NCERT Solutions Class 8 Maths Chapter 2 Power Play

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 8 · M AT H S

NCERT Solutions

Chapter 2: Power Play

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

19 – 47 25 96 English

Solutions, notes, sample papers & more at 75 pages

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 2: Power Play
Chapter 2 of Ganita Prakash Grade 8 Part I begins with a sheet of paper 0.001 cm thick and asks how thick it
becomes after 46 folds — the answer is more than 7,00,000 km. From that one question the chapter builds
exponential notation, the laws of exponents, zero and negative powers, and scientific notation, and then
uses them to talk about ants, stars and the age of the Earth.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) 19 – 47

SECTIONS QUESTIONS

25 96

MEDIUM

English

In-text Questions — Page 19
2.1 Experiencing the Power Play …

Q1 How many times can you fold it over and over?

With ordinary paper, about 6 or 7 times — then it stops folding, whatever your strength.
The reason is that folding does two things at once, and they work against each other:

the thickness doubles each time — after k folds it is 2k times the original;
the length is halved each time — after k folds it is 1/2k of the original.

After 7 folds: thickness × 27 = 128 times

length ÷ 27 = 1/128 of the sheet

So a sheet that started 20 cm long and 0.01 cm thick is now about 0.16 cm long and 1.28 cm
thick — a stubby block, taller than it is wide. There is no material left to bend round the fold.

Why it happens: Estu's "7 times" is not a rule about paper, it is a rule about
doubling. Every fold multiplies the thickness by 2, and repeated multiplication grows
far faster than our hands can keep up with. Roxie's idea is right in principle —
thinner paper starts smaller, so it takes a few more folds to reach the same block. In
2002 a student folded a very long, very thin roll of paper 12 times.

Page 1 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Try This: Fold a newspaper sheet and a tissue and count. Then measure the folded
stack with a ruler and check that the thickness really is doubling.

Q2 Say you can fold a sheet of paper as many times as you wish. What would its
thickness be after 30 folds? Make a guess.

Guess first, then check. Most people guess a few metres. The true answer is about 10.7 km —
the height at which aeroplanes fly.

Thickness after 30 folds = 0.001 cm × 230

230 = 1,07,37,41,824

= 0.001 × 1,07,37,41,824 cm

= 10,73,741.824 cm

= 10.737 km (about 10.7 km)

Why it happens: the guess feels wrong because we imagine 30 folds as 30 additions
of a thin layer. But each fold multiplies by 2. Ten folds already multiply by 210 = 1024,
so three sets of ten folds multiply by about 1024 × 1024 × 1024 — more than a
hundred crore.

Tip: A quick way to feel the size: 210 ≈ 103. So 230 = (210)3 ≈ (103)3 = 109, a hundred
crore. That estimate takes a few seconds and is right to within 8%.

In-text Questions — Page 20

Page 2 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

2.1 Experiencing the Power Play …

MATH TALK

Q1 Now, what do you think the thickness would be after 30 folds? 45 folds? Make a
guess.

After 30 folds, about 10.7 km. After 45 folds, about 3,51,844 km — very nearly the distance to
the Moon (3,84,400 km).

30 folds: 0.001 cm × 230 = 10,73,741.8 cm = 10.737 km

45 folds: 0.001 cm × 245 = 3,51,84,37,20,888 cm ≈ 3,51,844 km

Use the table on this page to see how quickly the jumps grow. From fold 17 (131 cm) to fold 26
the thickness goes from about a person's height to 671 m — taller than most buildings in the
country. Nine more folds after that and it is 343 km, higher than the space station.

Why it happens: the gap between one fold and the next is itself doubling. Fold 45
adds as much thickness as folds 1 to 44 put together, because 245 = 244 + 244. That is
the signature of multiplicative growth: the last step is always as big as everything
before it.

Q2 Fill the table below. [Folds 18–26, 27–30 and 31–45]

Every entry is 0.001 cm × 2n, i.e. double the row above. Here are the values, converted to a
sensible unit at each stage.

FOLD THICKNESS FOLD THICKNESS FOLD THICKNESS

18 ≈ 262 cm 21 ≈ 21.0 m 24 ≈ 167.8 m

19 ≈ 524 cm 22 ≈ 41.9 m 25 ≈ 335.5 m

20 ≈ 10.4 m 23 ≈ 83.9 m 26 ≈ 671 m

Page 3 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

co m
m.
FOLD THICKNESS FOLD THICKNESS

m ≈ 1.3 km as e
27
.co
29
a g l
≈ 5.4 km

se m
g l a
a
28 ≈ 2.7 km 30 ≈ 10.7 km

m
.co
FOLD THICKNESS FOLD THICKNESS FOLD THICKNESS

ag
asem ≈ 687 km
gl
31 ≈ 21.5 km 36 41 ≈ 21,990 km

32 ≈ 43.0 km
a
37 ≈ 1,374 km 42 ≈ 43,980 km

co m
m.
33 ≈ 85.9 km 38 ≈ 2,749 km 43 ≈ 87,961 km

m≈ 171.8 km as e ≈ 1,75,922 km
34
.co
39 ≈ 5,498 km
a
44
g l
se m
g l a
a
35 ≈ 343.6 km 40 ≈ 10,995 km 45 ≈ 3,51,844 km

One more fold gives 46: ≈ 7,03,687 km — past the Moon.
m a s
m .co agl
l a se
Check it yourself: you never need a calculator for the next row. Just double. And you
ag
can jump ten rows at a time by multiplying by 1024, because 210 = 1024.

co m
m .
as e
om
In-text.cQuestions — Page 21
a g l
se m
g l a
2.1 Experiencing the Power Play …

a
se m
com a
Notice the change in thickness after two folds. By how much does it increase?
l
Q1

. a g
m
gl ase
a

It becomes 4 times as thick.

co m
m .
as e
com ÷ 0.016 = 4
Fold 4 → 0.016 cm, Fold 6 → 0.064 cm

.0.064 a g l
se m
g l a
a Check with another pair: Fold 4 → 0.016, Fold 5 → 0.032, so one fold doubles
c
m .
Two folds: 2 × 2 = 4 times
m a s e
m . co agl
l a se
The book's next line asks you to check that 3 folds increase the thickness 8 times (= 2 × 2 × 2).
agfold 12 (4.096 cm): 4.096 ÷ 0.512 = 8. ✓
Take fold 9 (0.512 cm) and

co m
m .
m ase
.co


a g l Page 4 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: the factor depends only on how many folds you make, never on
where you start. Each fold multiplies by 2, so k folds multiply by 2k — from fold 3 to
fold 6, or from fold 30 to fold 33, the increase is the same 8 times. That is exactly
what the rule 2a ÷ 2b = 2a–b is saying, and it is why the table of "times increased by"
on this page reads 1024 in every row.

In-text Questions — Page 22
2.2 Exponential Notation and Operations

Q1 Which expression describes the thickness of a sheet of paper after it is folded 10
times? The initial thickness is represented by the letter-number v. (i) 10v (ii) 10 + v
(iii) 2 × 10 × v (iv) 2¹⁰ (v) 2¹⁰v (vi) 10²v

(v) 210v

Start: v

After 1 fold: v × 2

After 2 folds: v × 2 × 2 = v × 22

…

After 10 folds: v × 2 × 2 × … × 2 (ten 2s) = 210v = 1024v

Why the others fail:

10v and 2 × 10 × v multiply by a fixed number of times folded, not by 2 each time — that is
linear thinking.
10 + v adds; folding never adds.
210 forgets the starting thickness altogether — it is a plain number, not a thickness.
102v = 100v; but the base must be 2 (the doubling), and the exponent must be 10 (the
number of folds). Here they have been swapped.

Why it happens: in na, the base n is what is repeated and the exponent a is how
many times. Folding repeats "× 2" ten times, so the base is 2 and the exponent is 10
— never the other way round.

Page 5 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q2 Express the number 32400 as a product of its prime factors and represent the prime
factors in their exponential form.

Divide repeatedly by the smallest prime that goes in, as in the factor tree on this page.

32400 ÷ 2 = 16200

16200 ÷ 2 = 8100

8100 ÷ 2 = 4050

4050 ÷ 2 = 2025

2025 ÷ 5 = 405
405 ÷ 5 = 81

81 ÷ 3 = 27 ÷ 3 = 9 ÷ 3 = 3 ÷ 3 = 1

32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3

= 24 × 52 × 34

Check: 24 = 16, 52 = 25, 34 = 81, and 16 × 25 × 81 = 400 × 81 = 32400. ✓

Why it happens: exponential form is not just shorter — it is the form in which the
number's structure is visible. Reading 24 × 52 × 34 you can see at once that 32400 is a
perfect square (every exponent is even), that it is divisible by 16, by 25 and by 81, and
that its square root is 22 × 5 × 32 = 180.

Q3 What is (– 1)⁵ ? Is it positive or negative? What about (– 1)⁵⁶ ?

(–1)5 = (–1)(–1)(–1)(–1)(–1)

= [(–1)(–1)] × [(–1)(–1)] × (–1)

= 1 × 1 × (–1) = –1 (negative)

Page 6 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

(–1)56 = 28 pairs of (–1)(–1)

= 1 × 1 × … × 1 = 1 (positive)

Why it happens: the minus signs cancel two at a time, because (–1) × (–1) = +1. So
only the parity of the exponent matters. An even power of a negative number is
positive (all the minus signs pair up); an odd power is negative (one minus sign is left
over). Here 5 is odd and 56 is even.

Tip: this works for any negative base: (–3)4 = 81 but (–3)3 = –27. Be careful with
brackets, though — (–3)2 = 9 while –32 = –9, because in the second one only the 3 is
squared.

Q4 What is 0², 0⁵ ? What is 0ⁿ ?

02 = 0 × 0 = 0

05 = 0 × 0 × 0 × 0 × 0 = 0

0n = 0, for every counting number n

Why it happens: a product is zero the moment one factor is zero. Multiply 0 by itself
any number of times and there is nothing to build up with — unlike doubling, which
grows, or halving, which shrinks but never dies, repeated multiplication by 0
collapses at the first step and stays there.

Did you know? The one case left out is 00. The rule n0 = 1 needs n ≠ 0, and the
pattern 0n = 0 would want the answer to be 0. The two patterns disagree, so 00 is left
undefined. This is also the reason the chapter insists n ≠ 0 in na ÷ nb.

Page 7 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q5 Is (– 2)⁴ = 16? Verify.

Yes.

(–2)4 = (–2) × (–2) × (–2) × (–2)
= [(–2) × (–2)] × [(–2) × (–2)]

=4×4

= 16

Why it happens: 4 is even, so the four minus signs form two pairs and every pair
gives +1. In general (–n)a = na when a is even, and (–n)a = –na when a is odd.
Compare with the previous question: (–2)3 = –8, not 8.

Figure it Out — Pages 22–23
2.2 Exponential Notation and Operations

Q1 Express the following in exponential form: (i) 6 × 6 × 6 × 6 (ii) y × y (iii) b × b × b × b (iv)
5 × 5 × 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d

Count how many times each factor is repeated; that count becomes its exponent.

EXPANDED FORM EXPONENTIAL FORM

(i) 6×6×6×6 64 ( = 1296)

(ii) y×y y2

(iii) b×b×b×b b4

(iv) 5×5×7×7×7 52 × 73 ( = 8575)

(v) 2×2×a×a 22a2 ( = 4a2)

(vi) a×a×a×c×c×c×c×d a3c4d

Page 8 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: different bases keep their own exponents; you may not add them.

m l a se
In (iv) there are five factors altogether, but they are not all the same number, so the
o
answer is.c g
52 × 73 and not 55 or 355. In (vi), d appears once, so itsaexponent
e m is 1 —

g l asd1 is simply written as d.
a
and

. com ag
m
Express each of the following as aeproduct
s
a
of powers of their prime factors in

a l (iii) 540 (iv) 3600
Q2
exponential form. (i) 648 (ii)g405

co m
m.

as e
co2m× 324 = 2 × 2 × 162 = 2 × 2 × 2 × 81
(i) 648.=
a g l
m
ase3 4
agl
= 2 × 3 (8 × 81 = 648 ✓)

m a s
m .co agl
se
(ii) 405 = 5 × 81

g l a
= 34 × 5 (81 × 5 = 405 ✓)
a

co m
m .
as e
om
(iii) 540 = 2 × 270 = 2 × 2 × 135 = 4 × 27 × 5

= 2.2c× 33 × 5 (4 × 27 × 5 = 540 ✓) a g l
m
ase
agl
se m
(iv) 3600 = 36 × 100 = (22 × 32) × (22 × 52)
com g l a
m . a
ase
= 24 × 32 × 52 (16 × 9 × 25 = 3600 ✓)

agl
Why it happens: every number has exactly one prime factorisation, so the
co m
m .
e
exponential form is a fingerprint. Notice (iv): all three exponents are even, which is
m l as
.co g
another way of saying 3600 is a perfect square — indeed 602, since 60 = 22 × 3 × 5.
m a
l a se
ag Tip: break the number into two convenient factors first (3600 = 36 × 100), factorise
.c
s e m
a
each, then collect equal bases using na × nb = na+b. It is much faster than dividing by
m
2 eleven times.
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q3 Write the numerical value of each of the following: (i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴ (iv)
(– 3)² × (– 5)² (v) 3² × 10⁴ (vi) (– 2)⁵ × (– 10)⁶

(i) 2 × 103 = 2 × 1000 = 2000

(ii) 72 × 23 = 49 × 8 = 392

(iii) 3 × 44 = 3 × 256 = 768

(iv) (–3)2 × (–5)2 = 9 × 25 = 225

(v) 32 × 104 = 9 × 10000 = 90000

(vi) (–2)5 × (–10)6 = (–32) × 10,00,000 = –3,20,00,000

Why it happens: in (iv) both exponents are even, so both minus signs disappear and
the answer is positive. In (vi) the exponents are 5 (odd) and 6 (even): the first factor
stays negative, the second turns positive, and a negative times a positive is negative.
You can also see it as (–3)2 × (–5)2 = (–3 × –5)2 = 152 = 225, using ma × na = (mn)a.

Check it yourself: –3,20,00,000 is –3.2 × 107 in the standard form you meet later in
this chapter.

In-text Questions — Page 23
2.2 Exponential Notation and Operations · The Stones that Shine …

Q1 Three daughters with curious eyes, / Each got three baskets — a kingly prize. / Each
basket had three silver keys, / Each opens three big rooms with ease. / Each room
had tables — one, two, three, / With three bright necklaces on each, you see. / Each
necklace had three diamonds so fine… / Can you count these stones that shine?
[Hint: Find out the number of baskets and rooms.]

37 = 2187 diamonds.
Follow the riddle level by level. Each line multiplies the count by 3.

Page 10 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

LEVEL COUNT VALUE

daughters 3 31 = 3

baskets 3×3 32 = 9

keys 3×3×3 33 = 27

rooms 3×3×3×3 34 = 81

tables 35 243

necklaces 36 729

diamonds 37 2187

Why it happens: the riddle has seven "each … three" steps, one per level. Every step
multiplies by 3, so the total is 3 multiplied by itself 7 times. The exponent is simply a
count of the levels — which is why exponential notation is the natural language for
anything that branches.

Q2 How many rooms were there altogether?

34 = 81 rooms.

3 daughters × 3 baskets each = 3 × 3 = 9 baskets

9 baskets × 3 keys each = 27 keys

27 keys × 3 rooms each = 81 rooms

and 3 × 3 = 9, 9 × 3 = 27, 27 × 3 = 81 = 34

Why it happens: rooms sit four branchings below the king — daughters, baskets,
keys, rooms — so the exponent is 4. Counting the levels is faster and safer than
drawing all 81 rooms, and it is the only method that still works when the levels
number 20 instead of 4.

Page 11 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q3 How many diamonds were there in total? Can we find out by just one multiplication
using the products above?

Yes — 2187 diamonds, from a single multiplication.

Number of diamonds = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 37
Split the seven 3s as four and three:

37 = (3 × 3 × 3 × 3) × (3 × 3 × 3)

= 34 × 33
= 81 × 27

= 2187

We had already reached 81 while counting the rooms, and 27 while counting the keys. So one
multiplication, 81 × 27, finishes the job.

Why it happens: this is the first appearance of the product rule. Seven copies of 3
can be split into a group of 4 and a group of 3 in the multiplication, because
multiplication lets you regroup freely. The exponents therefore add: 34 × 33 = 34+3 =
37. In general na × nb = na+b.

In-text Questions — Page 24
2.2 Exponential Notation and Operations

MATH TALK

Q1 3⁷ can also be written as 3² × 3⁵. Can you reason out why?

Because seven 3s can be split into two 3s and five 3s, and it makes no difference where you put
the brackets.

Page 12 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

37 = 3 × 3 × 3 × 3 × 3 × 3 × 3

= (3 × 3) × (3 × 3 × 3 × 3 × 3)

= 32 × 35

Check: 9 × 243 = 2187 = 37 ✓

Every split of 7 into two counting numbers works equally well: 31 × 36, 33 × 34, 34 × 33. All of
them give 2187.

Why it happens: multiplication is associative — regrouping the factors never
changes the product. So a string of a + b equal factors can always be cut into a
group of a and a group of b. That single fact is the whole content of na × nb = na+b:
the exponents add because the counts of factors add.

Q2 Write the product p⁴ × p⁶ in exponential form.

p4 × p6 = (p × p × p × p) × (p × p × p × p × p × p)
= p multiplied by itself 4 + 6 = 10 times

= p10

Why it happens: nothing about the argument used the fact that the base was a
number. p is a letter-number standing for any value, and the count of factors is still 4
+ 6. This is how the chapter passes from particular cases such as 34 × 33 to the
general rule na × nb = na+b, where a and b are counting numbers.

Tip: the rule needs the same base. p4 × q6 cannot be shortened at all; only p4 × p6
can.

Page 13 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

co m
m.
Use this observation to compute the following. (i) 2⁹ (ii) 5⁷ (iii) 4⁶
e
Q3

m l as
.co a g
a s em
agleach exponent into two parts you already know, then multiply once.
Split

co m
(i) 29 = 24 × 25 = 16 × 32 = 512
e m . ag
(also 23 × 26 = 8 × 64 = 512)
g l as
a

co m
(ii) 57 = 53 × 54 = 125 × 625 = 78125
em.
m l as
.co
(also 52 × 55 = 25 × 3125 = 78125)
m a g
l a se
a g
(iii) 46 = 43 × 43 = 64 × 64 = 4096
m a s
m .co agl
se
(also 42 × 44 = 16 × 256 = 4096)

g l a
a
m
Why it happens: the rule turns nine multiplications into one. You only need to

. co
m
remember a few small powers — 24, 25, 53, 54, 43 — and every larger power of the

as e
om you know best. l
same base is within one multiplication of them. Since the split is your choice, pick the
. cpieces a g
m
ase
two

agl
se m
com g l a
. a
Q4 Is 2¹⁰ also equal to (2⁵)² ? Write it as a product.
m
ase
ANSWER agl
m
Yes.

. co
em
m l as
.co g
210 = (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
m = (25) × (25) a
l a se
ag
.c
m
= (25)2
m a s e
. co agl
Check: 32 × 32 = 1024 = 210 ✓
e m
g l as
a
The same ten 2s can also be grouped in fives of two: 210 = (22)5 = 4 × 4 × 4 × 4 × 4 = 1024. Both
groupings are legal, so (25)2 = (22)5 = 210.
co m
m .
m ase
.co


a g l Page 14 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: (na)b says "take b copies of a block of a factors". Altogether that is
a × b factors — and a × b does not care about the order of a and b. That is why (na)b
= (nb)a = nab. Note the contrast with the product rule: multiplying powers adds
exponents, raising a power to a power multiplies them.

Q5 Write the following expressions as a power of a power in at least two different
ways: (i) 8⁶ (ii) 7¹⁵ (iii) 9¹⁴ (iv) 5⁸

Split the exponent into two factors (that keeps the base), or rewrite the base as a power (that
changes the base).

SAME BASE — SPLIT THE EXPONENT NEW BASE — SPLIT THE BASE

(i) 86 (82)3 = (83)2 8 = 23, so 86 = (23)6 = 218 = (29)2

(ii) 715 (73)5 = (75)3 15 = 15 × 1, so also (715)1

(iii) 914 (92)7 = (97)2 9 = 32, so 914 = (32)14 = 328 = (34)7 = (37)4

(iv) 58 (52)4 = (54)2 = ((52)2)2 5 is prime, so the base cannot be split

Why it happens: the number of ways you can write nm as a power of a power is
exactly the number of ways of factorising m — 15 = 3 × 5 = 5 × 3 gives two, while a
prime exponent gives essentially none. If the base is itself a power, as with 8 = 23 and
9 = 32, you get extra choices, because (23)6 = 218 reopens the whole game with the
new exponent 18.

In-text Questions — Page 25
2.2 Exponential Notation and Operations · Magical Pond

Q1 In the middle of a beautiful, magical pond lies a bright pink lotus. The number of
lotuses doubles every day in this pond. After 30 days, the pond is completely
covered with lotuses. On which day was the pond half full?

On day 29.

Page 15 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Number of lotuses on day 30 = full pond

Each day the number doubles, so day 30 = 2 × (day 29)

day 29 = half of day 30 = half the pond

Why it happens: this is the most surprising fact about doubling, and the reason it is
worth a whole page. With 29 days gone the pond looks only half covered — it seems
there is plenty of time left. But everything that took 29 days to build is matched in
the single day that follows. Exponential growth stays invisible for a long time and
then finishes in one step.

Did you know? Turn the question round: on which day was the pond one-quarter
covered? Day 28. One-eighth? Day 27. Going back is halving, exactly as going
forward is doubling.

Q2 If the pond is completely covered by lotuses on the 30th day, how much of it is
covered by lotuses on the 29th day?

Half the pond is covered on day 29.

lotuses on day 30 = 2 × lotuses on day 29

so lotuses on day 29 = (1/2) × lotuses on day 30
= half the full pond

Why it happens: you do not need to know how many lotuses "full" means. The
doubling rule fixes the ratio between consecutive days, and the ratio alone answers
the question. This is the same reasoning as "the thickness after 10 folds is 1024
times the thickness before, wherever you start".

Page 16 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q3 Write the number of lotuses (in exponential form) when the pond was — (i) fully
covered (ii) half covered

The pond starts with one lotus on day 0, and doubles each day.

DAY 0 1 2 3 … 29 30

Lotuses 1 = 20 2 = 21 4 = 22 8 = 23 … 229 230

(i) fully covered = 230 lotuses ( = 1,07,37,41,824)

(ii) half covered = 229 lotuses ( = 53,68,70,912)

Why it happens: the exponent is a count of days. Writing 230 instead of
1,07,37,41,824 keeps the structure in view — you can see at once that the answer to
(ii) is one power lower, because 230 ÷ 2 = 229.

Q4 There is another pond in which the number of lotuses triples every day. When both
the ponds had no flowers, Damayanti placed a lotus in the doubling pond. After 4
days, she took all the lotuses from there and put them in the tripling pond. How
many lotuses will be in the tripling pond after 4 more days?

24 × 34 = 16 × 81 = 1296 lotuses.

After the first 4 days (doubling):

1 × 2 × 2 × 2 × 2 = 24 = 16 lotuses

These 16 go into the tripling pond. After 4 more days:

24 × 3 × 3 × 3 × 3 = 24 × 34

= 16 × 81 = 1296

Page 17 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: each pond simply multiplies whatever is in it by its own factor, once
per day. Four days of doubling contribute a factor 24, four days of tripling contribute
34, and the two factors multiply. The starting single lotus contributes a factor of 1
and disappears from the working.

Q5 What if Damayanti had changed the order in which she placed the flowers in the
lakes? How many lotuses would be there?

Exactly the same — 1296 lotuses.

Tripling first, then doubling:

1 × 34 × 24 = 81 × 16 = 1296
Doubling first, then tripling:

1 × 24 × 34 = 16 × 81 = 1296

Why it happens: the total is a product of eight factors — four 2s and four 3s — and
multiplication does not care about order. So the order of the ponds is irrelevant. This
is worth noticing: for additive processes the order can matter a great deal, but for
repeated multiplication it never does.

Q6 Can this product be expressed as an exponent mⁿ, where m and n are some
counting numbers?

Yes — 64.

1 × 34 × 24 = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2)

Regroup one 3 with one 2 at a time:

= (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2)

= (3 × 2)4 = 64 = 1296 ✓

Page 18 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

So m = 6 and n = 4. The general rule this demonstrates is
co m
e m.
m l as
.co
ma × na = (mn)a, where a is a counting number.
m a g
l a se
a g
Why it happens: the rule works only because the two powers have the same

o m
. c ag
exponent. Four 3s and four 2s can be paired off exactly, with nothing left over. With

s e
35 × 24 the pairing would leave a spare m3, giving 64 × 3 — still simplified, but not a
a
single power.
agl

co m
m.
Check it yourself: use the rule on 25 × 55. It equals (2 × 5)5 = 105 = 1,00,000 — far

m as e
.co
easier than 32 × 3125.
a g l
se m
g l a
a
s
Use this observation to compute the value of 2⁵ × 5⁵.
m a
Q7

m .co agl
l a se
g

a
m
25 × 55 = (2 × 5)5
. co
e m
as
= 105

. com (one lakh) a g l
m
ase
= 1,00,000

agl Check the long way: 25 = 32, 55 = 3125, and 32 × 3125 = 1,00,000. ✓
se m
com g l a
m . a
e
as This is exactly why any number ending in zeros can be
Why it happens: pairing each 2 with a 5 makes five 10s, and powers of 10 are the

g l
a — the trailing zeros count the 2–5 pairs in its prime
numbers we read most easily.
split off as a power of 10
factorisation.
co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 19 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q8 Simplify 10⁴/5⁴ and write it in exponential form.

104 ÷ 54 = (10 × 10 × 10 × 10) ÷ (5 × 5 × 5 × 5)
Pair each 10 with a 5:

= (10 ÷ 5) × (10 ÷ 5) × (10 ÷ 5) × (10 ÷ 5)

= (10 ÷ 5)4 = 24 = 16

Check: 104 = 10000, 54 = 625, and 10000 ÷ 625 = 16. ✓

In general: ma ÷ na = (m ÷ n)a (n ≠ 0)

Why it happens: this is the division twin of the previous rule, and it works for the
same reason — the equal exponents let every factor on top be paired with exactly
one factor below. Read from right to left it is just as useful: (m/n)a can be split as
ma/na.

In-text Questions — Page 26
2.2 Exponential Notation and Operations · How Many Combinations

Q1 Estu has 4 dresses and 3 caps. How many different ways can Estu combine the
dresses and caps?

12 ways.

Counting by caps: for each of the 3 caps there are 4 dresses

4 + 4 + 4 = 4 × 3 = 12

Counting by dresses: for each of the 4 dresses there are 3 caps

3 + 3 + 3 + 3 = 3 × 4 = 12

Page 20 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: the two counts agree because they count the same 12 pairs, once
row-wise and once column-wise. This is the multiplication principle: when a choice
is made in stages and the stages do not affect one another, multiply the numbers of
options. It is the idea the whole password section rests on.

Cap 1 Dress 1
Dress 2
Cap 2 Dress 3 3 × 4 = 12 outfits
Dress 4
Cap 3

4 dresses each

Every cap joins to every dress, so the lines number 3 × 4 = 12 — one line per outfit.

Q2 Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie
dress up? [Hint: Try drawing a diagram like the one above.]

7 × 2 × 3 = 42 ways.

Choose a dress: 7 ways

For each of those, choose a hat: 2 ways → 7 × 2 = 14 dress-and-hat pairs

For each of those 14, choose shoes: 3 ways → 14 × 3 = 42

Why it happens: add a third stage and you multiply once more. Each of the 14
dress-and-hat choices splits into 3 branches for the shoes, so the count triples.
Building the answer stage by stage is what makes the rule reliable — you never have
to picture all 42 outfits at once.

Page 21 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Tip: the order of the stages does not matter: 2 × 3 × 7 and 3 × 7 × 2 also give 42.
Choose whichever order makes the arithmetic easiest.

Q3 Estu and Roxie came across a safe containing old stamps and coins that their great-
grandfather had collected. It was secured with a 5-digit password. Since nobody
knew the password, they had no option except to try every password until it
opened. They were unlucky and the lock only opened with the last password, after
they had tried all possible combinations. How many passwords did they end up
checking?

105 = 1,00,000 passwords (one lakh).
Follow the book's advice and shrink the problem first.

LOCK REASONING PASSWORDS

1-digit 0 to 9 10 = 101

2-digit 10 first digits × 10 second digits 100 = 102

3-digit each of the 100 × 10 choices for the third 1000 = 103

5-digit 10 × 10 × 10 × 10 × 10 1,00,000 = 105

These are exactly the numbers 00000, 00001, 00002, …, 99998, 99999 — every whole number up
to 99,999 written with five digits.

Why it happens: each extra slot multiplies the count by 10, because it multiplies
every existing password into 10 new ones. The exponent in 105 is therefore the
number of slots, not the number of digits available. This is why adding one character
to a password makes it ten times harder to break, while adding one to its value
makes no difference at all.

Did you know? At one password per second it would take 1,00,000 seconds — a
little over a day of non-stop trying. Estu's worry about losing the whole vacation was
not far off.

In-text Questions — Pages 27–28

Page 22 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

2.2 Exponential Notation and Operations · 2.3 The Other Side of Powers

TRY THIS

Q1 How many 5-digit passwords are possible?

105 = 1,00,000.

Each of the 5 slots has 10 choices (0 to 9)

10 × 10 × 10 × 10 × 10 = 105
= 1,00,000 passwords

The list runs 00000, 00001, 00002, …, 00010, 00011, …, 00100, …, 00999, …, 30456, …, 99998,
99999.

Why it happens: writing every number from 0 to 99,999 with exactly five digits —
padding with leading zeros — matches passwords to numbers one for one. There
are 99,999 – 0 + 1 = 1,00,000 such numbers, which confirms the count independently
of the multiplication principle.

Q2 Estu says, “Next time, I will buy a lock that has 6 slots with the letters A to Z. I feel it
is safer.” How many passwords are possible with such a lock?

266 = 30,89,15,776 — about 30.9 crore, or 3.09 × 108.

26 × 26 × 26 × 26 × 26 × 26 = 266

262 = 676

263 = 17,576

266 = (263)2 = 17,576 × 17,576
= 30,89,15,776

Estu is right. Compared with the 1,00,000 of the 5-digit lock, this is more than 3000 times as
many passwords.

Page 23 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: two things improved at once. The base rose from 10 to 26 (more

m l a se
symbols per slot) and the exponent rose from 5 to 6 (one more slot). Of the two,
o
m .cexponent is usually the stronger move: 267 would bea26g times bigger
sewhereas widening the alphabet by one letter to 276 multiplies the count by
raising the

l a
g about 1.26.
aonly
again,

. c om ag
m
Tip: (263)2 is much less work than sixemultiplications
s
a
— the power-of-a-power rule
paying its way. agl

co m
se m.
o m l a
g of Vidisha in Madhya
.c
Think about how many combinations are possible in different contexts. Some
a
Q3

m
se Pradesh is 464001. The Pincode of Zemabawk in Mizoram is 796017. (ii) Mobile
examples are — (i) Pincodes of places in India — The Pincode
a
agl numbers. (iii) Vehicle registration numbers. Try to find out how these numbers or
codes are allotted/generated.
m a s
m .co agl
l a se
ag
Work out the raw count first, then see how the real system spends it.

com
CODE STRUCTURE POSSIBLE CODES
m .
m as e
.co
Pincode 6 digits, 0–9
a g l
106 = 10,00,000

s m
eMobile
gl a
a
number 10 digits, but the first must be 6, 7, 8 or 9 4 × 109 = 4,00,00,00,000

se m
com a
262 × 102 × 262 × 104 ≈ 4.57 × 1011
l
Vehicle number 2 letters + 2 digits + 2 letters + 4 digits

. a g
m
ase
agl
The real systems are not random. In a Pincode the first digit names one of the postal zones (4 =
central India, so Vidisha; 7 = eastern India, so Zemabawk), the first two digits fix the sub-zone or
state circle, the third fixes the sorting district, and the last three name the individual post office.

co m
.
Vehicle numbers start with the state code (MP, MZ, TN …) and then the RTO district number.

em
as
Mobile numbers begin with an operator-and-circle block.
m l
.co a
m Why it happens: a code that carries meaning is easier to route and to check, but
g
l a se
ag meaning costs capacity — once the first digit is fixed by geography, only 105
.c
Pincodes remain for that zone. Every real coding system settles somewhere between
s e m
m a
. co agl
"pure counting" (maximum codes) and "readable structure" (fewer codes, fewer
e m
as
mistakes).

a g l

co m
m .
m ase
.co


a g l Page 24 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q4 What is 2¹⁰⁰ ÷ 2²⁵ in powers of 2?

275.

2100 ÷ 225 = 2100 – 25 = 275

Why subtraction is the right operation: writing it out, there are 100 twos on top and 25 twos
below. Each two below cancels one above, and 25 cancellations leave 100 – 25 = 75 twos.

In general: na ÷ nb = na–b, where n ≠ 0 and a > b

Why it happens: division undoes multiplication, so where the product rule added
exponents the quotient rule must subtract them. The chapter reaches this rule by
halving a line of length 24 units: 24 ÷ 23 = 21. The condition n ≠ 0 is essential, since
you cannot cancel zeros.

In-text Questions — Page 28
2.3 The Other Side of Powers

MATH TALK

Q1 Why can’t n be 0?

Because the quotient rule would then ask us to divide by 0, which has no meaning.

na ÷ nb = na–b requires dividing by nb

If n = 0, then nb = 0
and division by 0 is not defined

The clearest case is a = b. Then the rule claims 0a ÷ 0a = 00, that is 0 ÷ 0 = 00. But 0 ÷ 0 has no
single value — any number multiplied by 0 gives 0, so there is nothing to choose. Hence 00 is
left undefined and the rules are stated with n ≠ 0.

Page 25 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: notice what is being protected. The rules of exponents are
shorthand for cancelling equal factors, and cancelling only works when the factor is
not zero — exactly the same reason you may cancel the 3s in (3 × 5)/(3 × 7) but may
not cancel the 0s in (0 × 5)/(0 × 7). Every "n ≠ 0" in this chapter is that one restriction,
showing up again and again.

In-text Questions — Page 29
2.3 The Other Side of Powers

MATH TALK

Q1 Can we write 10³ = 1/10⁻³ ?

Yes.

1 ÷ 10–3 = 1 ÷ (1/103)

= 1 × 103 (dividing by a fraction means multiplying by its reciprocal)

= 103

The same argument gives 72 = 1/7–2 and 4a = 1/4–a. In general

n–a = 1/na and na = 1/n–a, where n ≠ 0

Why it happens: a negative exponent means "reciprocal", and taking the reciprocal
twice returns you to where you started. So the minus sign in an exponent can always
be moved by shifting the power across the fraction bar — from the denominator to
the numerator, or back. This is the single most useful habit in the whole topic: a
power crossing the bar changes the sign of its exponent.

Page 26 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q2 We had required a and b to be counting numbers. Can a and b be any integers? Will
the generalised forms still hold true?

Yes — once n0 = 1 and n–a = 1/na are defined as they are, all three rules hold for every integer a
and b (with n ≠ 0).
Check the product rule where one exponent is negative:

2–4 × 27 = (1/24) × 27 = 27/24 = 23

and the rule predicts 2–4+7 = 23 ✓

3–2 × 3–5 = (1/32) × (1/35) = 1/37 = 3–7

and the rule predicts 3–2–5 = 3–7 ✓

(13–2)–3 = 1/(13–2)3 = 1/13–6 = 136

and the rule predicts 13(–2)×(–3) = 136 ✓

Why it happens: the definitions were not chosen at random — they were chosen so
that the rules would survive. The chapter finds n0 by insisting that 24 ÷ 24 = 24–4
must still be true, and finds n–a by insisting that 24 ÷ 25 = 24–5 must still be true.
Extending a rule and then defining the new symbols so the rule keeps working is a
move mathematics uses constantly.

Page 27 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q3 Write equivalent forms of the following. (i) 2⁻⁴ (ii) 10⁻⁵ (iii) (– 7)⁻² (iv) (– 5)⁻³ (v) 10⁻¹⁰⁰

GIVEN EQUIVALENT FORM VALUE

(i) 2–4 1/24 1/16 = 0.0625

(ii) 10–5 1/105 1/1,00,000 = 0.00001

(iii) (–7)–2 1/(–7)2 1/49 (positive)

(iv) (–5)–3 1/(–5)3 –1/125 (negative)

(v) 10–100 1/10100 a decimal point, 99 zeros, then 1

Why it happens: the minus in the exponent flips the number over; it does not make
the number negative. That is why (iii) is positive — its sign is decided by the even
exponent 2, not by the minus in front of it — while (iv) is negative because 3 is odd. A
negative exponent always produces a number between 0 and 1 when the base is
bigger than 1.

Did you know? 10100 is called a googol, so 10–100 is one googolth — smaller than
the size of an atom compared with the size of the universe.

Q4 Simplify and write the answers in exponential form. (i) 2⁻⁴ × 2⁷ (ii) 3² × 3⁻⁵ × 3⁶ (iii) p³ ×
p⁻¹⁰ (iv) 2⁴ × (– 4)⁻² (v) 8ᵖ × 8ᵠ

Same base each time, so add the exponents.

(i) 2–4 × 27 = 2–4+7 = 23 ( = 8)

(ii) 32 × 3–5 × 36 = 32–5+6 = 33 ( = 27)

(iii) p3 × p–10 = p3–10 = p–7 ( = 1/p7)

(v) 8p × 8q = 8p+q

Page 28 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

Part (iv) needs one extra step, because the bases 2 and –4 are different. Rewrite –4 as a power of
co m
2, remembering that the exponent is even so the sign disappears:
e m.
m l as
m .co a g
l a se
(iv) (–4)–2 = 1/(–4)2 = 1/16 = 2–4
g
a24 × (–4)–2 = 24 × 2–4 = 24–4 = 20 = 1

com
m . ag
l a se
Why it happens: (iii) is worth a second look — a negative answer for the exponent is
ag 1/p7, a number smaller than 1 when p > 1.
perfectly acceptable. p–7 is simply
Nothing has gone wrong; the exponent has just crossed zero, exactly as the power
co m
m.
line on the next page shows.

m as e
.co a g l
a s em
agl Questions — Page 30
In-text

s
2.3 The Other Side of Powers · Power Lines
m a
m .co agl
l
How many times larger than 4⁻² is 4²?
a se
g
Q1
a
m

. co
44 = 256 times.
e m
m l as
.co a g
a s e4m2 ÷ 4–2 = 42–(–2) = 42+2 = 44 = 256
agl
se m
com
Check with the values on the power line: 42 = 16 and 4–2 = 1/16, so 16 ÷ (1/16) = 16 × 16 = 256. ✓
g l a
m . a
a e
s line the two numbers sit four steps apart — 4–2, 4–1,
ag l
Why it happens: on the power
0 1 2 4
4 , 4 , 4 — and each step multiplies by 4. Four steps therefore multiply by 4 . The
power line turns "how many times larger" into "how many steps apart", which is the
co m
m .
e
whole point of arranging powers along a line: a ratio of values becomes a difference
m l as
.co g
of exponents.
m a
l a se
ag
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q2 Use the power line for 7 to answer the following questions. 2,401 × 49 = 49³ = 343 ×
2,401 = 16,807/49 = 7/343 = 16,807/8,23,543 = 1,17,649 × 1/343 = 1/343 × 1/343 =

First read each number off the power line as a power of 7, then just add or subtract the
exponents.

POWER 7 –4 7 –3 7 –2 7– 70 71 72 73 74 75 76 77
1

Value 1/2401 1/343 1/49 1/7 1 7 49 343 2401 16807 117649 823543

QUESTION IN POWERS OF 7 ANSWER

2,401 × 49 74 × 72 = 74+2 76 = 1,17,649

493 (72)3 = 72×3 76 = 1,17,649

343 × 2,401 73 × 74 = 73+4 77 = 8,23,543

16,807 ÷ 49 75 ÷ 72 = 75–2 73 = 343

7 ÷ 343 71 ÷ 73 = 71–3 7–2 = 1/49

16,807 ÷ 8,23,543 75 ÷ 77 = 75–7 7–2 = 1/49

1,17,649 × 1/343 76 × 7–3 = 76–3 73 = 343

1/343 × 1/343 7–3 × 7–3 = 7–3–3 7–6 = 1/1,17,649

Why it happens: notice how much arithmetic vanished. 343 × 2,401 is an awkward
multiplication of six-figure size, but as 73 × 74 it is the sum 3 + 4. Turning
multiplication into addition by working with exponents is the idea behind
logarithms, slide rules and the whole language of scientific notation later in this
chapter.

Check it yourself: the fourth and sixth rows give the same answer, 7–2. That is not a
coincidence — in both, the numerator is two steps below the denominator on the
power line.

Page 30 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

In-text Questions — Page 31
2.4 Powers of 10

Q1 Write these numbers in the same way: (i) 172, (ii) 5642, (iii) 6374.

Each digit is multiplied by the power of 10 belonging to its place. The units place is 100 = 1.

(i) 172 = (1 × 100) + (7 × 10) + 2

= (1 × 102) + (7 × 101) + (2 × 100)

(ii) 5642 = (5 × 1000) + (6 × 100) + (4 × 10) + 2

= (5 × 103) + (6 × 102) + (4 × 101) + (2 × 100)

(iii) 6374 = (6 × 1000) + (3 × 100) + (7 × 10) + 4

= (6 × 103) + (3 × 102) + (7 × 101) + (4 × 100)

Why it happens: our numerals are a place-value system built on 10, so the
expanded form is really a sum of powers of 10. Reading right to left the exponents
run 0, 1, 2, 3, … — which is exactly why the units place needs 100 = 1. Without the
convention n0 = 1 the pattern would break at the last digit.

Q2 How can we write 561.903?

Continue the same pattern to the right of the decimal point, where the exponents go on
decreasing: 2, 1, 0, then –1, –2, –3.

561.903 = (5 × 100) + (6 × 10) + 1 + (9 × 1/10) + (0 × 1/100) + (3 × 1/1000)

Page 31 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

= (5 × 102) + (6 × 101) + (1 × 100) + (9 × 10–1) + (0 × 10–2) + (3 × 10–3)

DIGIT 5 6 1 . 9 0 3

Place value 102 101 100 10–1 10–2 10–3

Why it happens: the decimal point is not a break in the system — it only marks
where 100 sits. Each step to the right divides the place value by 10, which is precisely
the step 100 → 10–1 → 10–2 on the power line. Negative exponents are what make
the whole of the decimal system one single idea instead of two.

Q3 Write the large-number facts we read just before in this form. [(i) The Sun is
30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy. (ii) The
number of stars in our galaxy is 1,00,00,00,00,000. (iii) The mass of the Earth is
59,76,00,00,00,00,00,00,00,00,00,000 kg.]

Count the digits after the leading figures; that count is the exponent.

(i) 30,00,00,00,00,00,00,00,00,000 m

= 3 followed by 20 zeros

= 3 × 1020 m

(ii) 1,00,00,00,00,000 stars

= 1 followed by 11 zeros

= 1 × 1011 (one hundred arab / one hundred billion stars)

(iii) 59,76,00,00,00,00,00,00,00,00,00,000 kg

= 5976 followed by 21 zeros

= 5976 × 1021 = 5.976 × 1024 kg

Page 32 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: (iii) shows why standard form insists that the coefficient lie
between 1 and 10. Written as 5976 × 1021, 59.76 × 1023 or 5.976 × 1024 the number is
the same, but only the last form lets you compare it with another number at a
glance, because then the exponent alone tells you the size.

Tip: to convert, move the decimal point until exactly one non-zero digit is left in front
of it, and count the moves. Moving left raises the exponent; moving right lowers it.

In-text Questions — Page 32
2.4 Powers of 10 · Scientific Notation

Q1 Can you say which of the three distances is the smallest?

The Sun–Earth distance, 1.496 × 1011 m.

DISTANCE STANDARD FORM EXPONENT

Sun to Saturn 1.4335 × 1012 m 12

Saturn to Uranus 1.439 × 1012 m 12

Sun to Earth 1.496 × 1011 m 11

1011 is 10 times smaller than 1012

1.496 × 1011 ≈ 0.15 × 1012, which is less than 1.4335 × 1012

Why it happens: compare exponents first. All three coefficients are close to 1.4, so
the coefficients decide nothing — the exponent does everything. Written in full the
three numbers are 14,33,50,00,00,000 m, 14,39,00,00,00,000 m and
1,49,60,00,00,000 m, and you would have to count commas to see which is shortest.
That counting is exactly what the exponent does for you.

Page 33 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

co m
m.
Did you know? Of the first two, Saturn–Uranus (1.439 × 1012) is very slightly the

m as e
.co l
larger — here the coefficients do decide, because the exponents tie.
a g
se m
g l a
a
Q2 The number line below shows the distance between the Sun and Saturn (1.4335 ×

o m
. c ag
10¹² m). On the number line below, mark the relative position of the Earth. The
distance between the Sun and the m
s e Earth is 1.496 × 10¹¹ m.

a gla

Find what fraction of the Sun–Saturn distance the Earth sits at.
co m
em.
Earth's.c om ÷ Saturn's distance g l as
em
distance
a
a s
a gl = (1.496 × 1011) ÷ (1.4335 × 1012)

s
= (1.496 ÷ 1.4335) × 1011–12
m a
= 1.0436 × 10–1
em
.co agl
a s
aglof the way along
≈ 0.104, that is about one-tenth

co m
.
So the Earth's mark goes very close to the Sun — about a tenth of the line from the left end.

e m
m l as
m .co a g
l a se Earth
ag
se m
com g l a
Sun
m . Saturn a
a e
s is about 0.104 of the way from the
a g l Earth
Sun to Saturn

co m
m .
The whole line is 1.4335 × 10¹² m. The Earth mark is roughly one-tenth of the way from the Sun.

m as e
.co a g l
se m Why it happens: dividing two numbers in standard form is easy — divide the
g l a
a coefficients and subtract the exponents. The subtraction 11 – 12 = –1 immediately
c
m .
says "about a tenth", before you touch the coefficients at all. That is the practical
m a s e
. co agl
value of the notation: the rough answer arrives first, the precision afterwards.
e m
g l as
a

co m
m .
m ase
.co


a g l Page 34 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q3 Express the following numbers in standard form. (i) 59,853 (ii) 65,950 (iii) 34,30,000
(iv) 70,04,00,00,000

Move the decimal point so that exactly one non-zero digit stands before it; the number of places
moved is the exponent.

NUMBER PLACES MOVED STANDARD FORM

(i) 59,853 4 5.9853 × 104

(ii) 65,950 4 6.595 × 104

(iii) 34,30,000 6 3.43 × 106

(iv) 70,04,00,00,000 10 7.004 × 1010

Why it happens: moving the decimal point one place to the left divides the number
by 10, so it must be balanced by multiplying by 101. Moving it k places left therefore
costs a factor 10k. Trailing zeros are dropped from the coefficient but the internal
zeros are not: in (iv) the 0s of 7.004 are between significant digits and carry
information, while the ten trailing zeros are wholly accounted for by 1010.

Check it yourself: the exponent is always one less than the number of digits in a
whole number. 59,853 has 5 digits, so the exponent is 4; 70,04,00,00,000 has 11
digits, so the exponent is 10.

In-text Questions — Pages 33–34
2.5 Did You Ever Wonder?

Q1 What would be the worth (in rupees) of the donated jaggery? What would be the
worth (in rupees) of the donated wheat?

First write down the relationships, then put numbers into them.

Page 35 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Worth of jaggery (₹) = Roxie's weight in kg × cost of 1 kg jaggery

Worth of wheat (₹) = Estu's weight in kg × cost of 1 kg wheat

Nothing can be computed until the four unknowns are estimated. Reasonable assumptions for
a 13-year-old and an 11-year-old, at present prices:

Roxie ≈ 45 kg, jaggery ≈ ₹70 per kg

Worth of jaggery = 45 × 70 = ₹3150

Estu ≈ 50 kg, wheat ≈ ₹50 per kg

Worth of wheat = 50 × 50 = ₹2500

Why it happens: the mathematics here is one multiplication; the real work is
modelling — deciding which quantities matter and how they are related. Once the
relationship is written down, missing values can be estimated and the answer
follows. Your figures may differ from these and still be perfectly correct, so long as
the assumptions are sensible and stated.

Q2 Make necessary and reasonable assumptions for the unknowns and find the
answers. Remember, Roxie is 13 years old and Estu is 11 years old.

State each assumption, then compute. Here is one complete set.

UNKNOWN ASSUMPTION REASON

Roxie's weight (13 yrs) 45 kg typical for that age

Estu's weight (11 yrs) 50 kg the book's own figure

Jaggery ₹70 / kg local market rate

Wheat ₹50 / kg local market rate

Page 36 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Jaggery: 45 × 70 = ₹3150

Wheat: 50 × 50 = ₹2500

Together ≈ ₹5650, that is about ₹5.7 × 103

Why it happens: the answer is only as good as the assumptions, so give the result
to the accuracy the assumptions deserve — "about ₹5000–6000" is honest,
"₹5650.00" pretends to a precision nobody has. This is the same idea as the
coefficient in scientific notation: how many digits you write should reflect how well
you know the number.

Q3 Roxie wonders, “Instead of jaggery if we use 1-rupee coins, how many coins are
needed to equal my weight?”. How can we find out?

Relate the quantities first:

Number of coins = Roxie's weight ÷ weight of one 1-rupee coin

Roxie's weight is known (≈ 45 kg). The weight of one coin is not — so measure it. A single coin is
too light for a kitchen balance, so weigh 100 coins together and divide by 100. That gives about
4 g per coin.

45 kg = 45,000 g

Number of coins ≈ 45,000 ÷ 4

= 11,250 coins ≈ 1.1 × 104

Why it happens: weighing 100 coins instead of one is a real measuring technique,
not a trick. Any error in reading the balance is spread over 100 coins, so the error in
the per-coin figure is 100 times smaller. Whenever a single item is too small to
measure, measure many and divide.

Page 37 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q4 Would the number of coins be in hundreds, thousands, lakhs, crores, or even more?
Make an instinctive guess.

In thousands — around ten thousand.

A coin weighs a few grams, so roughly 250 coins make 1 kg

250 × 45 ≈ 11,000 coins

Order of magnitude: about 104

Lakhs would need each coin to weigh less than half a gram — lighter than a small paper clip.
Hundreds would need each coin to weigh over 100 g, heavier than a mobile phone. Both are
clearly wrong, and you can rule them out before doing any arithmetic.

Why it happens: a good guess is not a wild one. Fix the two extremes by asking
what the answer would force the coin to weigh, discard the impossible ones, and you
are usually left with the right power of 10. Getting the exponent right is most of the
battle; the coefficient can wait.

Q5 Find the answer by making necessary and reasonable assumptions and
approximations for the unknowns. Remember, we are not looking for an exact
answer but a reasonably close estimate.

Assume: Roxie's weight = 45 kg = 45,000 g

Assume: one 1-rupee coin = 4 g

Number of coins = 45,000 ÷ 4 = 11,250

≈ 1.1 × 104 coins, worth ₹11,250

How sensitive is this? If the coin actually weighs 3.5 g the answer is about 12,900; if it weighs 5 g
it is 9,000. Every reasonable assumption lands between 9,000 and 13,000 — so the answer is
"about ten thousand coins", and that is as much as the data can support.

Page 38 of 75

ase
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

Why it happens: testing how the answer moves when an assumption is changed is
. com
m so the
the way to tell how far to trust it. Here the answer never leaves the 10s4erange,
co m a
gldistinction —
. a
em
order of magnitude is solid even though the exact figure is not. That
a s
agl
solid exponent, soft coefficient — is the practical heart of this section.

. c om ag
Estu asks, “What if we use 5-rupeem
a s e coins or 10-rupee notes instead? How much
money could it be?” Make an linstinctive
Q6

ag
guess first. Then find out (make necessary
and reasonable assumptions about the unknown details and find the answers).

com
e m.
as

m l
.co a g
Guess: the 5-rupee coins should be worth more than the 1-rupee coins, and the notes far more
m
l a se
than either — a note weighs almost nothing.

g
a ITEM ASSUMED WEIGHT NUMBER IN 45 KG VALUE

m a s
1-rupee coin 4g
m .co 11,250 ₹11,250
agl
l a se
5-rupee coin 6g
a g 7,500 ₹37,500

com
10-rupee note 1g 45,000 ₹4,50,000

m .
m ase
. co
5-rupee coins: 45,000 g ÷ 6 g = 7,500 coins × ₹5 = ₹37,500
m agl
l a se10-rupee notes: 45,000 g ÷ 1 g = 45,000 notes × ₹10 = ₹4,50,000
ag
se m
o m l a
m .c Notes win overwhelmingly because that ratio is ag
Why it happens: the value of a weight of money is (value per piece) ÷ (weight per

a se
piece), multiplied by the total weight.
l
enormous for paper — ₹10
ag per gram against ₹1 per 4 grams. Roughly, notes give 40
times as much money per kilogram as 1-rupee coins.

co m
m .
o m l a se
Try This: repeat with ₹500 notes. The same 45 kg becomes about ₹2.25 crore — a

m agon the paper.
.cjump of 50 times, purely because of the number printed
l a se
ag
.c
s e m
m a
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 39 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q7 Estu says, “When I become an adult, I would like to donate notebooks worth my
weight every year”. Roxie says, “When I grow up, I would like to do annadāna
(offering grains or meals) worth my weight every year”. How many people might
benefit from each of these offerings in a year? Again, guess first before finding out.

Take adult weights of about 60 kg each.

Notebooks. One notebook of 100 pages ≈ 200 g
60 kg = 60,000 g → 60,000 ÷ 200 = 300 notebooks

If each student receives 5 notebooks:

300 ÷ 5 = 60 students a year

Annadāna. One meal uses about 150 g of grain

60,000 ÷ 150 = 400 meals
= 400 people once, or about 1 person every day for a year, or 100 families of 4

Why it happens: both answers turn on "how much does one unit weigh", and the
two units are wildly different in mass — a notebook is worth more than a meal in
weight, so fewer notebooks are donated but each helps a student for months.
Notice also that these are yearly gifts: over a 40-year working life the notebooks
alone reach about 2400 students.

Did you know? Tulābhāra, offering goods equal to one's own weight, is a very old
practice and is still followed in many places in Southern India. It is a token of
gratitude and it supports the community.

Q8 Roxie and Estu overheard someone saying — “We did pādayātra for about 400 km to
reach this place! We arrived early this morning.” How long ago would they have
started their journey?

About 12 to 13 days ago — roughly two weeks.

Page 40 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Assume a comfortable walking speed = 4 km per hour

Assume 8 hours of walking a day (with rests and nights off)

Distance covered per day = 4 × 8 = 32 km

Days needed = 400 ÷ 32 = 12.5 days

Test the assumptions. At 3 km/h for 6 h a day it is 400 ÷ 18 ≈ 22 days; at 5 km/h for 10 h a day it
is 400 ÷ 50 = 8 days. So the honest answer is "somewhere between one and three weeks, most
likely about a fortnight".

Why it happens: a pādayātra is not a race, so the daily distance — not the speed —
is the quantity that really controls the answer. Once you fix a plausible 30 km or so a
day, the arithmetic is a single division. Guessing before calculating is worth doing
here: most people's first guess is far too short, because we judge distance by how
long a bus takes.

Did you know? Pādayātra is the traditional practice of walking long distances as part
of a religious or spiritual pursuit. Ajmer Sharif Dargah Ziyarat, Pandharpur Wari,
Kānwar Yatra, Sabarimala Yatra, Sammed Shikharji Yatra and the Lumbini to Sarnath
Yatra are some well-known examples.

In-text Questions — Pages 35–36
2.5 Did You Ever Wonder? · Linear Growth vs. Exponential Growth

Q1 How many times can a person circumnavigate (go around the world) the Earth in
their lifetime if they walk non-stop? Consider the distance around the Earth as
40,000 km.

About 18 times, on reasonable assumptions.

Page 41 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Assume walking at 5 km/h for 8 hours a day = 40 km a day

Assume walking for 50 years of a lifetime

Distance in a year = 40 × 365 = 14,600 km

Distance in 50 years = 14,600 × 50 = 7,30,000 km

Number of trips = 7,30,000 ÷ 40,000 ≈ 18 times

In standard form: 7.3 × 105 km walked, against 4 × 104 km per trip.

Why it happens: the answer is startlingly small for a lifetime of walking, and that is
the lesson. Walking is linear — the distance grows by a fixed 40 km a day, never
faster. Fifty years of it add up to less than twenty trips round the world, while the
paper on page 19 passed the Moon in 46 steps. This is the contrast the next page
names.

Did you know? Before modern transport, merchants, sages and scholars really did
walk thousands of kilometres across deserts, mountains and rivers to reach distant
parts of the world.

Q2 Roxie tells Estu about a science-fiction novel she is reading where they build a
ladder to reach the moon, “... I wonder if we actually had a ladder like that, how
many steps would it have?”. What do you think? Make an instinctive guess first.

Guess before calculating. Most first guesses are lakhs; the true answer is about 192 crore
steps.

Page 42 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Distance to the Moon = 3,84,400 km

Assume a gap of 20 cm between steps

3,84,400 km = 3,84,400 × 1,00,000 cm = 3.844 × 1010 cm

Number of steps = 3.844 × 1010 ÷ 20

= 1.922 × 109

= 1,92,20,00,000 steps

Why it happens: we judge such questions from ladders we have seen, which have
twenty or thirty steps, so a jump to 109 feels absurd. Working in powers of 10 keeps
it manageable: cm and km differ by 105, and that single factor is what our intuition
drops.

Q3 Would the number of steps be in thousands, lakhs, crores, or even more?

In crores — 192 crore and 20 lakh steps, that is 1 billion 922 million steps.

1,92,20,00,000 = 1.922 × 109

1 crore = 107, so this is 192.2 crore

1 arab = 109, so this is about 1.9 arab

A quick way to reach the same conclusion without exact arithmetic: the Moon is roughly 4 × 1010
cm away, and 20 cm is 2 × 101 cm, so the count is about (4 ÷ 2) × 1010–1 = 2 × 109.

Why it happens: this rough check uses only the rules of this chapter — divide the
coefficients, subtract the exponents. It gives the right power of 10 in one line, which
is exactly what the question asks for. The exact figure adds precision but changes
nothing about the answer "crores".

Page 43 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

co m
m.
We have to find out how many 20 cm make 3,84,400 km.
e
Q4

m l as
m .co a g
a se

a g l
3,84,400 km

co m
= 3,84,400 × 1000 m (1 km = 1000 m)
e m . ag
g l as
a
= 3,84,400 × 1000 × 100 cm (1 m = 100 cm)

= 3,84,400 × 105 cm

co m
m.
= 3.844 × 105 × 105 cm = 3.844 × 1010 cm

m as e
.co a g l
a s em of 20 cm lengths = 3.844 × 1010 ÷ (2 × 101)
gl
Number
a = 1.922 × 109
m a s
.co agl
= 1,92,20,00,000
se m
g l a
a
Why it happens: the unit change from km to cm is itself a power of 10 — a factor of

o m
105 — so the whole conversion is one exponent addition, 5 + 5 = 10. Doing it this way
c
.
mor gaining a zero
s e
removes the commonest mistake in such problems, which is losing

. com in the middle.
somewhere
a gla
m
ase
agl
se m
com a
Can you come up with some examples of linear growth and of exponential growth?
l
Q5

. a g
m
gl ase
a

Linear growth adds a fixed amount each step; exponential growth multiplies by a fixed
factor each step.
co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 44 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

LINEAR GROWTH (ADDITIVE) EXPONENTIAL GROWTH (MULTIPLICATIVE)

Steps up a ladder: + 20 cm each time Folding paper: × 2 each time

Pocket money saved at ₹50 a week Money at compound interest

Filling a tank at a steady rate Lotuses in the magical pond: × 2 a day

Distance walked at a steady speed Bacteria dividing in two every 20 minutes

Numbering the pages of a book Passwords as slots are added: × 10 each slot

Height of a stack, one book at a time Diamonds in "The Stones that Shine": × 3 each level

Linear after n steps: start + n × d

Exponential after n steps: start × rn

Why it happens: the two behave very differently over time. Linear growth crosses
any given level once, at a predictable moment. Exponential growth is small for a long
while and then overtakes any linear process, however fast, because a fixed multiplier
compounds. 46 folds beat 192 crore steps for exactly this reason.

In-text Questions — Pages 37–38
2.5 Did You Ever Wonder? · Getting a Sense for Large Numbers

Q1 The estimated global population of starlings is around 1.3 arab/1.3 billion (_________).
[Fill in the blank in scientific notation.]

1.3 × 109

1 arab = 1,00,00,00,000 = 109

1 billion = 1,000,000,000 = 109

So 1.3 arab = 1.3 billion = 1.3 × 109 starlings

The row is headed 109, and the coefficient 1.3 lies between 1 and 10, so the entry is in proper
standard form.

Page 45 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: arab and billion are names for the same number, 109, reached by
two different naming systems — the Indian one counting in hundreds (lakh, crore,
arab) and the international one counting in thousands (thousand, million, billion).
Powers of 10 let you write the quantity once and read it in either system.

Did you know? A starling murmuration is a mesmerising aerial display of thousands
of starlings flying in synchronised, swirling patterns, often described as a
"choreographed dance".

Q2 With a global human population of about 8 × 10⁹ and about 4 × 10⁵ African
elephants, can we say that there are nearly 20,000 people for every African
elephant?

Yes.

People per elephant = (8 × 109) ÷ (4 × 105)

= (8 ÷ 4) × 109–5

= 2 × 104

= 20,000 people per elephant

Why it happens: dividing numbers in standard form splits into two easy halves —
divide the coefficients (8 ÷ 4 = 2) and subtract the exponents (9 – 5 = 4). Neither half
needs long division. Compare doing 8,00,00,00,000 ÷ 4,00,000 by hand, where the
only real difficulty is keeping count of the zeros; the exponent does that counting for
you.

Tip: when the coefficients divide awkwardly, round them. (8.2 × 109) ÷ (4.15 × 105) is
still about 2 × 104 — the answer to the nearest power of 10 barely moves.

Page 46 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q3 The estimated mosquito population worldwide (2023) is 11 neel/110 trillion (________).
[Fill in the blank in scientific notation.]

1.1 × 1014

1 neel = 1013, so 11 neel = 11 × 1013

1 trillion = 1012, so 110 trillion = 110 × 1012

11 × 1013 = 110 × 1012 = 1,10,00,00,00,00,00,000

In standard form = 1.1 × 1014

Compare with the Antarctic krill in the same row, 50 neel / 500 trillion = 5 × 1014 — nearly five
times as many.

Why it happens: 11 × 1013 and 110 × 1012 are both correct values but neither is in
standard form, because the coefficient must satisfy 1 ≤ x < 10. Shifting the decimal
point one place left in 11 costs a factor of 10, which is paid back by raising the
exponent from 13 to 14.

In-text Questions — Pages 38–39

Page 47 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

2.5 Did You Ever Wonder? · Getting a Sense for Large Numbers

Q1 Calculate and write the answer using scientific notation: (i) How many ants are
there for every human in the world?

Ants ≈ 20 padma / 20 quadrillion = 2 × 1016

Humans ≈ 8 × 109 (2025)

Ants per human = (2 × 1016) ÷ (8 × 109)

= 0.25 × 107

= 2.5 × 106 ants per person

That is about 25 lakh ants for each of us. Using the sharper figure 8.2 × 109 for the human
population gives 2.4 × 106 — the same to one significant figure, which is all the data deserves.

Why it happens: 0.25 × 107 is a correct value but not standard form, since the
coefficient must be at least 1. Moving the decimal point one place right multiplies by
10, so the exponent must drop by 1: 0.25 × 107 = 2.5 × 106. Watch for this whenever
the top coefficient is smaller than the bottom one.

Page 48 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

co m
m.
Calculate and write the answer using scientific notation: (ii) If a flock of starlings
e
Q2

m l as
.co
contains 10,000 birds, how many flocks could there be in the world?

a g
se m
g l a
a

Starlings ≈ 1.3 × 109
co m
em . ag
as
Birds per flock = 10,000 = 104
a g l
Number of flocks = (1.3 × 109) ÷ 104
co m
em.
m
= 1.3 × 109–4
l as
m .co5 a g
l a se
= 1.3 × 10 flocks (about 1,30,000)

a g
a s
com kind of division there is. It also shows agl
Why it happens: dividing by a pure power of 10 leaves the coefficient untouched
and only lowers the exponent — the .fastest
a s em form: 1.3 was between 1 and 10 to begin with
gl
why the answer is already in standard
and nothing changed it. a

co m
m .
m as e
.co a g l
Calculate and write the answer using scientific notation: (iii) If each tree had about
m 10⁴ leaves, find the total number of leaves on all the trees in the world.
Q3

l a se
ag
se m
com g l a
m . a
ase
agl
Trees ≈ 30 kharab / 3 trillion = 3 × 1012

Leaves per tree = 104

co m
m .
m as e
.co
Total leaves = (3 × 1012) × 104
a g l
a s em = 3 × 10 12+4

agl
= 3 × 1016 leaves
.c
s e m
m a
e m . co
That is 30 padma leaves — about the same order as the number of ants on Earth (2 × 1016), agl
g l as
a
which is a pleasant thing to notice.

co m
m .
m ase
.co


a g l Page 49 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: multiplying in standard form is the mirror of dividing — multiply
the coefficients, add the exponents. Here the coefficient stays 3 because the other
factor is a bare power of 10.

Q4 Calculate and write the answer using scientific notation: (iv) If you stacked sheets
of paper on top of each other, how many would you need to reach the Moon?

Distance to the Moon = 3,84,400 km = 3.844 × 1010 cm

Thickness of one sheet = 0.001 cm = 10–3 cm

Number of sheets = (3.844 × 1010) ÷ 10–3

= 3.844 × 1010–(–3)

= 3.844 × 1010+3

= 3.844 × 1013 sheets

About 38 lakh crore sheets — and yet a single sheet folded 46 times covers the same distance,
because 246 ≈ 7 × 1013.

Why it happens: dividing by 10–3 is the same as multiplying by 103, which is why the
exponent went up. Subtracting a negative number adds, and this is where that rule
earns its keep. It is also the neatest summary of the whole chapter: stacking 3.8 ×
1013 sheets is linear, folding one sheet 46 times is exponential, and both end at the
Moon.

In-text Questions — Page 39
2.5 Did You Ever Wonder? · A different way to say your age!

Q1 “I’m ______ hours old!” said Roxie. Make an estimate before finding this number.

About 1,16,160 hours — roughly 1.2 × 105 hours.

Page 50 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Roxie said she is 4840 days old

1 day = 24 hours

Hours = 4840 × 24

= 4840 × 20 + 4840 × 4

= 96,800 + 19,360

= 1,16,160 hours ≈ 1.16 × 105

Estimate first: 4840 ≈ 5 × 103 and 24 ≈ 2.4 × 101, so the answer is about 12 × 104 = 1.2 × 105. The
estimate is within 4% of the exact value.

Why it happens: the same age looks utterly different in different units — 13 years,
4840 days, 1.16 lakh hours. The quantity has not changed at all; only the size of the
unit has. Choosing a smaller unit multiplies the number, and here the unit shrank by
a factor of 24.

Q2 “I am 69,70,710 … old”. What could this number mean? Find out!

It is Roxie's age in minutes.

Test the units one by one, starting from 4840 days.

In hours: 4840 × 24 = 1,16,160 — far too small

In minutes: 1,16,160 × 60 = 69,69,600 — matches!

In seconds: 69,69,600 × 60 = 41,81,76,000 — far too big

The book's 69,70,710 is 1110 minutes more than 69,69,600 — about 18½ hours, which is simply
the part-day that has passed since she counted her 4840 days. So the answer is 69,70,710
minutes ≈ 7 × 106 minutes.

Page 51 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: you can identify an unknown unit purely from the size of the
number — its power of 10 is a fingerprint. 105 means hours, 107 means minutes, 108
would mean seconds. That is exactly the skill the whole "powers of 10" table on the
following pages is training.

Q3 Estu: “I am 4070 days old today. Can you find out my date of birth?”

Count backwards 4070 days from today.

4070 ÷ 365.25 = 11.14 years

= 11 years and about 0.14 × 365 ≈ 51 days

So Estu was born about 11 years and 7 weeks ago — which agrees with the book telling us he
is 11 years old. To get the exact date, subtract in stages:

1. Go back 11 years from today. That accounts for 11 × 365 + 3 leap days = 4018 days.
2. 4070 – 4018 = 52 days still to go back. Subtract 52 more days from that date.

Worked out from 31 August 2026, this gives 10 July 2015. Do the same from your own "today"
and you will get a date about 11 years and 7 weeks earlier.

Why it happens: years are not a whole number of days, so 365.25 (three ordinary
years and one leap year) is the right average to divide by. Using 365 flat would put
the answer out by about 11 days over 11 years — small, but enough to name the
wrong date.

Q4 If you have lived for a million seconds, how old would you be?

About 11½ days old — not even a fortnight.

Page 52 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

1 day = 24 × 60 × 60 = 86,400 seconds ≈ 8.64 × 104 s

106 ÷ (8.64 × 104)

= (1 ÷ 8.64) × 106–4

= 0.1157 × 102
= 11.57 days (11 days and about 14 hours)

Why it happens: "a million" sounds like a large age, and it is a large number — but a
second is a very small unit. Size in mathematics is always number × unit, and
forgetting the unit is what makes such quantities feel wrong. For contrast, a billion
seconds is 109 ÷ 8.64 × 104 ≈ 11,574 days ≈ 31.7 years, which really is an age.

Check it yourself: the jump from 106 to 109 seconds is a factor of 1000 — from a
fortnight to a lifetime. That single comparison is the best illustration in the chapter
of how fast powers of 10 climb.

In-text Questions — Page 40

Page 53 of 75

as e
Class 8 Maths Chapter 2 Power Play
a g l AglaSem · NCERT Solutions

2.5 Did You Ever Wonder?
co m
e m.
m l as
.co
10⁵ seconds ≈ 1.16 days and 10⁶ seconds ≈ 11.57 days. Think of some events or
a g
em
Q1

a s
phenomena whose time is of the order of (i) 10⁵ seconds and (ii) 10⁶ seconds. Write

a gl them in scientific notation.

co m
. ag

e m
g l as
a
ORDER EVENT OR PHENOMENON TIME IN SCIENTIFIC
NOTATION

co m
m.
105 s One full day and night (24 h) 8.64 × 104 s

m
(about a day)
as e
.co A train from Delhi to Chennai, about 33 hours a g l
m
ase
1.2 × 105 s

agl A cyclone crossing a coastline, about a day ≈ 1 × 105 s

m a s
.co agl
A five-day Test match (5 × 6 h of play) 1.1 × 105 s

se m
g l a
a
106 s A hen's egg hatching, 21 days 1.8 × 106 s
(about 12

m
days)

. c o
2.5 × 10 sem
m as
One phase of the Moon, new moon to new moon, 29.5 6

. co days
a g l
e m
g l as
a
Recovering from a viral fever, about 10 days 8.6 × 105 s

se m
com a
A school examination period of two weeks 1.2 × 106 s

. a g l
m
ase
Useful conversions:
agl
1 day = 8.64 × 104 s
co m
m .
e
1 week = 6.05 × 105 s
m l as
m .1comonth ≈ 2.6 × 106 s a g
l a se
ag 1 year ≈ 3.15 × 107 s
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 75

Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: "of the order of 105" does not mean exactly 1,00,000 seconds — it
means anything from about 3 × 104 to 3 × 105, which is why a 24-hour day and a 33-
hour train journey both qualify. That looseness is deliberate. When quantities span
twenty powers of 10, as in this chapter, sorting them into the right power is far more
informative than pinning down the coefficient.

In-text Questions — Pages 41–42
2.5 Did You Ever Wonder?

TRY THIS

Q1 A fossil of Kelenken Guillermoi, a type of terror bird, is dated to 15 million years ago
( ≈_______________ seconds). [Fill in the blank.]

≈ 4.7 × 1014 seconds

1 year ≈ 3.15 × 107 seconds

15 million years = 1.5 × 107 years

Time = (1.5 × 107) × (3.15 × 107)

= (1.5 × 3.15) × 107+7

= 4.725 × 1014

≈ 4.7 × 1014 seconds

This sits in the 1014-second row, which the table gives as ≈ 3.17 million years — and 15 ÷ 3.17 ≈
4.7, matching the coefficient exactly.

Why it happens: the row heading itself is a conversion factor. Once you know 1014 s
≈ 3.17 million years, any span in that row can be converted by a single division of
coefficients, with no need to touch the exponent. Reading a table this way is much
faster than multiplying out year lengths every time.

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Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Q2 Plants on land started 47 crore/470 million years ago ( ≈ _______________ seconds). [Fill
in the blank.]

≈ 1.5 × 1016 seconds

470 million years = 4.7 × 108 years

1 year ≈ 3.15 × 107 seconds

Time = (4.7 × 108) × (3.15 × 107)

= 14.8 × 1015

= 1.48 × 1016 ≈ 1.5 × 1016 seconds

Cross-check with the row heading: 1016 s ≈ 31.7 crore years, and 47 ÷ 31.7 ≈ 1.5. ✓

Why it happens: notice the intermediate 14.8 × 1015. It is right but not standard, so
the decimal point moves one place left and the exponent rises by 1. Every
multiplication in standard form may need this final adjustment, because two
coefficients under 10 can multiply to something over 10.

Q3 Calculate and write the answer using scientific notation: (i) If one star is counted
every second, how long would it take to count all the stars in the universe? Answer
in terms of the number of seconds using scientific notation.

Stars in the observable universe ≈ 2 × 1023
Rate = 1 star per second

Time = 2 × 1023 ÷ 1

= 2 × 1023 seconds

How long is that? Convert to years:

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Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

(2 × 1023) ÷ (3.15 × 107)

≈ 0.63 × 1016 = 6.3 × 1015 years

The universe itself is about 1.38 × 1010 years old, so counting the stars one per second would
take roughly 4.6 lakh times the age of the universe.

Why it happens: the number of seconds is easy — one per star — but it means
nothing until it is compared with something we know. Turning 2 × 1023 seconds into
"far longer than the universe has existed" is the whole purpose of this section: very
large quantities are beyond experience, so we relate and compare them with
quantities we are familiar with.

Q4 Calculate and write the answer using scientific notation: (ii) If one could drink a
glass of water (200 ml) every 10 seconds, how long would it take to finish the entire
volume of water on Earth?

Drops of water on Earth ≈ 2 × 1025, at 16 drops per millilitre

Volume = (2 × 1025) ÷ 16 ml

= 0.125 × 1025 = 1.25 × 1024 ml

Glasses = 1.25 × 1024 ÷ 200

= 0.00625 × 1024 = 6.25 × 1021 glasses

Time = 6.25 × 1021 × 10 s

= 6.25 × 1022 seconds

In years that is about 2 × 1015 years — again far beyond the age of the universe.

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Class 8 Maths Chapter 2 Power Play AglaSem · NCERT Solutions

Why it happens: the chain has three steps — drops to millilitres, millilitres to
glasses, glasses to seconds — and each one is a division or multiplication in
standard form, so the exponents simply run 25 → 24 → 21 → 22. Keeping the
working in this notation is what makes a three-step calculation on 25-digit numbers
something you can do on one line.

Did you know? 1018 seconds ago the universe did not exist according to modern
physics, so an answer of 1022 seconds is not a length of time anyone could ever
spend. These questions are thought experiments, meant to build a feel for scale.

In-text Questions — Page 43
2.5 A Pinch of History

Q1 What does the first part of each name denote? [million (10⁶), billion (10⁹), trillion
(10¹²), quadrillion (10¹⁵), quintillion (10¹⁸), sextillion (10²¹), septillion (10²⁴), octillion
(10²⁷), nonillion (10³⁰), decillion (10³³)]

It is a counting word, and it tells you how many times 1000 is multiplied into 1000.

NAME FIRST PART MEANS BUILT AS VALUE

million 1 (Latin mille, thousand) 1000 × 1000 106

billion bi = 2 1000 × 10002 109

trillion tri = 3 1000 × 10003 1012

quadrillion quadri = 4 1000 × 10004 1015

quintillion quint = 5 1000 × 10005 1018

decillion dec = 10 1000 × 100010 1033

If the first part stands for k, the name equals 1000 × 1000k = 103 × 103k = 103k+3

Check: k = 3 (trillion) gives 109+3 = 1012 ✓

Page 58 of 75

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Class 8 Maths Chapter 2 Power Play
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Why it happens: the whole family is generated by one rule — each name is 1000

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times the one before, so the exponents rise in steps of 3. The Indian names work the
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same way.cbut step by 100 instead: lakh 105, crore 107, arab 109,akharab
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s, epadma 1015, shankh 1017, maha shankh 1019 — exponents rising by 2 each
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a s emcentury BCE) the mathematician Arjuna
Did you know? In the Lalitavistara (first

aglway to a tallakshana, 10 . A Pali grammar treatise
names odd powers of ten all the 53

of Kāccāyana goes to 10140, called asaṅkhyeya.

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Q2
s em The currency note with the highest denomination in India currently is 2000 rupees.

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The record is the 1 sextillion pengő note
ait was never issued.
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printed in Hungary in 1946, though

NOTE FACE VALUE
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IN STANDARD FORM

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₹2000 (India) 2000 2 × 103

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100 trillion Zimbabwean dollars (2009) 1,00,000,000,000,000 1014

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1 sextillion pengő (Hungary, 1946) 1 followed by 21 zeros 1021

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Pengő note ÷ ₹2000 note = 1021 ÷ (2 × 103)
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= 0.5 × 1018 = 5 × 10 17times the face value

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Why it happens: such notes appear when prices are themselves growing
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a s em exponentially — during hyperinflation prices multiply, they do not merely add, so the

agl government must keep adding zeros to keep up. The Zimbabwean 100 trillion dollar
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note was worth about $30 when it was printed, which shows how quickly the printed

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number and the real value part company.

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a g l Page 59 of 75