This MCQ module is based on: Large Numbers, Comparisons & Chapter Exercises
TOPIC 6 OF 22
Large Numbers, Comparisons & Chapter Exercises
🎓 Class 8
Mathematics
CBSE
Theory
Ch 2 — Power Play (Exponents and Powers)
⏱ ~35 min
🌐 Language: [gtranslate]
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Targeting Class 8 level in Algebra, with Basic difficulty.
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Chapter 2: Power Play
Class 8 Mathematics — Ganita Prakash Part-I | Part 3 of 3: Large Numbers, History & Exercises
2.5 Did You Ever Wonder?
Large numbers appear naturally when we try to measure real-world quantities. The key skill is: guess first, then calculate using estimation and approximation.
Activity: The Problem
Nanjundappa wants to donate jaggery equal to Roxie's weight (13 years old, ~45 kg) and wheat equal to Estu's weight (11 years old, ~50 kg).
What if instead of jaggery, we use 1-rupee coins equal to Roxie's weight?
Step 1 — Guess first: Would it be hundreds, thousands, lakhs, or crores of coins?
Step 2 — Calculate:
- Weight of a 1-rupee coin ≈ 4.85 g ≈ 5 g
- Roxie's weight ≈ 45 kg = 45,000 g
- Number of coins = \(\frac{45000}{5} = 9000\) coins
- Value = 9000 × ₹1 = ₹9,000
Only about 9,000 coins — far less than a lakh! (Most people overestimate because coins feel light individually but are heavy in bulk.)
How Many Steps to the Moon?
Linear growth: ~192 crore steps to reach the Moon at 20 cm per step
Estimation: Steps to the Moon
Distance to Moon = 3,84,400 km = \(3.844 \times 10^8\) m = \(3.844 \times 10^{10}\) cm
Step height = 20 cm
Steps needed = \(\dfrac{3.844 \times 10^{10}}{20} = \dfrac{3.844 \times 10^{10}}{2 \times 10^1} = 1.922 \times 10^9\)
= 192 crore and 20 lakh steps with linear growth
vs just 46 paper folds with exponential growth — that's the difference!
Getting a Sense for Large Numbers
Powers of 10 let us compare quantities across wildly different scales. Here is a tour of the world's populations — from almost extinct species to all the ants on Earth!
10⁰
Northern White Rhino
Only 2 remaining (2 × 10⁰) — both females, Kenya
10¹
Hainan Gibbon
~42 alive (≈ 4 × 10¹) as of 2024
10²
Kākāpō (Parrot)
242 alive (≈ 2 × 10²) mid-2025
10³
Komodo Dragon
< 3000 (≈ 3 × 10³), all in Indonesia
10⁴
Maned Wolf
> 17,000 (≈ 1.7 × 10⁴), mostly Brazil
10⁵
African Elephant
~4.15 lakh (≈ 4 × 10⁵) as of 2018
10⁶
American Alligator
~50 lakh / 5 million (5 × 10⁶)
10⁷
Camel (global)
~3.5 crore / 35 million (3.5 × 10⁷)
10⁸
Water Buffalo
> 20 crore / 200 million (2 × 10⁸)
10⁹
Starling + Humans
Starlings ≈ 1.3 × 10⁹; Humans ≈ 8.2 × 10⁹
10¹⁰
Chicken
~33 billion (3.3 × 10¹⁰) at any time
10¹²
Trees (global)
~3 trillion (3 × 10¹²) — 2023 estimate
10¹⁴
Mosquito / Krill
Mosquitoes ≈ 1.1 × 10¹⁴; Krill ≈ 5 × 10¹⁴
10¹⁵
Beetle / Earthworm
Each ≈ 1 × 10¹⁵ (1 quadrillion)
10¹⁶
Ants
~20 quadrillion (2 × 10¹⁶) — ants alone outweigh all wild birds + mammals!
10²³
Stars in Universe
≈ 2 × 10²³ observable stars
A staggering fact — ants (2 × 10¹⁶) collectively outweigh all wild vertebrates
A Different Way to Say Your Age — Time in Seconds
Roxie says "I am 4840 days old today." Can you find her age in hours? In seconds? These numbers are best expressed using scientific notation and powers of 10.
Quick Estimation
Roxie is 13 years old. Her age in seconds:
\(13 \times 365.25 \times 24 \times 3600 \approx 13 \times 3.156 \times 10^7 \approx 4.1 \times 10^8\) seconds
If you have lived for a million seconds (\(10^6\) s): \(10^6 \div 3600 \approx 277.8\) hours ≈ 11.6 days.
| Power | ≈ Real time | Real-world example |
|---|---|---|
| 10⁰ = 1 s | 1 second | A ball thrown up takes ~2–3 s to return |
| 10¹ ≈ 10 s | 10 seconds | Blood circulation takes 10–20 s; waiting at a traffic signal |
| 10² ≈ 1.6 min | 100 seconds | Making tea: 5–10 min (≈ 4–8 × 10² s); light from Sun: ~500 s |
| 10³ ≈ 17 min | 1,000 seconds | Satellites in low Earth orbit: 90 min (≈ 5.5 × 10³ s) per revolution |
| 10⁴ ≈ 2.7 hr | 10,000 s | Digesting a meal: 2–4 hours; lifespan of a mayfly: ~1 day (≈ 9 × 10⁴ s) |
| 10⁵ ≈ 1.2 days | 1 lakh seconds | A long weekend; a cross-country train journey |
| 10⁶ ≈ 11.6 days | 10 lakh seconds | A school term break; time for a spacecraft to reach the Moon |
| 10⁷ ≈ 3.8 months | 1 crore seconds | Mangalyaan's journey to Mars: 298 days (≈ 2.65 × 10⁷ s) |
| 10⁸ ≈ 3.2 years | 10 crore seconds | Typical dog lifespan: 3–15 years |
| 10⁹ ≈ 31.7 years | 1 arab seconds | Halley's Comet: 75–79 year orbit (≈ 2.4 × 10⁹ s); Neptune's year ≈ 5.2 × 10⁹ s |
| 10¹⁰ ≈ 317 years | Chola dynasty ruled for >900 years (≈ 3 × 10¹⁰ s) | |
| 10¹¹ ≈ 3,170 years | Oldest known living tree: ~5,000 years (≈ 1.57 × 10¹¹ s) | |
| 10¹² ≈ 31,700 years | Early Homo sapiens appeared 2–3 lakh years ago (≈ 7–9 × 10¹² s) | |
| 10¹⁵ ≈ 3.17 crore yr | Himalayas formed: 5.5 crore years ago (≈ 1.7 × 10¹⁵ s); dinosaurs extinct 6.6 crore years ago (≈ 2 × 10¹⁵ s) | |
| 10¹⁷ ≈ 3.17 billion yr | Earth is 4.5 billion years old; Universe formed 13.8 billion years ago (≈ 4.4 × 10¹⁷ s) |
Remarkable: \(10^9\) seconds ≈ 31.7 years — roughly half a human lifespan. Yet \(10^{18}\) seconds ago the universe did not even exist according to modern physics! Exponential notation spans human life to cosmic time in just a handful of digits.
2.5 A Pinch of History
Ancient Indian Number Names
In the Lalitavistara, a Buddhist treatise from the 1st century BCE, number-names for odd powers of 10 appear up to 10⁵³ in a dialogue between the mathematician Arjuna and Prince Gautama (the Bodhisattva):
"Hundred kotis are called an ayuta (10⁹), hundred ayutas a niyuta (10¹¹)... hundred sarvajnas a vibhutangama (10⁵¹), a hundred vibhutangamas is a tallakshana (10⁵³)."
Mahaviracharya's Ganita-sara-sangraha lists up to 10²³. The Jaina treatise Amalasiddhi gives names for every power of 10 up to 10⁹⁶. A Pali grammar lists up to 10¹⁴⁰ (named asaṅkhyeya)!
Indian Naming System
| Name | Value | Exponential |
|---|---|---|
| Lakh | 1,00,000 | \(10^5\) |
| Crore | 1,00,00,000 | \(10^7\) |
| Arab | 1,00,00,00,000 | \(10^9\) |
| Kharab | 10¹¹ | \(10^{11}\) |
| Neel | 10¹³ | \(10^{13}\) |
| Padma | 10¹⁵ | \(10^{15}\) |
| Shankh | 10¹⁷ | \(10^{17}\) |
| Maha Shankh | 10¹⁹ | \(10^{19}\) |
Pattern: each unit = 100 × previous
International Naming System
| Name | Value | Exponential |
|---|---|---|
| Million | 1,000,000 | \(10^6\) |
| Billion | 10⁹ | \(10^9\) |
| Trillion | 10¹² | \(10^{12}\) |
| Quadrillion | 10¹⁵ | \(10^{15}\) |
| Quintillion | 10¹⁸ | \(10^{18}\) |
| Sextillion | 10²¹ | \(10^{21}\) |
| Septillion | 10²⁴ | \(10^{24}\) |
| Decillion | 10³³ | \(10^{33}\) |
Pattern: each unit = 1000 × previous
and
\(10^{100}\) is called a googol. The estimated number of atoms in the universe is only \(10^{78}\) to \(10^{82}\) — far less than a googol!
\(10^{\text{googol}} = 10^{(10^{100})}\) is called a googolplex. Writing it out — even one digit per atom in the universe — is physically impossible.
The highest denomination currency note ever printed: 1 sextillion pengő (\(10^{21}\)) in Hungary, 1946 — never issued. Zimbabwe's 100 trillion dollar note (\(10^{14}\) Zimbabwe dollars) was worth about $30 when printed in 2009!
Figure it Out — All Questions (Pages 44–45)
Q1. Find the units digit in the value of \(2^{224} \div 4^{32}\).
Hint: \(4 = 2^2\)
\(4^{32} = (2^2)^{32} = 2^{64}\)
\(2^{224} \div 2^{64} = 2^{224-64} = 2^{160}\)
Units digit of powers of 2 cycle as 2, 4, 8, 6 (period 4).
\(160 \div 4 = 40\) with remainder 0. Remainder 0 → same as remainder 4 → units digit = 6.
Q2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would there be after 40 days?
Each day a new container (of 5 bottles) is added. After 40 days: \(5 \times 40 = 200\) bottles.
Note: This is linear growth (adding 5 each day), not exponential. The answer is 200.
Q3. Write \(64^3\), \(192^8\), and \(32^{-5}\) each as a product of two or more powers in three different ways.
(i) \(64^3\): Note \(64 = 2^6\), so \(64^3 = 2^{18}\)
Three ways: \((2^9)^2\) | \((2^6)^3\) | \((2^3)^6\) | \(2^{10} \times 2^8\) etc.
(ii) \(192^8\): \(192 = 64 \times 3 = 2^6 \times 3\), so \(192^8 = 2^{48} \times 3^8\)
Three ways: \((2^{48} \times 3^8)\) | \((2^{24})^2 \times 3^8\) | \(2^{48} \times (3^4)^2\)
(iii) \(32^{-5}\): \(32 = 2^5\), so \(32^{-5} = (2^5)^{-5} = 2^{-25}\)
Three ways: \(2^{-25}\) | \((2^{-5})^5\) | \(2^{-12} \times 2^{-13}\)
Q4. Examine each statement — is it 'Always True', 'Only Sometimes True', or 'Never True'?
(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 prime).
(i) Only Sometimes True. E.g. \(64 = 4^3 = 8^2\) ✓, but \(8 = 2^3\) is not a perfect square.
(ii) Always True. \(n^4 = (n^2)^2\) — the 4th power is always a perfect square.
(iii) Always True. \(n^5 \div n^3 = n^2\), which is always an integer when n is an integer. So \(n^5\) is always divisible by \(n^3\).
(iv) Always True. \(a^3 \times b^3 = (ab)^3\) — always a perfect cube (by Law 3).
(v) Always True (for prime q). \(q^{46} = (q^{23})^2\) → square; \(q^{46} = (q^{11.5})^4\)? No — but \(q^{46} = (q^{46/4})\) needs integer. Let's check: 46/4 is not integer, but \(q^{46} = (q^{23})^2\) is a square. Is it a 4th power? Only if \(4 | 46\) — no. Is it a 6th power? \(q^{46} = (q^{46/6})\) — not integer. Hmm — actually q⁴⁶: for 4th power need exponent divisible by 4 (46/4 not integer). For 6th power need 6|46 (no). So the answer is Never True for a prime q (since 4 ∤ 46 and 6 ∤ 46).
Q5. Simplify and write in 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) \(m^5 n^{12} (mn)^9\)
(i) \(10^{-2+(-5)} = 10^{-7}\)
(ii) \(5^{7-4} = 5^3 = 125\)
(iii) \(9^{-7-4} = 9^{-11}\)
(iv) \(13^{(-2)\times(-3)} = 13^6\)
(v) \(m^5 n^{12} \times m^9 n^9 = m^{5+9} \times n^{12+9} = m^{14} n^{21}\)
Q6. If \(12^2 = 144\), what is (i) \((1.2)^2\) (ii) \((0.12)^2\) (iii) \((0.012)^2\) (iv) \(120^2\)?
(i) \((1.2)^2 = (12 \times 10^{-1})^2 = 144 \times 10^{-2} = \mathbf{1.44}\)
(ii) \((0.12)^2 = (12 \times 10^{-2})^2 = 144 \times 10^{-4} = \mathbf{0.0144}\)
(iii) \((0.012)^2 = (12 \times 10^{-3})^2 = 144 \times 10^{-6} = \mathbf{0.000144}\)
(iv) \(120^2 = (12 \times 10)^2 = 144 \times 100 = \mathbf{14400}\)
Q7. Circle the numbers that are the same: \(2^4 \times 3^6\) \(6^4 \times 3^2\) \(6^{10}\) \(18^2 \times 6^2\) \(6^{24}\)
Compute each in terms of primes:
\(2^4 \times 3^6 = 16 \times 729 = 11664\)
\(6^4 \times 3^2 = 1296 \times 9 = 11664\) ← same!
Check: \(6^4 \times 3^2 = (2\times3)^4 \times 3^2 = 2^4 \times 3^4 \times 3^2 = 2^4 \times 3^6\) ✓
\(18^2 \times 6^2 = (18 \times 6)^2 = 108^2 = 11664\) ← same!
Check: \(18^2 \times 6^2 = (2\times3^2)^2 \times (2\times3)^2 = 2^2\times3^4\times2^2\times3^2 = 2^4\times3^6\) ✓
\(6^{10} = 60{,}466{,}176\) — different. \(6^{24}\) — much larger, different.
Answer: \(\boxed{2^4 \times 3^6\text{,}\quad 6^4 \times 3^2\text{, and}\quad 18^2 \times 6^2}\) are all equal to 11,664.
Q8. Identify the greater number in each pair: (i) \(4^3\) or \(3^4\) (ii) \(2^8\) or \(8^2\) (iii) \(100^2\) or \(2^{100}\)
(i) \(4^3 = 64\) vs \(3^4 = 81\). Greater: \(3^4 = 81\)
(ii) \(2^8 = 256\) vs \(8^2 = 64\). Greater: \(2^8 = 256\)
(iii) \(100^2 = 10{,}000 = 10^4\) vs \(2^{100} \approx 1.27 \times 10^{30}\). Greater: \(2^{100}\) — by an enormous margin!
Q9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID code using digits 0–9. How many digits should the code consist of?
8.5 billion = \(8.5 \times 10^9 \approx 10^{10}\)
A code with \(n\) digits can produce \(10^n\) unique IDs.
We need \(10^n \geq 8.5 \times 10^9\)
\(10^{10} = 10{,}000{,}000{,}000 \geq 8{,}500{,}000{,}000\) ✓
So the code must be at least 10 digits long.
Q10. 64 is both a square number (\(8^2\)) and a cube number (\(4^3\)). Are there other such numbers? How can we describe them in general? [Math Talk]
Numbers that are both perfect squares and perfect cubes must be perfect 6th powers!
Reason: If \(n = a^2 = b^3\), then \(n = k^6\) for some integer \(k\) (since lcm(2,3) = 6).
Examples: \(1^6 = 1\), \(2^6 = 64\), \(3^6 = 729\), \(4^6 = 4096\), \(5^6 = 15625\), ...
General form: \(n = k^6\) for any positive integer k.
Q11. A digital locker has an alphanumeric passcode of length 5 (digits 0–9 and letters A–Z). How many such codes are possible?
Each position can be any of 10 digits + 26 letters = 36 options.
For 5 positions: \(36^5 = 36 \times 36 \times 36 \times 36 \times 36\)
\(36^2 = 1296\), \(36^3 = 46{,}656\), \(36^4 = 16{,}79{,}616\), \(36^5 = 6{,}04{,}66{,}176\)
= 6,04,66,176 ≈ \(6 \times 10^7\) possible codes.
Q12. The worldwide population of sheep (2024) is about \(10^9\), and goats also about \(10^9\). What is the total population of sheep and goats?
(i) \(20^9\) (ii) \(10^{11}\) (iii) \(10^{10}\) (iv) \(10^{18}\) (v) \(2 \times 10^9\) (vi) \(10^9 + 10^9\)
Total = \(10^9 + 10^9 = 2 \times 10^9\).
Correct options: (v) and (vi) — they are equal to each other.
Note: (i) \(20^9 \neq 2 \times 10^9\); (ii) and (iii) are far larger; (iv) is multiplication, not addition.
Q13. Calculate and write in scientific notation: (i) Each of 8.2 × 10⁹ people has 30 clothing pieces — total pieces? (ii) 100 million bee colonies, 50,000 bees each — total honeybees? (iii) Human body has ~38 trillion bacteria; total for all humans? (iv) Total time eating in a lifetime (in seconds).
(i) \(8.2 \times 10^9 \times 30 = 8.2 \times 30 \times 10^9 = 246 \times 10^9 = \mathbf{2.46 \times 10^{11}}\)
(ii) \(100\ \text{million} = 10^8\) colonies. \(10^8 \times 5 \times 10^4 = 5 \times 10^{12}\) bees = \(5 \times 10^{12}\)
(iii) \(38 \times 10^{12} \times 8.2 \times 10^9 = 38 \times 8.2 \times 10^{21} = 311.6 \times 10^{21} = \mathbf{3.116 \times 10^{23}}\)
(iv) Assume: eat 3 times/day × 20 min each = 60 min/day = 1 hr/day; lifetime ≈ 75 years.
Total = \(75 \times 365 \times 3600\) s \(\approx 75 \times 3.6 \times 10^3 \times 3.65 \times 10^2 \approx 75 \times 1.31 \times 10^6 \approx 9.9 \times 10^7\) seconds = ≈ \(10^8\) seconds
Q14. [Try This] What was the date 1 arab (1 billion) seconds ago?
1 billion seconds = \(10^9\) s
\(10^9 \div (365.25 \times 24 \times 3600) \approx \frac{10^9}{3.156 \times 10^7} \approx 31.7\) years
From 2026: 2026 − 31.7 ≈ around the year 1994
More precisely: \(10^9\) seconds = 31 years, 251 days, 13 hours, 34 minutes, 8 seconds ago from today (April 10, 2026) ≈ early August 1994.
Chapter Summary
Definition
\(n^a = n \times n \times \ldots \times n\) (a times); \(n^{-a} = \frac{1}{n^a}\)
Product Law
\(n^a \times n^b = n^{a+b}\)
Power of Power
\((n^a)^b = n^{ab}\)
Division Law
\(n^a \div n^b = n^{a-b}\) (n ≠ 0)
Same Exponent
\(m^a \times n^a = (mn)^a\) and \(m^a \div n^a = (m/n)^a\)
Zero & Negative
\(n^0 = 1\) (n ≠ 0) | \(n^{-a} = \frac{1}{n^a}\)
Scientific Notation
\(x \times 10^y\) where \(1 \leq x < 10\), y is any integer
Growth Types
Linear (additive) vs Exponential (multiplicative) — exponential grows far faster
Puzzle Time: Tremendous in Ten!
Play with a partner. In 10 seconds, write a number or expression using only the digits 0–9 and arithmetic operations. Whoever writes the larger number wins the round!
Round 1: Roxie wrote \(10^{13}\). Estu wrote \(999{,}999 \times 999{,}999\).
Roxie's number is \(10^{13}\). Estu's is less than \((10^6)^2 = 10^{12}\). Roxie wins!
Round 2: Roxie wrote \(4 \times 10^{1000}\). Estu wrote \((10^{1{,}000{,}000}) \times 9000 = 9 \times 10^{1{,}000{,}003}\).
Estu's number has exponent 1,000,003 vs Roxie's exponent 1,000. Estu wins Round 2!
Try these rule variants:
(i) Exponents not allowed, only addition (ii) Exponents not allowed, addition + multiplication (iii) Exponents allowed, only addition (iv) Exponents allowed, any operation
Interactive: Estimation Calculator
Order of Magnitude Explorer
Enter two quantities to compare their orders of magnitude
Competency-Based Questions
Q1. A species of fern produces \(10^6\) spores per plant. Each spore that germinates produces another plant with \(10^6\) spores. If exactly 1 spore from each plant germinates, how many spores exist after 3 generations? Express in scientific notation. L1 Remember
Generation 1: \(10^6\) spores. Generation 2: \(10^6 \times 10^6 = 10^{12}\) spores. Generation 3: \(10^{12} \times 10^6 = 10^{18}\) spores.
Answer: \(\mathbf{10^{18}}\) spores (1 quintillion).
Q2. The number of trees globally is \(3 \times 10^{12}\). If each tree has \(2 \times 10^5\) leaves, find the total number of leaves on Earth. Express in scientific notation. L2 Understand
Total leaves = \(3 \times 10^{12} \times 2 \times 10^5 = 6 \times 10^{17}\)
= \(6 \times 10^{17}\) leaves (600 quadrillion).
Q3. An international student exchange programme sends students in the ratio of human population to African elephant population (about 8 × 10⁹ humans, 4 × 10⁵ elephants). How many humans correspond to every 1 elephant? Use scientific notation. L3 Apply
\(\frac{8 \times 10^9}{4 \times 10^5} = \frac{8}{4} \times 10^{9-5} = 2 \times 10^4 = 20{,}000\)
There are approximately 20,000 humans for every African elephant.
Q4. Meera claims: "Since \(2^{10} = 1024 \approx 10^3\), we can say \(2^{100} \approx 10^{30}\)." Evaluate her reasoning and determine whether it leads to an over-estimate or under-estimate. L4 Analyse
Meera's reasoning: \(2^{100} = (2^{10})^{10} \approx (10^3)^{10} = 10^{30}\)
Actual: \(2^{10} = 1024 > 10^3\), so \((1024)^{10} > (10^3)^{10} = 10^{30}\)
Therefore \(2^{100} > 10^{30}\) — Meera's estimate is an under-estimate.
More precisely: \(2^{100} \approx 1.268 \times 10^{30}\). So her approximation is close but slightly low.
Q5. Design a system to give every grain of sand on Earth (\(\approx 10^{21}\) grains) a unique numeric ID using digits 0–9. What is the minimum number of digits needed? If instead you used alphanumeric codes (36 symbols), how many symbols suffice? Compare and explain which system is more efficient. L6 Create
Numeric: Need \(10^n \geq 10^{21}\), so n ≥ 22 digits.
Alphanumeric (36 symbols): Need \(36^k \geq 10^{21}\).
\(\log_{36}(10^{21}) = \frac{21}{\log_{10}36} = \frac{21}{1.556} \approx 13.5\). So k ≥ 14 symbols.
The alphanumeric system is more efficient — uses 14 symbols vs 22 digits — because each symbol carries more information (\(\log_2 36 \approx 5.2\) bits vs \(\log_2 10 \approx 3.3\) bits per digit).
Assertion–Reason Questions
Options: (A) Both true, Reason is correct explanation. (B) Both true, Reason is NOT correct explanation. (C) Assertion true, Reason false. (D) Assertion false, Reason true.
ARQ 1.
Assertion: Numbers that are both perfect squares and perfect cubes must be perfect 6th powers.
Reason: The LCM of 2 (for square) and 3 (for cube) is 6, so the exponent must be a multiple of 6.
(A) — Both true. If \(n = a^2 = b^3\), the prime factorisation of n must have all exponents divisible by both 2 and 3, hence divisible by lcm(2,3) = 6. So n is a perfect 6th power. ✓
ARQ 2.
Assertion: Ants (\(\approx 2 \times 10^{16}\)) are more numerous than all stars in the observable universe (\(\approx 2 \times 10^{23}\)).
Reason: When comparing numbers in scientific notation, the one with the higher exponent is larger.
(D) — Assertion is false: Stars (\(2 \times 10^{23}\)) far outnumber ants (\(2 \times 10^{16}\)) — stars are \(10^7\) times more numerous! The Reason is true and is a valid general rule for scientific notation comparison. The Assertion wrongly states ants are more numerous than stars.
ARQ 3.
Assertion: In scientific notation, changing the exponent from 7 to 8 causes a 10× change, but changing the coefficient from 2 to 3 causes only a 1.5× change.
Reason: The exponent represents the order of magnitude — a change of 1 in the exponent multiplies the value by 10.
(A) — Both true. \(2 \times 10^8 \div 2 \times 10^7 = 10\) (exponent change → 10× change). \(3 \times 10^7 \div 2 \times 10^7 = 1.5\) (coefficient change → 1.5× change). The Reason correctly explains why the exponent is more significant than the coefficient. ✓
Frequently Asked Questions — Power Play (Exponents and Powers)
What is Large Numbers, Comparisons & Chapter Exercises in NCERT Class 8 Mathematics?
Large Numbers, Comparisons & Chapter Exercises is a key concept covered in NCERT Class 8 Mathematics, Chapter 2: Power Play (Exponents and Powers). This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Large Numbers, Comparisons & Chapter Exercises step by step?
To solve problems on Large Numbers, Comparisons & Chapter Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 2: Power Play (Exponents and Powers)?
The essential formulas of Chapter 2 (Power Play (Exponents and Powers)) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Large Numbers, Comparisons & Chapter Exercises important for the Class 8 board exam?
Large Numbers, Comparisons & Chapter Exercises is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Large Numbers, Comparisons & Chapter Exercises?
Common mistakes in Large Numbers, Comparisons & Chapter Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Large Numbers, Comparisons & Chapter Exercises?
End-of-chapter NCERT exercises for Large Numbers, Comparisons & Chapter Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.
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