Class 8 Ganita Prakash Chapter 2 Power Play Solutions

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Class 8 Ganita Prakash · Chapter 2
Power Play

Page-wise textbook solutions (pages 19–47) · Every step · Figures · Common mistakes · Tips

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✔ Understood 0/81

🗝️ Key ideas of this chapter

• \(n^a\) means \(n\) multiplied by itself \(a\) times. \(n\) is the base, \(a\) the exponent.
• Laws: \(n^a \times n^b = n^{a+b}\), \(\;n^a \div n^b = n^{a-b}\), \(\;(n^a)^b = n^{ab}\), \(\;m^a \times n^a = (mn)^a\), \(\;\dfrac{m^a}{n^a} = \left(\dfrac{m}{n}\right)^a\).
• \(n^0 = 1\) and \(n^{-a} = \dfrac{1}{n^a}\) (when \(n \ne 0\)).
• Scientific form: \(x \times 10^y\) with \(1 \le x \lt 10\). The exponent \(y\) matters most.
• Linear growth = repeated adding; exponential growth = repeated multiplying (much faster).

📖 Textbook Page 19

📖 Page 19Activity

Take a big sheet of paper and keep folding it in half. How many times can you fold it?

1Try with a newspaper, a notebook page and a tissue.

2Each fold doubles the number of layers: 2, 4, 8, 16, 32, 64, 128 …

3After 7 folds there are 128 layers — too thick and too small to fold again by hand.

✅ Answer: Usually only about 7 or 8 times, whatever the paper.

💡 Tip: It gets hard quickly because the layers double every time — that is exponential growth.

📖 Page 19Guess

Suppose you could fold a sheet as many times as you like. Guess its thickness after 30 folds. (Thickness of the sheet = 0.001 cm.)

1Thickness after \(n\) folds \(= 0.001 \times 2^n\) cm.

2\(2^{30} = 1\,073\,741\,824\)

3\(0.001 \times 1\,073\,741\,824 \approx 10\,73\,742\) cm \(\approx 10.7\) km

Thickness doubles each fold: 0.001 cm × 2ⁿ. Small at first, then huge!

✅ Answer: About 10.7 km — the height at which aeroplanes fly!

📖 Textbook Page 20

📖 Page 20Math Talk

Guess the thickness after 30 folds and after 45 folds.

130 folds: \(0.001 \times 2^{30}\) cm \(\approx 10.7\) km.

245 folds: \(0.001 \times 2^{45}\) cm \(\approx 3\,51\,844\) km.

3The Moon is about 3,84,400 km away — 45 folds almost reach it, and 46 folds go past it!

✅ Answer: 30 folds ≈ 10.7 km; 45 folds ≈ 3.5 lakh km.

📖 Page 20Table

Fill in the thickness for folds 18 to 45.

1Each fold doubles the thickness: multiply the previous value by 2.

2100 cm = 1 m and 1000 m = 1 km. Change units when numbers get big.

3Example: fold 20 ≈ 10.5 m, so fold 21 ≈ 21 m, fold 22 ≈ 42 m.

Fold	Thickness	Fold	Thickness	Fold	Thickness
18	≈ 262 cm	28	≈ 2.7 km	38	≈ 2,748.8 km
19	≈ 524 cm	29	≈ 5.4 km	39	≈ 5,497.6 km
20	≈ 10.5 m	30	≈ 10.7 km	40	≈ 10,995.1 km
21	≈ 21.0 m	31	≈ 21.5 km	41	≈ 21,990.2 km
22	≈ 41.9 m	32	≈ 42.9 km	42	≈ 43,980.5 km
23	≈ 83.9 m	33	≈ 85.9 km	43	≈ 87,960.9 km
24	≈ 167.8 m	34	≈ 171.8 km	44	≈ 175,921.9 km
25	≈ 335.5 m	35	≈ 343.6 km	45	≈ 351,843.7 km
26	≈ 671.1 m	36	≈ 687.2 km
27	≈ 1.3 km	37	≈ 1,374.4 km

✅ Answer: See the completed table (values rounded).

⚠️ Common mistake: Adding instead of doubling. Each fold MULTIPLIES by 2.

📖 Textbook Page 21

📖 Page 21Think

Look at fold 4 (0.016 cm) and fold 6 (0.064 cm). By how much does the thickness grow after two folds?

1\(0.064 \div 0.016 = 4\)

2Two folds = doubling twice \(= 2 \times 2 = 4\) times.

✅ Answer: It becomes 4 times thicker.

📖 Page 21Check

Is it true that after any 3 folds the thickness becomes 8 times?

1Fold 4 → Fold 7: \(0.128 \div 0.016 = 8\) ✔

2Fold 10 → Fold 13: \(8.192 \div 1.024 = 8\) ✔

3Three doublings \(= 2 \times 2 \times 2 = 2^3 = 8\).

✅ Answer: Yes, it always becomes 8 times.

💡 Tip: After any 10 folds it becomes 2¹⁰ = 1024 times.

📖 Textbook Page 22

📖 Page 22Choose

The starting thickness is \(v\). Which expression gives the thickness after 10 folds?
(i) \(10v\) (ii) \(10 + v\) (iii) \(2 \times 10 \times v\) (iv) \(2^{10}\) (v) \(2^{10}v\) (vi) \(10^2 v\)

1Each fold multiplies the thickness by 2.

210 folds → multiply by 2 ten times \(= 2^{10}\).

3Thickness \(= 2^{10} \times v = 2^{10}v\).

✅ Answer: (v) \(2^{10}v\)

⚠️ Common mistake: Choosing (iii) 2 × 10 × v. That is only 20 times, not 1024 times.

📖 Page 22Think

What is \((-1)^5\)? Is it positive or negative? What about \((-1)^{56}\)?

1\((-1) \times (-1) = +1\). So every PAIR of (−1)s gives +1.

2\((-1)^5\): 2 pairs and one (−1) left over → \(-1\).

3\((-1)^{56}\): 56 is even → 28 pairs → \(+1\).

✅ Answer: \((-1)^5 = -1\) (negative); \((-1)^{56} = 1\) (positive).

💡 Tip: Negative base: even power → positive, odd power → negative.

📖 Page 22Verify

Is \((-2)^4 = 16\)?

1\((-2)^4 = (-2) \times (-2) \times (-2) \times (-2)\)

2\(= 4 \times 4 = 16\)

✅ Answer: Yes, it is 16.

⚠️ Common mistake: Confusing (−2)⁴ = 16 with −2⁴ = −16. The bracket matters!

📖 Page 22Think

What is \(0^2\)? \(0^5\)? \(0^n\)?

1Zero multiplied by anything is zero.

2\(0^2 = 0 \times 0 = 0\), \(0^5 = 0\).

✅ Answer: All are 0. \(0^n = 0\) for any counting number \(n\).

📖 Page 22Worked

Write 32400 as a product of prime factors in exponential form.

1\(32400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 3 \times 3 \times 3 \times 3\)

2\(= 2^4 \times 3^4 \times 5^2\)

✅ Answer: \(32400 = 2^4 \times 3^4 \times 5^2\)

📖 Page 22Figure it Out · Q1

Write in exponential form:
(i) \(6\times6\times6\times6\) (ii) \(y \times y\) (iii) \(b\times b\times b\times b\) (iv) \(5\times5\times7\times7\times7\) (v) \(2\times2\times a\times a\) (vi) \(a\times a\times a\times c\times c\times c\times c\times d\)

1Count how many times each base is multiplied.

2(i) \(6^4\) (ii) \(y^2\) (iii) \(b^4\)

3(iv) \(5^2 \times 7^3\) (v) \(2^2 \times a^2\) (or \((2a)^2\))

4(vi) \(a^3 c^4 d\)

✅ Answer: (i) \(6^4\) (ii) \(y^2\) (iii) \(b^4\) (iv) \(5^2 \times 7^3\) (v) \(2^2a^2\) (vi) \(a^3c^4d\)

⚠️ Common mistake: Writing d¹ is fine, but do not forget d — it appears once.

📖 Textbook Page 23

📖 Page 23Figure it Out · Q2

Write as a product of powers of prime factors: (i) 648 (ii) 405 (iii) 540 (iv) 3600

1(i) \(648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 = 2^3 \times 3^4\)

2(ii) \(405 = 3 \times 3 \times 3 \times 3 \times 5 = 3^4 \times 5\)

3(iii) \(540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^3 \times 5\)

4(iv) \(3600 = 36 \times 100 = 2^2 \cdot 3^2 \times 2^2 \cdot 5^2 = 2^4 \times 3^2 \times 5^2\)

✅ Answer: (i) \(2^3 \times 3^4\) (ii) \(3^4 \times 5\) (iii) \(2^2 \times 3^3 \times 5\) (iv) \(2^4 \times 3^2 \times 5^2\)

💡 Tip: Split into easy parts first: 3600 = 36 × 100.

📖 Page 23Figure it Out · Q3

Find the value:
(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) \((-2)^5 \times (-10)^6\)

1(i) \(2 \times 1000 = 2000\)

2(ii) \(49 \times 8 = 392\)

3(iii) \(3 \times 256 = 768\)

4(iv) \(9 \times 25 = 225\)

5(v) \(9 \times 10000 = 90\,000\)

6(vi) \((-32) \times 1\,000\,000 = -32\,000\,000\)

✅ Answer: (i) 2000 (ii) 392 (iii) 768 (iv) 225 (v) 90,000 (vi) −3,20,00,000

⚠️ Common mistake: In (vi), (−2)⁵ is negative (odd power) but (−10)⁶ is positive (even power).

📖 Page 23Puzzle

The Stones that Shine: 3 daughters, each with 3 baskets, each basket with 3 keys, each key opens 3 rooms. How many rooms are there altogether?

1Daughters = 3

2Baskets = \(3 \times 3 = 9\)

3Keys = \(9 \times 3 = 27\)

4Rooms = \(27 \times 3 = 81 = 3^4\)

Each level is 3 times the one above it. Diamonds = 3⁷ = 2187.

✅ Answer: 81 rooms \((3^4)\).

📖 Page 23Puzzle

Each room has 3 tables, each table 3 necklaces, each necklace 3 diamonds. How many diamonds are there? Can you find it with just one multiplication using the products above?

1Diamonds \(= 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^7\)

2We already know \(3^4 = 81\) and \(3^3 = 27\).

3\(3^7 = 3^4 \times 3^3 = 81 \times 27 = 2187\)

✅ Answer: 2187 diamonds — found by one multiplication 81 × 27.

📖 Textbook Page 24

📖 Page 24Reason

\(3^7\) can also be written as \(3^2 \times 3^5\). Why?

1\(3^7 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3\)

2Group as \((3 \times 3) \times (3 \times 3 \times 3 \times 3 \times 3)\)

3\(= 3^2 \times 3^5\), and \(2 + 5 = 7\).

✅ Answer: Because seven 3s can be split into two 3s and five 3s: \(3^2 \times 3^5 = 3^{2+5} = 3^7\).

📖 Page 24Math Talk

Use \(n^a \times n^b = n^{a+b}\) to compute (i) \(2^9\) (ii) \(5^7\) (iii) \(4^6\).

1(i) \(2^9 = 2^4 \times 2^5 = 16 \times 32 = 512\)

2(ii) \(5^7 = 5^3 \times 5^4 = 125 \times 625 = 78\,125\)

3(iii) \(4^6 = 4^3 \times 4^3 = 64 \times 64 = 4096\)

✅ Answer: (i) 512 (ii) 78,125 (iii) 4096

💡 Tip: Split the power into two parts whose values you already know.

📖 Page 24Think

Is \(2^{10}\) also equal to \((2^5)^2\)? Write it as a product.

1\(2^{10} = (2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2 \times 2)\)

2\(= 2^5 \times 2^5 = (2^5)^2\)

3Check: \(32 \times 32 = 1024\) ✔

✅ Answer: Yes, \(2^{10} = (2^5)^2 = (2^2)^5\).

📖 Page 24Practice

Write as a power of a power in at least two ways: (i) \(8^6\) (ii) \(7^{15}\) (iii) \(9^{14}\) (iv) \(5^8\)

1Use \((n^a)^b = n^{a \times b}\). Split the exponent into two factors.

2(i) \(8^6 = (8^2)^3 = (8^3)^2\); also \(8^6 = 2^{18} = (2^9)^2\)

3(ii) \(7^{15} = (7^3)^5 = (7^5)^3\)

4(iii) \(9^{14} = (9^2)^7 = (9^7)^2\); also \(9^{14} = 3^{28} = (3^4)^7\)

5(iv) \(5^8 = (5^2)^4 = (5^4)^2\)

✅ Answer: (i) \((8^2)^3, (8^3)^2\) (ii) \((7^3)^5, (7^5)^3\) (iii) \((9^2)^7, (9^7)^2\) (iv) \((5^2)^4, (5^4)^2\)

⚠️ Common mistake: Adding exponents here. For a power of a power we MULTIPLY: (8²)³ = 8⁶, not 8⁵.

📖 Textbook Page 25

📖 Page 25Puzzle

Magical Pond: the number of lotuses doubles every day and the pond is full on day 30. On which day was it half full?

1Think backwards. From day 29 to day 30 the lotuses double.

2So on day 29 there were half as many lotuses → half full.

Working backwards: full on day 30, so half full on day 29.

✅ Answer: Day 29 (not day 15!).

⚠️ Common mistake: Saying day 15 because 15 is half of 30. Doubling is not adding.

📖 Page 25Practice

Write the number of lotuses (in exponential form) when the pond was (i) fully covered (ii) half covered.

1Start: 1 lotus. After each day it is multiplied by 2.

2After 30 days: \(1 \times 2^{30} = 2^{30}\)

3Half of \(2^{30}\) is \(2^{30} \div 2 = 2^{29}\)

✅ Answer: (i) \(2^{30}\) (ii) \(2^{29}\)

📖 Page 25Think

Damayanti puts 1 lotus in the doubling pond for 4 days, then moves all of them to the tripling pond for 4 more days. How many lotuses? What if she uses the ponds in the other order?

1After 4 days of doubling: \(1 \times 2^4 = 16\)

2After 4 days of tripling: \(2^4 \times 3^4 = 16 \times 81 = 1296\)

3Other order: \(3^4 \times 2^4 = 81 \times 16 = 1296\) — the same!

4Regroup: \((3 \times 2)^4 = 6^4 = 1296\).

✅ Answer: 1296 lotuses either way, because \(2^4 \times 3^4 = 6^4\).

📖 Page 25Practice

Use \(m^a \times n^a = (mn)^a\) to find \(2^5 \times 5^5\).

1\(2^5 \times 5^5 = (2 \times 5)^5 = 10^5\)

2\(= 1\,00\,000\)

✅ Answer: \(10^5 = 1,00,000\)

💡 Tip: Pair 2s with 5s to make 10s — very quick!

📖 Page 25Practice

Simplify \(\dfrac{10^4}{5^4}\) and write it in exponential form.

1\(\dfrac{10^4}{5^4} = \left(\dfrac{10}{5}\right)^4\)

2\(= 2^4 = 16\)

✅ Answer: \(2^4\) (= 16)

📖 Textbook Page 26

📖 Page 26Count

Estu has 4 dresses and 3 caps. How many different outfits can he make?

1For each cap there are 4 dress choices.

2\(4 + 4 + 4 = 3 \times 4 = 12\)

For every cap there are 4 dresses: 4 + 4 + 4 = 3 × 4 = 12.

✅ Answer: 12 outfits.

📖 Page 26Count

Roxie has 7 dresses, 2 hats and 3 pairs of shoes. In how many ways can she dress up?

1Each dress can go with each hat: \(7 \times 2 = 14\) dress-hat pairs.

2Each pair can go with each of the 3 shoes: \(14 \times 3 = 42\).

✅ Answer: \(7 \times 2 \times 3 = 42\) ways.

💡 Tip: When choices are made one after another, MULTIPLY the number of choices.

📖 Page 26Count

A safe has a 5-digit password (digits 0–9). Estu and Roxie tried every possible password. How many did they check?

1Start simple: a 2-digit lock has 10 × 10 = 100 passwords (00 to 99).

2A 3-digit lock: 100 × 10 = 1000 passwords.

3A 5-digit lock: \(10 \times 10 \times 10 \times 10 \times 10 = 10^5\).

Each wheel has 10 choices, and the choices multiply.

✅ Answer: \(10^5 = 1,00,000\) passwords (00000 to 99999).

📖 Textbook Page 27

📖 Page 27Count

Estu wants a lock with 6 slots, each showing a letter A to Z. How many passwords are possible?

1Each slot has 26 choices.

2Total \(= 26 \times 26 \times 26 \times 26 \times 26 \times 26 = 26^6\)

3\(26^6 = 30\,89\,15\,776\)

✅ Answer: \(26^6 = 30,89,15,776\) — about 3000 times more than the 5-digit lock.

📖 Page 27Try This

Think about how many codes are possible for (i) Indian pincodes (ii) mobile numbers (iii) vehicle numbers.

1(i) A pincode has 6 digits → at most \(10^6\) = 10 lakh codes. (The first digit shows the region, so not all are used.)

2(ii) A mobile number has 10 digits → at most \(10^{10}\) numbers. (It usually starts with 6, 7, 8 or 9.)

3(iii) A number like KA 01 AB 1234 has state letters, a 2-digit district code, 2 letters and 4 digits. For one district: \(26 \times 26 \times 10^4 = 67\,60\,000\) numbers.

✅ Answer: Codes multiply quickly — that is why short codes can label huge numbers of things.

📖 Page 27Practice

What is \(2^{100} \div 2^{25}\) in powers of 2?

1Use \(n^a \div n^b = n^{a-b}\).

2\(2^{100} \div 2^{25} = 2^{100-25} = 2^{75}\)

✅ Answer: \(2^{75}\)

⚠️ Common mistake: Dividing the exponents (100 ÷ 25 = 4). We SUBTRACT exponents when dividing.

📖 Textbook Page 28

📖 Page 28Math Talk

In \(n^a \div n^b = n^{a-b}\), why can't \(n\) be 0?

1If \(n = 0\), then \(n^b = 0\).

2Dividing by 0 has no meaning (it is not defined).

✅ Answer: Because we would be dividing by zero, which is not allowed.

📖 Page 28Think

Why is \(2^0 = 1\)?

1\(2^4 \div 2^4 = 16 \div 16 = 1\)

2Using the rule: \(2^4 \div 2^4 = 2^{4-4} = 2^0\)

3So \(2^0\) must be 1.

✅ Answer: \(x^0 = 1\) for any \(x \ne 0\).

📖 Page 28Think

A line of 16 units is halved 5 times. How long is it? Write it with a power of 2.

1\(2^4 \div 2^5 = 2^{4-5} = 2^{-1}\)

2Halving 16 → 8 → 4 → 2 → 1 → ½

3So \(2^{-1} = \tfrac{1}{2}\).

Each halving divides by 2, so the power of 2 goes down by 1 — even below zero.

✅ Answer: \(\tfrac{1}{2}\) unit \(= 2^{-1}\).

📖 Textbook Page 29

📖 Page 29Think

Can we write \(10^3 = \dfrac{1}{10^{-3}}\)?

1\(10^{-3} = \dfrac{1}{10^3}\)

2\(\dfrac{1}{10^{-3}} = 1 \div \dfrac{1}{10^3} = 1 \times 10^3 = 10^3\)

✅ Answer: Yes. In general \(n^a = \dfrac{1}{n^{-a}}\) (\(n \ne 0\)).

📖 Page 29Think

We used these rules only for counting-number exponents. Do they work for any integers (including negative ones and 0)?

1Check: \(2^3 \times 2^{-5} = 8 \times \tfrac{1}{32} = \tfrac{1}{4} = 2^{-2}\), and \(3 + (-5) = -2\) ✔

2Check: \((2^{-1})^2 = \tfrac{1}{4} = 2^{-2}\), and \(-1 \times 2 = -2\) ✔

✅ Answer: Yes. All the rules work for any integer exponents (base not 0).

📖 Page 29Practice

Write equivalent forms: (i) \(2^{-4}\) (ii) \(10^{-5}\) (iii) \((-7)^{-2}\) (iv) \((-5)^{-3}\) (v) \(10^{-100}\)

1Use \(n^{-a} = \dfrac{1}{n^a}\).

2(i) \(\dfrac{1}{2^4} = \dfrac{1}{16}\)

3(ii) \(\dfrac{1}{10^5} = \dfrac{1}{100000}\)

4(iii) \(\dfrac{1}{(-7)^2} = \dfrac{1}{49}\)

5(iv) \(\dfrac{1}{(-5)^3} = -\dfrac{1}{125}\)

6(v) \(\dfrac{1}{10^{100}}\)

✅ Answer: (i) \(\tfrac{1}{16}\) (ii) \(\tfrac{1}{100000}\) (iii) \(\tfrac{1}{49}\) (iv) \(-\tfrac{1}{125}\) (v) \(\tfrac{1}{10^{100}}\)

⚠️ Common mistake: Thinking a negative exponent makes the number negative. 2⁻⁴ is a small positive number (1/16).

📖 Page 29Practice

Simplify 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) \(8^p \times 8^q\)

1(i) \(2^{-4+7} = 2^3\)

2(ii) \(3^{2-5+6} = 3^3\)

3(iii) \(p^{3-10} = p^{-7}\)

4(iv) \((-4)^{-2} = \dfrac{1}{16} = 2^{-4}\), so \(2^4 \times 2^{-4} = 2^0 = 1\)

5(v) \(8^{p+q}\)

✅ Answer: (i) \(2^3\) (ii) \(3^3\) (iii) \(p^{-7}\) (iv) \(2^0 = 1\) (v) \(8^{p+q}\)

📖 Textbook Page 30

📖 Page 30Power line

Using the power line of 4: how many times larger than \(4^{-2}\) is \(4^2\)?

Power line of 4: each step up multiplies by 4.

1\(4^2 \div 4^{-2} = 4^{2-(-2)} = 4^4\)

2\(4^4 = 256\). Check: \(16 \div \tfrac{1}{16} = 256\) ✔

✅ Answer: \(4^4 = 256\) times.

📖 Page 30Power line

Use the power line of 7 to answer:
(a) \(2401 \times 49\) (b) \(49^3\) (c) \(343 \times 2401\) (d) \(16807 \div 49\) (e) \(7 \div 343\) (f) \(16807 \div 823543\) (g) \(117649 \times \tfrac{1}{343}\) (h) \(\tfrac{1}{343} \times \tfrac{1}{343}\)

Power line of 7.

1Read each number as a power of 7: \(7 = 7^1, 49 = 7^2, 343 = 7^3, 2401 = 7^4, 16807 = 7^5, 117649 = 7^6, 823543 = 7^7\).

2(a) \(7^4 \times 7^2 = 7^6 = 117649\)

3(b) \((7^2)^3 = 7^6 = 117649\)

4(c) \(7^3 \times 7^4 = 7^7 = 823543\)

5(d) \(7^5 \div 7^2 = 7^3 = 343\)

6(e) \(7^1 \div 7^3 = 7^{-2} = \tfrac{1}{49}\)

7(f) \(7^5 \div 7^7 = 7^{-2} = \tfrac{1}{49}\)

8(g) \(7^6 \times 7^{-3} = 7^3 = 343\)

9(h) \(7^{-3} \times 7^{-3} = 7^{-6}\)

✅ Answer: (a) \(7^6\) (b) \(7^6\) (c) \(7^7\) (d) \(7^3\) (e) \(7^{-2}\) (f) \(7^{-2}\) (g) \(7^3\) (h) \(7^{-6}\)

📖 Page 30Practice

Write using powers of 10 (like \(47561 = 4 \times 10^4 + 7 \times 10^3 + \dots\)): (i) 172 (ii) 5642 (iii) 6374

1(i) \(172 = 1 \times 10^2 + 7 \times 10^1 + 2 \times 10^0\)

2(ii) \(5642 = 5 \times 10^3 + 6 \times 10^2 + 4 \times 10^1 + 2 \times 10^0\)

3(iii) \(6374 = 6 \times 10^3 + 3 \times 10^2 + 7 \times 10^1 + 4 \times 10^0\)

✅ Answer: See the steps.

💡 Tip: The units place is 10⁰ = 1. After the decimal point the powers become 10⁻¹, 10⁻², …

📖 Textbook Page 31

📖 Page 31Practice

Write these large-number facts in scientific form:
(i) Sun to centre of the Milky Way: 30,00,00,00,00,00,00,00,00,000 m
(ii) Stars in our galaxy: 1,00,00,00,00,000
(iii) Mass of the Earth: 59,76,00,00,00,00,00,00,00,00,00,000 kg

1Put the decimal point after the first digit, then count how many places it moved.

2(i) 3 followed by 20 zeros → \(3 \times 10^{20}\) m

3(ii) 1 followed by 11 zeros → \(1 \times 10^{11}\)

4(iii) 25 digits in all → \(5.976 \times 10^{24}\) kg

✅ Answer: (i) \(3 \times 10^{20}\) m (ii) \(10^{11}\) (iii) \(5.976 \times 10^{24}\) kg

⚠️ Common mistake: Counting commas instead of digits. Count the digits after the first one.

📖 Textbook Page 32

📖 Page 32Compare

Sun–Saturn \(= 1.4335 \times 10^{12}\) m, Saturn–Uranus \(= 1.439 \times 10^{12}\) m, Sun–Earth \(= 1.496 \times 10^{11}\) m. Which is the smallest?

1First compare the powers of 10: 10¹¹ is smaller than 10¹².

2So Sun–Earth \((1.496 \times 10^{11})\) is the smallest, even though 1.496 looks bigger.

✅ Answer: Sun to Earth is the smallest distance.

💡 Tip: In scientific form, the exponent matters most. Compare exponents first, then the numbers in front.

📖 Page 32Number line

On a line from the Sun to Saturn \((1.4335 \times 10^{12}\) m\()\), mark the Earth \((1.496 \times 10^{11}\) m\()\).

1\(\dfrac{1.496 \times 10^{11}}{1.4335 \times 10^{12}} = \dfrac{1.496}{14.335} \approx 0.104\)

2So Earth is about \(\tfrac{1}{10}\) of the way from the Sun.

1.496 × 10¹¹ is about one-tenth of 1.4335 × 10¹², so Earth sits about 1/10 of the way from the Sun.

✅ Answer: Mark Earth about one-tenth of the way from the Sun.

📖 Page 32Practice

Write in standard form: (i) 59,853 (ii) 65,950 (iii) 34,30,000 (iv) 70,04,00,00,000

1(i) \(59853 = 5.9853 \times 10^4\)

2(ii) \(65950 = 6.595 \times 10^4\)

3(iii) \(3430000 = 3.43 \times 10^6\)

4(iv) \(70040000000 = 7.004 \times 10^{10}\)

✅ 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 59.853 × 10³. The first number must be at least 1 and less than 10.

📖 Textbook Page 33

📖 Page 33Estimate

Tulābhāra: what is the worth of jaggery equal to Roxie's weight, and wheat equal to Estu's weight?

1Assume Roxie = 45 kg and jaggery = ₹70 per kg.

2Jaggery worth \(= 45 \times 70 = 3150\) rupees

3Assume Estu = 50 kg and wheat = ₹50 per kg.

4Wheat worth \(= 50 \times 50 = 2500\) rupees

✅ Answer: About ₹3150 of jaggery and ₹2500 of wheat (answers change with your assumptions).

💡 Tip: Always write your assumptions clearly. Different assumptions → different but reasonable answers.

📖 Textbook Page 34

📖 Page 34Estimate

How many 1-rupee coins would equal Roxie's weight? Guess first: hundreds, thousands, lakhs or crores?

1Assume Roxie = 45 kg = 45,000 g.

2Assume one 1-rupee coin weighs about 3 g (you can weigh 10 coins together and divide by 10).

3Coins \(= 45000 \div 3 = 15\,000\)

✅ Answer: About 15,000 coins — in the thousands (₹15,000).

📖 Page 34Estimate

What if we use 5-rupee coins or 10-rupee notes instead?

15-rupee coin ≈ 6 g → \(45000 \div 6 = 7500\) coins → \(7500 \times 5 = 37\,500\) rupees

210-rupee note ≈ 1 g → \(45000 \div 1 = 45\,000\) notes → \(45000 \times 10 = 4\,50\,000\) rupees

✅ Answer: About ₹37,500 in 5-rupee coins, and about ₹4.5 lakh in 10-rupee notes.

💡 Tip: Notes are very light, so the same weight is worth much more money.

📖 Page 34Math Talk

Estu wants to donate notebooks equal to his weight every year; Roxie wants to do annadāna (food) equal to her weight. How many people might benefit?

1Notebooks: assume Estu = 50 kg and a notebook ≈ 200 g → 50,000 ÷ 200 = 250 notebooks.

2If each child gets 5 notebooks → 250 ÷ 5 = 50 children each year.

3Annadāna: assume 45 kg of rice and one meal uses about 150 g of rice → 45,000 ÷ 150 = 300 meals.

✅ Answer: About 50 children (notebooks) and about 300 meals (annadāna) each year.

📖 Page 34Estimate

Pilgrims walked about 400 km and arrived this morning. How long ago did they start?

1Assume walking speed ≈ 5 km per hour.

2Assume they walk about 8 hours a day → 40 km per day.

3Days \(= 400 \div 40 = 10\)

✅ Answer: About 10 days ago.

📖 Textbook Page 35

📖 Page 35Estimate

If a person walked non-stop, how many times could they go around the Earth (40,000 km) in a lifetime?

1Assume speed 5 km/h, walking 24 hours a day → 120 km per day.

2In a year: \(120 \times 365 = 43\,800\) km ≈ 1 trip around the Earth.

3In about 70 years of walking → about 75 trips.

✅ Answer: About 70–80 times (if they could walk non-stop, which nobody can!).

📖 Page 35Estimate

A ladder to the Moon (3,84,400 km) with steps 20 cm apart: how many steps? Guess first.

1\(3\,84\,400 \text{ km} = 3\,84\,400 \times 1\,00\,000 \text{ cm} = 3.844 \times 10^{10}\) cm

2Steps \(= 3.844 \times 10^{10} \div 20 = 1.922 \times 10^{9}\)

To reach the Moon (3,84,400 km): adding is slow, multiplying is super fast.

✅ Answer: 1,92,20,00,000 steps (about 192 crore).

📖 Textbook Page 36

📖 Page 36Think

Give some examples of linear growth and of exponential growth.

1Linear (add the same amount each time): saving ₹100 every week; a candle burning 1 cm each hour; a taxi fare rising ₹15 per km.

2Exponential (multiply by the same amount each time): bacteria doubling every 20 minutes; a message forwarded to 3 friends each, who forward to 3 more; money with compound interest.

✅ Answer: Linear = repeated adding. Exponential = repeated multiplying.

📖 Textbook Page 37

📖 Page 37Practice

Write the starling population, about 1.3 arab (1.3 billion), in scientific form.

11 arab = 1 billion \(= 10^9\)

21.3 arab \(= 1.3 \times 10^9\)

✅ Answer: \(1.3 \times 10^9\)

📖 Textbook Page 38

📖 Page 38Think

With about \(8 \times 10^9\) people and \(4 \times 10^5\) African elephants, are there nearly 20,000 people for every elephant?

1\(\dfrac{8 \times 10^9}{4 \times 10^5} = 2 \times 10^{9-5} = 2 \times 10^4\)

2\(2 \times 10^4 = 20\,000\)

✅ Answer: Yes, about 20,000 people per African elephant.

📖 Page 38Practice

Mosquitoes: about 11 neel (110 trillion). Write in scientific form.

1110 trillion \(= 110 \times 10^{12} = 1.1 \times 10^{14}\)

✅ Answer: \(1.1 \times 10^{14}\)

📖 Page 38Calculate

(i) Ants per human? (ants \(2 \times 10^{16}\), humans \(8.2 \times 10^9\))
(ii) Starling flocks of 10,000 birds?
(iii) Leaves on all trees if each tree has \(10^4\) leaves? (trees \(3 \times 10^{12}\))
(iv) Sheets of paper (0.001 cm thick) to reach the Moon?

1(i) \(\dfrac{2 \times 10^{16}}{8.2 \times 10^9} \approx 0.244 \times 10^7 = 2.4 \times 10^6\) (about 24 lakh ants per person)

2(ii) \(\dfrac{1.3 \times 10^9}{10^4} = 1.3 \times 10^5\) flocks

3(iii) \(3 \times 10^{12} \times 10^4 = 3 \times 10^{16}\) leaves

4(iv) Distance \(= 3.844 \times 10^{10}\) cm; sheets \(= \dfrac{3.844 \times 10^{10}}{10^{-3}} = 3.844 \times 10^{13}\)

✅ Answer: (i) \(\approx 2.4 \times 10^6\) (ii) \(1.3 \times 10^5\) (iii) \(3 \times 10^{16}\) (iv) \(3.844 \times 10^{13}\)

💡 Tip: Divide the front numbers, and subtract the powers of 10.

📖 Textbook Page 39

📖 Page 39Think

Roxie is 4840 days old. How many hours old is she? Estimate first.

1Estimate: 5000 × 24 ≈ 1,20,000 hours.

2Exact: \(4840 \times 24 = 1\,16\,160\) hours.

✅ Answer: 1,16,160 hours (≈ \(1.16 \times 10^5\)).

📖 Page 39Think

Roxie says "I am 69,70,710 … old". What could this number mean?

1Try minutes: \(4840 \times 24 \times 60 = 69\,69\,600\) minutes.

269,70,710 is very close to this (about 18 more hours).

✅ Answer: It is her age in minutes.

📖 Page 39Think

Estu is 4070 days old today. Find his date of birth.

1\(4070 \div 365 \approx 11.15\) → about 11 years and a few weeks.

211 years = 4015 or 4016 days (counting 2 or 3 leap years). 4070 − 4017 ≈ 53 days more.

3Example: if today is 3 October 2026, go back 11 years → 3 October 2015, then back about 52 more days.

✅ Answer: About 11 years and 52 days ago — e.g. 12 August 2015 if today is 3 October 2026.

💡 Tip: Change ‘today’ to the real date when you solve it.

📖 Page 39Think

If you have lived for a million seconds, how old are you?

11 day \(= 24 \times 60 \times 60 = 86\,400\) seconds.

2\(10^6 \div 86\,400 \approx 11.6\) days.

✅ Answer: About 11½ days old!

📖 Textbook Page 40

📖 Page 40Think

\(10^5\) seconds ≈ 1.16 days and \(10^6\) seconds ≈ 11.57 days. Give events of these sizes in scientific notation.

1Order \(10^5\): one day and night \(= 86\,400 \approx 8.6 \times 10^4\) s; a 2-day weekend \(\approx 1.7 \times 10^5\) s.

2Order \(10^6\): a 2-week school holiday \(\approx 1.2 \times 10^6\) s; one Moon cycle (29.5 days) \(\approx 2.5 \times 10^6\) s.

✅ Answer: Any events lasting about a day (10⁵ s) or about 1–3 weeks (10⁶ s).

📖 Textbook Page 41

📖 Page 41Calculate

A terror-bird fossil is 15 million years old. Write this in seconds.

11 year \(\approx 3.15 \times 10^7\) s

2\(1.5 \times 10^7 \times 3.15 \times 10^7 \approx 4.7 \times 10^{14}\) s

✅ Answer: \(\approx 4.7 \times 10^{14}\) seconds

📖 Textbook Page 42

📖 Page 42Calculate

Land plants began 470 million years ago. Write this in seconds.

1\(4.7 \times 10^8 \times 3.15 \times 10^7 \approx 14.8 \times 10^{15} = 1.48 \times 10^{16}\) s

✅ Answer: \(\approx 1.5 \times 10^{16}\) seconds

📖 Page 42Try This

(i) Counting one star every second, how long to count all \(2 \times 10^{23}\) stars?
(ii) Drinking one 200 ml glass every 10 s, how long to finish all water on Earth (\(2 \times 10^{25}\) drops, 16 drops per ml)?

1(i) \(2 \times 10^{23}\) stars → \(2 \times 10^{23}\) seconds (about \(6 \times 10^{15}\) years!)

2(ii) Water \(= \dfrac{2 \times 10^{25}}{16} = 1.25 \times 10^{24}\) ml

3Glasses \(= \dfrac{1.25 \times 10^{24}}{200} = 6.25 \times 10^{21}\)

4Time \(= 6.25 \times 10^{21} \times 10 = 6.25 \times 10^{22}\) s

✅ Answer: (i) \(2 \times 10^{23}\) s (ii) \(6.25 \times 10^{22}\) s

📖 Textbook Page 43

📖 Page 43Think

Million, billion, trillion, quadrillion … What does the first part of each name tell us?

1bi = 2, tri = 3, quadr = 4, quint = 5, sext = 6, sept = 7, oct = 8, non = 9, dec = 10.

2Billion \(= 10^9 = 1000 \times 1000^2\); trillion \(= 10^{12} = 1000 \times 1000^3\).

3The prefix tells how many more groups of ‘thousand’ come after one thousand.

✅ Answer: The prefix is a number n, and the name means \(10^{3n+3}\).

📖 Textbook Page 44

📖 Page 44Figure it Out · Q1

Find the units digit of \(2^{224} \div 4^{32}\). [Hint: \(4 = 2^2\)]

1\(4^{32} = (2^2)^{32} = 2^{64}\)

2\(2^{224} \div 2^{64} = 2^{160}\)

3Units digits of powers of 2 repeat: 2, 4, 8, 6, 2, 4, 8, 6 …

4160 ÷ 4 = 40 exactly, so \(2^{160}\) ends like \(2^4\) → 6.

✅ Answer: 6

📖 Page 44Figure it Out · Q2

A container has 5 bottles. Every day a new container is brought in. How many bottles after 40 days?

140 new containers \(\times\) 5 bottles \(= 200\)

2\(200 = 2 \times 10^2\)

3This is linear growth (adding 5 each day), not exponential.

✅ Answer: 200 bottles (205 if you also count the container that was there at the start).

⚠️ Common mistake: Thinking it is 5⁴⁰. Nothing is multiplying here — we just add 5 each day.

📖 Page 44Figure it Out · Q3

Write as a product of two or more powers in three ways: (i) \(64^3\) (ii) \(192^8\) (iii) \(32^{-5}\)

1(i) \(64^3 = 2^{18}\): \(2^{10} \times 2^8\), \(\;4^5 \times 4^4\), \(\;8^3 \times 8^3\)

2(ii) \(192 = 2^6 \times 3\), so \(192^8 = 2^{48} \times 3^8\); also \(2^{40} \times 6^8\), \(\;8^{16} \times 3^8\)

3(iii) \(32^{-5} = 2^{-25}\): \(2^{-10} \times 2^{-15}\), \(\;2^{-5} \times 2^{-20}\), \(\;4^{-12} \times 2^{-1}\)

✅ Answer: Many answers are possible — check that the exponents add up correctly.

📖 Page 44Figure it Out · Q4

Always true, 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 its cube.
(iv) The product of two cube numbers is a cube.
(v) \(q^{46}\) is both a 4th power and a 6th power (q prime).

1(i) Sometimes: \(64 = 4^3 = 8^2\) ✔, but \(8 = 2^3\) is not a square.

2(ii) Always: \(n^4 = (n^2)^2\).

3(iii) Always: \(n^5 \div n^3 = n^2\), a whole number (n ≠ 0).

4(iv) Always: \(a^3 \times b^3 = (ab)^3\).

5(v) Never: 46 is not a multiple of 4 or of 6, and q is prime.

✅ Answer: (i) Sometimes (ii) Always (iii) Always (iv) Always (v) Never

📖 Page 44Figure it Out · Q5

Simplify: (i) \(10^{-2} \times 10^{-5}\) (ii) \(5^7 \div 5^4\) (iii) \(9^{-7} \div 9^4\) (iv) \((13^{-2})^{-3}\) (v) \(m^5 n^{12} (mn)^9\)

1(i) \(10^{-2-5} = 10^{-7}\)

2(ii) \(5^{7-4} = 5^3\)

3(iii) \(9^{-7-4} = 9^{-11}\)

4(iv) \(13^{(-2)\times(-3)} = 13^6\)

5(v) \(m^5 n^{12} m^9 n^9 = m^{14} n^{21}\)

✅ Answer: (i) \(10^{-7}\) (ii) \(5^3\) (iii) \(9^{-11}\) (iv) \(13^6\) (v) \(m^{14}n^{21}\)

⚠️ Common mistake: In (iv), (−2) × (−3) = +6, not −6.

📖 Page 44Figure it Out · Q6

If \(12^2 = 144\), find (i) \((1.2)^2\) (ii) \((0.12)^2\) (iii) \((0.012)^2\) (iv) \(120^2\)

1Decimal places in the square = 2 × decimal places in the number.

2(i) 1 place → 2 places: 1.44

3(ii) 2 places → 4 places: 0.0144

4(iii) 3 places → 6 places: 0.000144

5(iv) one zero → two zeros: 14400

✅ Answer: (i) 1.44 (ii) 0.0144 (iii) 0.000144 (iv) 14,400

📖 Textbook Page 45

📖 Page 45Figure it Out · Q7

Which of these are the same? \(2^4 \times 3^6\), \(6^4 \times 3^2\), \(6^{10}\), \(18^2 \times 6^2\), \(6^{24}\)

1Write each in powers of 2 and 3.

2\(6^4 \times 3^2 = 2^4 3^4 \times 3^2 = 2^4 \times 3^6\) ✔

3\(18^2 \times 6^2 = (2 \cdot 3^2)^2 (2 \cdot 3)^2 = 2^2 3^4 \cdot 2^2 3^2 = 2^4 \times 3^6\) ✔

4\(6^{10} = 2^{10}3^{10}\) and \(6^{24} = 2^{24}3^{24}\) — different.

✅ Answer: \(2^4 \times 3^6\), \(6^4 \times 3^2\) and \(18^2 \times 6^2\) are the same.

📖 Page 45Figure it Out · Q8

Which is greater? (i) \(4^3\) or \(3^4\) (ii) \(2^8\) or \(8^2\) (iii) \(100^2\) or \(2^{100}\)

1(i) \(4^3 = 64\), \(3^4 = 81\) → \(3^4\)

2(ii) \(2^8 = 256\), \(8^2 = 64\) → \(2^8\)

3(iii) \(100^2 = 10^4\); \(2^{100} = (2^{10})^{10} \gt (10^3)^{10} = 10^{30}\) → \(2^{100}\)

✅ Answer: (i) \(3^4\) (ii) \(2^8\) (iii) \(2^{100}\)

💡 Tip: A big exponent usually beats a big base.

📖 Page 45Figure it Out · Q9

A dairy makes 8.5 billion milk packets a year. Each needs a unique code using digits 0–9. How many digits must the code have?

1Packets \(= 8.5 \times 10^9\).

2A code of n digits gives \(10^n\) different codes.

3\(10^9 = 1\) billion is too few; \(10^{10} = 10\) billion is enough.

✅ Answer: At least 10 digits.

📖 Page 45Figure it Out · Q10

64 is a square \((8^2)\) and a cube \((4^3)\). Are there other such numbers? Describe them.

1A number that is both a square and a cube must be a 6th power.

2\(n^6 = (n^3)^2 = (n^2)^3\)

3Examples: \(1^6 = 1, 2^6 = 64, 3^6 = 729, 4^6 = 4096\)

✅ Answer: Yes, infinitely many: all 6th powers \(n^6\).

📖 Page 45Figure it Out · Q11

A digital locker uses 5-character codes with digits and letters (like G89P0, BRJKW, 003AZ). How many codes are possible?

1Each place: 10 digits + 26 letters = 36 choices.

2Codes \(= 36^5 = 6\,04\,66\,176\)

✅ Answer: \(36^5 = 6,04,66,176\)

📖 Page 45Figure it Out · Q12

Sheep \(\approx 10^9\) and goats \(\approx 10^9\). Total? (i) \(20^9\) (ii) \(10^{11}\) (iii) \(10^{10}\) (iv) \(10^{18}\) (v) \(2 \times 10^9\) (vi) \(10^9 + 10^9\)

1We ADD: \(10^9 + 10^9 = 2 \times 10^9\).

2\(10^{18}\) would be \(10^9 \times 10^9\) — that is multiplying, not adding.

✅ Answer: (v) and (vi) — both mean 2 × 10⁹ (2 billion).

⚠️ Common mistake: Adding exponents when adding numbers. Exponents add only when we multiply powers.

📖 Page 45Figure it Out · Q13

Give answers in scientific notation:
(i) 30 clothes per person for the world
(ii) 100 million bee colonies × 50,000 bees
(iii) 38 trillion bacteria per person, for all humans
(iv) Time spent eating in a lifetime, in seconds

1(i) \(8.2 \times 10^9 \times 30 = 246 \times 10^9 = 2.46 \times 10^{11}\)

2(ii) \(10^8 \times 5 \times 10^4 = 5 \times 10^{12}\)

3(iii) \(3.8 \times 10^{13} \times 8.2 \times 10^9 \approx 31.2 \times 10^{22} = 3.1 \times 10^{23}\)

4(iv) Assume 1 hour a day for 70 years: \(3600 \times 365 \times 70 = 9.198 \times 10^7\) s

✅ Answer: (i) \(2.46 \times 10^{11}\) (ii) \(5 \times 10^{12}\) (iii) \(\approx 3.1 \times 10^{23}\) (iv) \(\approx 9.2 \times 10^7\) s

📖 Page 45Figure it Out · Q14

What was the date 1 arab (1 billion) seconds ago?

1\(10^9 \div 86\,400 \approx 11\,574\) days

2\(11\,574 \div 365.25 \approx 31.7\) years

3Example: from 3 October 2026, go back 31 years 8 months → around 25 January 1995.

✅ Answer: About 31.7 years ago — around late January 1995 if today is 3 October 2026.

📖 Textbook Page 47

📖 Page 47Game: Tremendous in Ten

Round 1: Roxie wrote 10000000000000, Estu wrote \(999999 \times 999999\). Round 2: Roxie wrote \(10^{1000} + 10^{1000} + 10^{1000} + 10^{1000}\), Estu wrote \(10^{1000000} \times 9000\). Who wins each round?

1Round 1: Roxie's number \(= 10^{13}\). Estu's \(\lt 10^6 \times 10^6 = 10^{12}\). Roxie wins.

2Round 2: Roxie's \(= 4 \times 10^{1000}\).

3Estu's \(= 9 \times 10^{3} \times 10^{1000000} = 9 \times 10^{1000003}\) — far bigger. Estu wins.

✅ Answer: Round 1: Roxie. Round 2: Estu.

💡 Tip: To win, use exponents — a tower of powers grows fastest!

🌍 Where do we use powers?

• Writing huge and tiny numbers in science: distances in space, sizes of atoms and cells.
• Passwords, PINs and codes: the number of choices multiplies.
• Money growing with compound interest, population growth, spread of viruses.

🚀 Link to higher classes

• Class 9: laws of exponents for real numbers, \(a^{1/n}\) (roots as powers).
• Class 10–11: geometric progressions, compound interest, exponential growth.
• Class 11–12: logarithms (the reverse of powers).

⬆ Top 🏠 Class 8 Hub

🔑 keytoenjoylearningmaths.blogspot.com · Solutions written in our own words, based on NCERT Ganita Prakash Class 8 (Part 1)