Chapter 2 Power Play (Class 8 - Latest Maths NCERT (Ganita Prakash I) Solutions)

To Find:

Exponential form of the given repeated multiplications.

Solution:

(i) $6 \times 6 \times 6 \times 6$: Here, the base is $6$ and it is multiplied $4$ times. Thus, the exponential form is $6^4$.

(ii) $y \times y$: Here, the base is $y$ and it is multiplied $2$ times. Thus, the exponential form is $y^2$.

(iii) $b \times b \times b \times b$: Here, the base is $b$ and it is multiplied $4$ times. Thus, the exponential form is $b^4$.

(iv) $5 \times 5 \times 7 \times 7 \times 7$: Here, $5$ is multiplied $2$ times and $7$ is multiplied $3$ times. Thus, the exponential form is $5^2 \times 7^3$.

(v) $2 \times 2 \times a \times a$: Here, $2$ is multiplied $2$ times and $a$ is multiplied $2$ times. Thus, the exponential form is $2^2 \times a^2$.

(vi) $a \times a \times a \times c \times c \times c \times c \times d$: Here, $a$ is multiplied $3$ times, $c$ is multiplied $4$ times, and $d$ is multiplied $1$ time. Thus, the exponential form is $a^3 \times c^4 \times d^1$ (or $a^3 c^4 d$).

To Find:

The prime factorisation and exponential representation of the given numbers.

Solution for (i) $648$:

$\begin{array}{c|cc} 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Prime factors: $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

Exponential form: $2^3 \times 3^4$

Solution for (ii) $405$:

$\begin{array}{c|cc} 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Prime factors: $3 \times 3 \times 3 \times 3 \times 5$

Exponential form: $3^4 \times 5^1$

Solution for (iii) $540$:

$\begin{array}{c|cc} 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Prime factors: $2 \times 2 \times 3 \times 3 \times 3 \times 5$

Exponential form: $2^2 \times 3^3 \times 5^1$

Solution for (iv) $3600$:

$\begin{array}{c|cc} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Prime factors: $2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$

Exponential form: $2^4 \times 3^2 \times 5^2$

To Find:

The simplified numerical value (product) of the expressions.

Solution:

(i) $2 \times 10^3$: $2 \times (10 \times 10 \times 10) = 2 \times 1000 = \mathbf{2000}$

(ii) $7^2 \times 2^3$: $(7 \times 7) \times (2 \times 2 \times 2) = 49 \times 8 = \mathbf{392}$

(iii) $3 \times 4^4$: $3 \times (4 \times 4 \times 4 \times 4) = 3 \times 256 = \mathbf{768}$

(iv) $(-3)^2 \times (-5)^2$: $(-3 \times -3) \times (-5 \times -5) = 9 \times 25 = \mathbf{225}$

(v) $3^2 \times 10^4$: $(3 \times 3) \times (10 \times 10 \times 10 \times 10) = 9 \times 10000 = \mathbf{90000}$

(vi) $(-2)^5 \times (-10)^6$: $(-32) \times (1000000) = \mathbf{-32000000}$

Conclusion:

The numerical values for the expressions are $2000$, $392$, $768$, $225$, $90000$, and $-32000000$ respectively.

Given:

Expression: $2^{224} \div 4^{32}$

To Find:

The units digit of the result.

Solution:

First, we simplify the expression using the laws of exponents. We convert $4^{32}$ to base $2$:

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

Now, the original expression becomes:

$2^{224} \div 2^{64} = 2^{224 - 64}$

$2^{224 - 64} = 2^{160}$

To find the units digit of $2^{160}$, we look at the cyclicity of powers of $2$:

$2^1 = 2$

$2^2 = 4$

$2^3 = 8$

$2^4 = 16$ (Ends in $6$)

The units digits repeat in the cycle: $2, 4, 8, 6$. The length of the cycle is $4$.

We divide the exponent $160$ by the cycle length $4$:

$160 \div 4 = 40$ with a remainder of $0$.

A remainder of $0$ means the units digit corresponds to the $4^{th}$ position in the cycle, which is $6$.

Conclusion:

The units digit of $2^{224} \div 4^{32}$ is $6$.

Given:

Bottles per container = $5$

Containers per day = $1$

Total time = $40$ days

To Find:

Total number of bottles after $40$ days.

Solution:

First, find the total number of containers brought in $40$ days:

$\text{Total Containers} = 1 \text{ container/day} \times 40 \text{ days} = 40 \text{ containers}$

Now, find the total number of bottles:

$\text{Total Bottles} = \text{Total Containers} \times \text{Bottles per container}$

$\text{Total Bottles} = 40 \times 5$

$\text{Total Bottles} = 200$

Conclusion:

There would be $200$ bottles after $40$ days.

(i) $64^3$

We know $64 = 2^6$. So, $64^3 = (2^6)^3 = 2^{18}$.

Way 1: $2^{10} \times 2^8$

Way 2: $64^1 \times 64^2$

Way 3: $(2^2)^9 = 4^9 = 4^4 \times 4^5$

(ii) $192^8$

Way 1: $192^4 \times 192^4$

Way 2: $192^{10} \times 192^{-2}$

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

(iii) $32^{-5}$

We know $32 = 2^5$. So, $32^{-5} = (2^5)^{-5} = 2^{-25}$.

Way 1: $2^{-10} \times 2^{-15}$

Way 2: $32^{-2} \times 32^{-3}$

Way 3: $2^5 \times 2^{-30}$

(i) Cube numbers are also square numbers.

Only Sometimes True. For example, $64$ is a cube ($4^3$) and also a square ($8^2$). However, $8$ is a cube ($2^3$) but is not a square number.

(ii) Fourth powers are also square numbers.

Always True. Any number raised to the power of $4$ can be written as $(x^2)^2$. Since it is expressed as something squared, it is always a square number.

(iii) The fifth power of a number is divisible by the cube of that number.

Always True. Using the law of exponents $a^m \div a^n = a^{m-n}$, we get $q^5 \div q^3 = q^2$. For any integer $q$ (where $q \neq 0$), $q^2$ is an integer, so $q^5$ is divisible by $q^3$.

(iv) The product of two cube numbers is a cube number.

Always True. Let the numbers be $a^3$ and $b^3$. Their product is $a^3 \times b^3 = (ab)^3$. Since the result can be written as a cube, the statement is always true.

(v) $q^{46}$ is both a $4$th power and a $6$th power ($q$ is a prime number).

Never True. For $q^{46}$ to be a $4$th power, the exponent $46$ must be divisible by $4$. $46 \div 4 = 11.5$. For it to be a $6$th power, $46$ must be divisible by $6$. $46 \div 6 = 7.66$. Since $46$ is not divisible by $4$ or $6$, it cannot be a $4$th or $6$th power of $q$.

(i) $10^{-2} \times 10^{-5}$

We use the law of exponents: $a^m \times a^n = a^{m+n}$

$10^{-2} \times 10^{-5} = 10^{(-2) + (-5)}$

$= 10^{-2 - 5}$

$= 10^{-7}$

(ii) $5^7 \div 5^4$

We use the law of exponents: $a^m \div a^n = a^{m-n}$

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

$= 5^3$

(iii) $9^{-7} \div 9^4$

We use the law of exponents: $a^m \div a^n = a^{m-n}$

$9^{-7} \div 9^4 = 9^{-7 - 4}$

$= 9^{-11}$

(iv) $(13^{-2})^{-3}$

We use the law of exponents: $(a^m)^n = a^{m \times n}$

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

$= 13^6$

(v) $m^5 n^{12} (mn)^9$

First, we apply the power of a product rule: $(ab)^n = a^n b^n$

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

Now, we group the like bases and use $a^m \times a^n = a^{m+n}$:

$= (m^5 \times m^9) \times (n^{12} \times n^9)$

$= m^{5+9} \times n^{12+9}$

$= m^{14} n^{21}$

Conclusion:

The simplified exponential forms are: (i) $10^{-7}$, (ii) $5^3$, (iii) $9^{-11}$, (iv) $13^6$, and (v) $m^{14} n^{21}$.

Given:

$12^2 = 144$

Solution:

(i) $(1.2)^2$:

We can write $1.2$ as $\frac{12}{10}$.

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

(ii) $(0.12)^2$:

We can write $0.12$ as $\frac{12}{100}$.

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

(iii) $(0.012)^2$:

We can write $0.012$ as $\frac{12}{1000}$.

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

(iv) $120^2$:

We can write $120$ as $12 \times 10$.

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

Conclusion:

The values are: (i) $1.44$, (ii) $0.0144$, (iii) $0.000144$, and (iv) $14400$.

Analysis:

We will convert all the expressions to their prime factors (base $2$ and $3$) to compare them.

1. $2^4 \times 3^6$ is already in prime factor form.

2. $6^4 \times 3^2$:

$= 2^4 \times 3^4 \times 3^2 = 2^4 \times 3^{4+2} = 2^4 \times 3^6$

3. $6^{10}$:

$6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10}$ (Not the same)

4. $18^2 \times 6^2$:

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

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

$= 2^4 \times 3^6$

5. $6^{24}$ is obviously much larger.

Conclusion:

The numbers that are the same are: $2^4 \times 3^6$, $6^4 \times 3^2$, and $18^2 \times 6^2$.

(i) $4^3$ or $3^4$:

$4^3 = 4 \times 4 \times 4 = 64$

$3^4 = 3 \times 3 \times 3 \times 3 = 81$

Since $81 > 64$, $3^4$ is the greater number.

(ii) $2^8$ or $8^2$:

$2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256$

$8^2 = 8 \times 8 = 64$

Since $256 > 64$, $2^8$ is the greater number.

(iii) $100^2$ or $2^{100}$:

$100^2 = 100 \times 100 = 10,000$

For $2^{100}$, we know that $2^{10} = 1024$.

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

$(1024)^{10}$ is obviously much larger than $10,000$.

Therefore, $2^{100}$ is the greater number.

Given:

Total packets to be produced = $8.5$ billion.

In the Indian number system, $1$ billion equals $100$ crore ($10^9$).

So, total packets = $8.5 \times 10^9 = 8,500,000,000$.

Digits available for the code = $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ (Total $10$ digits).

To Find:

The minimum number of digits ($n$) required for the unique ID code.

Solution:

If an ID code consists of $n$ digits, the total number of unique combinations possible is given by the formula $10^n$.

We need to find the smallest integer $n$ such that:

$10^n \geq 8,500,000,000$

Let us check the powers of $10$:

If $n = 9$:

$10^9 = 1,000,000,000$ ($1$ billion)

Since $1,000,000,000 < 8,500,000,000$, a $9$-digit code is insufficient.

If $n = 10$:

$10^{10} = 10,000,000,000$ ($10$ billion)

Since $10,000,000,000 \geq 8,500,000,000$, a $10$-digit code can accommodate all packets.

Conclusion:

The unique ID code should consist of at least $10$ digits.

Solution:

Yes, there are many other numbers that are both perfect squares and perfect cubes. These numbers are called sixth powers.

Reasoning:

A number is a perfect square if it can be written as $x^2$.

A number is a perfect cube if it can be written as $y^3$.

For a number to be both, its prime factors must have exponents that are divisible by both $2$ and $3$. The smallest common multiple of $2$ and $3$ is $6$. Therefore, any number of the form $n^6$ (where $n$ is a natural number) will be both a square and a cube.

Examples:

1. For $n = 1$: $1^6 = 1$ (Square of $1$, Cube of $1$)

2. For $n = 2$: $2^6 = 64$ (Square of $8$, Cube of $4$)

3. For $n = 3$: $3^6 = 729$ (Square of $27$, Cube of $9$)

4. For $n = 4$: $4^6 = 4096$ (Square of $64$, Cube of $16$)

Conclusion:

Numbers that are both squares and cubes can be generally described as sixth powers of natural numbers, represented as $n^6$.

Given:

Length of the passcode = $5$ characters.

Types of characters allowed: Digits ($0-9$) and Letters ($A-Z$).

To Find:

Total number of possible codes.

Solution:

First, calculate the total number of available characters for each position:

Number of digits ($0, 1, ..., 9$) = $10$

Number of letters ($A, B, ..., Z$) = $26$

Total characters per position = $10 + 26 = 36$

Since the passcode has a length of $5$, and each position can be filled by any of the $36$ characters independently, the total number of combinations is:

$\text{Total Codes} = 36 \times 36 \times 36 \times 36 \times 36 = 36^5$

$\text{Total Codes} = 6,04,66,176$

Conclusion:

The total number of possible passcode combinations is $6,04,66,176$ (Six crore four lakh sixty-six thousand one hundred and seventy-six).

Given:

Population of sheep $\approx 10^9$

Population of goats $\approx 10^9$

Solution:

Total population = Population of sheep + Population of goats

Total population = $10^9 + 10^9$

Since both terms are identical, we can write this as:

Total population = $2 \times 10^9$

Conclusion:

Comparing with the given options, both (v) $2 \times 10^9$ and (vi) $10^9 + 10^9$ are correct representations of the total population.

Note: For these calculations, we assume the current world population to be approximately $8$ billion ($8 \times 10^9$).

(i) Total pieces of clothing:

$\text{Total clothing} = (\text{Population}) \times (\text{Clothing per person})$

$\text{Total clothing} = (8 \times 10^9) \times 30$

$\text{Total clothing} = 240 \times 10^9$

In scientific notation: $2.4 \times 10^{11}$

(ii) Total honeybees:

$\text{Bee colonies} = 100 \text{ million} = 10^8$

$\text{Bees per colony} = 50,000 = 5 \times 10^4$

$\text{Total bees} = 10^8 \times (5 \times 10^4)$

In scientific notation: $5 \times 10^{12}$

(iii) Total bacterial population in humans:

$\text{Bacteria per human} = 38 \text{ trillion} = 38 \times 10^{12}$

$\text{Total humans} = 8 \times 10^9$

$\text{Total bacteria} = (38 \times 10^{12}) \times (8 \times 10^9)$

$\text{Total bacteria} = 304 \times 10^{21}$

In scientific notation: $3.04 \times 10^{23}$

(iv) Total time spent eating in a lifetime:

Assume average lifetime = $80$ years and time spent eating per day = $1.5$ hours.

$\text{Total seconds} = 80 \text{ years} \times 365.25 \text{ days} \times 1.5 \text{ hours} $$ \times 3600 \text{ seconds}$

$\text{Total seconds} \approx 29220 \text{ days} \times 5400 \text{ seconds/day}$

$\text{Total seconds} \approx 157,788,000$

In scientific notation: $1.58 \times 10^8$

Given:

Time interval = $1 \text{ billion seconds} = 1,000,000,000 \text{ s}$

Solution:

We need to convert seconds into years, days, hours, and minutes.

1. Seconds to Minutes: $1,000,000,000 \div 60 \approx 16,666,666.67 \text{ minutes}$

2. Minutes to Hours: $16,666,666.67 \div 60 \approx 277,777.78 \text{ hours}$

3. Hours to Days: $277,777.78 \div 24 \approx 11,574.07 \text{ days}$

4. Days to Years: $11,574.07 \div 365.25$ (including leap years) $\approx 31.688 \text{ years}$

Breaking down the time:

$0.688$ of a year is approximately $0.688 \times 12 \approx 8.25$ months.

So, $1$ billion seconds is approximately $31$ years and $8$ months.

Calculation:

If today's date is taken as June 2024:

Subtracting $31$ years: June 1993

Subtracting $8$ months from June 1993: October 1992

Conclusion:

One billion seconds ago, the date was approximately in October 1992 (assuming the current date is in mid-2024).