Figure it Out — Powers, Combinatorics and Scientific Notation — Power Play

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12 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q1
Find out the units digit in the value of 22²⁴ ÷ 4³². [Hint: 4 = 2²]
Hint. Rewrite both sides with base 2, subtract the exponents, then look only at the units digit.
Step 1 — Rewrite 4³² with base 2. 4³² = (2²)³² = 2⁶⁴.

Step 2 — Divide. 2²²⁴ ÷ 2⁶⁴ = 2²²⁴⁻⁶⁴ = 2¹⁶⁰.

Step 3 — Find the units digit of 2¹⁶⁰. Powers of 2 cycle through units digits 2, 4, 8, 6 with period 4. Since 160 is a multiple of 4, 2¹⁶⁰ has the same units digit as 2⁴ = 16.

✦ Answer: The units digit is 6.
Where students slip. Trying to compute 2¹⁶⁰ in full — only the 4-cycle pattern of units digits (2,4,8,6) is needed, matched to where the exponent falls in that cycle.
22 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q2
There are 5 bottles in a container. Every day, a new container (also with 5 bottles) is brought in. How many bottles would there be after 40 days?
Hint. Work out how many containers accumulate in 40 days, then multiply by bottles per container.
Step 1 — Count the containers after 40 days. One new container arrives each day for 40 days, so there are 40 containers.

Step 2 — Multiply by bottles per container. 40 × 5 = 200.

Step 3 — Write in exponential/scientific form. 200 = 2 × 10².

✦ Answer: 200 = 2 × 10² bottles.
Where students slip. Assuming the number of bottles doubles or grows exponentially each day — this is steady day-by-day accumulation (addition of a fixed amount daily), not repeated doubling.
33 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q3
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
Hint. Try splitting the exponent into a sum, and also try rewriting the base itself as a product of powers.
Step 1 — (i) 64³. Since 64 = 2⁶, 64³ = 2¹⁸. Some valid splits: 2¹⁰ × 2⁸; 4⁵ × 4⁴ (since 4³=2⁶, this checks as (2²)⁹=2¹⁸); 8³ × 8³ (since 8⁶=2¹⁸).

Step 2 — (ii) 192⁸. Since 192 = 2⁶ × 3, 192⁸ = 2⁴⁸ × 3⁸. Some valid splits: 2⁴⁸ × 3⁸; 2⁴⁰ × 2⁸ × 3⁸; 2⁴⁰ × 6⁸ (since 6⁸=2⁸×3⁸).

Step 3 — (iii) 32⁻⁵. Since 32 = 2⁵, 32⁻⁵ = 2⁻²⁵. Some valid splits: 2⁻¹⁰ × 2⁻¹⁵; 2⁻⁵ × 2⁻²⁰; 4⁻¹² × 2⁻¹.

✦ Answer: (i) e.g. 2¹⁰ × 2⁸, 4⁵ × 4⁴, 8³ × 8³ (ii) e.g. 2⁴⁸ × 3⁸, 2⁴⁰ × 2⁸ × 3⁸, 2⁴⁰ × 6⁸ (iii) e.g. 2⁻¹⁰ × 2⁻¹⁵, 2⁻⁵ × 2⁻²⁰, 4⁻¹² × 2⁻¹.
Where students slip. Forgetting that the exponents in a valid split must add up to the original exponent (or, for a base-change split, that the new base's own exponent relationship must be tracked correctly).
44 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q4
Examine each statement 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 (q is a prime number).
Hint. Try to write each general form (n³, n⁴, n⁵, n₁³×n₂³, q⁴⁶) as a power of 2 to check the square/cube/4th/6th-power condition directly on the exponent.
Step 1 — (i) Cube numbers (n³) are squares only when 3 itself is a multiple of an even number pattern — specifically only when n is itself a perfect square, so n³ = (n^(3/2))² only works for numbers whose exponent structure allows it; concretely this is only true for numbers of the form n⁶ = (n²)³ = (n³)². Only Sometimes True.

Step 2 — (ii) Fourth powers (n⁴) can always be written as (n²)², a perfect square. Always True.

Step 3 — (iii) The fifth power n⁵ = n³ × n², so n⁵ is always divisible by n³. Always True.

Step 4 — (iv) The product of two cube numbers n₁³ × n₂³ = (n₁ × n₂)³, always another cube number. Always True.

Step 5 — (v) For q⁴⁶ to be a 4th power, 46 must be divisible by 4 (it isn't: 46/4 = 11.5); for it to be a 6th power, 46 must be divisible by 6 (it isn't: 46/6 ≈ 7.67). Never True.

✦ Answer: (i) Only Sometimes True (only for numbers that are 6th powers) (ii) Always True (iii) Always True (iv) Always True (v) Never True.
Where students slip. Assuming (i) is Always True by analogy with (ii) — being a cube doesn't automatically make a number a square; only numbers whose exponent is itself a multiple of 6 satisfy both conditions.
53 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q5
Simplify and write these in exponential form: (i) 10⁻² × 10⁻⁵ (ii) 5⁷ ÷ 5⁴ (iii) 9⁻⁷ ÷ 9⁴ (iv) (13⁻²)⁻³ (v) m⁵n¹²(mn)⁹
Hint. Use add-exponents for same-base multiplication, subtract-exponents for division, and multiply-exponents for a power of a power.
Step 1 — (i) Add exponents: −2 + (−5) = −7, so 10⁻² × 10⁻⁵ = 10⁻⁷.

Step 2 — (ii) Subtract exponents: 7 − 4 = 3, so 5⁷ ÷ 5⁴ = 5³.

Step 3 — (iii) Subtract exponents: −7 − 4 = −11, so 9⁻⁷ ÷ 9⁴ = 9⁻¹¹.

Step 4 — (iv) Multiply exponents: −2 × −3 = 6, so (13⁻²)⁻³ = 13⁶.

Step 5 — (v) Expand (mn)⁹ = m⁹n⁹, then combine with m⁵n¹²: m⁵⁺⁹n¹²⁺⁹ = m¹⁴n²¹.

✦ Answer: (i) 10⁻⁷ (ii) 5³ (iii) 9⁻¹¹ (iv) 13⁶ (v) m¹⁴n²¹.
Where students slip. Forgetting to expand (mn)⁹ into m⁹n⁹ before combining like bases in (v) — the two different letters' exponents must be added separately, not combined as if they were the same base.
63 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q6
If 12² = 144, what is (i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²?
Hint. Each number here is 12 shifted by a power of 10 — track how squaring affects that shift.
Step 1 — (i) 1.2 = 12 ÷ 10, so (1.2)² = 144 ÷ 10² = 144 ÷ 100 = 1.44.

Step 2 — (ii) 0.12 = 12 ÷ 100, so (0.12)² = 144 ÷ 100² = 144 ÷ 10000 = 0.0144.

Step 3 — (iii) 0.012 = 12 ÷ 1000, so (0.012)² = 144 ÷ 1000² = 144 ÷ 1000000 = 0.000144.

Step 4 — (iv) 120 = 12 × 10, so 120² = 144 × 10² = 144 × 100 = 14400.

✦ Answer: (i) 1.44 (ii) 0.0144 (iii) 0.000144 (iv) 14400.
Where students slip. Shifting the decimal point in the answer by the same number of places as in the question — since squaring doubles the shift, the decimal point actually moves twice as many places as in the original number.
72 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q7
Circle the numbers that are the same: 2⁴ × 3⁶, 6⁴ × 3², 6¹⁰, 18² × 6², 6²⁴
Hint. Convert everything to a product of powers of 2 and 3 to compare directly.
Step 1 — Convert 2⁴ × 3⁶ — already in base form.

Step 2 — Convert 6⁴ × 3². 6⁴ = 2⁴ × 3⁴, so 6⁴ × 3² = 2⁴ × 3⁶ — matches.

Step 3 — Convert 6¹⁰. 6¹⁰ = 2¹⁰ × 3¹⁰ — does not match 2⁴ × 3⁶.

Step 4 — Convert 18² × 6². 18 = 2 × 3², so 18² = 2² × 3⁴; combined with 6² = 2² × 3², total = 2⁴ × 3⁶ — matches.

Step 5 — Convert 6²⁴. 6²⁴ = 2²⁴ × 3²⁴ — does not match.

✦ Answer: 2⁴ × 3⁶, 6⁴ × 3², and 18² × 6² are all equal; 6¹⁰ and 6²⁴ are not.
Where students slip. Comparing the numbers by rough size instead of breaking each one into its prime factors — converting every expression to powers of 2 and 3 makes the matches exact and checkable.
82 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q8
Identify the greater number in each of the following: (i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰
Hint. Evaluate the smaller-looking pair directly; for the last one, compare their sizes using powers of the same base where possible.
Step 1 — (i) 4³ = 64 and 3⁴ = 81, so 3⁴ is greater.

Step 2 — (ii) 2⁸ = 256 and 8² = 64, so 2⁸ is greater.

Step 3 — (iii) 100² = 10⁴, while 2¹⁰⁰ is vastly larger (2¹⁰⁰ is far beyond 10³⁰), so 2¹⁰⁰ is greater.

✦ Answer: (i) 3⁴ (ii) 2⁸ (iii) 2¹⁰⁰ — all greater than the other option in each pair.
Where students slip. Assuming the number with the larger base is always greater — 8² looks bigger than 2⁸ at a glance, but the one with the larger exponent usually wins once both are actually evaluated.
93 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q9
A dairy plans to produce 8.5 billion packets of milk in a year and wants a unique ID code for each packet, using digits 0-9. How many digits should the code consist of?
Hint. Find the smallest n such that 10ⁿ is at least as large as the number of packets.
Step 1 — Write the number of packets as a power of 10 for comparison. 8.5 billion = 8,500,000,000.

Step 2 — Compare against powers of 10. 10⁹ = 1,000,000,000 is smaller than 8.5 billion, but 10¹⁰ = 10,000,000,000 is larger.

Step 3 — Conclude the minimum number of digits. Since an n-digit code gives 10ⁿ possible codes, and n = 9 isn't enough while n = 10 is, at least 10 digits are needed.

✦ Answer: At least 10 digits.
Where students slip. Using the number of digits in '8,500,000,000' itself (10 digits) as the answer without checking — this happens to be correct here, but the actual requirement is that 10ⁿ must be at least the packet count, which needs to be verified rather than assumed.
103 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q10
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?
Hint. Think about what condition an exponent must satisfy to be both a multiple of 2 and a multiple of 3.
Step 1 — Recall why 64 works. 64 = 2⁶, and 6 is a multiple of both 2 (so it's a square, (2³)²) and 3 (so it's a cube, (2²)³).

Step 2 — Generalise the condition. Any number of the form n⁶ is both a square ((n³)²) and a cube ((n²)³), since 6 is divisible by both 2 and 3.

Step 3 — Give more examples. 1⁶ = 1, 2⁶ = 64, 3⁶ = 729, 4⁶ = 4096, and so on — infinitely many.

✦ Answer: Yes, there are infinitely many such numbers — they are exactly the 6th powers (n⁶), such as 1, 64, 729, 4096, ...
Where students slip. Searching for more square-and-cube numbers by trial and error — recognising that the exponent needs to be a multiple of 6 (the LCM of 2 and 3) generates every such number directly.
112 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q11
A digital locker has an alphanumeric passcode of length 5 (digits 0-9 and letters A-Z). How many such codes are possible?
Hint. Count the total number of available characters first, then raise it to the power of the code length.
Step 1 — Count the available characters. 10 digits + 26 letters = 36 possible characters for each position.

Step 2 — Raise to the power of the code length. Each of the 5 positions independently has 36 choices, so the total is 36⁵.

✦ Answer: 36⁵ possible codes.
Where students slip. Using only 26 or only 10 as the base — the passcode allows both digits and letters at every position, so the base must be the combined count of 36.
122 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q12
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
Hint. This is a straightforward addition of two equal quantities, not a multiplication of exponents.
Step 1 — Add the two populations. 10⁹ (sheep) + 10⁹ (goats), since the two populations are separate totals that must be combined, not multiplied.

Step 2 — Recognise this equals doubling one quantity. 10⁹ + 10⁹ = 2 × 10⁹, which is the same value however it's written.

✦ Answer: (vi) 10⁹ + 10⁹, which is the same value as (v) 2 × 10⁹.
Where students slip. Multiplying the two populations (giving 10¹⁸) instead of adding them — the question asks for the total (combined) population, which is a sum, not a product.
134 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q13
Calculate and write the answer in scientific notation: (i) If each person in the world (~8.2 × 10⁹) had 30 pieces of clothing, find the total 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) Estimate the total time spent eating in a lifetime, in seconds (assume 70 years and 1 hour of eating per day).
Hint. These repeat the earlier thought-experiment calculations — convert every quantity to scientific notation first, then multiply.
Step 1 — (i) 8.2 × 10⁹ × 30 = 246 × 10⁹ = 2.46 × 10¹¹.

Step 2 — (ii) 10⁸ colonies × 5 × 10⁴ bees = 5 × 10¹².

Step 3 — (iii) 3.8 × 10¹³ cells/person × 8.2 × 10⁹ people = 31.16 × 10²² = 3.116 × 10²³.

Step 4 — (iv) 3600 seconds/day × 365 × 70 days = 91,980,000 = 9.198 × 10⁷ seconds, since converting to seconds first keeps every factor in the same unit before multiplying.

✦ Answer: (i) 2.46 × 10¹¹ pieces of clothing (ii) 5.0 × 10¹² honeybees (iii) 3.116 × 10²³ bacterial cells (iv) 9.198 × 10⁷ seconds.
Where students slip. Leaving intermediate coefficients above 10 (like 246 or 31.16) instead of converting them back into standard scientific-notation form by folding the extra factor of 10 into the exponent.
143 marksGanita Prakash Cl-8 Part 1, Figure it Out (§2.6), Q14
What was the date 1 arab (1 billion) seconds ago?
Hint. Convert a billion seconds into years by dividing by the number of seconds in a year, then count back from today.
Step 1 — Find the number of seconds in a year. 365 × 24 × 60 × 60 ≈ 3.15 × 10⁷ seconds.

Step 2 — Divide a billion seconds by this. 10⁹ ÷ (3.15 × 10⁷) ≈ 31.7 years, since dividing total seconds by seconds-per-year converts the count directly into years.

Step 3 — State the result relative to today. A billion seconds is about 31.7 years, or roughly 11,574 days.

✦ Answer: About 31.7 years (roughly 11,574 days) before today.
Where students slip. Confusing a billion seconds with a billion minutes, hours, or days — a billion seconds is only about 32 years, a far smaller span than a billion of any larger time unit.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp102.pdf). Questions are scattered as 'Math Talk'/'Try This' prompts through the running text, plus two formal 'Figure it Out' blocks. Every answer here is checked against the book's own printed answer key at the end of the chapter.. Questions are referenced from the NCERT textbook for identification.

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