Key Concept: Cubes, Cube Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that 1729 can be expressed as the sum of two cubes in two different ways. This is true because:

$1729 = 12^3 + 1^3 = 1728 + 1$

and

$1729 = 10^3 + 9^3 = 1000 + 729$

The reason claims that Ramanujan discovered that 1729 is the smallest number with this property. This is also true, and it directly explains why 1729 is significant in mathematics. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Sum of Cubes

a) $2^3 + 16^3$

[Solution Description]

4104 can be expressed as the sum of two cubes in two different ways: $2^3 + 16^3$ and $9^3 + 15^3$. Therefore, the correct expression is $2^3 + 16^3$.

Key Concept: Properties of Numbers

b) It is the smallest number expressible as the sum of two cubes in two different ways.

[Solution Description]

The number 1729 is unique because it is the smallest number that can be expressed as a sum of two cubes in two different ways: $12^3 + 1^3$ and $10^3 + 9^3$. This property makes it known as the Hardy-Ramanujan Number.

Key Concept: Hardy-Ramanujan Number, Sum of Cubes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To verify whether the assertion and reason are correct, we need to check if 4104 can indeed be expressed as the sum of two cubes in two different ways. According to the syllabus, 4104 is a Hardy-Ramanujan Number and can be expressed as $2^3 + 16^3$ and $9^3 + 15^3$. Let's verify this:

First, calculate $2^3 + 16^3$:

$2^3 = 8$

$16^3 = 4096$

$8 + 4096 = 4104$

Next, calculate $9^3 + 15^3$:

$9^3 = 729$

$15^3 = 3375$

$729 + 3375 = 4104$

Both calculations confirm that 4104 can be expressed as the sum of two cubes in two different ways. Therefore, the Assertion is true, and the Reason is also true. Moreover, the Reason correctly explains why 4104 is a Hardy-Ramanujan Number.

Key Concept: Hardy-Ramanujan Number, Sum of Cubes

b) Two ways

[Solution Description]

4104 is a Hardy-Ramanujan Number, which means it can be expressed as the sum of two cubes in two different ways. The two pairs of cubes are:

$4104 = 16^3 + 2^3 = 4096 + 8$

and

$4104 = 15^3 + 9^3 = 3375 + 729.$

Therefore, 4104 can be expressed as the sum of two cubes in two ways.

Key Concept: Hardy-Ramanujan Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that 1729 is the smallest Hardy-Ramanujan Number, which is true. The reason given is that it can be expressed as the sum of two cubes in two different ways, which is also true. Moreover, the reason correctly explains why 1729 is considered a Hardy-Ramanujan Number because this property defines such numbers. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Cubes

c) 27

[Solution Description]

To find the number of small cubes, calculate the volume of the larger cube and divide it by the volume of one small cube. The volume of the larger cube is $3^3 = 27$ cubic cm. The volume of one small cube is $1^3 = 1$ cubic cm. Therefore, the number of small cubes needed is $27 / 1 = 27$.

Key Concept: Concept of Cubes in Geometry

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The volume of a cube is given by $V = s^3$, where $s$ is the side length. For a cube with side length 2 cm, the volume is $V = 2^3 = 8$ cubic cm. Similarly, the volume of a smaller cube with side length 1 cm is $V = 1^3 = 1$ cubic cm. Therefore, 8 smaller cubes are needed to make a larger cube of side 2 cm. Both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Cubes, Perfect Cubes

b) 64

[Solution Description]

A perfect cube is a number that can be expressed as $n^3$ where $n$ is an integer. Checking the options:

- 64: $4^3 = 64$, so it is a perfect cube.

- 50: There is no integer $n$ such that $n^3 = 50$.

- 125: $5^3 = 125$, so it is a perfect cube.

- 28: There is no integer $n$ such that $n^3 = 28$.

Therefore, both 64 and 125 are perfect cubes. However, since we need to choose one option, we select 64.

Key Concept: Cubes and Cube Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the number 1729 can be expressed as the sum of two cubes in two different ways. This is true because:

$1729 = 12^3 + 1^3 = 10^3 + 9^3$

The reason states that 1729 is the smallest number that can be expressed as the sum of two cubes in two different ways. This is also true, as it is known as the Hardy-Ramanujan Number due to this unique property.

Since both the assertion and reason are true, and the reason explains why 1729 is significant, the correct option is (a).

Key Concept: Perfect Cubes

a) 8

[Solution Description]

A perfect cube is a number that can be expressed as the cube of an integer. Let's check each option:

- 8: $2^3 = 8$ (Perfect cube)

- 14: Not a perfect cube

- 16: Not a perfect cube

- 20: Not a perfect cube

Therefore, 8 is a perfect cube.

Key Concept: Cube Numbers, Perfect Cubes

b) 1728

[Solution Description]

The product of two perfect cubes is given by multiplying their values. Here, $27 = 3^3$ and $64 = 4^3$. The product is:

$27 \times 64 = 3^3 \times 4^3 = (3 \times 4)^3 = 12^3 = 1728$

Thus, the product is 1728.

Key Concept: Cubes and Their Prime Factors

c) $2^3 \times 3^3$

[Solution Description]

First, find the cube of 6: $6^3 = 216$. Now, find the prime factorisation of 216.

$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3$

Thus, the prime factorisation of the cube of 6 is $2^3 \times 3^3$.

Key Concept: Smallest Multiple that is a Perfect Cube

d) 180

[Solution Description]

For a number to be a perfect cube, each prime in its factorisation must appear three times. The given factorisation is $2 \times 3 \times 5^2$. To make it a perfect cube:

We need two more 2's: $2^3$.

We need two more 3's: $3^3$.

We need one more 5: $5^3$.

The smallest number to multiply by is $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$.

Key Concept: Patterns in Cubes

d) 6

[Solution Description]

Observing the pattern, the number of consecutive odd numbers required to sum up to $n^3$ is equal to $n$. Therefore, for $6^3$, we need 6 consecutive odd numbers. This follows from the pattern where $1^3$ uses 1 odd number, $2^3$ uses 2 odd numbers, and so on.

Key Concept: Volume of Cube

c) 27

[Solution Description]

The volume of a cube with side length $s$ is calculated using the formula:

$\text{Volume} = s^3$

First, calculate the volume of the smaller cube with side 1 cm:

$\text{Volume of small cube} = 1^3 = 1 \text{ cm}^3$

Next, calculate the volume of the larger cube with side 3 cm:

$\text{Volume of large cube} = 3^3 = 27 \text{ cm}^3$

To find out how many small cubes fit into the larger cube, divide the volume of the larger cube by the volume of the smaller cube:

$\frac{27 \text{ cm}^3}{1 \text{ cm}^3} = 27$

Therefore, 27 cubes of side 1 cm will make a cube of side 3 cm.

Key Concept: Perfect Cubes

c) 64

[Solution Description]

To find the cube of 4, we multiply 4 by itself three times: $4^3 = 4 \times 4 \times 4$. First, calculate $4 \times 4 = 16$. Then, multiply the result by 4 again: $16 \times 4 = 64$. Therefore, the cube of 4 is 64.

Key Concept: Perfect Cubes

b) 27

[Solution Description]

A perfect cube is a number that can be expressed as $n^3$ where $n$ is an integer. Let's check each option:

- 16: $2^3 = 8$ and $3^3 = 27$, so 16 is not a perfect cube.

- 27: $3^3 = 27$, so 27 is a perfect cube.

- 35: Not a perfect cube as it falls between $3^3 = 27$ and $4^3 = 64$.

- 48: Not a perfect cube as it falls between $3^3 = 27$ and $4^3 = 64$.

Therefore, the correct answer is 27.

Key Concept: Perfect Cubes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine whether 1000 is a perfect cube, we need to check its prime factorisation. The number 1000 can be expressed as $10^3$, which implies it is a perfect cube. Let's break down the prime factors:

$1000 = 10 \times 10 \times 10 = (2 \times 5)^3 = 2^3 \times 5^3$

Here, both 2 and 5 appear three times in their respective groups, confirming that 1000 is indeed a perfect cube. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Perfect Cubes, Prime Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine whether the assertion and reason are true, we first perform the prime factorization of 27000.

Step 1: Factorize 27000 into its prime factors.

$27000 = 27 \times 1000 = 3^3 \times 10^3 = 3^3 \times (2 \times 5)^3 = 2^3 \times 3^3 \times 5^3$

Step 2: Analyze the prime factors.

The prime factors of 27000 are $2^3 \times 3^3 \times 5^3$. Each prime factor (2, 3, and 5) appears exactly three times in the factorization.

Step 3: Determine if 27000 is a perfect cube.

Since each prime factor appears three times, 27000 is indeed a perfect cube. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Cube Roots

d) 50

[Solution Description]

To find the cube root of 125000, we first express it as $125 \times 1000$. We know that $5^3 = 125$ and $10^3 = 1000$. So, the cube root of 125000 is $5 \times 10 = 50$.

Key Concept: Cubes, Cube Roots

c) 5

[Solution Description]

The smallest perfect cube greater than 100 is $5^3 = 125$, and the largest perfect cube less than 1000 is $9^3 = 729$. The cubes between 100 and 1000 are $5^3, 6^3, 7^3, 8^3,$ and $9^3$. Thus, there are 5 perfect cubes in this range.

Key Concept: Cube Numbers

b) 1728

[Solution Description]

To find the cube of 12, we multiply 12 by itself three times. This can be calculated as follows:

Step 1: Multiply 12 by 12.

$12 \times 12 = 144$

Step 2: Multiply the result by 12 again.

$144 \times 12 = 1728$

Therefore, the cube of 12 is 1728.

Key Concept: Cube Numbers

a) 64

[Solution Description]

A cube number is obtained by multiplying a number by itself three times. Let's check each option:

1. $64$ is $4^3 = 4 \times 4 \times 4 = 64$.

2. $100$ is not a cube number.

3. $99$ is not a cube number.

4. $50$ is not a cube number.

Therefore, $64$ is a cube number.

Key Concept: Perfect Cubes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To check whether 216 is a perfect cube, we need to find if there exists an integer whose cube is 216. Let's calculate the cube of 6: $6^3 = 6 × 6 × 6 = 216$. Since 216 is the cube of 6, which is an integer, it is indeed a perfect cube. Similarly, a perfect cube is defined as the cube of an integer, so the reason correctly explains why 216 is a perfect cube.

Key Concept: Finding Perfect Cubes

b) 343

[Solution Description]

To find which number is a perfect cube between 200 and 500, we need to check the cubes of integers within this range.

Let's calculate the cubes of numbers around this range:

$5^3 = 125$, $6^3 = 216$, $7^3 = 343$, $8^3 = 512$.

Among these, 343 falls within the range of 200 to 500.

Therefore, 343 is the correct answer.

Key Concept: Perfect cubes, properties of cubes

d) The cube of an even number is even, and the cube of an odd number is odd.

[Solution Description]

From the syllabus, it is observed that the cube of an even number is always even, and the cube of an odd number is always odd. For example, $2^3 = 8$ (even) and $3^3 = 27$ (odd). Therefore, the statement "The cube of an even number is even, and the cube of an odd number is odd" is correct.

Key Concept: Properties of Cube Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let us consider an even number, say 2. When we cube it, $2^3 = 8$, which is even. Similarly, for another even number, say 4, $4^3 = 64$, which is also even. This shows that the cube of an even number is always even. The reason states that the product of any even number multiplied by itself three times will always be even, which is true because multiplying an even number by itself any number of times (including three) results in an even number. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Properties of Cube Numbers

c) 1000

[Solution Description]

The smallest two-digit even number is 10. To find its cube, we calculate:

$10^3 = 10 \times 10 \times 10 = 1000$

Therefore, the cube of the smallest two-digit even number is 1000.

Key Concept: Perfect Cube Identification

a) 216

[Solution Description]

To determine if a number is a perfect cube, we check if it can be expressed as $n^3$ where $n$ is an integer. Let's evaluate each option:

1. 216: $6^3 = 216$, so 216 is a perfect cube.

2. 100: There is no integer $n$ such that $n^3 = 100$, so 100 is not a perfect cube.

3. 144: There is no integer $n$ such that $n^3 = 144$, so 144 is not a perfect cube.

4. 225: There is no integer $n$ such that $n^3 = 225$, so 225 is not a perfect cube.

Therefore, the correct answer is 216.

Key Concept: Cube of an even number

b) 2744

[Solution Description]

To find the cube of 14, we calculate $14^3$.

Step 1: Multiply 14 by itself: $14 \times 14 = 196$.

Step 2: Multiply the result by 14 again: $196 \times 14 = 2744$.

Therefore, the cube of 14 is 2744.

Key Concept: Cube of an odd number is odd

b) The cube of $m$ is odd

[Solution Description]

Let $m$ be an odd number. By definition, an odd number can be expressed as $m = 2k + 1$, where $k$ is an integer. The cube of $m$ is given by:

$m^3 = (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1$

The expression $8k^3 + 12k^2 + 6k$ is even since each term is divisible by 2. Adding 1 to an even number results in an odd number. Therefore, the cube of an odd number is always odd.

Key Concept: Cubes, Even Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let’s analyze the assertion and reason step by step:

1. **Assertion Analysis**: To check if the cube of an even number is even, let’s take an even number represented as $2k$, where $k$ is an integer. The cube of this number is $(2k)^3 = 8k^3$. Since $8k^3$ is divisible by 2, it is even. Thus, the assertion is true.

2. **Reason Analysis**: The reason states that an even number can be expressed as $2k$, where $k$ is an integer. This is a fundamental property of even numbers and is correct.

3. **Explanation Link**: The reason correctly explains why the cube of an even number is even because expressing the even number as $2k$ leads to the cube being divisible by 2.

Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Cube of an odd number is odd

a) 125

[Solution Description]

To determine which number is the cube of an odd number, let's check each option:

- 125: $\sqrt[3]{125} = 5$, which is odd.

- 216: $\sqrt[3]{216} = 6$, which is even.

- 343: $\sqrt[3]{343} = 7$, which is odd.

- 512: $\sqrt[3]{512} = 8$, which is even.

Both 125 and 343 are cubes of odd numbers, but only 125 appears in the options.

Key Concept: Cube of an odd number

b) It is odd

[Solution Description]

The cube of an odd number is always odd. An odd number multiplied by an odd number results in another odd number, and multiplying that result again by an odd number will still give an odd number.

Key Concept: Cube of an odd number

b) 1

[Solution Description]

Let $m$ be an odd integer. We know that any odd integer can be written as $m = 2k + 1$, where $k$ is an integer. Now, let's compute $m^3$:

$m^3 = (2k + 1)^3$

Expanding the cube:

$m^3 = 8k^3 + 12k^2 + 6k + 1$

Now, let's find the remainder when $m^3$ is divided by 4. Notice that $8k^3$ and $12k^2$ are both divisible by 4, so they leave no remainder. The term $6k$ is divisible by 2 but not necessarily by 4, depending on the value of $k$. However, since we are interested in the remainder when divided by 4, we can write $6k$ as $4k + 2k$. Here, $4k$ is divisible by 4, leaving no remainder, and $2k$ leaves a remainder of 2 if $k$ is odd or 0 if $k$ is even. Finally, adding 1 gives us:

$\text{Remainder} = 0 + 0 + (0 \text{ or } 2) + 1 = 1 \text{ or } 3$

But since $m$ is odd, $m^3$ is also odd, so the remainder must be odd. Therefore, the possible remainders are 1 or 3. In this case, since the question asks for a specific remainder, we consider the general form and conclude that the remainder is 1.

Key Concept: Pattern in the Last Digits of Cubes

c) 2

[Solution Description]

To find the one’s digit of the cube of 8888, observe that the number ends with an 8. According to the pattern, the cube of any number ending with 8 will end with a 2 since $8^3 = 512$.

Key Concept: Pattern in the Last Digits of Cubes

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

Let us consider the assertion and reason step by step.

**Assertion**: The cube of any number ending with 3 will always end with 7. To verify this, let's take an example: $3^3 = 27$, which ends with 7. Similarly, $13^3 = 2197$, which also ends with 7. This pattern holds true for all numbers ending with 3. Therefore, the assertion is true.

**Reason**: The one’s digit of a cube depends only on the one’s digit of the original number and not on the other digits. This is a fundamental property of cubes. For example, the cube of 4 ($4^3 = 64$) and the cube of 14 ($14^3 = 2744$) both end with 4 because they share the same one’s digit. Hence, the reason is also true.

However, the reason does not explain why the cube of a number ending with 3 specifically ends with 7. It only explains how the one’s digit of the cube is determined based on the one’s digit of the original number. Therefore, the reason is correct but not the explanation for the assertion.

Key Concept: Pattern in the Last Digits of Cubes

a) 1

[Solution Description]

To find the one’s digit of the cube of 3331, observe that the one’s digit of a number determines the one’s digit of its cube. The number 3331 ends with 1. From the pattern, we know that the cube of any number ending with 1 will also end with 1. Therefore, the one’s digit of $3331^3$ is 1.

Key Concept: Observing the units digit of cube numbers

b) 4

[Solution Description]

To determine the unit’s digit of the cube of a number ending with 4, note that $4^3 = 64$. Since 64 ends with 4, the unit’s digit of the cube of such a number will be 4.

Key Concept: Observing the units digit of cube numbers

d) 9

[Solution Description]

To determine the units digit of the cube of a number ending with 9, consider the cube of 9. $9^3 = 729$. The units digit of 729 is 9. Therefore, the units digit of the cube of any number ending with 9 will be 9.

Key Concept: Observing the units digit of cube numbers

b) 3

[Solution Description]

To find the unit’s digit of the cube of 777, observe that the unit’s digit of 777 is 7. Now, we know that the cube of a number ending with 7 will have a unit’s digit of 3 because $7^3 = 343$, which ends with 3.

Key Concept: Cubes ending in 1

b) 1

[Solution Description]

To find the one's digit of the cube of a number ending with 1, we observe that when any number ending with 1 is cubed, the result always ends with 1. For example, $11^3 = 1331$, which ends with 1. Therefore, the one's digit of the cube of such a number is always 1.

Key Concept: Cubes and their one’s digit

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let us consider a number ending with 3, say 3 itself. The cube of 3 is $3^3 = 27$, which ends with 7. Now, let's take another number ending with 3, such as 13. The cube of 13 is $13^3 = 2197$, which also ends with 7. Similarly, for 23, the cube is $23^3 = 12167$, which again ends with 7. This shows that the cube of any number ending with 3 will always end with 7. The reason given is true because the one’s digit of the cube of a number depends solely on the one’s digit of the original number. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Cube numbers ending in 3

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let us consider a general number ending with 3, say $n = 10k + 3$, where $k$ is any integer. Now, let's find the cube of $n$:

$n^3 = (10k + 3)^3 = 1000k^3 + 900k^2 + 270k + 27$

Observing the last term, 27, we can see that the unit's digit of $n^3$ is determined solely by this term. Since 27 ends with 7, the cube of any number ending with 3 will also end with 7. Therefore, the Assertion is true. The Reason correctly explains that cubing the digit 3 results in 27, which ensures the unit's digit of the cube is 7. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Adding Consecutive Odd Numbers, Cubes

b) 109

[Solution Description] First, determine the number of consecutive odd numbers whose sum is $1000$. Since $1000 = 10^3$, there are $10$ consecutive odd numbers. The first term can be calculated using the formula for the nth odd number: $a_n = 2n - 1$. For $10$ consecutive odd numbers, the first term is $2(1) - 1 = 1$, but since this is not the case here, we use the general pattern where the starting odd number for $k$ terms is $k(k-1)+1$. For $k=10$, the starting odd number is $91$. The last odd number is $91 + 2(10-1) = 109$.

Key Concept: Adding Consecutive Odd Numbers

b) 125

[Solution Description]

The first 5 consecutive odd numbers starting from 21 are 21, 23, 25, 27, and 29. Summing them up:

$21 + 23 + 25 + 27 + 29 = 125$

According to the pattern, $5^3 = 125$. Hence, the sum is 125.

Key Concept: Adding Consecutive Odd Numbers, Cubes

b) 13

[Solution Description] The sum of four consecutive odd numbers equals $4^3 = 64$. To find the starting odd number, let the first term be $x$. Then, the next three odd numbers are $x+2$, $x+4$, and $x+6$. Adding them up: $x + (x+2) + (x+4) + (x+6) = 4x + 12 = 64$. Solving for $x$, we get $4x = 52$ and $x = 13$. So, the starting odd number is $13$.

Key Concept: Sum of Odd Numbers

b) 216

[Solution Description]

According to the given pattern, the sum of the first $n$ consecutive odd numbers is equal to $n^3$. Here, $n = 6$. Thus, the sum is $6^3 = 216$.

Key Concept: Relationship between cube numbers and sum of odd numbers

b) 43

[Solution Description]

To find the first odd number in the sequence summing to $7^3 = 343$, observe that the sequence starts at $n^2 - n + 1$. For $n = 7$, this gives:

$7^2 - 7 + 1 = 49 - 7 + 1 = 43$

Thus, the first odd number is 43.

Key Concept: Relationship between cube numbers and sum of odd numbers

d) 9

[Solution Description]

The pattern shows that for any integer $n$, the sum of $n$ consecutive odd numbers equals $n^3$. Therefore, to express $9^3$ as a sum of consecutive odd numbers, we need $9$ such numbers.

Key Concept: Prime Factorisation and Perfect Cubes

a) 27

[Solution Description]

For a number to become a perfect cube when multiplied by 8, it must satisfy the condition that each prime factor in its prime factorisation, when combined with those of 8, appears three times. The prime factorisation of 8 is $2^3$. Let's evaluate each option:

Option 1: 27 Prime factorisation: $27 = 3^3$. When multiplied by 8 ($2^3$), the result is $2^3 \times 3^3$, which is a perfect cube.

Option 2: 50 Prime factorisation: $50 = 2 \times 5^2$. When multiplied by 8 ($2^3$), the result is $2^4 \times 5^2$, which is not a perfect cube.

Option 3: 100 Prime factorisation: $100 = 2^2 \times 5^2$. When multiplied by 8 ($2^3$), the result is $2^5 \times 5^2$, which is not a perfect cube.

Option 4: 125 Prime factorisation: $125 = 5^3$. When multiplied by 8 ($2^3$), the result is $2^3 \times 5^3$, which is a perfect cube.

Both 27 and 125 satisfy the condition. However, since only one option needs to be selected, we choose 27 as it is the first correct option.

Key Concept: Cubes and Their Prime Factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine whether 1728 is a perfect cube, we need to analyze its prime factorisation. Let's break down 1728 into its prime factors:

$1728 = 12^3 = (2 \times 2 \times 3)^3 = 2^3 \times 2^3 \times 3^3$

Here, each prime factor (2 and 3) appears exactly three times in the prime factorisation of 1728. This satisfies the condition for a number to be a perfect cube, which states that each prime factor must appear three times in its prime factorisation. Therefore, both the Assertion and the Reason are true, and the Reason correctly explains why the Assertion is true.

Key Concept: Cubes and Their Prime Factors

b) 216000

[Solution Description]

Given the prime factorisation $2^2 \times 3^1 \times 5^1$, we find the cube by raising each prime factor to the power of 3.

Step 1: Cube each prime factor.

$(2^2)^3 = 2^{6}, \quad (3^1)^3 = 3^{3}, \quad (5^1)^3 = 5^{3}$

Step 2: Multiply them together.

$2^{6} \times 3^{3} \times 5^{3} = 64 \times 27 \times 125 = 216000$

Therefore, the cube of the number is 216000.

Key Concept: Prime factorization of cubes, Perfect cubes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine whether 27000 is a perfect cube, we need to perform its prime factorization. Let's break down 27000 into its prime factors:

$27000 = 27 \times 1000 = 3^3 \times 10^3 = 3^3 \times (2 \times 5)^3 = 2^3 \times 3^3 \times 5^3$

From the above factorization, we can see that each prime factor (2, 3, and 5) appears exactly three times. According to the definition of a perfect cube, if each prime factor in the prime factorization of a number appears three times, then the number is a perfect cube. Therefore, 27000 is indeed a perfect cube. Thus, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Smallest Multiple to Make a Perfect Cube

c) 7

[Solution Description]

First, find the prime factorization of 392:

$392 = 2 \times 2 \times 2 \times 7 \times 7 = 2^3 \times 7^2$

For 392 to be a perfect cube, each prime factor must appear in multiples of three. Here, 7 appears twice, so we need one more 7. Hence, the smallest number to multiply is 7.

Key Concept: Smallest Multiple That Is a Perfect Cube

c) 7

[Solution Description]

The prime factorization of 392 is: $392 = 2 \times 2 \times 2 \times 7 \times 7$.

The prime factor 7 does not appear in a group of three. Therefore, 392 is not a perfect cube. To make it a cube, we need one more 7.

Hence, the smallest natural number by which 392 should be multiplied to make a perfect cube is 7.

Key Concept: Perfect Cube, Prime Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine whether 27000 is a perfect cube, we first perform its prime factorisation.

$27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5$

Here, each prime factor (2, 3, and 5) appears exactly three times. According to the rule, if each prime factor in the prime factorisation of a number appears three times, then the number is a perfect cube. Therefore, 27000 is indeed a perfect cube. This confirms that both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Prime Factorisation and Perfect Cubes

c) 27

[Solution Description]

Check each number’s prime factorisation:

64: $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 = (2^2)^3$

100: $100 = 2 \times 2 \times 5 \times 5$, not all factors appear three times.

27: $27 = 3 \times 3 \times 3 = 3^3$

18: $18 = 2 \times 3 \times 3$, not all factors appear three times.

Thus, both 64 and 27 are perfect cubes. However, since only one option can be correct, we choose 27 as it directly fits the criteria without further manipulation.

Key Concept: Perfect Cube

d) 2028

[Solution Description]

To determine which number is not a perfect cube, we perform prime factorisation of each option.

Option 1: 1331

Step 1: Divide 1331 by 11 repeatedly:

$1331 \div 11 = 121$, $121 \div 11 = 11$, $11 \div 11 = 1$.

So, $1331 = 11^3$, which is a perfect cube.

Option 2: 2744

Step 1: Divide 2744 by 2 repeatedly:

$2744 \div 2 = 1372$, $1372 \div 2 = 686$, $686 \div 2 = 343$.

Step 2: Now, divide 343 by 7 repeatedly:

$343 \div 7 = 49$, $49 \div 7 = 7$, $7 \div 7 = 1$.

So, $2744 = 2^3 × 7^3$, which is a perfect cube.

Option 3: 2197

Step 1: Divide 2197 by 13 repeatedly:

$2197 \div 13 = 169$, $169 \div 13 = 13$, $13 \div 13 = 1$.

So, $2197 = 13^3$, which is a perfect cube.

Option 4: 2028

Step 1: Divide 2028 by 2 repeatedly:

$2028 \div 2 = 1014$, $1014 \div 2 = 507$.

Step 2: Now, divide 507 by 3:

$507 \div 3 = 169$.

Step 3: Finally, divide 169 by 13:

$169 \div 13 = 13$, $13 \div 13 = 1$.

So, $2028 = 2^2 × 3 × 13^2$. Since the exponents of 2 and 3 are not multiples of three, 2028 is not a perfect cube.

Key Concept: Sum of Consecutive Odd Numbers

c) 8

[Solution Description]

To find how many consecutive odd numbers are needed to obtain the sum as $8^3$, we observe the pattern:

The sum of $n$ consecutive odd numbers starting from 1 is equal to $n^3$. Thus, to get $8^3$, we need 8 consecutive odd numbers.

Therefore, the correct answer is 8.

Key Concept: Difference of cubes

b) 127

[Solution Description]

Using the given pattern:

$n^3 - (n-1)^3 = 1 + n \times (n-1) \times 3$

For $n = 7$,

$7^3 - 6^3 = 1 + 7 \times 6 \times 3$

Calculate step by step:

$7 \times 6 = 42$

$42 \times 3 = 126$

Add 1: $126 + 1 = 127$

Therefore, $7^3 - 6^3 = 127$.

Key Concept: Adding consecutive odd numbers

c) 6

[Solution Description]

To find how many consecutive odd numbers are needed to get the sum as $6^3$, we observe the pattern:

$1 = 1^3$ (1 odd number)

$3 + 5 = 8 = 2^3$ (2 odd numbers)

$7 + 9 + 11 = 27 = 3^3$ (3 odd numbers)

Thus, for $n^3$, we need $n$ consecutive odd numbers. Therefore, for $6^3$, we need 6 consecutive odd numbers.

Key Concept: Adding consecutive odd numbers, Sum of cubes

d) 13

[Solution Description]

To find the number of consecutive odd numbers needed to achieve the sum equal to $7^3$, we use the pattern observed in the syllabus. The sequence starts from $15$. Let the number of terms be $n$. The last term in the sequence can be expressed as $15 + 2(n - 1)$. The sum of the sequence is given by:

$n \times \frac{15 + 15 + 2(n - 1)}{2} = 7^3$

Simplifying the equation:

$n \times (15 + n - 1) = 343$

$n(n + 14) = 343$

Solving for $n$:

$n^2 + 14n - 343 = 0$

Using the quadratic formula:

$n = \frac{-14 \pm \sqrt{196 + 1372}}{2} = \frac{-14 \pm \sqrt{1568}}{2}$

$n = \frac{-14 \pm 39.6}{2}$

Taking the positive value:

$n \approx \frac{25.6}{2} \approx 12.8$

Since the number of terms must be an integer, we round up to $13$.

Key Concept: Sum of Consecutive Odd Numbers

b) 125

[Solution Description]

The first 5 consecutive odd numbers starting from 21 are 21, 23, 25, 27, and 29. Adding them: $21 + 23 + 25 + 27 + 29 = 125$.

Key Concept: Sum of Consecutive Odd Numbers

d) Assertion is false, but Reason is true.

[Solution Description]

Let's verify the assertion and reason step by step. According to the pattern, the sum of the first n consecutive odd numbers is equal to $n^3$. For example,

- Sum of the first 1 odd number: $1 = 1^3$

- Sum of the first 2 odd numbers: $3 + 5 = 8 = 2^3$

- Sum of the first 3 odd numbers: $7 + 9 + 11 = 27 = 3^3$

This pattern continues for any positive integer n. Therefore, the sum of the first 6 consecutive odd numbers should be $6^3 = 216$. Now let's calculate the sum of the first 6 odd numbers:

1. The sequence of the first 6 odd numbers is: 1, 3, 5, 7, 9, 11.

2. Sum = $1 + 3 + 5 + 7 + 9 + 11 = 36$.

Here, we observe that the sum of the first 6 odd numbers is 36, which is not equal to $6^3 = 216$. Hence, the assertion is false. However, the reason is true because the sum of the first n consecutive odd numbers is indeed $n^3$ as per the given pattern. Thus, the correct option is d).

Key Concept: Difference Between Consecutive Cube Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The formula for the difference between the cubes of two consecutive numbers, $n^3 - (n-1)^3$, can be simplified to $3n^2 - 3n + 1$. This matches the given pattern: $1 + n \times (n-1) \times 3$. Therefore, the assertion is true. The reason explains how the formula is derived from the observed pattern, making it the correct explanation of the assertion.

Key Concept: Difference Between Consecutive Cube Numbers

c) 127

[Solution Description] Substitute $n = 7$ into the formula: $7^3 - 6^3 = 1 + 7 \times 6 \times 3$. Calculate step by step: $7 \times 6 = 42$, then $42 \times 3 = 126$. Finally, add 1 to get $127$.

Key Concept: Difference Between Consecutive Cube Numbers

c) 397

[Solution Description]

First, identify the values of $n$ and $(n-1)$. Here, $n = 12$ and $(n-1) = 11$. Plug these values into the given formula:

$12^3 - 11^3 = 1 + 12 \times 11 \times 3$

Next, calculate the multiplication part:

$12 \times 11 = 132$

Then multiply by 3:

$132 \times 3 = 396$

Finally, add 1 to get the result:

$1 + 396 = 397$

Therefore, $12^3 - 11^3 = 397$.

Key Concept: Cube Numbers, Prime Factorization

a) 5

[Solution Description]

The volume of one cuboid is $12 \times 18 \times 25$. First, find the prime factorization of each dimension: $12 = 2^2 \times 3$, $18 = 2 \times 3^2$, and $25 = 5^2$. The combined prime factorization is $2^3 \times 3^3 \times 5^2$. To make this a perfect cube, we need an additional $5$. Therefore, we need $5$ cuboids to form a perfect cube.

Key Concept: Cubes and their Prime Factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine if 1728 is a perfect cube, we need to check its prime factorisation. Let's break down 1728 into its prime factors:

$1728 = 12^3 = (2^2 \times 3)^3 = 2^{6} \times 3^{3}$

Here, the prime factors are 2 and 3. Both 2 and 3 appear in multiples of 3 (i.e., $2^6$ and $3^3$). Since each prime factor in the prime factorisation of 1728 appears three times, 1728 is indeed a perfect cube. Therefore, both the Assertion and Reason are true, and the Reason correctly explains why 1728 is a perfect cube.

Key Concept: Cube Numbers in Prime Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that if every prime factor in the prime factorization of a number appears exactly three times, then the number is a perfect cube. This is true because a perfect cube is defined as a number that can be expressed as the cube of an integer. When a number is cubed, each prime factor in its prime factorization is raised to the power of three. For example, consider the number 8. Its prime factorization is $2 \times 2 \times 2 = 2^3$, which is a perfect cube since it can be written as $2^3$. The reason given is also correct, stating that a number is a perfect cube if and only if each prime factor in its prime factorization is raised to a power that is a multiple of three. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Perfect Cube, Prime Factorization, Smallest Multiple

d) No, multiply by 98

[Solution Description]

First, perform the prime factorization of 9604:

$9604 = 2 \times 2 \times 7 \times 7 \times 7 \times 7$

This can be written as:

$9604 = 2^2 \times 7^4$

The prime factor 2 appears twice, and 7 appears four times. To make them appear in groups of three, we need to multiply by $2 \times 7^2 = 98$.

Key Concept: Perfect Cube

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To check if 512 is a perfect cube, we perform its prime factorisation. The prime factorisation of 512 is $2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2$ or $2^9$. Here, the prime factor 2 appears nine times, which is a multiple of three. Therefore, 512 is a perfect cube. The assertion is true. The reason states that in the prime factorisation of 512, each prime factor appears three times. This is also true because 2 appears nine times, which is a multiple of three. Moreover, the reason correctly explains why 512 is a perfect cube. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Checking whether a number is a perfect cube using prime factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To check if 1728 is a perfect cube, we need to perform its prime factorisation and verify if each prime factor appears three times. Let's start by factorising 1728:

$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

Grouping the factors into triplets:

$1728 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$

Each prime factor (2 and 3) appears exactly three times in the factorisation. Therefore, 1728 is indeed a perfect cube. Specifically, it can be expressed as $12^3$ because $(2 \times 2 \times 3)^3 = 12^3$.

Thus, both the Assertion and Reason are true, and the Reason correctly explains why 1728 is a perfect cube.

Key Concept: Smallest Multiple That is a Perfect Cube

c) Assertion is true, but Reason is false.

[Solution Description]

First, let us find the prime factorisation of 108:

$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$

For a number to be a perfect cube, all the exponents in its prime factorisation must be multiples of 3. Here, the exponent of 2 is 2, which is not a multiple of 3. Therefore, we need one more 2 to make the exponent of 2 equal to 3. Thus, the smallest natural number by which 108 must be multiplied to make it a perfect cube is 2. The assertion is true. However, the reason states that we need two more 2's, which is incorrect because only one more 2 is required. Hence, the reason is false.

Key Concept: Smallest Multiple That is a Perfect Cube

b) 30

[Solution Description]

First, we need to find the prime factorization of 7200.

$7200 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$.

To make it a perfect cube, each prime factor must appear in groups of three. Here, the prime factor 2 appears five times, the prime factor 3 appears two times, and the prime factor 5 appears two times. So, we need one more 2, one more 3, and one more 5 to make all prime factors appear in groups of three. Therefore, the smallest natural number by which 7200 must be multiplied is $2 \times 3 \times 5 = 30$.

Key Concept: Smallest Multiple That is a Perfect Cube

a) 2

[Solution Description]

Find the prime factors of 500. Prime factorisation of 500 is $500 = 2 × 2 × 5 × 5 × 5$. Here, 2 appears only twice, and 5 appears three times. To make it a perfect cube, we need one more 2, so multiply by 2. The product is $500 × 2 = 1000$, which is a perfect cube.

Key Concept: Smallest multiple that is a perfect cube

d) 25

[Solution Description]

To determine if 1080 is a perfect cube, we first perform its prime factorisation. The prime factorisation of 1080 is:

$1080 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$

We observe that the prime factor 5 appears only once, which is not a multiple of three. To make it a perfect cube, we need two more 5s. Therefore, multiplying 1080 by $5 \times 5 = 25$ gives:

$1080 \times 25 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 = 27000$

Hence, the smallest natural number required is 25.

Key Concept: Smallest multiple that is a perfect cube

a) 2

[Solution Description]

To determine if 500 is a perfect cube, we first perform its prime factorisation. The prime factorisation of 500 is:

$500 = 2 \times 2 \times 5 \times 5 \times 5$

We observe that the prime factor 2 appears only twice, which is not a multiple of three. To make it a perfect cube, we need one more 2. Therefore, multiplying 500 by 2 gives:

$500 \times 2 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 1000$

Hence, the smallest natural number required is 2.

Key Concept: Prime Factorization and Perfect Cubes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

First, let's find the prime factorization of 72. We can write $72 = 2 \times 2 \times 2 \times 3 \times 3$. Here, 2 appears three times, but 3 appears only twice. To make 72 a perfect cube, we need to have each prime factor appear three times. Therefore, we need to multiply 72 by one more 3, i.e., $72 \times 3 = 216$, which is $6^3$ and a perfect cube. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Finding the Smallest Number to Divide

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let us first find the prime factorization of 128.

$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$

For a number to be a perfect cube, each prime factor in its prime factorization must appear in multiples of three. Here, the prime number 2 appears seven times. To make it a perfect cube, we need to divide by 2 so that the count of 2 becomes six, which is a multiple of three.

$128 \div 2 = 64 = 4^3$

Thus, the smallest number by which 128 must be divided to obtain a perfect cube is indeed 2. Both the Assertion and Reason are true, and the Reason correctly explains why dividing by 2 makes 128 a perfect cube.

Key Concept: Finding the Smallest Number to Divide

b) 4

[Solution Description]

To determine if 256 is a perfect cube, we first factorize it into its prime factors:

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

For a number to be a perfect cube, all prime factors must appear in groups of three. In this case, $2^8$ has two extra $2$'s (since $8 \div 3 = 2$ with a remainder of 2).

Therefore, to make 256 a perfect cube, we need to divide it by $2^2 = 4$.

The quotient will be $256 \div 4 = 64$, which is a perfect cube ($4^3$).

Hence, the smallest natural number by which 256 should be divided to make it a perfect cube is 4.

Key Concept: Finding the Smallest Number to Divide

b) 3

[Solution Description]

First, we factorize 192 into its prime factors:

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

For a number to be a perfect cube, all prime factors must appear in groups of three. Here, the prime factor 3 appears only once, and it needs to be removed to form a perfect cube.

Additionally, $2^6$ can be expressed as $(2^2)^3$, but we still have an extra 3.

Therefore, dividing 192 by 3 will give us $192 \div 3 = 64$, which is a perfect cube ($4^3$).

Hence, the smallest natural number by which 192 should be divided to make it a perfect cube is 3.

Key Concept: Cube Root Calculation

b) 3

[Solution Description]

To find the cube root of 27, we need to determine a number that, when multiplied by itself three times, gives 27. We know that $3 \times 3 \times 3 = 27$. Therefore, $\sqrt[3]{27} = 3$.

Key Concept: Cube Root Calculation

b) 6

[Solution Description]

To find the cube root of 216, we need to determine a number that, when multiplied by itself three times, gives 216. We know that $6 \times 6 \times 6 = 216$. Therefore, $\sqrt[3]{216} = 6$.

Key Concept: Cube Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the cube root of 27000 using the prime factorisation method:

First, we factorise 27000 into its prime factors:

$27000 = 27 \times 1000 = 3^3 \times 10^3 = 3^3 \times (2 \times 5)^3 = 2^3 \times 3^3 \times 5^3$

Now, the cube root of 27000 is given by:

$\sqrt[3]{27000} = \sqrt[3]{2^3 \times 3^3 \times 5^3} = 2 \times 3 \times 5 = 30$

Therefore, the cube root of 27000 is indeed 30.

Key Concept: Cube Root Interpretation

c) $\sqrt[3]{27} = 3$

[Solution Description]

The cube root of a number is the value that, when multiplied by itself three times, gives the original number. From the syllabus, we know that $3^3 = 27$, so $\sqrt[3]{27} = 3$ is a true statement.

Key Concept: Cube Roots, Real-World Application

b) 0.9 m

[Solution Description]

First, convert the volume from liters to cubic meters since $1 \text{ liter} = 0.001 \text{ m}^3$. Thus, $729 \text{ liters} = 0.729 \text{ m}^3$. The volume of a cube is given by $V = s^3$, where $s$ is the length of the side. So:

$s^3 = 0.729$

Taking the cube root of both sides:

$s = \sqrt[3]{0.729}$

We know that $0.9^3 = 0.729$, so:

$s = 0.9 \text{ m}$

Therefore, the length of each edge of the container is $0.9 \text{ m}$.

Key Concept: Cube Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To solve this, we need to verify both the assertion and the reason. The assertion states that the cube root of 27 is 3. This can be checked by calculating $3^3$. If $3^3 = 27$, then the assertion is true. Similarly, the reason states that $3^3 = 27$, which is a mathematical fact. Since the assertion and reason are both true, and the reason correctly explains the assertion, the correct answer is option a).

Key Concept: Cube Root Application

c) 7 cm

[Solution Description]

The volume of a cube is given by the formula $V = s^3$, where $s$ is the length of the side. To find the side length, take the cube root of the volume: $\sqrt[3]{343}$. Since $7^3 = 343$, the side length is 7 cm.

Key Concept: Cube Root Definition

c) 3

[Solution Description]

The cube root of a number is a value that, when multiplied by itself three times, gives the original number. To find $\sqrt[3]{27}$, we look for a number that, when cubed, equals 27. Since $3^3 = 27$, the cube root of 27 is 3.

Key Concept: Cube Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the cube root of 125, we need to determine a number whose cube is 125. We know that $5^3 = 125$. Therefore, the cube root of 125 is indeed 5. The reason provided states that the cube of 5 is 125, which is also true. Moreover, this reason correctly explains why the assertion is true. Hence, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Prime Factorization, Cube Roots

c) 56

[Solution Description]

To find the cube root of 175616 using prime factorization:

First, factorize 175616 into its prime factors.

175616 = 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 7 $\times$ 7 $\times$ 7 $\times$ 7 $\times$ 7 $\times$ 7 = $2^6 \times 7^6$.

Group the prime factors in triples: $(2^3 \times 2^3 \times 7^3 \times 7^3)$.

The cube root is $2 \times 2 \times 7 \times 7 = 56$.

Therefore, the cube root of 175616 is 56.

Key Concept: Cube Roots Using Prime Factorization

c) 4

[Solution Description]

First, find the prime factorization of 64.

$64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^3 × 2^3$

Therefore, the cube root of 64 is $2 × 2 = 4$.

Key Concept: Cube Roots, Prime Factorization

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To find the cube root of 5832 using prime factorization, first, we factorize 5832 into its prime factors.

$5832 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^3 \times 3^6$

Since each prime factor appears in multiples of three, we can take one factor from each group to find the cube root.

$\sqrt[3]{5832} = 2 \times 3 \times 3 = 18$

The Assertion states that the cube root of 5832 is 18, which is correct. However, the Reason states that the prime factorization of 5832 is $2^3 \times 3^6$, which is also correct. The prime factorization does support the calculation of the cube root, but it is not the direct explanation for why the cube root is 18.

Therefore, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Cube Root by Prime Factorisation

b) 30

[Solution Description]

To find the cube root of 27000, we use the prime factorisation method. First, factorise 27000 into its prime factors:

$27000 = 27 \times 1000 = 3^3 \times 10^3$

Now, express 10 as its prime factors:

$10 = 2 \times 5$

So,

$27000 = 3^3 \times (2 \times 5)^3 = 3^3 \times 2^3 \times 5^3$

Therefore, the cube root of 27000 is:

$\sqrt[3]{27000} = 3 \times 2 \times 5 = 30$

Key Concept: Cube Roots, Real-World Application

c) 30

[Solution Description]

The volume of a cube is given by the formula $V = s^3$, where $s$ is the length of one side. To find the side length, we need to take the cube root of the volume.

Given that the volume is 27000 cubic meters, we calculate $\sqrt[3]{27000}$.

Using the prime factorisation method, we find that $27000 = 3 \times 3 \times 3 \times 10 \times 10 \times 10 = 3^3 \times 10^3$.

Therefore, $\sqrt[3]{27000} = 3 \times 10 = 30$.

So, the length of one side of the tank is 30 meters.

Key Concept: Cube Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the cube root of 216, we need to identify a number which when multiplied three times gives 216. Here, $6 \times 6 \times 6 = 216$. Therefore, the cube root of 216 is 6. The reason provided is correct as it correctly explains that $6^3$ equals 216.

Key Concept: Cube Roots, Prime Factorisation

b) 42

[Solution Description]

To find the cube root of 74088, we use the prime factorisation method. First, we express 74088 as a product of its prime factors:

$74088 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7$.

This can be written as $74088 = 2^3 \times 3^3 \times 7^3$.

Taking the cube root of both sides, we get $\sqrt[3]{74088} = 2 \times 3 \times 7 = 42$.

Therefore, the cube root of 74088 is 42.