ch2 POWER PLAY figure it out CLASS 8

Q1

Find the units digit in the value of $\;2^{224}\div 4^{32}$. (Hint: $4=2^2$.)

Solution and explanation

Write everything as powers of 2:

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

So

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

Units digits of powers of 2 cycle every 4: $2,4,8,6,$ then repeat. We find the position in the cycle:

$$160 \bmod 4 = 0,$$

so $2^{160}$ is at the 4th position of the cycle → units digit = 6.

Answer: 6

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Q2

There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?

Solution and explanation

Each container has 5 bottles. After 40 days there are 40 containers (one per day), so total bottles:

$$40 \times 5 = 200.$$

Answer: 200 bottles

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Q3

Write the given number as the product of two or more powers in three different ways.
(i) $64^3$ \qquad (ii) $192^8$ \qquad (iii) $32^{-5}$

Solution and explanation

We use prime-factor forms and power rules $(a^b)^c=a^{bc}$.

(i) $64^3$
$64=2^6$. So

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

Other equivalent ways:

$$64^3=(8^2)^3=8^6,\qquad 64^3=(4^3)^3=4^9.$$

(All three equal $2^{18}$.)

Three forms: $2^{18},\;8^{6},\;4^{9}.$

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(ii) $192^8$
Prime factorise $192$:

$$192=64\times 3 = 2^6\times 3.$$

So

$$192^8=(2^6\cdot3)^8 =2^{48}\cdot 3^8.$$

Other ways (same exponents grouped differently):

$$192^8=(2^{24})^2\cdot 3^8,\qquad 192^8=(2^{16})^3\cdot 3^8.$$

Three forms: $2^{48}\cdot3^8,\; (2^{24})^2\cdot3^8,\; (2^{16})^3\cdot3^8.$

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(iii) $32^{-5}$
$32=2^5$. So

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

Other forms:

$$32^{-5}=(2^{-5})^{5},\qquad 32^{-5}=\frac{1}{32^{5}}.$$

Three forms: $2^{-25},\; (2^{-5})^{5},\; \dfrac{1}{32^{5}}.$

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Q4

Examine each statement and say if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain.

(i) Cube numbers are also square numbers.

• A number that is both a cube and a square must be both $n^2$ and $m^3$. Such numbers are 6th powers ($k^6$). Example: $64=2^6$ is both. But $8=2^3$ is not a square.
  Answer: Only Sometimes True (exactly those numbers that are 6th powers).

(ii) Fourth powers are also square numbers.

• $x^4=(x^2)^2$, so every 4th power is a square (square of $x^2$).
  Answer: Always True.

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

• $n^5 = n^3\cdot n^2$, so yes $n^5$ is divisible by $n^3$ for any integer $n$.
  Answer: Always True.

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

• If numbers are $a^3$ and $b^3$, product $a^3b^3=(ab)^3$, which is a cube.
  Answer: Always True.

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

• For a power $q^{46}$ to be a 4th power, exponent 46 must be divisible by 4; to be a 6th power, 46 must be divisible by 6. 46 is divisible by neither 4 nor 6 (46 ÷ 4 = 11.5; 46 ÷ 6 ≈ 7.67). So it is neither.
  Answer: Never True.

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Q5

Simplify and write these in exponential form.

(i) $10^{-2}\times 10^{-5}$
(ii) $5^7\div5^4$
(iii) $9^{-7}\div9^4$
(iv) $(13^{-2})^{-3}$
(v) $m^5 n^{12}(mn)^9$

Solution and explanation (use exponent laws)

• Law used: $a^m a^n = a^{m+n}$, $a^m\div a^n = a^{m-n}$, $(a^m)^n=a^{mn}$, $(ab)^n=a^n b^n$.

(i) $10^{-2}\times 10^{-5} = 10^{-2 + (-5)} = 10^{-7}.$

(ii) $5^7\div5^4 = 5^{7-4} = 5^{3} = 125.$

(iii) $9^{-7}\div9^4 = 9^{-7-4} = 9^{-11} = \dfrac{1}{9^{11}}.$

(iv) $(13^{-2})^{-3} = 13^{(-2)\cdot(-3)} = 13^{6}.$

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

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Q6

If $12^2=144$, find:

(i) $(1.2)^2$
(ii) $(0.12)^2$
(iii) $(0.012)^2$
(iv) $120^2$

Solution and explanation (scale by powers of 10)

$(1.2) = 12/10$. So

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

$(0.12)=12/100$. So

$$(0.12)^2=\frac{12^2}{100^2}=\frac{144}{10,000}=0.0144.$$

$(0.012)=12/1000$. So

$$(0.012)^2=\frac{12^2}{1000^2}=\frac{144}{1,000,000}=0.000144.$$

$120=12\times10$. So

$$120^2=(12\times10)^2=12^2\times10^2=144\times100=14,400.$$

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

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Q7

Circle the numbers that are the same —
$2^{4}\times3^{6},\;6^{4}\times3^{2},\;6^{10},\;18^{2}\times6^{2},\;6^{24}.$

Solution and explanation

Compute prime factor form:

• $2^{4}\cdot3^{6}$ is already $2^4 3^6$.

• $6^{4}\cdot3^{2} = (2\cdot3)^{4}\cdot3^2 = 2^{4}3^{4}\cdot3^{2} = 2^{4}3^{6}$. → same as first.

• $6^{10} = 2^{10}3^{10}$ → different.

• $18^{2}\cdot6^{2} = (2\cdot3^2)^2 \cdot (2\cdot3)^2 = (2^2 3^4)\cdot(2^2 3^2) = 2^{4}3^{6}$. → same as first.

• $6^{24}=2^{24}3^{24}$ → different.

So the equal ones are:
$\;2^{4}\cdot3^{6},\;6^{4}\cdot3^{2},\;18^{2}\cdot6^{2}.$

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Q8

Identify the greater number in each of the following—

(i) $4^3$ or $3^4$
(ii) $2^8$ or $8^2$
(iii) $100^2$ or $2^{100}$

Solution and explanation

(i) $4^3=64$. $3^4=81$. So $3^4$ is greater.
Answer: $3^4$.

(ii) $2^8=256$. $8^2=64$. So $2^8$ is greater.
Answer: $2^8$.

(iii) $100^2=10{,}000$. $2^{100}$ is extremely large (approx $1.27\times10^{30}$). So $2^{100}$ is far greater.
Answer: $2^{100}$.

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Q9

A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID for each packet using digits 0–9. How many digits should the code consist of?

Solution and explanation

Number of packets = $8.5$ billion $= 8.5\times10^9$.

A code of length $n$ (digits 0–9) has $10^n$ possible codes. We need the smallest $n$ with $10^n\ge 8.5\times10^9$.

Check:

• $10^9 = 1\times10^9$ (too small).

• $10^{10} = 1\times10^{10} = 10{,}000{,}000{,}000$ (which is $>8.5\times10^9$).

Therefore $n=10$ digits are needed.

Answer: 10 digits

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Q10

64 is a square number ($8^2$) and a cube number ($4^3$). Are there other numbers that are both squares and cubes? Describe them in general.

Solution and explanation

A number that is both a perfect square and a perfect cube must be a power that is simultaneously of the form $a^2$ and $b^3$. This happens exactly when the exponent is a common multiple of 2 and 3, i.e. a multiple of $\mathrm{lcm}(2,3)=6$.

Thus numbers of the form $k^6$ (6th powers) are both squares and cubes:

$$k^6 = (k^3)^2 = (k^2)^3.$$

Examples: $1^6=1,\;2^6=64,\;3^6=729,\;4^6=4096$, etc.

Answer: Yes. Exactly the sixth powers $k^6$.

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Q11

A digital locker has an alphanumeric passcode of length 5 (digits 0–9 and letters A–Z). How many such codes are possible?

Solution and explanation

Assume uppercase letters only → 26 letters + 10 digits = 36 possible characters per slot. For 5 positions, total possibilities:

$$36^5.$$

Compute value (optional): $36^2=1296,\;36^3=46,656,\;36^4=1,679,616,\;36^5=60,466,176.$

Answer: $36^5 = 60,466,176$ possible codes.

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Q12

The worldwide population of sheep (2024) is about $10^9$, and goats are about the same. What is the total population of sheep and goats?

Solution and explanation

Sheep ≈ $10^9$ and goats ≈ $10^9$. Sum:

$$10^9 + 10^9 = 2\times10^9.$$

Answer: $2\times10^9$

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Q13

Calculate and write the answer in scientific notation:

(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population in all humans in the world (use world pop ≈ $8.2\times10^9$).
(iv) Total time spent eating in a lifetime in seconds. (Make and state reasonable assumptions.)

Solution and explanation

We show method + result in scientific notation.

(i) Let world population ≈ $8.2\times10^9$ people. Pieces per person = 30.
Total $=30\times 8.2\times10^9 =246\times10^9 =2.46\times10^{11}.$

Answer (i): $\boxed{2.46\times10^{11}}$ pieces of clothing (using 8.2 billion people).

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(ii) Colonies = $100$ million = $1.0\times10^8$. Bees per colony = 50,000 = $5.0\times10^4$.
Total bees $=(1.0\times10^8)\times(5.0\times10^4)=5.0\times10^{12}.$

Answer (ii): $\boxed{5.0\times10^{12}}$ honeybees.

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(iii) Bacteria per person ≈ 38 trillion = $3.8\times10^{13}$. World pop ≈ $8.2\times10^9$.
Total bacteria ≈ $(3.8\times10^{13})\times(8.2\times10^9)$.

Multiply coefficients: $3.8\times8.2=31.16$. Add exponents: $13+9=22$. So

$$31.16\times10^{22}=3.116\times10^{23}.$$

Answer (iii): $\boxed{3.116\times10^{23}}$ (approx).

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(iv) Total time spent eating in a lifetime (example model & result).
We must assume average daily time spent eating and an expected lifetime. One reasonable model:

• Eating time ≈ 1.5 hours/day = $1.5\times60\times60 = 5400$ seconds/day.

• Lifetime ≈ 70 years. Use 1 year ≈ 365 days (ignore leap days for a rough estimate).

Total seconds $=5400$ s/day $\times 70\times365$ days
First compute days: $70\times365=25,550$ days.
Now multiply: $25,550\times5,400$.

We compute carefully:

• $25,550\times5,400 = 25,550\times(54\times100) = (25,550\times54)\times100.$

• $25,550\times54 = 25,550\times(50+4) = 25,550\times50 + 25,550\times4.$

• $25,550\times50 = 1,277,500$.

• $25,550\times4 = 102,200$.

• Sum = $1,277,500 + 102,200 = 1,379,700.$
  Multiply by 100 → $137,970,000$ seconds.

So about $1.3797\times10^8$ seconds.

Answer (iv): ≈ $\boxed{1.38\times10^{8}\ \text{seconds}}$ (using 1.5 h/day and 70 years).
(If you pick different assumptions — e.g. 1 hour/day or 80 years — change numbers accordingly; show assumptions when giving to students.)

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Q14

What was the date 1 arab / 1 billion seconds ago?

Solution and explanation

1 arab = 1 billion seconds = $10^9$ seconds.

Convert $10^9$ seconds to days:

$$1\,\text{day}=86400\ \text{seconds}. \quad \text{Compute } \frac{10^9}{86400}.$$

We compute exact integer days and remainder (division):

• $86400\times 11{,}574 = 86400\times 11{,}574 = 999{,}993{,}600$ seconds. (Check by multiplication.)

• Remainder: $10^9 - 999{,}993{,}600 = 4{,}400$ seconds.

So $10^9$ seconds = 11,574 days + 4,400 seconds.

Convert 4,400 seconds to hours/minutes:

• $4,400 \div 3{,}600 = 1$ hour remainder $800$ seconds.

• $800$ seconds = $13$ minutes $20$ seconds.

So $10^9$ seconds = 11,574 days, 1 hour, 13 minutes, 20 seconds.

Now convert days to years roughly:

$$11,574\ \text{days}\div365\approx31.7\ \text{years}.$$

So about 31.7 years ago.

• If you want a calendar date, subtract ~31 years and 259 days (approx) from today. For example, if "today" were mid-2025, one billion seconds earlier would be around 1993 (roughly 31 years and 8–9 months earlier). Always state the reference date when giving an exact calendar date.

Answer: Approximately 11,574 days ≈ 31.7 years ago (i.e. ~31 years, 8 months, 1 hour, 13 min, 20 s earlier).

Powers and Exponents – Questions & Step-by-Step Solutions

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1. Simplify: \( 2^3 \times 2^4 \)

View Solution

Using the law \( a^m \times a^n = a^{m+n} \):
\( 2^3 \times 2^4 = 2^{3+4} = 2^7 = 128 \)

2. Simplify: \( 5^6 \div 5^2 \)

View Solution

Using \( a^m \div a^n = a^{m-n} \):
\( 5^6 \div 5^2 = 5^{6-2} = 5^4 = 625 \)

3. Simplify: \( (3^2)^4 \)

View Solution

Using \( (a^m)^n = a^{m \times n} \):
\( (3^2)^4 = 3^{2 \times 4} = 3^8 = 6561 \)

4. Write 81 as a power of 3.

View Solution

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

5. Simplify: \( 2^{-3} \)

View Solution

Negative exponent rule: \( a^{-n} = \frac{1}{a^n} \)
\( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \)

6. Simplify: \( 10^0 \)

View Solution

Any non-zero number raised to the power 0 is 1.
\( 10^0 = 1 \)

7. Express \( 1/125 \) as a power of 5.

View Solution

\( 125 = 5^3 \), so \( 1/125 = 5^{-3} \)

8. Simplify: \( (2^3 \times 3^2)^2 \)

View Solution

First simplify inside the bracket:
\( 2^3 = 8, \; 3^2 = 9 \) → \( 8 \times 9 = 72 \)
Then square: \( 72^2 = 5184 \)

9. Simplify: \( (4^2 \div 4)^3 \)

View Solution

Inside: \( 4^2 \div 4 = 4^{2-1} = 4^1 = 4 \)
Then: \( 4^3 = 64 \)

10. Evaluate: \( (5^{-1})^{-3} \)

View Solution

Multiply powers: \( (5^{-1})^{-3} = 5^{(-1)\times(-3)} = 5^3 = 125 \)

11. Simplify: \( 9^{3/2} \)

View Solution

Fractional exponent: \( a^{m/n} = \sqrt[n]{a^m} \)
\( 9^{3/2} = (\sqrt{9})^3 = 3^3 = 27 \)

12. Simplify: \( 27^{2/3} \)

View Solution

\( 27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9 \)

13. Simplify: \( (16^{1/2})^3 \)

View Solution

\( 16^{1/2} = 4 \), so \( 4^3 = 64 \)

14. If \( 2^x = 32 \), find \( x \).

View Solution

\( 32 = 2^5 \) → Therefore, \( x = 5 \)

Chapter 2 – Power Play: Case-Based & Assertion–Reasoning Questions Case-Based Study Questions with Answer Keys

 

Chapter 2 – Power Play: Case-Based & Assertion–Reasoning Questions

Case-Based Study Questions with Answer Keys

Case Study 1 — Magical Pond & Doubling
A magical pond has 1 lotus flower on Day 1. Each day, the number of lotus flowers doubles. On Day 30, the pond is completely covered.
Q1.1 On which day is the pond half-covered?
Q1.2 Write the number of lotuses on Day 30 in exponential form.
Q1.3 If each lotus takes up 0.5 m², what is the total area covered by lotuses on Day 30?

Q1: On which day is the pond half-covered?
Answer: Day 29 (one day before full coverage).

Q2: Write the number of lotuses on Day 30 in exponential form.
Answer: 2²⁹ lotuses on Day 29, 2³⁰ lotuses on Day 30.

Q3: If each lotus takes up 0.5 m², what is the total area covered by lotuses on Day 30?
Answer: 2³⁰ × 0.5 m² = 536,870,912 m².

Case Study 2 — Paper Folding to the Moon

Q1: Write the thickness after 10 folds in cm using exponents.
Answer: 0.001 × 2¹⁰ = 1.024 cm.

Q2: Find the thickness after 46 folds in km.
Answer: 0.001 × 2⁴⁶ cm ≈ 7,036,874 km.

Q3: Compare exponential growth in folding to linear growth of stacking 46 sheets without folding.
Answer: Exponential growth reaches Moon in 46 folds; linear stacking = only 0.046 cm thickness.

Case Study 3 — Roxie’s Tulābhāra Donation

Q1: Find the worth of the donation.
Answer: ₹3150.

Q2: If 1-rupee coins replace the jaggery (mass = 4 g each), find the number of coins needed.
Answer: 11,250 coins.

Q3: Write your answer in scientific notation.
Answer: 1.125 × 10⁴ coins.

Case Study 4 — Locks and Combinations

Q1: How many different codes are possible?
Answer: 10⁵ = 100,000 codes.

Q2: If the lock instead uses 5 characters (A–Z letters only), how many codes are possible?
Answer: 26⁵ = 11,881,376 codes.

Q3: Which is more secure and why?
Answer: Letter-based is more secure (more possibilities).

Case Study 5 — Earth to Moon Steps vs Folding

Q1: Find number of steps needed.
Answer: 1,922,000,000 steps.

Q2: Compare with number of folds needed (paper example).
Answer: Folding paper = only 46 folds to reach Moon’s distance.

Q3: Explain the difference in terms of linear vs exponential growth.
Answer: Linear = constant addition; exponential = rapid multiplication.

Assertion and Reasoning Questions with Answer Keys

(A) Both true and R explains A
(B) Both true but R does not explain A
(C) A true but R false
(D) A false but R true

1. A: Doubling every day is an example of exponential growth.
   R: In exponential growth, the quantity increases by a fixed multiple each time.
   Answer: A
2. A: 2¹⁰ = 1024 means 2 multiplied by itself 10 times equals 1024.
   R: The base tells how many times to multiply the exponent by itself.
   Answer: C
3. A: Scientific notation expresses numbers as x × 10ⁿ with 1 ≤ x < 10.
   R: This makes it easier to write and compare very large or small numbers.
   Answer: A
4. A: In the magical pond, the pond is one-quarter full on Day 28.
   R: Each day’s lotus count is double the previous day’s.
   Answer: A
5. A: Linear growth can overtake exponential growth if given enough time.
   R: In exponential growth, increase per step decreases over time.
   Answer: D
6. A: a^m × aⁿ = a^m+n for all integers m and n.
   R: Multiplying powers with the same base adds the exponents.
   Answer: A
7. A: a^-n = 1/aⁿ for a ≠ 0.
   R: Negative exponents represent reciprocals.
   Answer: A
8. A: The number 64 is both a perfect square and a perfect cube.
   R: 64 = 8² and 64 = 4³.
   Answer: A
9. A: The cube of any even number is odd.
   R: Cube of even = even × even × even = even.
   Answer: D
10. A: The largest 3-digit power of 2 is 2⁹.
    R: 2⁹ = 512 and 2¹⁰ = 1024.
    Answer: A
11. A: Multiplying by 10³ increases a number’s value by 100.
    R: 10³ = 1000, so it multiplies the number by 1000.
    Answer: D
12. A: (a^m)ⁿ = a^mn holds for all integers m, n.
    R: Raising a power to another power multiplies exponents.
    Answer: A
13. A: The number of 6-letter passwords from A–Z is 26⁶.
    R: Each letter has 26 choices, independent of others.
    Answer: A
14. A: Linear growth means adding the same amount each step.
    R: Linear growth is faster than exponential growth for small step sizes.
    Answer: B
15. A: In scientific notation, the exponent tells the number of decimal shifts.
    R: Shifting decimal right means negative exponent.
    Answer: C

Class-8 Ganita prakash ASSERTION & REASONING Maths ch-2 Power play with answer key

 Class-8 Ganita prakash ASSERTION & REASONING Maths ch-2 Power play

Assertion and Reasoning Questions with Answer Keys

Assertion and Reasoning Questions with Answer Keys

(A) Both true and R explains A
(B) Both true but R does not explain A
(C) A true but R false
(D) A false but R true

1. A: Doubling every day is an example of exponential growth.
   R: In exponential growth, the quantity increases by a fixed multiple each time.
   Answer: A

2. A: $2^{10} = 1024$ means 2 multiplied by itself 10 times equals 1024.
   R: The base tells how many times to multiply the exponent by itself.
   Answer: C

3. A: Scientific notation expresses numbers as $x \times 10^n$ with $1 \le x < 10$.
   R: This makes it easier to write and compare very large or small numbers.
   Answer: A

4. A: In the magical pond, the pond is one-quarter full on Day 28.
   R: Each day’s lotus count is double the previous day’s.
   Answer: A

5. A: Linear growth can overtake exponential growth if given enough time.
   R: In exponential growth, increase per step decreases over time.
   Answer: D

6. A: $a^m \times a^n = a^{m+n}$ for all integers m and n.
   R: Multiplying powers with the same base adds the exponents.
   Answer: A

7. A: $a^{-n} = 1/a^n$ for $a \neq 0$.
   R: Negative exponents represent reciprocals.
   Answer: A

8. A: The number 64 is both a perfect square and a perfect cube.
   R: $64 = 8^2$ and $64 = 4^3$.
   Answer: A

9. A: The cube of any even number is odd.
   R: Cube of even = even × even × even = even.
   Answer: D

10. A: The largest 3-digit power of 2 is $2^9$.
    R: $2^9 = 512$ and $2^{10} = 1024$.
    Answer: A

11. A: Multiplying by $10^3$ increases a number’s value by 100.
    R: $10^3 = 1000$, so it multiplies the number by 1000.
    Answer: D

12. A: $(a^m)^n = a^{mn}$ holds for all integers m, n.
    R: Raising a power to another power multiplies exponents.
    Answer: A

13. A: The number of 6-letter passwords from A–Z is $26^6$.
    R: Each letter has 26 choices, independent of others.
    Answer: A

14. A: Linear growth means adding the same amount each step.
    R: Linear growth is faster than exponential growth for small step sizes.
    Answer: B

15. A: In scientific notation, the exponent tells the number of decimal shifts.
    R: Shifting decimal right means negative exponent.
    Answer: C

Class-8 Ganita prakash ASSERTION & REASONING Maths ch-1 a square and a cube

Class-8 Ganita prakash ASSERTION & REASONING Maths ch-1 a square and a cube

Assertion & Reasoning Questions (15)

For each question:

• A = Assertion

• R = Reason
  Choose the correct option:
  

  A. Both A and R are true, and R is the correct explanation of A.
  B. Both A and R are true, but R is not the correct explanation of A.
  C. A is true, but R is false.
  D. A is false, but R is true.

• 
  

1. (A): All perfect squares have an odd number of factors.
   (R): They have one factor that is repeated in the middle.
   Answer: Both A and R are true, and R is the correct explanation.

2. (A): If a number ends in 2, 3, 7, or 8, it can never be a perfect square.
   (R): The units digit of a perfect square is always 0, 1, 4, 5, 6, or 9.
   Answer: Both A and R are true, and R is the correct explanation.

3. (A): 36 is a perfect square.
   (R): It can be expressed as the product of two identical numbers, 6 × 6.
   Answer: Both A and R are true, and R is the correct explanation.

4. (A): Every perfect square can be written as the sum of consecutive odd numbers starting from 1.
   (R): The sum of the first n odd numbers equals n².
   Answer: Both A and R are true, and R is the correct explanation.

5. (A): All prime numbers are perfect squares.
   (R): Prime numbers have exactly two factors: 1 and itself.
   Answer: A is false, R is true.

6. (A): The cube of a number ending in 0 will always have exactly two zeros at the end.
   (R): Cubes can have multiple zeros, depending on the number of 10s in the factors.
   Answer: A is false, R is true.

7. (A): 729 is a perfect cube.
   (R): It can be written as 9 × 9 × 9.
   Answer: Both A and R are true, and R is the correct explanation.

8. (A): A number is a perfect cube if all prime factors occur in triples in its prime factorisation.
   (R): Triplet grouping of prime factors ensures equal grouping in three identical sets.
   Answer: Both A and R are true, and R is the correct explanation.

9. (A): 1729 is called the Hardy–Ramanujan number.
   (R): It is the smallest number expressible as the sum of two cubes in two different ways.
   Answer: Both A and R are true, and R is the correct explanation.

10. (A): The cube of an even number is always even.
    (R): Multiplying an even number three times still results in an even number.
    Answer: Both A and R are true, and R is the correct explanation.

11. (A): The square root of a perfect square is always a natural number.
    (R): A perfect square is the product of an integer with itself.
    Answer: Both A and R are true, and R is the correct explanation.

12. (A): A cube can end in exactly two zeros.
    (R): Each zero in the cube comes from a factor of 10³.
    Answer: Both A and R are false.

13. (A): 500 is a perfect cube.
    (R): Its prime factors cannot be arranged into triplets of identical factors.
    Answer: A is false, R is true.

14. (A): Every perfect square is also a perfect cube.
    (R): Both are powers of integers.
    Answer: A is false, R is true.

15. (A): The positive square root of 64 is 8.
    (R): 8 × 8 = 64.
    Answer: Both A and R are true, and R is the correct explanation.