = 81 × 9 = 729.
(iv) \( \frac{2187}{27} \) = \( \frac{3^{7}}{3^{3}} \) = 3⁴, and \( \frac{1}{243} \) = \( \frac{1}{3^{5}} \) = 3⁻⁵.
In simple words: Along the power line each step right multiplies by 3 and each step left divides by 3.

Teacher's Note:
1. Here the line is read left to right with increasing exponents.
2. Know the powers of 3 up to 3⁷ = 2187.

Question 8. A ladder is to be built to a height of 1000 km, with a gap of 25 cm between consecutive steps. A sheet of paper 0.01 cm thick is folded again and again so that its thickness doubles with every fold.
(i) Find the number of steps the ladder needs. Give the answer in standard form.
(ii) Find the least number of folds after which the folded paper would be at least 1000 km thick.
(iii) Compare the two answers and explain the difference between linear (additive) growth and exponential (multiplicative) growth, giving one further everyday example of each.
Answer:
(i) 1000 km = 1000 × 1000 × 100 cm = 10⁸ cm.
Number of steps = \( \frac{10^{8}}{25} \) = 40,00,000 = 4 × 10⁶ steps.
(ii) Thickness after n folds = 0.01 × 2ⁿ cm. We need 0.01 × 2ⁿ ≥ 10⁸, that is 2ⁿ ≥ 10¹⁰.
2³³ ≈ 8.59 × 10⁹, which is less than 10¹⁰, while 2³⁴ ≈ 1.72 × 10¹⁰, which is more. So 34 folds are needed.
(iii) The same height needs 40 lakh steps but only 34 folds. In linear growth a fixed amount (25 cm) is added at every stage. In exponential growth the quantity is multiplied by a fixed number (2) at every stage, so each stage adds as much as everything before it.
Examples: filling a tank at 5 litres a minute is linear growth; money that doubles every year, or a rumour passed on by everyone who hears it, is exponential growth.
In simple words: Adding 25 cm needs 4 × 10⁶ steps, but doubling needs only 34 folds.

Teacher's Note:
1. Convert km to cm before dividing.
2. 2¹⁰ ≈ 10³ helps estimate: 2³³ ≈ 8.6 × 10⁹.

Question 9. The rules for exponents are first written for counting numbers and then extended.
(i) Using only the rule \( n^{a} \div n^{b} = n^{a-b} \), justify that n⁰ = 1 for every n ≠ 0, and explain why n = 0 must be excluded.
(ii) Using the same rule with a < b, justify that \( n^{-a} = \frac{1}{n^{a}} \), taking 2³ ÷ 2⁷ as your worked example.
(iii) Show, by writing out the factors, that \( (n^{a})^{b} = n^{ab} = (n^{b})^{a} \), using (2³)⁴ as the example.
(iv) Use these results to simplify \( \frac{m^{6} \times m^{-2}}{m^{4}} \) and state the value obtained.
Answer:
(i) Take a = b: \( n^{a} \div n^{a} = n^{a-a} = n^{0} \). But a non-zero number divided by itself is 1. Hence n⁰ = 1.
If n = 0, the left side is \( \frac{0}{0} \), which has no meaning. So 0⁰ is left undefined and the rule needs n ≠ 0.
(ii) By the rule, 2³ ÷ 2⁷ = 2³⁻⁷ = 2⁻⁴.
Writing the same division in full: \( \frac{2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{2^{4}} \).
Comparing, 2⁻⁴ = \( \frac{1}{2^{4}} \) = \( \frac{1}{16} \). The same argument with any n ≠ 0 gives \( n^{-a} = \frac{1}{n^{a}} \).
(iii) (2³)⁴ = 2³ × 2³ × 2³ × 2³, which is four groups of three factors, that is 12 factors of 2, so it equals 2¹².
Grouping the same 12 factors as three groups of four gives (2⁴)³. Hence \( (n^{a})^{b} = n^{ab} = (n^{b})^{a} \).
(iv) \( \frac{m^{6} \times m^{-2}}{m^{4}} = \frac{m^{6-2}}{m^{4}} = m^{4-4} = m^{0} = 1 \) (m ≠ 0). The value is 1.
In simple words: The zero and negative exponent meanings are chosen so that the division rule keeps working.

Teacher's Note:
1. Every new meaning comes from an old rule.
2. Expanded products are the proof behind each law.

Question 10. The chapter compares the sizes of some huge populations: there are about 8 × 10⁹ humans, about 2 × 10¹⁶ ants, about 3 × 10¹² trees, and about 10²¹ grains of sand on all the beaches and deserts of the Earth.
(i) Find the number of ants per human, in standard form.
(ii) Find the number of sand grains available per ant, in standard form.
(iii) If each tree carries about 10⁴ leaves, find the total number of leaves on all the trees of the world, in standard form.
(iv) Arrange the four given quantities in increasing order and state, for each consecutive pair, roughly how many powers of 10 separate them.
Answer:
(i) Ants per human = \( \frac{2 \times 10^{16}}{8 \times 10^{9}} \) = 0.25 × 10⁷ = 2.5 × 10⁶, about 25 lakh ants for every human.
(ii) Sand grains per ant = \( \frac{10^{21}}{2 \times 10^{16}} \) = 0.5 × 10⁵ = 5 × 10⁴, about 50,000 grains for every ant.
(iii) Total leaves = 3 × 10¹² × 10⁴ = 3 × 10¹⁶ leaves.
(iv) Increasing order: 8 × 10⁹ (humans) < 3 × 10¹² (trees) < 2 × 10¹⁶ (ants) < 10²¹ (sand grains).
Trees are about 375 times the number of humans, a gap of between two and three powers of 10.
Ants are about 6.7 × 10³ times the number of trees, a gap of nearly four powers of 10.
Sand grains are 5 × 10⁴ times the number of ants, a gap of nearly five powers of 10.
Reading the exponents alone (9, 12, 16, 21) is enough to order these quantities at a glance.
In simple words: Divide coefficients and subtract exponents to compare huge counts; the exponents give the order at once.

Teacher's Note:
1. Adjust 0.25 × 10⁷ to 2.5 × 10⁶ for standard form.
2. Compare exponents first when ordering.

Case Study Based Questions

Question 11. A doubling pocket-money plan. Meera asks her grandfather for pocket money for a month. Her grandfather makes an unusual offer: “I will give you ₹2 on the first day, ₹4 on the second day, ₹8 on the third day, and on every following day I will give you double of what I gave you the day before.” Meera thinks this is a poor deal, because ₹2 is very little money, and she agrees at once. Her brother warns her that she has made a far better bargain than she realises, because the daily amount is growing multiplicatively and not by a fixed amount. By the end of the first week the daily amount has already crossed ₹100.
(i) Write the amount Meera gets on the first, second and third day as powers of 2.
(ii) How much does she get on the 5th day?
(iii) Write an expression in exponential form for the amount she gets on the nth day.
(iv) On which day does the daily amount first cross ₹10,000?
Answer:
(i) Day 1: ₹2 = 2¹; Day 2: ₹4 = 2²; Day 3: ₹8 = 2³.
(ii) On the 5th day she gets 2⁵ = ₹32.
(iii) On the nth day the amount is ₹2ⁿ, since the day number equals the exponent.
(iv) 2¹³ = 8192 and 2¹⁴ = 16384. As 8192 < 10000 < 16384, the daily amount first crosses ₹10,000 on Day 14.
In simple words: The amount on day n is 2ⁿ rupees, which passes ₹10,000 on day 14.

Teacher's Note:
1. The exponent matches the day number here.
2. Doubling seems small at first but grows enormously within a fortnight.

Question 12. A message that spreads. A school posts a notice about a cleanliness drive. On Day 1 the head girl sends the notice to 3 students. On Day 2, each of those 3 students sends it to 3 fresh students, and this continues: every student who receives the notice on a day passes it to 3 new students the very next day. Nobody receives the notice twice. The teacher in charge is surprised to see that within a week the notice has reached more students than the school has, and she uses the situation in the mathematics class as an example of multiplicative growth.
(i) How many students receive the notice on Day 1 and on Day 2?
(ii) Write the number of fresh students receiving the notice on Day n in exponential form.
(iii) Find the number of fresh receivers on Day 4.
(iv) On which day do the fresh receivers first cross 1000?
Answer:
(i) Day 1: 3 students; Day 2: 3 × 3 = 9 students.
(ii) Fresh receivers on Day n = 3ⁿ, because the number is multiplied by 3 every day.
(iii) Day 4: 3⁴ = 81 students.
(iv) 3⁶ = 729 and 3⁷ = 2187. Since 729 < 1000 < 2187, the fresh receivers first cross 1000 on Day 7.
In simple words: Each day three times as many new students get the notice, so Day n has 3ⁿ fresh receivers.

Teacher's Note:
1. 'Fresh' receivers means only that day's new students.
2. Find the first power of 3 above 1000.

Question 13. Bacteria in a dish. In a school laboratory, a dish is prepared with a nutrient jelly and 50 bacteria are placed on it. The teacher explains that in these warm conditions the colony doubles every hour, so the count after each hour is twice the count of the hour before. The students record the count every hour for a whole day and plot it. The first few readings look small and the curve seems almost flat, but after some hours the numbers shoot up so steeply that the graph paper is not tall enough. The teacher says this is exactly the multiplicative growth seen in the paper-folding story.
(i) How many bacteria are there after 3 hours?
(ii) Write the count after n hours in exponential form.
(iii) Find the count after 6 hours.
(iv) Find the count after 10 hours, and write it in standard form.
Answer:
(i) After 3 hours: 50 × 2³ = 50 × 8 = 400 bacteria.
(ii) After n hours the count is 50 × 2ⁿ.
(iii) After 6 hours: 50 × 2⁶ = 50 × 64 = 3200 bacteria.
(iv) After 10 hours: 50 × 2¹⁰ = 50 × 1024 = 51200 bacteria, which in standard form is 5.12 × 10⁴.
In simple words: The starting 50 is multiplied by 2 once for every hour.

Teacher's Note:
1. Do not forget the starting count of 50.
2. Exponential curves look flat at first and then shoot up.

Question 14. Distances in the Solar System. Astronomers must handle distances far beyond everyday experience, so they write them in standard form. The distance between the Sun and the Earth is 1.496 × 10¹¹ m. The distance between the Sun and Saturn is 1.4335 × 10¹² m, and the distance between Saturn and Uranus is 1.439 × 10¹² m. Our Sun itself lies about 3 × 10²⁰ m from the centre of the Milky Way galaxy. Written out in full, these numbers carry so many zeroes that a single miscounted zero changes the answer ten times over, which is why the exponent is written separately.
(i) Which of the first three distances is the smallest? Give a reason.
(ii) Write the Sun to Earth distance in the usual form, without powers.
(iii) About how many times farther from the Sun is Saturn than the Earth?
(iv) Express the Sun’s distance from the galaxy centre in kilometres, in standard form.
Answer:
(i) The Sun to Earth distance, 1.496 × 10¹¹ m, is the smallest, because its exponent 11 is less than the exponent 12 of the other two; the exponent is compared first.
(ii) 1.496 × 10¹¹ m = 1,49,60,00,00,000 m.
(iii) \( \frac{1.4335 \times 10^{12}}{1.496 \times 10^{11}} \) ≈ 9.58, so Saturn is nearly 10 times as far from the Sun as the Earth is.
(iv) 1 km = 10³ m, so 3 × 10²⁰ m = \( \frac{3 \times 10^{20}}{10^{3}} \) km = 3 × 10¹⁷ km.
In simple words: Compare exponents first, and divide by 10³ to change metres into kilometres.

Teacher's Note:
1. 10¹² ÷ 10¹¹ = 10, then adjust using the coefficients.
2. Moving the decimal 11 places gives the usual form.

Question 15. Computer memory counts in powers of two. A computer stores information in bytes, and memory sizes are counted in powers of 2 because each memory address is built from switches that are either off or on. One kilobyte (KB) is defined as 2¹⁰ bytes, one megabyte (MB) as 2²⁰ bytes and one gigabyte (GB) as 2³⁰ bytes. A shopkeeper tells a customer that a 4 GB pen drive holds “about four billion” bytes. The customer, who has just studied powers in school, decides to check the shopkeeper’s claim for herself using the laws of exponents instead of multiplying out long strings of twos.
(i) How many bytes are there in 1 MB? Give the value as a number.
(ii) Using the laws of exponents, find how many kilobytes make 1 GB, as a power of 2.
(iii) Express the number of bytes in 4 GB as a power of 2.
(iv) Write that number in standard form, correct to two decimal places, and say whether the shopkeeper’s claim is fair.
Answer:
(i) 1 MB = 2²⁰ bytes = 10,48,576 bytes.
(ii) Number of KB in 1 GB = \( \frac{2^{30}}{2^{10}} \) = 2³⁰⁻¹⁰ = 2²⁰ kilobytes.
(iii) 4 GB = 4 × 2³⁰ = 2² × 2³⁰ = 2³² bytes.
(iv) 2³² = 4,29,49,67,296 ≈ 4.29 × 10⁹ bytes. This is just over four billion, so the shopkeeper’s claim is fair.
In simple words: 4 GB is 2³² bytes, about 4.29 × 10⁹, which is roughly four billion.

Teacher's Note:
1. One billion = 10⁹.
2. Write 4 as 2² before using the product law.

Question 16. Things too small to see. A science magazine lists the sizes of some very small things. The diameter of a typical virus is about 1.2 × 10⁻⁷ m, the length of a typical bacterium is about 2 × 10⁻⁶ m, and the thickness of a human hair is about 6 × 10⁻⁵ m. The magazine says that negative powers of 10 are used for such lengths for exactly the same reason that positive powers are used for the distances to the planets: it is the exponent, and not the leading digit, that first tells us how big or how small the quantity really is.
(i) Write the diameter of the virus in the usual decimal form, in metres.
(ii) Arrange the three lengths in increasing order.
(iii) How many times thicker than the virus is a human hair?
(iv) How many bacteria, laid end to end, would stretch across the thickness of one hair?
Answer:
(i) 1.2 × 10⁻⁷ m = 0.00000012 m (the decimal point moves 7 places to the left).
(ii) Increasing order: virus (1.2 × 10⁻⁷ m) < bacterium (2 × 10⁻⁶ m) < hair (6 × 10⁻⁵ m). The more negative the exponent, the smaller the length.
(iii) \( \frac{6 \times 10^{-5}}{1.2 \times 10^{-7}} \) = 5 × 10^{−5 − (−7)} = 5 × 10² = 500, so a hair is 500 times thicker than a virus.
(iv) \( \frac{6 \times 10^{-5}}{2 \times 10^{-6}} \) = 3 × 10¹ = 30, so about 30 bacteria laid end to end stretch across one hair.
In simple words: With negative powers, the more negative exponent belongs to the smaller length.

Teacher's Note:
1. Subtracting a negative exponent means adding.
2. −7 is less than −6, which is less than −5.

Question 17. Tulābhāra at a village temple. In the practice of Tulābhāra, a devotee donates goods equal to his or her own weight. At a village temple, a donor of weight 60 kg offers rice costing ₹55 per kilogram. Her son then asks a different question: instead of rice, how many 5-rupee coins would be needed to balance the same 60 kg? Nobody knows the mass of a coin, so the family weighs one on a kitchen scale and finds it to be about 6 g. They first make an instinctive guess, then describe the relationship between the quantities, make this reasonable assumption, and only then calculate.
(i) Find the worth of the donated rice in rupees.
(ii) Find the number of 5-rupee coins needed to balance 60 kg, and write it as a power of 10.
(iii) Find the total value of those coins, in standard form.
(iv) List, in order, the steps of estimation the family followed.
Answer:
(i) Worth of rice = 60 × 55. Since 60 × 50 = 3000 and 60 × 5 = 300, the worth is ₹3300.
(ii) 60 kg = 60 × 1000 = 60000 g. Number of coins = \( \frac{60000}{6} \) = 10000 = 10⁴ coins.
(iii) Total value = 10⁴ × 5 = 50000 rupees = ₹5 × 10⁴.
(iv) First make an instinctive guess without calculating; then describe the relationship between the quantities (number of coins = total mass ÷ mass of one coin); then make reasonable assumptions or measurements for the unknowns (here, 6 g per coin); finally compute and check how close the first guess was.
In simple words: 60 kg of 6 g coins is 10⁴ coins, worth ₹50,000.

Teacher's Note:
1. Change kg to g before dividing by 6 g.
2. Estimation follows the steps guess, relate, assume, compute.

Question 18. Measuring time with powers of ten. The chapter places everyday and cosmic time spans on a scale of powers of ten seconds. Light takes about 8 minutes to reach the Earth from the Sun, which is of the order of 10² seconds. A satellite in a low orbit takes about 90 minutes for one round of the Earth, of the order of 10³ seconds. Going further up the scale, 10⁵ seconds is a little more than a day, 10⁶ seconds is under a fortnight, and 10⁹ seconds is about one-third of a long human life. Each step up the ladder multiplies the time by ten.
(i) How many seconds are there in one day? Write the answer in standard form.
(ii) About how many days are there in 10⁶ seconds?
(iii) A person who has lived for 10⁹ seconds is about how many years old?
(iv) A film runs for 2 hours. Which power of 10 seconds is its order of magnitude?
Answer:
(i) Seconds in a day = 24 × 60 × 60 = 86400 = 8.64 × 10⁴ seconds.
(ii) \( \frac{10^{6}}{86400} \) ≈ 11.6, so 10⁶ seconds is about 11.6 days, a little under a fortnight.
(iii) 10⁹ seconds = \( \frac{10^{9}}{8.64 \times 10^{4} \times 365.25} \) ≈ 31.7 years, so the person is about 32 years old.
(iv) 2 hours = 7200 seconds = 7.2 × 10³ seconds, so the power of 10 is 10³ seconds, the same as one orbit of a low-orbit satellite (5400 s).
In simple words: Divide by the seconds in a day or a year to turn big powers of 10 into days or years.

Teacher's Note:
1. Use 365.25 days in a year to allow for leap years.
2. In this chapter's scale, the order is read from the exponent in standard form.

Question 19. The game “Tremendous in Ten”. Rhea and Dev play a game. In ten seconds each writes a number or an expression using only the digits 0 to 9 and arithmetic signs, and the larger value wins the round. In Round 1, Rhea writes 10²⁰ and Dev writes 99999 × 99999 × 99999. In Round 2, Rhea writes 10⁵⁰⁰ + 10⁵⁰⁰ + 10⁵⁰⁰ + 10⁵⁰⁰ + 10⁵⁰⁰ while Dev writes 8000 × 10⁴⁹⁰. Dev complains that Rhea only wins because she is allowed to use exponents, and that his numbers “look” much bigger because they have far more digits written down.
(i) Show that Dev’s Round 1 number is less than 10¹⁵, without multiplying it out.
(ii) Who wins Round 1? Justify using powers of 10.
(iii) Write each player’s Round 2 number in standard form.
(iv) Who wins Round 2, and what does this tell you about comparing the coefficient and the exponent of a number in standard form?
Answer:
(i) Each factor 99999 is less than 10⁵. So the product of three such factors is less than 10⁵ × 10⁵ × 10⁵ = (10⁵)³ = 10¹⁵.
(ii) Rhea wins Round 1. Her number is 10²⁰, and 10²⁰ > 10¹⁵ > Dev’s number; in fact 10²⁰ is more than 10⁵ times Dev’s number.
(iii) Rhea: 10⁵⁰⁰ added five times = 5 × 10⁵⁰⁰. Dev: 8000 × 10⁴⁹⁰ = 8 × 10³ × 10⁴⁹⁰ = 8 × 10⁴⁹³.
(iv) Rhea wins Round 2, because her exponent 500 is larger than Dev’s 493, even though her coefficient 5 is smaller than his 8. Dividing, \( \frac{5 \times 10^{500}}{8 \times 10^{493}} \) = 0.625 × 10⁷ = 6.25 × 10⁶, so Rhea’s number is over 60 lakh times Dev’s.
When two numbers are in standard form, compare the exponents first; the coefficient matters only when the exponents are equal.
In simple words: The exponent decides which number is bigger; the coefficient matters only for equal exponents.

Teacher's Note:
1. Adding equal powers multiplies by the count: 5 × 10⁵⁰⁰, not 10²⁵⁰⁰.
2. Bound each factor by a power of 10 to compare quickly.

Question 20. Names for the powers of ten. Indian texts named very high powers of ten long ago. The Lalitavistara, a Buddhist treatise of the first century BCE, records a dialogue in which a hundred kotis make an ayuta (10⁹), a hundred ayutas a niyuta (10¹¹), a hundred niyutas a kankara (10¹³), and the list rises to a tallakshana (10⁵³). Mahaviracharya lists 24 terms up to 10²³, and a Pali treatise of Kaccayana names a number as high as 10¹⁴⁰. In modern Indian usage a hundred arab is a kharab (10¹¹) and a hundred kharab is a neel. The number 10¹⁰⁰ is called a googol, while the atoms in the universe are estimated at 10⁷⁸ to 10⁸².
(i) How many zeroes follow the digit 1 when a tallakshana is written out in full?
(ii) By what factor does each named step rise in the Lalitavistara list? Answer as a power of 10.
(iii) Express a neel as a power of 10, and name the Lalitavistara term equal to it.
(iv) How many times larger is a googol than the largest estimate for the number of atoms in the universe?
Answer:
(i) A tallakshana is 10⁵³, so it is written as 1 followed by 53 zeroes.
(ii) A hundred kotis make an ayuta, a hundred ayutas a niyuta, and so on, so each step is 100 = 10² times the previous one. The exponents 9, 11, 13 rise by 2 each time, which is why only odd powers of ten up to 10⁵³ get names.
(iii) A neel = 10² × 10¹¹ = 10¹³. The Lalitavistara term of the same value is the kankara.
(iv) The largest estimate is 10⁸², so \( \frac{10^{100}}{10^{82}} \) = 10¹⁰⁰⁻⁸² = 10¹⁸. A googol is 10¹⁸ times that estimate, far larger than any count of real objects.
In simple words: Each named step is 10² times the last, and a googol is 10¹⁸ times the number of atoms.

Teacher's Note:
1. The exponent of 10 equals the number of zeros.
2. Dividing powers of 10 means subtracting exponents.