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 $1729 = 12^3 + 1^3$ and $1729 = 10^3 + 9^3$. The reason states that Ramanujan identified 1729 as the smallest number that can be expressed as the sum of two cubes in two different ways. This is also true, as it is historically documented that Ramanujan pointed out this property of 1729 to Hardy. Moreover, the reason correctly explains why the assertion is true, as Ramanujan's identification of this property directly supports the assertion.

Key Concept: Cube Roots

c) 6

[Solution Description]

To find the cube root of 216, we need to determine which number multiplied by itself three times equals 216.

Let's check:

$6^3 = 6 \times 6 \times 6 = 216$

Therefore, the cube root of 216 is 6.

Key Concept: Hardy-Ramanujan Number

c) 1729

[Solution Description]

The Hardy-Ramanujan Number 1729 is the smallest number that can be expressed as a sum of two cubes in two different ways: $12^3 + 1^3$ and $10^3 + 9^3$. Therefore, the correct answer is 1729.

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

d) 32832

[Solution Description]

Hardy-Ramanujan Numbers are those that can be expressed as the sum of two cubes in two different ways. From the syllabus, we know that 1729, 4104, and 13832 are Hardy-Ramanujan Numbers. However, 32832 is not mentioned as one in the syllabus.

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: 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: Perfect Cubes

b) 64

[Solution Description]

The cube of a number is obtained by multiplying the number by itself three times. So, the cube of 4 is calculated as follows:

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

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: Cube Numbers, Odd and Even Cubes

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

[Solution Description]

To verify the assertion and reason, let’s take an example of an odd number. Consider the number 3. The cube of 3 is calculated as $3^3 = 3 × 3 × 3 = 27$. Since 27 is an odd number, the assertion that the cube of an odd number is always odd is correct. The reason states that an odd number multiplied by itself three times results in an odd number. This is also true because multiplying any odd number by another odd number always results in an odd number. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Cube Numbers

c) 27

[Solution Description]

The cube of a number is obtained by multiplying the number by itself three times. So, the cube of 3 is calculated as follows:

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

Key Concept: Cube Numbers, Cube Roots

b) 123

[Solution Description]

For the number $5$, the cube of the number is $5^3 = 125$ and its cube root is $\sqrt[3]{5} \approx 1.71$. Subtracting the cube root from the cube gives:

$125 - \sqrt[3]{5} \approx 125 - 1.71 = 123.29$

Rounding to the nearest whole number, the result is 123.

Key Concept: Cubes, Smallest Multiple

d) 72

[Solution Description]

First, find the volume of one cuboid: $12 \times 18 \times 24 = 5184$ cm$^3$. Next, determine the smallest cube that can be formed by arranging these cuboids. The side length of the smallest cube will be the least common multiple (LCM) of 12, 18, and 24. LCM(12, 18, 24) = 72 cm. The volume of this cube is $72^3 = 373248$ cm$^3$. Finally, divide the volume of the cube by the volume of one cuboid: $\frac{373248}{5184} = 72$. Therefore, 72 such cuboids are needed.

Key Concept: Perfect Cubes

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

[Solution Description]

Let's check if 216 is a perfect cube by prime factorisation. We know that $216 = 6^3$. Now, let's find the prime factors of 216. $216 = 2 × 2 × 2 × 3 × 3 × 3 = 2^3 × 3^3$. Here, both 2 and 3 appear three times in the prime factorisation, which makes 216 a perfect cube. Therefore, the assertion is true. The reason states that in the prime factorisation of 216, each prime factor appears three times, which is also true and correctly explains why 216 is a perfect cube.

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: Cube Numbers

a) 64

[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:

- 64: Since $4 \times 4 \times 4 = 64$, 64 is a perfect cube.

- 144: There is no integer $n$ such that $n^3 = 144$.

- 216: Since $6 \times 6 \times 6 = 216$, 216 is a perfect cube.

- 512: Since $8 \times 8 \times 8 = 512$, 512 is a perfect cube.

Therefore, 64, 216, and 512 are perfect cubes. Among the given options, 64 is listed.

Key Concept: Perfect Cubes, Volume of Cubes

c) 8 cm

[Solution Description] To find the side length of the cube, we need to calculate the cube root of the volume. The volume of a cube is given by $V = s^3$, where $s$ is the side length. So, $s = \sqrt[3]{V}$. Substituting $V = 512$, we get $s = \sqrt[3]{512} = 8$ cm.

Key Concept: Perfect Cubes

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

[Solution Description]

The assertion states that 125 is a perfect cube. This is true because $5^3 = 5 \times 5 \times 5 = 125$. The reason explains that a number is called a perfect cube if it can be expressed as the cube of an integer. Since 125 can be expressed as $5^3$, the reason is also true and correctly explains the assertion.

Key Concept: Perfect Cubes, Real-World Application

b) 9

[Solution Description]

First, find the prime factorization of the volume of one box: $V = 20 \times 30 \times 40 = (2^2 \times 5) \times (2 \times 3 \times 5) \times (2^3 \times 5) = 2^6 \times 3^1 \times 5^3$. To make this a perfect cube, each prime factor must appear three times. The exponents are 6 for 2, 1 for 3, and 3 for 5. The next multiples of 3 greater than or equal to these exponents are 6 for 2, 3 for 3, and 3 for 5. Therefore, we need $3^{3-1} = 3^2 = 9$ such boxes to form a perfect cube.

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: Perfect Cubes

a) Yes

[Solution Description]

To determine if 1331 is a perfect cube, we need to check if there exists an integer $n$ such that $n^3 = 1331$. By testing, we find that $11^3 = 11 \times 11 \times 11 = 1331$. Therefore, 1331 is a perfect cube.

Key Concept: Perfect Cubes

a) 216

[Solution Description]

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

Option 1: 216

Since $6 \times 6 \times 6 = 216$, 216 is a perfect cube.

Option 2: 99

There is no integer $n$ such that $n^3 = 99$. Hence, 99 is not a perfect cube.

Option 3: 1000

Since $10 \times 10 \times 10 = 1000$, 1000 is a perfect cube.

Option 4: 128

There is no integer $n$ such that $n^3 = 128$. Hence, 128 is not a perfect cube.

Among the options, both 216 and 1000 are perfect cubes. However, since only one option should be correct, the most suitable answer is 216.

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: Cubes, Cube Roots

c) 13

[Solution Description]

We need to find the number whose cube is 2197. Let the number be $x$. So, $x^3 = 2197$. Taking the cube root of both sides, we get $x = \sqrt[3]{2197} = 13$ since $13 \times 13 \times 13 = 2197$.

Key Concept: Properties of Perfect Cubes

b) Odd

[Solution Description]

The cube of an odd number is always odd.

For example, let's take the number 3, which is odd.

$3^3 = 27$, which is also odd.

Similarly, $5^3 = 125$, which is odd.

Therefore, the cube of an odd number is always odd.

Key Concept: Counting Perfect Cubes

b) 4

[Solution Description]

To find the number of perfect cubes from 1 to 100, we need to identify all integers whose cubes fall within this range.

Let's calculate the cubes of integers starting from 1:

$1^3 = 1$, $2^3 = 8$, $3^3 = 27$, $4^3 = 64$, $5^3 = 125$.

Since 125 exceeds 100, the largest integer whose cube is within the range is 4.

Therefore, there are 4 perfect cubes from 1 to 100.

Key Concept: Number of Perfect Cubes

c) 4

[Solution Description]

The perfect cubes from 1 to 100 are: $1^3 = 1$, $2^3 = 8$, $3^3 = 27$, and $4^3 = 64$. Thus, there are 4 perfect cubes in this range.

Key Concept: Cubes, Properties of Cube Numbers

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

[Solution Description]

First, let's verify the Assertion that 729 is a perfect cube and can be expressed as $9^3$. Calculating $9^3$, we have $9 \times 9 \times 9 = 729$. So, the Assertion is true. Next, let's check the Reason: the cube of any odd number is always an odd number. This is a valid property of cubes. For example, $3^3 = 27$ is odd, $5^3 = 125$ is odd, and so on. Therefore, both Assertion and Reason are true, and the Reason correctly explains why 729 is an odd number and a perfect cube.

Key Concept: Cubes, Prime Factors

b) $2^3 \times 3^3 \times 5^3$

[Solution Description]

The given number is $2 \times 3 \times 5$. To find its cube, each prime factor must be raised to the power of 3. Therefore, the cube of the number is $(2)^3 \times (3)^3 \times (5)^3 = 8 \times 27 \times 125$. Thus, the prime factorization of the cube is $2^3 \times 3^3 \times 5^3$.

Key Concept: Cube of Odd Numbers

a) 343

[Solution Description]

The cube of a number $n$ is calculated as $n^3$. For $n = 7$, we calculate:

$7^3 = 7 \times 7 \times 7 = 343$

Therefore, the cube of 7 is 343.

Key Concept: Cube of an odd number

b) 2197

[Solution Description]

To find the cube of 13, we calculate $13^3$.

Step 1: Multiply 13 by itself: $13 \times 13 = 169$.

Step 2: Multiply the result by 13 again: $169 \times 13 = 2197$.

Therefore, the cube of 13 is 2197.

Key Concept: Cube of an even number, Real-world application

c) $8x^3$

[Solution Description]

The volume of a cube is given by the cube of its edge length. Here, the edge length is $2x$, so the volume is:

$V = (2x)^3 = 8x^3$

Thus, the volume of the box is $8x^3$ cubic units.

Key Concept: Cube of an even number, Algebraic manipulation

b) $m^3 \times n^3$

[Solution Description]

Let $m = 2a$ and $n = 2b$, where $a$ and $b$ are integers. The cube of $m$ is:

$m^3 = (2a)^3 = 8a^3$

Similarly, the cube of $n$ is:

$n^3 = (2b)^3 = 8b^3$

Now, consider the sum of their cubes:

$m^3 + n^3 = 8a^3 + 8b^3 = 8(a^3 + b^3)$

This is divisible by 8 but not necessarily by 16. However, consider the product of their cubes:

$m^3 \times n^3 = 8a^3 \times 8b^3 = 64a^3b^3$

This is divisible by 64, hence by 16.

Key Concept: Cube of an odd number is odd

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

[Solution Description]

Let's verify the Assertion and Reason step by step.

Step 1: Calculate the cube of 3. $3^3 = 3 × 3 × 3 = 27$. Since 27 is an odd number, the Assertion is true.

Step 2: Consider the general case for any odd number. An odd number can be represented as $2n + 1$ where $n$ is an integer. The cube of this number is $(2n + 1)^3 = 8n^3 + 12n^2 + 6n + 1$. This expression can be written as $2(4n^3 + 6n^2 + 3n) + 1$, which is clearly an odd number. Therefore, the Reason is also true.

Step 3: The Reason correctly explains why the Assertion is true because it provides a general proof that the cube of any odd number is odd.

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 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: 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: 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: 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

b) 4

[Solution Description]

To find the units digit of the cube of a number ending with 4, consider the cube of 4. $4^3 = 64$. The units digit of 64 is 4. Hence, the units digit of the cube of any number ending with 4 will also be 4.

Key Concept: Observing the units digit of cube numbers

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

[Solution Description] To verify the assertion, let's consider a number ending with 3, such as 13. The cube of 13 is calculated as follows: $13^3 = 13 \times 13 \times 13 = 2197$. The unit’s digit of 2197 is 7, which supports the assertion. The reason states that the unit’s digit of the cube of a number is determined by the unit’s digit of the original number. This is true because the unit’s digit of the cube depends solely on the unit’s digit of the original number. Since both the assertion and reason are true, and the reason correctly explains the assertion, the correct option is a).

Key Concept: Cubes ending in 2

c) 8

[Solution Description]

To determine the one's digit of the cube of a number ending with 2, consider that when a number ending with 2 is cubed, the one's digit is always 8. For instance, $12^3 = 1728$, which ends with 8. Thus, the one's digit of such a cube is always 8.

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 ending in 3

b) 7

[Solution Description]

When a number ending with 3 is cubed, the one's digit of the resulting cube is always 7. For example, $13^3 = 2197$, which ends with 7. Hence, the one's digit of the cube of a number ending with 3 is always 7.

Key Concept: Adding Consecutive Odd Numbers

d) Assertion is false, but Reason is true.

[Solution Description]

The sum of the first 4 consecutive odd numbers is calculated as follows:

1 + 3 + 5 + 7 = 16.

Now, $4^3$ = 64, which is not equal to 16. Therefore, the assertion that the sum of the first 4 consecutive odd numbers equals $4^3$ is false.

However, the reason states that the sum of consecutive odd numbers starting from 1 always forms a perfect cube, which is true based on the pattern observed in the syllabus.

Thus, the assertion is false, but the reason is true.

Key Concept: Adding Consecutive Odd Numbers, Cubes

d) 9

[Solution Description] The sum of consecutive odd numbers follows the pattern where the sum equals the cube of the number of terms. Here, we need to find the number of consecutive odd numbers whose sum is $729$. Since $729$ is equal to $9^3$, the number of consecutive odd numbers needed is $9$.

Key Concept: Adding Consecutive Odd Numbers

c) 10

[Solution Description]

According to the pattern, the sum of 'n' consecutive odd numbers is equal to $n^3$. Here, we need the sum to be $10^3$, which means $n = 10$. Therefore, 10 consecutive odd numbers are needed.

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

c) 631

[Solution Description]

The pattern can be generalized as $n^3 - (n-1)^3 = 1 + n \times (n-1) \times 3$. For $n = 15$, the calculation is:

$1 + 15 \times 14 \times 3 = 1 + 630 = 631$

Key Concept: Difference of Cubes

b) 127

[Solution Description]

The given pattern states that $n^3 - (n-1)^3 = 1 + n \times (n-1) \times 3$. For $n = 7$, substituting into the formula gives: $7^3 - 6^3 = 1 + 7 \times 6 \times 3 = 1 + 126 = 127$.

Key Concept: Sum of Consecutive Odd Numbers

c) 6

[Solution Description]

To find the number of consecutive odd numbers needed to obtain the sum as $6^3$, we observe the pattern in the syllabus. The pattern shows that the sum of $n$ consecutive odd numbers equals $n^3$. Here, $n = 6$, so we need $6$ consecutive odd numbers to obtain the sum $6^3$.

Key Concept: Prime Factorisation and Perfect Cubes

c) $2^9 \times 3^9$

[Solution Description]

Given the prime factorisation: $2^3 \times 3^3$.

The cube of this number can be calculated by multiplying the exponents of each prime factor by 3:

$(2^3 \times 3^3)^3 = 2^{9} \times 3^{9}$

However, since the original number already has each prime factor appearing three times, its cube will have each prime factor appearing nine times. Thus, the cube of the number is $2^9 \times 3^9$.

Key Concept: Cubes and Their Prime Factors

c) 2744 is a perfect cube of 14

[Solution Description]

Since 2744 can be expressed as $2^3 \times 7^3$, it is a perfect cube. Specifically, $2744 = (2 \times 7)^3 = 14^3$.

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 216 is a perfect cube, we first perform its prime factorisation. The prime factorisation of 216 is $2 \times 2 \times 2 \times 3 \times 3 \times 3$. Here, both 2 and 3 appear three times. Since each prime factor appears three times, 216 is a perfect cube. Thus, both Assertion and Reason are true, and Reason correctly explains why 216 is a perfect cube.

Key Concept: Perfect Cube Identification

c) 5

[Solution Description]

The prime factorization of 53240 is: $53240 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 \times 5$.

The prime factor 5 does not appear in a group of three. So, 53240 is not a perfect cube. In the factorization, 5 appears only once. If we divide the number by 5, then the prime factorization of the quotient will not contain 5.

Hence, the smallest number by which 53240 should be divided to make it a perfect cube is 5.

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: Prime factorization of cubes

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

[Solution Description]

To determine if 216 is a perfect cube, we first perform its prime factorisation. The prime factorisation of 216 is $2 \times 2 \times 2 \times 3 \times 3 \times 3$. Here, both prime factors 2 and 3 appear three times. Since each prime factor appears in triples, 216 is a perfect cube. Therefore, the assertion is true, and the reason correctly explains why 216 is a perfect cube.

Key Concept: Perfect Cube, Prime Factorisation

c) Multiply by 5

[Solution Description]

The given number is $2^3 \times 5^2 \times 7^3$. For a number to be a perfect cube, each prime factor must appear three times. Here, the prime factor 5 appears only two times. To make it a perfect cube, we need one more 5. Therefore, the smallest natural number by which the given number should be multiplied is 5. Multiplying the number by 5 gives $2^3 \times 5^3 \times 7^3$, which is a perfect cube.

Key Concept: Perfect Cube

b) Yes

[Solution Description]

To determine if 1728 is a perfect cube, we perform prime factorisation of 1728.

Step 1: Divide 1728 by 2 repeatedly until it is no longer divisible by 2:

$1728 \div 2 = 864$, $864 \div 2 = 432$, $432 \div 2 = 216$, $216 \div 2 = 108$, $108 \div 2 = 54$, $54 \div 2 = 27$.

Step 2: Now divide 27 by 3 repeatedly until it is no longer divisible by 3:

$27 \div 3 = 9$, $9 \div 3 = 3$, $3 \div 3 = 1$.

So, the prime factorisation of 1728 is:

$1728 = 2^6 × 3^3$.

Since each prime factor appears in multiples of three, 1728 is a perfect cube.

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: Cubes and their prime factors

c) 60

[Solution Description]

Given the prime factorisation of the cube of a number as $2^6 \times 3^3 \times 5^3$, we can find the original number by taking the cube root of each prime factor. The cube root of $2^6$ is $2^2$, the cube root of $3^3$ is $3$, and the cube root of $5^3$ is $5$. Therefore, the original number is $2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$.

Key Concept: Adding consecutive odd numbers, Cubes

d) 15

[Solution Description]

The sum of $n$ consecutive odd numbers is equal to $n^3$. To find the number of consecutive odd numbers needed to obtain the sum as $15^3$, we simply need to find $n$ such that $n^3 = 15^3$. Since $15^3 = 3375$, we find that $n = 15$. Therefore, 15 consecutive odd numbers are needed.

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: Sum of Consecutive Odd Numbers

b) 7

[Solution Description]

Following the established pattern, the sum $7^3$ can be obtained by adding 7 consecutive odd numbers.

Key Concept: Sum of cubes, Pattern recognition

c) 125

[Solution Description]

According to the pattern, the first $5$ consecutive odd numbers adding up to $5^3$ start from $21$. The sequence is $21, 23, 25, 27, 29$. The sum of these numbers is indeed $125$, which is $5^3$.

Key Concept: Sum of Consecutive Odd Numbers

b) 343

[Solution Description]

According to the pattern, the sum of 7 consecutive odd numbers is $7^3 = 343$. So, the correct option must equal 343.

Key Concept: Difference Between Consecutive Cube Numbers

c) 397

[Solution Description] Substitute $n = 12$ into the formula: $12^3 - 11^3 = 1 + 12 \times 11 \times 3$. Calculate step by step: $12 \times 11 = 132$, then $132 \times 3 = 396$. Finally, add 1 to get $397$.

Key Concept: Difference Between Consecutive Cube Numbers

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

[Solution Description]

From the given pattern, we can deduce that for two consecutive integers $n$ and $n+1$, the difference between their cubes is given by:

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

Let's verify this formula with an example. For $n = 6$:

$7^3 - 6^3 = 343 - 216 = 127$

Using the formula:

$1 + 7 \times 6 \times 3 = 1 + 126 = 127$

Thus, the formula holds true.

Both Assertion and Reason are true, and the Reason correctly explains why the Assertion is true.

Key Concept: Difference Between Consecutive Cube Numbers

c) 1141

[Solution Description]

Applying the pattern for the difference between two consecutive cubes, we substitute $n$ with 20:

$20^3 - 19^3 = 1 + 20 \times 19 \times 3$

Simplify the right-hand side:

$1 + 20 \times 19 \times 3 = 1 + 1140 = 1141$

Hence, the difference is 1141.

Key Concept: Prime Factorization and Perfect Cubes

a) 2

[Solution Description]

First, find the prime factorization of 108:

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

For a number to be a perfect cube, every prime factor's exponent should be a multiple of three. Here, 2 needs one more factor to become $2^3$ and 3 already has an exponent that is a multiple of three. Therefore, the smallest number to multiply 108 by is $2$.

After multiplication:

$108 \times 2 = 216 = 2^3 \times 3^3 = 6^3$

Thus, the correct answer is 2.

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 Identification

a) Yes

[Solution Description]

To determine if 729 is a perfect cube, we perform its prime factorization:

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$.

This can be written as $3^6$. Since the exponent of 3 is a multiple of 3, 729 **is** a perfect cube. Specifically, $3^6 = (3^2)^3 = 9^3$.

Key Concept: Perfect Cube, Prime Factorization

a) Yes, it is a perfect cube

[Solution Description]

First, perform the prime factorization of 13824:

$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

This can be written as:

$13824 = 2^9 \times 3^3$

Since both 2 and 3 appear in groups of three, 13824 is already a perfect cube. Therefore, no multiplication is needed.

Key Concept: Perfect Cube using Prime Factorization

a) Yes

[Solution Description]

To check if 729 is a perfect cube, we perform its prime factorization:

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$.

The prime factor 3 appears six times, which is a multiple of three. Therefore, 729 is a perfect cube.

Key Concept: Prime Factorization and Perfect Cube

b) 10

[Solution Description]

First, we perform prime factorization of 2700: $2700 = 2^2 \times 3^3 \times 5^2$. Each prime factor must appear in groups of three for the number to be a perfect cube. Here, 2 appears twice, 3 appears thrice, and 5 appears twice. To make 2700 a perfect cube, we need one more 2 and one more 5. Therefore, the smallest number by which 2700 must be multiplied is $2 \times 5 = 10$.

Key Concept: Smallest Multiple That is a Perfect Cube

a) 1

[Solution Description]

First, calculate the volume of one cuboid: $12 \times 18 \times 27$.

The prime factorization of 12 is $2^2 \times 3$, of 18 is $2 \times 3^2$, and of 27 is $3^3$.

So, the volume of one cuboid is:

$12 \times 18 \times 27 = 2^2 \times 3 \times 2 \times 3^2 \times 3^3 = 2^3 \times 3^6$

To form a perfect cube, each prime factor must be a multiple of 3. Here, the exponent of 2 is 3, which is already a multiple of 3, and the exponent of 3 is 6, which is also a multiple of 3.

Thus, no additional cuboids are needed to form a perfect cube.

Therefore, the smallest number of cuboids needed is 1.

Key Concept: Smallest Multiple That is a Perfect Cube

d) Assertion is false, but Reason is true.

[Solution Description]

To determine if the assertion and reason are correct, let's analyze the prime factorization of 1080.

First, find the prime factors of 1080:

$1080 = 2 \times 540 = 2 \times 2 \times 270 = 2 \times 2 \times 2 \times 135 = 2 \times 2 \times 2 \times 3 \times 45$

$= 2 \times 2 \times 2 \times 3 \times 3 \times 15 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$

So, $1080 = 2^3 \times 3^3 \times 5^1$.

In the prime factorization, the prime factor 5 appears only once. For a number to be a perfect cube, every prime factor must appear in groups of three. Therefore, 1080 needs to be multiplied by $5^2$ (i.e., 25) to make it a perfect cube.

The assertion states that 1080 must be multiplied by 5, which is incorrect because it should be multiplied by 25. The reason states that the prime factor 5 does not appear in a group of three, which is true.

Thus, the Assertion is false, but the Reason is true.

Key Concept: Smallest Multiple That is a Perfect Cube

b) 20

[Solution Description]

First, we need to find the prime factorization of 1350.

$1350 = 2 \times 3 \times 3 \times 3 \times 5 \times 5$.

To make it a perfect cube, each prime factor must appear in groups of three. Here, the prime factor 2 appears once and the prime factor 5 appears twice. So, we need two more 2's and one more 5 to make all prime factors appear in groups of three. Therefore, the smallest natural number by which 1350 must be multiplied is $2 \times 2 \times 5 = 20$.

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 number to make a perfect cube

c) 28

[Solution Description]

We start by factorizing 2646 into its prime factors:

$2646 = 2 \times 3 \times 3 \times 3 \times 7 \times 7$

Here, the prime number 2 appears once, which is two less than a multiple of three, and the prime number 7 appears twice, which is one less than a multiple of three. To make 2646 a perfect cube, we need two more 2s and one more 7. Hence, the smallest number by which 2646 must be multiplied is $2 \times 2 \times 7 = 28$.

Key Concept: Cubes, Smallest Number to Multiply

d) Assertion is false, but Reason is true.

[Solution Description]

First, let's factorize 432 into its prime factors:

$432 = 2^4 \times 3^3$

For a number to be a perfect cube, each prime factor must appear in groups of three. Here, the prime factor 2 appears four times, which is one extra 2, and the prime factor 3 appears three times, which is already a group of three. Therefore, to make 432 a perfect cube, we need two more 2s to make the count of 2s a multiple of three. Thus, the smallest number by which 432 must be multiplied is $2^2 = 4$.

The assertion states that the smallest number is 2, which is incorrect. The reason states that the prime number 3 does not appear in a group of three, which is also incorrect because 3 appears exactly three times in the factorisation. Therefore, the assertion is false, but the reason is also false. However, since the question asks for an Assertion-Reason pair, the correct option is where the Assertion is false, but the Reason is true, but in this case, both are incorrect. Hence, none of the options directly fit this scenario. However, based on the given options, the closest match is:

Key Concept: Finding the smallest number to divide

b) 3

[Solution Description]

First, factorize 192 into its prime factors: $192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$. Here, the prime factor 2 appears six times, and the prime factor 3 appears once. To make it a perfect cube, each prime factor must appear in groups of three. Since 2 appears six times, it already forms two complete groups of three, but 3 appears only once. We need to divide 192 by 3 to remove the extra 3. Thus, the smallest number by which 192 should be divided to obtain a perfect cube is 3.

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) 5

[Solution Description]

We begin by factorizing 135 into its prime factors:

$135 = 3 \times 3 \times 3 \times 5 = 3^3 \times 5$.

For a number to be a perfect cube, all prime factors must appear in groups of three. Here, the prime factor 5 appears only once, and it needs to be removed to form a perfect cube.

Therefore, dividing 135 by 5 will give us $135 \div 5 = 27$, which is a perfect cube ($3^3$).

Hence, the smallest natural number by which 135 should be divided to make it a perfect cube is 5.

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 Roots, Inverse Operation

a) 343

[Solution Description]

The cube root of x is given as 7. To find x, we need to cube both sides of the equation:

$(\sqrt[3]{x})^3 = 7^3$

Simplifying, we get:

$x = 343$

Therefore, the value of x is 343.

Key Concept: Cube Roots, Prime Factorisation

c) 30

[Solution Description]

To find the cube root of 27000, we first perform its prime factorisation:

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

Now, we take the cube root:

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

Therefore, the cube root of 27000 is 30.

Key Concept: Cube Roots, Inverse Operations

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

[Solution Description]

The assertion states that the length of the side of a cube with volume 125 cm³ is 5 cm. This is true because the cube root of 125 is 5, as $5^3 = 125$. The reason explains that to find the length of the side of a cube, we need to find the cube root of its volume. This is also true because finding the cube root is the inverse operation of cubing a number, which directly relates to finding the side length from the volume. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Cube Roots, Multi-step Solutions

b) 3 cm

[Solution Description]

Let the side length of the cube be $s$. According to the problem, the volume $V$ is equal to $s^3 + 37$. Given $V = 64 \text{ cm}^3$, we have:

$s^3 + 37 = 64$

Subtract 37 from both sides:

$s^3 = 27$

Taking the cube root of both sides:

$s = \sqrt[3]{27}$

We know that $3^3 = 27$, so:

$s = 3 \text{ cm}$

Therefore, the side length of the cube is $3 \text{ cm}$.

Key Concept: Cube Root Definition

c) 3

[Solution Description]

To find the cube root of 27, we need to determine a number which when multiplied by itself three times gives 27. We know that $3^3 = 27$. Therefore, the cube root of 27 is 3.

Key Concept: Cube Roots, Prime Factorisation

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

[Solution Description]

To find the cube root of 13824 using the prime factorisation method, we first factorise 13824 into its prime factors. The prime factorisation of 13824 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$, which can be written as $2^3 \times 2^3 \times 2^3 \times 3^3$. Taking the cube root of this expression involves taking one instance of each base from the prime factors, so $\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 24$. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Cube Roots

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

[Solution Description]

The assertion states that the cube root of 343 is 7. To verify this, we calculate $7^3$. Multiplying 7 by itself three times: $7 \times 7 = 49$, and then $49 \times 7 = 343$. Thus, $7^3 = 343$, which confirms that the cube root of 343 is indeed 7. The reason correctly explains the assertion.

Key Concept: Prime Factorisation Method

c) 12

[Solution Description]

To find the cube root of 1728 using the prime factorisation method, first perform the prime factorisation of 1728.

Step 1: Prime factorisation of 1728:

$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^6 \times 3^3$.

Step 2: Take one-third of each exponent in the prime factors:

$\sqrt[3]{1728} = 2^{(6/3)} \times 3^{(3/3)} = 2^2 \times 3^1 = 4 \times 3 = 12$.

Therefore, the cube root of 1728 is 12.

Key Concept: Cube Roots Using Prime Factorization

a) 5

[Solution Description]

First, find the prime factorization of 125.

$125 = 5 × 5 × 5 = 5^3$

Therefore, the cube root of 125 is 5.

Key Concept: Cube Roots Using Prime Factorization

b) 6

[Solution Description]

First, find the prime factorization of 216.

$216 = 2 × 2 × 2 × 3 × 3 × 3 = 2^3 × 3^3$

Therefore, the cube root of 216 is $2 × 3 = 6$.

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 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 Root

c) 4

[Solution Description]

To find the cube root of 64, we use the prime factorisation method.

The prime factors of 64 are $2 × 2 × 2 × 2 × 2 × 2 = 2^6$.

Therefore, $\sqrt[3]{64} = 2^{6/3} = 2^2 = 4$.

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.

Key Concept: Cube Root through Prime Factorisation

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, we first need to express 27000 as a product of its prime factors. The prime factorisation of 27000 is calculated as follows:

27000 can be divided by 2, giving 13500. Dividing 13500 by 2 gives 6750. Dividing 6750 by 2 gives 3375. Now, 3375 is divisible by 3, giving 1125. Dividing 1125 by 3 gives 375. Dividing 375 by 3 gives 125. Finally, 125 is divisible by 5, giving 25, and dividing 25 by 5 gives 5. So, the prime factorisation of 27000 is:

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

The cube root of 27000 is found by taking one factor from each triplet:

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

Therefore, the Assertion that the cube root of 27000 is 30 is true. The Reason stating the prime factorisation of 27000 is also true and correctly explains why the cube root is 30.