NCERT Solutions for Class 8 Maths Chapter 2 Power Play (PDF) – 2027

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

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

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).

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.

One more fold gives 46: ≈ 7,03,687 km — past the Moon.

It becomes 4 times as thick.

The book’s next line asks you to check that 3 folds increase the thickness 8 times (= 2 × 2 × 2). Take fold 9 (0.512 cm) and fold 12 (4.096 cm): 4.096 ÷ 0.512 = 8. ✓

(v) 2¹⁰v

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

Yes.

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

3⁷ = 2187 diamonds.

Follow the riddle level by level. Each line multiplies the count by 3.

3⁴ = 81 rooms.

Yes — 2187 diamonds, from a single multiplication.

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

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

Split each exponent into two parts you already know, then multiply once.

Yes.

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

On day 29.

Half the pond is covered on day 29.

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

2⁴ × 3⁴ = 16 × 81 = 1296 lotuses.

Exactly the same — 1296 lotuses.

Yes — 6⁴.

So m = 6 and n = 4. The general rule this demonstrates is

12 ways.

7 × 2 × 3 = 42 ways.

10⁵ = 1,00,000 passwords (one lakh).

Follow the book’s advice and shrink the problem first.

10⁵ = 1,00,000.

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

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

Work out the raw count first, then see how the real system spends it.

The real systems are not random. In a Pincode the first digit names one of the postal zones (4 = Western zone, which includes Madhya Pradesh, so Vidisha; 7 = Eastern zone, which includes Mizoram, 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. Vehicle numbers start with the state code (MP, MZ, TN …) and then the RTO district number. Mobile numbers begin with an operator-and-circle block.

2⁷⁵.

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

Yes.

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

Yes — once n⁰ = 1 and n^–a = 1/n^a 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:

Same base each time, so add the exponents.

(iv) The bases 2 and –4 differ, so first write (–4)^–2 as a power of 2 (the even exponent removes the minus sign):

4⁴ = 256 times.

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

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

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

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

The Sun–Earth distance, 1.496 × 10¹¹ m.

Find what fraction of the Sun–Saturn distance the Earth sits at.

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

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

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

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:

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

Relate the quantities first:

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.

In thousands — around ten thousand.

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

Take adult weights of about 60 kg each.

About 12 to 13 days ago — roughly two weeks.

About 66 times — roughly once a year — if a person could really walk non-stop for 60 years.

In standard form: about 2.6 × 10⁶ km walked, against 4 × 10⁴ km per trip.

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

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

Linear growth adds a fixed amount each step; exponential growth multiplies by a fixed factor each step.

1.3 × 10⁹

Yes.

1.1 × 10¹⁴

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

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.

It is Roxie’s age in minutes.

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.

Estu was born 11 years and about 52 days (7 weeks) before today.

Example: if today is 31 August 2026, going back 11 years gives 31 August 2015, and 52 days before that is 10 July 2015.

About 11½ days old — not even a fortnight.

≈ 4.7 × 10¹⁴ seconds

≈ 1.5 × 10¹⁶ seconds

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

The record is the 1 sextillion pengő note — 10²¹ pengő, also called 1 milliard bilpengő — printed in Hungary in 1946, though it was never issued.

The units digit is 6.

Now look at the units digits of the powers of 2. They repeat in a cycle of four:

200 bottles = 2 × 10².

Part (v) has two bases, so handle each separately.

Write each number as 12 times a power of 10, then square both parts.

Bring every expression down to powers of the primes 2 and 3.

So the three that are the same are 2⁴ × 3⁶, 6⁴ × 3² and 18² × 6², each equal to 11664.

10 digits.

A 10-digit code gives 1000 crore codes for 850 crore packets — comfortably more than needed, with room to spare.

Yes — infinitely many. They are exactly the sixth powers, n⁶.

36⁵ = 6,04,66,176 — about 6 crore, or 6.05 × 10⁷.

(vi) 10⁹ + 10⁹ — and this equals (v) 2 × 10⁹, so both express the correct total.

About 31 years and 8 months before today — 11,574 days ago.

So go back 31 years, 8 months and about 9 days from today’s date. Example: counting back from the start of 31 August 2026 gives about 10 pm on 22 December 1994.

Compare the two as powers of 10, without multiplying anything out.

Estu’s number is greater — overwhelmingly so.