NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 2 — Figure it Out (Pages 44–45)#

We first write $4^{32}$ as a power of 2:

$4^{32} = (2^2)^{32}$
$4^{32} = 2^{64}$

So the expression is a single power of 2:

$2^{224} \div 4^{32} = 2^{224} \div 2^{64}$
$2^{224} \div 4^{32} = 2^{224 - 64}$
$2^{224} \div 4^{32} = 2^{160}$

The units digits of the powers of 2 repeat in a cycle of 4:

Every power whose exponent is a multiple of 4 ends in 6. Since $160 = 4 \times 40$, $2^{160}$ ends in 6.

∴ The units digit is 6.

One container of 5 bottles comes in each day, so over 40 days, 40 containers come in:

Number of bottles $= 40 \times 5$
Number of bottles $= 200$

∴ There would be 200 bottles, that is $2 \times 10^2$.

This growth is additive: the same 5 bottles are added every day. Compare it with doubling every day, where 40 days would give a number with 13 digits.

Note: if the first container was already there before the 40 days began, there would be one more container, that is $41 \times 5 = 205$ bottles. NCERT’s answer is 200.

$64 = 2^6$, so:

$64^3 = (2^6)^3$
$64^3 = 2^{18}$

Splitting $2^{18}$ in different ways:

• $64^3 = 2^{10} \times 2^8$
• $64^3 = 4^5 \times 4^4$, because $4^9 = (2^2)^9 = 2^{18}$
• $64^3 = 8^3 \times 8^3$, because $8^6 = (2^3)^6 = 2^{18}$

$192 = 2^6 \times 3$, so:

$192^8 = (2^6 \times 3)^8$
$192^8 = 2^{48} \times 3^8$

Three ways:

• $192^8 = 2^{48} \times 3^8$
• $192^8 = 2^{40} \times 2^8 \times 3^8$
• $192^8 = 2^{40} \times 6^8$, because $2^8 \times 3^8 = (2 \times 3)^8$

$32 = 2^5$, so:

$32^{-5} = (2^5)^{-5}$
$32^{-5} = 2^{-25}$

Three ways:

• $32^{-5} = 2^{-10} \times 2^{-15}$
• $32^{-5} = 2^{-5} \times 2^{-20}$
• $32^{-5} = 4^{-12} \times 2^{-1}$, because $4^{-12} = 2^{-24}$

Only Sometimes True.

• $64 = 4^3$ is a cube, and also $64 = 8^2$, a square.
• $8 = 2^3$ is a cube, but 8 is not a square.

A cube is a square only when it is a sixth power: $n^6 = (n^2)^3 = (n^3)^2$.

Always True. Any fourth power can be written as the square of a square:

$n^4 = (n^2)^2$

For example, $3^4 = 81 = 9^2$.

Always True (for any number other than 0). Dividing leaves a whole power of the number:

$n^5 \div n^3 = n^{5 - 3}$
$n^5 \div n^3 = n^2$

For example, $2^5 = 32$ and $32 \div 2^3 = 4$.

Always True. For any two numbers $a$ and $b$:

$a^3 \times b^3 = (a \times b)^3$

For example, $8 \times 27 = 2^3 \times 3^3$, which is $6^3 = 216$.

Never True. Since $q$ is prime, $q^{46}$ is a 4th power only if 46 can be split into groups of 4, and a 6th power only if 46 can be split into groups of 6. But:

$46 \div 4 = 11$ remainder 2
$46 \div 6 = 7$ remainder 4

46 is divisible by neither 4 nor 6, so $q^{46}$ is not a 4th power or a 6th power.

$10^{-2} \times 10^{-5} = 10^{-2 + (-5)}$
$10^{-2} \times 10^{-5} = 10^{-7}$, that is $\dfrac{1}{10^7}$

$5^7 \div 5^4 = 5^{7 - 4}$
$5^7 \div 5^4 = 5^3$

$9^{-7} \div 9^4 = 9^{-7 - 4}$
$9^{-7} \div 9^4 = 9^{-11}$, that is $\dfrac{1}{9^{11}}$

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

First $(mn)^9 = m^9 n^9$, then we add the exponents of each base:

$m^5 n^{12} (mn)^9 = m^5 n^{12} \times m^9 n^9$
$m^5 n^{12} (mn)^9 = m^{5 + 9} \times n^{12 + 9}$
$m^5 n^{12} (mn)^9 = m^{14} n^{21}$

Each number is 12 multiplied or divided by a power of 10. When we square it, that power of 10 is squared too:

$(1.2)^2 = \left(\dfrac{12}{10}\right)^2$
$(1.2)^2 = \dfrac{144}{100}$
$(1.2)^2 = 1.44$

$(0.12)^2 = \left(\dfrac{12}{100}\right)^2$
$(0.12)^2 = \dfrac{144}{10000}$
$(0.12)^2 = 0.0144$

$(0.012)^2 = \left(\dfrac{12}{1000}\right)^2$
$(0.012)^2 = \dfrac{144}{1000000}$
$(0.012)^2 = 0.000144$

$120^2 = (12 \times 10)^2$
$120^2 = 144 \times 100$
$120^2 = 14400$

∴ (i) 1.44, (ii) 0.0144, (iii) 0.000144, (iv) 14400.

A quick rule: the square has twice as many decimal places as the number.

We write each number as a power of 2 times a power of 3, using $6 = 2 \times 3$ and $18 = 2 \times 3^2$:

$2^4 \times 3^6 = 2^4 \times 3^6$

$6^4 \times 3^2 = 2^4 \times 3^4 \times 3^2$
$6^4 \times 3^2 = 2^4 \times 3^6$

$6^{10} = 2^{10} \times 3^{10}$

$18^2 \times 6^2 = (2 \times 3^2)^2 \times (2 \times 3)^2$
$18^2 \times 6^2 = 2^2 \times 3^4 \times 2^2 \times 3^2$
$18^2 \times 6^2 = 2^4 \times 3^6$

$6^{24} = 2^{24} \times 3^{24}$

∴ $2^4 \times 3^6$, $6^4 \times 3^2$ and $18^2 \times 6^2$ are the same. Each equals $2^4 \times 3^6 = 11664$.

$4^3 = 64$
$3^4 = 81$

∴ $3^4$ is greater.

$2^8 = 256$
$8^2 = 64$

∴ $2^8$ is greater.

$100^2 = 10000$, which is $10^4$.

For $2^{100}$, we use $2^{10} = 1024$, which is more than 1000:

$2^{100} = (2^{10})^{10}$
$2^{100} = 1024^{10}$

So $2^{100}$ is more than $10^{30}$, far more than $10^4$.

∴ $2^{100}$ is greater.

Each place in the code can hold any of the 10 digits, so a code with $n$ digits gives $10^n$ different codes. We need at least as many codes as packets:

Number of packets $= 8.5$ billion
Number of packets $= 8.5 \times 10^9$

A 9-digit code gives only $10^9$ (1 billion) codes, which is too few. A 10-digit code gives $10^{10}$ (10 billion) codes, which is enough.

∴ The code should consist of 10 digits.

Yes, infinitely many. A number is both a square and a cube when it is a sixth power:

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

So $n^6$ is the square of $n^3$ and the cube of $n^2$.

Examples:

• $1^6 = 1$
• $2^6 = 64 = 8^2 = 4^3$
• $3^6 = 729 = 27^2 = 9^3$
• $4^6 = 4096 = 64^2 = 16^3$

∴ The numbers that are both squares and cubes are exactly the sixth powers: 1, 64, 729, 4096, 15625, …

Each of the 5 places can hold one of the 26 letters A–Z or one of the 10 digits 0–9:

Choices for each place $= 26 + 10 = 36$

The choice in each place does not depend on the others, so we multiply:

Number of codes $= 36 \times 36 \times 36 \times 36 \times 36$
Number of codes $= 36^5$
Number of codes $= 60466176$

∴ $36^5$, that is 60,466,176 codes are possible.

We add the two populations:

Total population $= 10^9 + 10^9$
Total population $= 2 \times 10^9$

∴ The total is $10^9 + 10^9 = 2 \times 10^9$, that is about 2 billion. Options (vi) and (v) are the same number, so both are correct; NCERT gives (vi).

The other options are much too large: $10^{10}$ would be ten times $10^9$, not two times, and $10^9 \times 10^9 = 10^{18}$ is what we get by multiplying, not adding.

Note: in the textbook, the first option is printed as “(ii) 20⁹”. It is option (i).

Total pieces of clothing $= 8.2 \times 10^9 \times 30$
Total pieces of clothing $= 246 \times 10^9$
Total pieces of clothing $= 2.46 \times 10^{11}$

100 million $= 10^8$ and $50000 = 5 \times 10^4$.

Number of honeybees $= 10^8 \times 5 \times 10^4$
Number of honeybees $= 5 \times 10^{12}$

38 trillion $= 38 \times 10^{12}$.

Bacterial population $= 38 \times 10^{12} \times 8.2 \times 10^9$
Bacterial population $= 311.6 \times 10^{21}$
Bacterial population $= 3.116 \times 10^{23}$

We assume a lifetime of 70 years and 1 hour of eating each day, and 1 hour $= 3600$ seconds.

Time spent eating $= 3600 \times 365 \times 70$
Time spent eating $= 91980000$ seconds
Time spent eating $= 9.198 \times 10^7$ seconds

That is roughly $9.2 \times 10^7$ seconds, or about 3 years spent eating.

1 arab is 1 billion, that is $10^9$. We change $10^9$ seconds into days:

Seconds in a day $= 24 \times 60 \times 60 = 86400$

Number of days $= \dfrac{10^9}{86400}$
Number of days $= 11574.07$ (about)

So 1 billion seconds is about 11,574 days, or about 31.7 years (more exactly 11,574 days, 1 hour, 46 minutes and 40 seconds).

To get the date, count back about 31 years and 8 months from today. For example, counting back from midnight at the start of 1 January 2026 gives 24 April 1994, at about 10:13 pm.