Key Concept: Perfect Squares, Sum of Odd Numbers

d) 50
[Solution Description]
To determine which number is not a perfect square, we check each option by verifying if it can be written as the sum of consecutive odd numbers starting from 1 or by confirming its non-membership in the list of perfect squares.
- $1 + 3 + 5 = 9$ (valid, so 9 is a perfect square).
- $1 + 3 + 5 + 7 = 16$ (valid, so 16 is a perfect square).
- The sum $1 + 3 + 5 + 7 + 9 + 11 + 13 = 49$ (valid, so 49 is a perfect square).
- The number 50 cannot be expressed as the sum of consecutive odd numbers starting from 1, nor is it listed among perfect squares.

Key Concept: Perfect Squares

c) 64
[Solution Description]
To find the square of 8, we multiply 8 by itself: $8 \times 8 = 64$. Thus, the square of 8 is 64.

Key Concept: Units Digit Patterns

c) 47
[Solution Description]
A perfect square can only end with digits 0, 1, 4, 5, 6, or 9. Therefore, any number ending with 2, 3, 7, or 8 is not a perfect square. Here, 47 ends with 7, so it cannot be a perfect square.

Key Concept: Difference Between Consecutive Squares

c) 23
[Solution Description]
The difference between consecutive squares is given by $(n+1)^2 - n^2 = 2n + 1$. For $n = 11$, the difference is $2(11) + 1 = 23$.

Key Concept: Square Numbers, Properties of Perfect Squares, Units Digit Patterns

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the sum of the first 15 odd natural numbers is a perfect square. Let's verify this by calculating the sum of the first 15 odd numbers:
$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 225$
We know that $15^2 = 225$, which confirms that 225 is indeed a perfect square. Thus, the assertion is true.
The reason states that the sum of the first $n$ odd natural numbers is $n^2$. This is a well-known property of square numbers, where:
$\sum_{k=1}^{n} (2k - 1) = n^2$
For example, $1 + 3 = 4 = 2^2$, $1 + 3 + 5 = 9 = 3^2$, and so on. Therefore, the reason is also true and correctly explains the assertion.

Key Concept: Prime Factorisation and Perfect Squares

c) Yes, because its prime factors can be paired identically.
[Solution Description]
The prime factorisation of 1156 is $2 \times 2 \times 17 \times 17$. Since these factors can be grouped into two identical groups ($2 \times 17$ and $2 \times 17$), 1156 is a perfect square. Specifically, $(2 \times 17)^2 = 34^2 = 1156$.

Key Concept: Patterns in Perfect Squares, Odd Number Summation

b) 10
[Solution Description]
The smallest three-digit perfect square is $100$ ($10^2$).
According to the pattern:
$1 + 3 + 5 + \ldots + (2n-1) = n^2$.
Here, $n^2 = 100 \Rightarrow n = 10$.
So, $10$ odd numbers must be added:
$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100$.
The correct answer is 10.

Key Concept: Perfect Squares, Prime Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, we first find the prime factorization of 2025:
$2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5 = (3 \times 3 \times 5)^2 = 45^2$
This shows that the prime factors can indeed be grouped into two identical pairs (four 3's and two 5's), confirming that 2025 is a perfect square. The reason correctly explains why the assertion is true, as it states the fundamental property of perfect squares regarding their prime factorization.

Key Concept: Perfect Squares and Units Digit

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that 49 is a perfect square, which is true since $7^2 = 49$. The reason claims that all numbers ending with the digit 9 are perfect squares. However, this is false because although some numbers ending with 9 are perfect squares (like 49), others like 19, 29, etc., are not perfect squares. Hence, the assertion is true but the reason is false.

Key Concept: Perfect Squares and Units Digit

c) 338
[Solution Description]
Perfect squares can only end with 0, 1, 4, 5, 6, or 9. Numbers ending with 2, 3, 7, or 8 cannot be perfect squares.

Key Concept: Estimating Square Roots

b) 44
[Solution Description]
To find $\sqrt{1936}$, we note that $40^2 = 1600$ and $50^2 = 2500$, so $\sqrt{1936}$ lies between 40 and 50. The last digit being 6 suggests the root could be 44 or 46. Calculating $44^2 = 1936$ confirms the answer.

Key Concept: Patterns in Units Place

c) 537
[Solution Description]
The units digit of a perfect square can only be $0, 1, 4, 5, 6,$ or $9$. Numbers ending with $2, 3, 7,$ or $8$ are not perfect squares. Among the options, only the number ending with $7$ does not fit this pattern.

Key Concept: Sum of Consecutive Odd Numbers, Finding the Next Square

a) 1296
[Solution Description]
The sum of the first $n$ odd numbers is given by $n^2 = 1225$. Therefore, $n = \sqrt{1225} = 35$. The next square is $(n + 1)^2 = 36^2$. To find $36^2$, we add the $(n+1)^{th}$ odd number to the previous sum. The $(n+1)^{th}$ odd number is $2(n + 1) - 1 = 2(36) - 1 = 71$. Thus, $36^2 = 35^2 + 71 = 1225 + 71 = 1296$.

Key Concept: Square Numbers, Patterns in Units Place, Consecutive Odd Numbers and Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, we calculate $45^2 = 2025$, which confirms that 2025 is indeed a perfect square. For the reason, the sum of the first $n$ odd numbers starting from 1 is given by $n^2$. Here, the sum of the first 45 odd numbers is $45^2 = 2025$. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Perfect Squares

c) 49
[Solution Description]
A perfect square is the square of a natural number. From the options, only 49 can be written as $7^2 = 49$.

Key Concept: Square Numbers Ending with 6, Patterns in Units Place

d) 2
[Solution Description]

If a perfect square ends in 6, then the number being squared must end in 4 or 6. The possible last two digits of such perfect squares are:

$16,\ 36,\ 56,\ 76,\ 96$

So, the possible tens digits are:

$1,\ 3,\ 5,\ 7,\ 9$

Therefore, the digit 2 cannot be the tens digit of a perfect square ending in 6.

Correct Option: D. 2

Key Concept: Zeros in Perfect Squares

d) 6
[Solution Description]
According to the properties, if a number has 3 zeros at the end, its square will have 6 zeros. Since 5000 has 3 zeros ($5000 = 5 \times 10^3$), squaring it gives $(5 \times 10^3)^2 = 25 \times 10^6 = 25,000,000$. Thus, the square of 5000 has 6 zeros at the end.

Key Concept: Units Place Property

c) 198
[Solution Description]
Perfect squares can only end with digits 0, 1, 4, 5, 6, or 9. The number ending with 8 is not a perfect square.

Key Concept: Patterns in Units Place, Non-Integer Squares

d) 6
[Solution Description]
A number with 3 zeros at the end can be written as $N = k \times 1000$, where $k$ is an integer not divisible by 10.
Squaring $N$ gives $N^2 = k^2 \times 1000^2 = k^2 \times 1,000,000$.
Since $k^2$ does not contribute additional zeros, the square will have exactly $6$ zeros.

Key Concept: Sum of Consecutive Odd Numbers

c) 1, 3, 5, 7, 9
[Solution Description]
A perfect square can be expressed as the sum of consecutive odd numbers starting from 1. Let's evaluate the options:
- Option 1: 5 + 7 + 9 = 21 (Not a perfect square)
- Option 2: 11 + 13 + 15 = 39 (Not a perfect square)
- Option 3: 1 + 3 + 5 + 7 + 9 = 25 (Which is $5^2$, a perfect square)
- Option 4: 3 + 5 + 7 + 9 = 24 (Not a perfect square)
Hence, the correct set is 1, 3, 5, 7, 9.

Key Concept: Differences Between Consecutive Squares, Largest Square Below 1000

b) 100
[Solution Description]
The difference between consecutive squares is given by $(n+1)^2 - n^2 = 2n + 1$.
Given $2n + 1 = 19$, solve for $n$:
$2n = 18 \Rightarrow n = 9$.
Thus, the larger square is $(9 + 1)^2 = 10^2 = 100$.

Key Concept: Square Numbers, Units Place Digits

c) Assertion is true, but Reason is false.
[Solution Description]
To verify if 2025 is a perfect square, let's analyze its properties:
1. **Units Place Digit:** The number 2025 ends with 5, which is one of the digits (0, 1, 4, 5, 6, 9) that perfect squares can end with. However, not all numbers ending with these digits are perfect squares. For example, 35 ends with 5 but is not a perfect square. Thus, ending with 5 does not guarantee that a number is a perfect square.
2. **Checking for Perfect Square:** Calculate the square root of 2025.
$45^2 = 2025$
Since $45^2$ equals 2025, 2025 is indeed a perfect square. Therefore, the Assertion is true.
3. **Reason Analysis:** The Reason states that a number ending with 5 is always a perfect square. This is false because numbers like 15 or 35 end with 5 but are not perfect squares. Hence, the Reason is incorrect.
Conclusion: The Assertion is true, but the Reason is false.

Key Concept: Perfect Squares and Odd Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify if 36 can be expressed as the sum of successive odd natural numbers, let's calculate the sum of the first $n$ odd numbers:
$1 + 3 + 5 + 7 + 9 + 11 = 36$. Here, $n = 6$ since there are 6 terms.
According to the property, the sum of the first $n$ odd numbers is $n^2$. Hence, $6^2 = 36$, which matches the assertion.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Square Numbers: Perfect Squares and Odd Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify if 16 is a perfect square, we subtract successive odd numbers starting from 1 until we reach or cross 0.
$16 - 1 = 15$
$15 - 3 = 12$
$12 - 5 = 7$
$7 - 7 = 0$
Since the result is 0, 16 is a perfect square. Also, $16 = 1 + 3 + 5 + 7$, which matches the definition given in the Reason. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Square Numbers and Triangular Numbers, Patterns in Perfect Squares

b) 25
[Solution Description]
The sequence follows the pattern where the sum of two consecutive triangular numbers gives a square number. The triangular numbers are $1, 3, 6, 10, 15, 21, \ldots$. The next pair is $10 + 15 = 25$, which is $5^2$.

Key Concept: Sum of Odd Numbers

d) 16
[Solution Description]
The sum of the first $n$ odd numbers is equal to $n^2$. Here, $n = 4$.
Step 1: Calculate $n^2 = 4^2 = 16$.
Therefore, the sum of the first 4 odd numbers is $16$.

Key Concept: Perfect Squares and Odd Numbers, Verification Method

a) 49
[Solution Description]
We apply the verification method to each option:
1) For 49: $49 - 1 = 48$, $48 - 3 = 45$, $45 - 5 = 40$, $40 - 7 = 33$, $33 - 9 = 24$, $24 - 11 = 13$, $13 - 13 = 0$. Since it reaches zero, 49 is a perfect square.
2) For 50: The process does not reach zero.
3) For 64: It also reaches zero after successive subtractions.
4) For 72: The process does not reach zero.
Both 49 and 64 are perfect squares, but since only one option is correct, we choose 49 as the primary example.

Key Concept: Difference Between Perfect Squares

b) 23
[Solution Description]
Calculate $12^2 = 144$ and $11^2 = 121$.
Difference: $144 - 121 = 23$.
According to the pattern, differences between consecutive squares are odd numbers. Here, $23$ is indeed an odd number.

Key Concept: Perfect Squares and Units Digit

c) 87654
[Solution Description]
A perfect square can only end with digits 0, 1, 4, 5, 6, or 9. Therefore, any number ending with 2, 3, 7, or 8 cannot be a perfect square.

Key Concept: Perfect Squares and Triangular Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The $5$-th triangular number is calculated using the formula for triangular numbers: $T_n = \frac{n(n+1)}{2}$. For $n=5$, $T_5 = \frac{5(5+1)}{2} = 15$. The $6$-th triangular number is $T_6 = \frac{6(6+1)}{2} = 21$. The sum of $T_5 + T_6 = 15 + 21 = 36$, which is indeed $6^2$.
The reason states a general property that the sum of the $n$-th and $(n+1)$-th triangular numbers is $(n+1)^2$. This is verified by the calculation above, as it holds for $n=5$. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Perfect Squares and Triangular Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The 5th triangular number is calculated as $T_5 = \frac{5 \times (5 + 1)}{2} = 15$. The 6th triangular number is $T_6 = \frac{6 \times (6 + 1)}{2} = 21$. Their sum is $15 + 21 = 36$, which is $6^2$, a perfect square. This matches the given pattern where $T_n + T_{n+1} = (n+1)^2$. Here, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Perfect Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that 144 is a perfect square, which is true since $12 \times 12 = 144$. The reason states that a perfect square can be expressed as the sum of consecutive odd numbers starting from 1, which is also true. For example, $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 144$. Moreover, the reason correctly explains why 144 is a perfect square by illustrating the property of perfect squares being sums of consecutive odd numbers.

Key Concept: Perfect Squares

d) 25
[Solution Description]
To find the square of 5, multiply 5 by itself: $5 \times 5 = 25$.

Key Concept: Square Numbers and Odd Number Summation

c) 15
[Solution Description]
A perfect square can be expressed as the sum of consecutive odd numbers starting from 1.
The sum of the first $n$ odd numbers is $n^2$.
Given the sum is 225, we need to find $n$ such that $n^2 = 225$.
Taking the square root of both sides: $n = \sqrt{225} = 15$.
Therefore, the first 15 odd numbers must be added to get 225.

Key Concept: Square Root Calculation

c) 12
[Solution Description]
We need to find a number $x$ such that $x^2 = 144$. From the given syllabus, we know that $12 \times 12 = 144$. Therefore, the positive square root of 144 is 12.

Key Concept: Identifying Perfect Squares

b) 1764
[Solution Description]
A number is a perfect square if all exponents in its prime factorization are even. Let's check each option by prime factorisation:
$2800 = 2^4 \times 5^2 \times 7^1$
The exponent of 7 is 1 (odd), so 2800 is not a perfect square.
$1764 = 2^2 \times 3^2 \times 7^2$
All exponents are even, so 1764 is a perfect square.

Key Concept: Square Roots of Perfect Squares

b) 34
[Solution Description]
To find the positive square root of 1156, we can use prime factorisation. First, factorise 1156 into its prime factors:
$1156 = 2 \times 2 \times 17 \times 17$
Now, group the prime factors into two identical groups:
$(2 \times 17) \times (2 \times 17)$
Hence, $\sqrt{1156} = 2 \times 17 = 34$.

Key Concept: Perfect Square Identification

a) 100
[Solution Description]
A perfect square is obtained by multiplying an integer by itself. Checking each option:
- 100: $10 \times 10 = 100$ (perfect square).
- 110: No integer multiplied by itself gives 110.
- 120: No integer multiplied by itself gives 120.
- 130: No integer multiplied by itself gives 130.
Thus, 100 is the only perfect square in the options.

Key Concept: Perfect Cubes and Prime Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if the assertion and reason are correct:
1. **Assertion**: Check if 216 is a perfect cube.
$6^3 = 6 \times 6 \times 6 = 216$, so 216 is indeed a perfect cube.
2. **Reason**: Analyze the prime factorization of 216.
$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3$. Here, both prime factors (2 and 3) appear three times, meaning they can be split into three identical groups. Thus, the reason correctly explains why 216 is a perfect cube.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Prime Factorisation of Cubes

c) $2^3 \times 3^3 \times 5^3 \times 7^3$
[Solution Description]
When cubing a number, each prime factor appears three times in the prime factorization of the cube. Given the prime factors $2 \times 3 \times 5 \times 7$, the cube will have:
$2^3 \times 3^3 \times 5^3 \times 7^3.$
Hence, the correct answer is $2^3 \times 3^3 \times 5^3 \times 7^3$.

Key Concept: Cube Roots, Consecutive Odd Numbers

b) 25
[Solution Description]
From the pattern given in the syllabus, the sum of $n$ consecutive odd numbers starting from $n^2 - n + 1$ equals $n^3$. For $n = 5$, the sequence starts at $5^2 - 5 + 1 = 21$:
$21 + 23 + 25 + 27 + 29 = 125 = 5^3$
The middle number is 25.

Key Concept: Perfect Cubes, Taxicab Numbers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that 4104 is the second smallest taxicab number. This is true because the smallest taxicab number is 1729, followed by 4104.
The reason claims that 4104 can be expressed as the sum of two cubes in two different ways: $15^3 + 9^3 = 3375 + 729 = 4104$ and $16^3 + 2^3 = 4096 + 8 = 4104$. This is also correct.
However, the reason does not explain why 4104 is the second smallest taxicab number; it only verifies that 4104 is a taxicab number. Thus, both statements are true, but the reason is not the correct explanation of the assertion.

Key Concept: Identifying Perfect Cubes

b) 27
[Solution Description]
A perfect cube is a number obtained by multiplying an integer by itself three times. Checking the options:
- 9 is not a perfect cube ($2^3 = 8$ and $3^3 = 27$).
- 27 is a perfect cube ($3 \times 3 \times 3 = 27$).
- 36 is not a perfect cube.
- 49 is not a perfect cube.
Thus, 27 is the perfect cube among the given options.

Key Concept: Prime Factorisation, Perfect Cubes

a) 216
[Solution Description]
To determine if a number is a perfect cube, its prime factors must form identical triplets. Let's analyze the options:
Option A: 216 = $2^3 \times 3^3$ → Perfect cube ($6^3$).
Option B: 500 = $2^2 \times 5^3$ → Not a perfect cube (exponent of 2 is not a multiple of 3).
Option C: 1000 = $2^3 \times 5^3$ → Perfect cube ($10^3$).
Option D: 800 = $2^5 \times 5^2$ → Not a perfect cube (exponents are not multiples of 3).
Thus, both A and C are correct, but since only one option can be chosen, we select the smallest valid choice.

Key Concept: Perfect Cubes and Prime Factorization

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 examine its prime factorization. Let's break it down step by step:
- Prime factorization of 216:
$216 = 6 \times 6 \times 6 = (2 \times 3)^3 = 2^3 \times 3^3$.
- Since the exponents of all prime factors (2 and 3) are multiples of 3, 216 is indeed a perfect cube.
- The assertion (A) states that 216 is a perfect cube, which is true.
- The reason (R) provides the correct prime factorization of 216, which also justifies why it is a perfect cube. Therefore, the reason correctly explains the assertion.

Key Concept: Perfect Cubes

c) 5
[Solution Description]
To find the cube root of 125, we need to find a number that when multiplied by itself three times gives 125. We can break it down:
Step 1: Find the prime factorization of 125. $125 = 5 \times 5 \times 5 = 5^3$.
Step 2: The cube root of $5^3$ is $\sqrt[3]{5^3} = 5$.
Therefore, the cube root of 125 is 5.

Key Concept: Perfect Cubes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
64 can be factorized as $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$. Since all the exponents are multiples of 3, they can be grouped into triplets ($2^3 \times 2^3$). Hence, 64 is a perfect cube and the Reason correctly explains the Assertion.

Key Concept: Consecutive Odd Numbers and Cubes

b) 6
[Solution Description]
From the given pattern in the syllabus, the sum of the first n consecutive odd numbers equals $n^3$.
Step 1: For $6^3 = 216$, the pattern shows that the sum of 6 consecutive odd numbers gives $6^3$.
Step 2: Example from the syllabus: $31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3$. There are 6 numbers here.
Therefore, 6 consecutive odd numbers must be added.

Key Concept: Prime Factorisation

c) 8
[Solution Description]
A number is a perfect cube if its prime factors can be grouped into triplets. For example, 8 can be written as $2 \times 2 \times 2 = 2^3$, making it a perfect cube.

Key Concept: Taxicab Numbers

c) 1729
[Solution Description]
The smallest such number is called a taxicab number. From the syllabus, the first taxicab number is 1729, which can be written as $1^3 + 12^3$ and $9^3 + 10^3$.

Key Concept: Prime Factorisation of Cubes

c) $2^3 \times 3^3$
[Solution Description]
The cube of a number has each prime factor raised to three times its original power. For $6 = 2 \times 3$, its cube is:
$6^3 = (2 \times 3)^3 = 2^3 \times 3^3$

Key Concept: Prime Factorisation of Cubes, Cube Roots

b) 90
[Solution Description]
To find the cube root of $N^3$, we first express $N$ in terms of its prime factors:
$N = 2 \times 3^2 \times 5$.
The cube of $N$ is:
$N^3 = (2 \times 3^2 \times 5)^3 = 2^3 \times 3^6 \times 5^3.$
Taking the cube root of $N^3$ gives:
$\sqrt[3]{N^3} = N = 2 \times 3^2 \times 5 = 90.$

Key Concept: Cubic Numbers: Properties of Cube Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let us verify the assertion and reason step by step:
1. The assertion states that the cube of a negative integer is always negative. For example, $(-2)^3 = -8$, $(-3)^3 = -27$, which are both negative.
2. The reason states that multiplying a number by itself three times preserves its sign. This is true because:
- Negative $\times$ Negative $\times$ Negative = Negative.
- Positive $\times$ Positive $\times$ Positive = Positive.
3. Therefore, the reason correctly explains why the cube of a negative integer is always negative.

Key Concept: Properties of Cube Numbers

b) 125
[Solution Description]
The cube of a number $n$ is given by $n^3$. Here, $n = 5$.
Step 1: Multiply 5 by itself three times:
$5^3 = 5 \times 5 \times 5$
Step 2: Calculate step-by-step:
$5 \times 5 = 25$
$25 \times 5 = 125$
So, $5^3 = 125$.

Key Concept: Cube Root

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The cube root of a number $n$ is the value that, when multiplied by itself three times, gives $n$. Here, $3 \times 3 \times 3 = 27$, so the cube root of 27 is indeed 3. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Cube Roots, Taxicab Numbers

c) Assertion is true, but Reason is false.
[Solution Description]
To verify if 4104 is a taxicab number, we check if it can be expressed as the sum of two cubes in two different ways. First, find the prime factorisation of 4104:
$4104 = 2^3 \times 3^3 \times 19$.
Now, express 4104 as sums of cubes:
1. $15^3 + 9^3 = 3375 + 729 = 4104$
2. $16^3 + 2^3 = 4096 + 8 = 4104$
Thus, 4104 is indeed a taxicab number. However, the Reason states that a taxicab number must have at least three distinct prime factors. While 4104 has three distinct prime factors (2, 3, 19), this is not a necessary condition for a taxicab number. For example, 1729 (another taxicab number) has only two distinct prime factors: $7 \times 13 \times 19$.
Therefore, Assertion is true, but Reason is false.

Key Concept: Cube Root and Prime Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if the assertion is true, let's find the prime factorization of 2744:
$2744 \div 2 = 1372$
$1372 \div 2 = 686$
$686 \div 2 = 343$
$343 \div 7 = 49$
$49 \div 7 = 7$
$7 \div 7 = 1$
So, the prime factorization is $2^3 \times 7^3$. Each prime factor appears three times, meaning they can indeed be split into three identical groups ($2^3$ and $7^3$). Therefore, 2744 is a perfect cube. The reason correctly explains the assertion.

Key Concept: Prime Factorization, Cube Roots

b) 18
[Solution Description]
To make the number a perfect cube after multiplying by 7, the exponents in its prime factorization must all be multiples of 3. The given prime factorization is $2^2 \times 3^1 \times 7^2$. Multiplying by 7 gives $2^2 \times 3^1 \times 7^3$. To make all exponents divisible by 3:
- For 2: $2^2$ needs another $2^1$ to become $2^3$.
- For 3: $3^1$ needs another $3^2$ to become $3^3$.
Thus, the smallest number to multiply is $2^1 \times 3^2 = 2 \times 9 = 18$.

Key Concept: Cube Root Calculation

d) 12
[Solution Description]
To find the cube root of $1728$, we perform prime factorization:
$1728 = 12 \times 12 \times 12 = 12^3$.
Therefore, $\sqrt[3]{1728} = 12$.

Key Concept: Taxicab Numbers, Perfect Cubes

a) 1729
[Solution Description]
The smallest taxicab number is 1729, known as the Hardy-Ramanujan Number. It can be expressed as:
$1729 = 1^3 + 12^3 = 9^3 + 10^3$.
The next taxicab numbers are larger (e.g., 4104, 13832), so 1729 is the correct answer.

Key Concept: Taxicab Numbers

b) 1729
[Solution Description]
The smallest taxicab number is 1729, as famously mentioned by Srinivasa Ramanujan. It can be expressed as:
$1729 = 1^3 + 12^3$ and $1729 = 9^3 + 10^3$.

Key Concept: Taxicab Numbers

c) 4104
[Solution Description]
The next two taxicab numbers after 1729 are 4104 and 13832. Here, 4104 can be expressed as:
$4104 = 2^3 + 16^3$ and $4104 = 9^3 + 15^3$.

Key Concept: Taxicab Numbers

c) 4104
[Solution Description]
After 1729, the next taxicab numbers are 4104 and 13832. Here, 4104 is one such taxicab number. It can be expressed as $2^3 + 16^3 = 4104$ and $9^3 + 15^3 = 4104$.

Key Concept: Taxicab Numbers

c) 3
[Solution Description]
There are three known taxicab numbers below 20000: 1729, 4104, and 13832.

Key Concept: Taxicab Numbers

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 number expressible as the sum of two cubes in two different ways, which is true as proven by Ramanujan. The reason provides the two valid cube pairs $(1^3 + 12^3)$ and $(9^3 + 10^3)$ that sum to 1729, correctly explaining why the assertion is true.

Key Concept: Taxicab Numbers

b) $(2, 16)$ and $(9, 15)$
[Solution Description]
The taxicab number $4104$ can be expressed as the sum of two cubes in two different ways:
$4104 = 2^3 + 16^3 = 9^3 + 15^3$
Calculating:
$2^3 + 16^3 = 8 + 4096 = 4104$
$9^3 + 15^3 = 729 + 3375 = 4104$
Therefore, the correct pairs are $(2, 16)$ and $(9, 15)$.

Key Concept: Perfect Cubes and Consecutive Odd Numbers

b) 64
[Solution Description]
The pattern given shows that the sum of consecutive odd numbers equals a perfect cube. For $4^3$, the corresponding sum is $13 + 15 + 17 + 19$. Adding these numbers:
$13 + 15 = 28$,
$28 + 17 = 45$,
$45 + 19 = 64$.
Thus, the sum is $64$.

Key Concept: Cube Roots

c) 9
[Solution Description]
We know that $9 \times 9 \times 9 = 729$. Therefore, $\sqrt[3]{729} = 9$.

Key Concept: Perfect Cubes via Prime Factorisation

b) Yes, because it can be written as $6^3$.
[Solution Description]
Prime factorize the number 216:
$216 = 6 \times 6 \times 6 = (2 \times 3) \times (2 \times 3) \times (2 \times 3) = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$
Since the factors can be grouped into three identical groups, 216 is a perfect cube.

Key Concept: Perfect Cubes and Consecutive Odd Numbers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To verify the assertion, let's analyze the given pattern:
- For $n=1$, the sum is $1 = 1^3$.
- For $n=2$, the sum is $3 + 5 = 8 = 2^3$.
- For $n=3$, the sum is $7 + 9 + 11 = 27 = 3^3$.
Observing the pattern, the sum of $n$ consecutive odd numbers (starting from the correct odd number) equals $n^3$. Here, there are $10$ terms ($n=10$) in the sum $91 + 93 + \ldots + 109$.
Calculating the mid-point of the range: $\frac{91 + 109}{2} = 100$.
Total sum = Mid-point $\times$ number of terms = $100 \times 10 = 1000 = 10^3$.
Thus, the sum is indeed a perfect cube.
The reason states that the sum of the first $n$ consecutive odd numbers starting from an appropriate odd number results in $n^3$. This is true as demonstrated by the pattern. However, the reason does not directly explain why the specific sum $91 + \ldots + 109$ equals $10^3$ but rather provides a general rule.
Therefore, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Third Differences in Cubic Numbers

d) 6
[Solution Description]
From the given data:
$19 - 7 = 12, \quad 37 - 19 = 18, \quad 61 - 37 = 24, \quad 91 - 61 = 30.$
The third differences are:
$18 - 12 = 6, \quad 24 - 18 = 6, \quad 30 - 24 = 6.$
The constant third difference is $6$.

Key Concept: Second Differences in Cubic Numbers

b) 18
[Solution Description]
Calculate the first differences:
$8 - 1 = 7, \quad 27 - 8 = 19, \quad 64 - 27 = 37.$
Now, compute the second difference:
$37 - 19 = 18.$

Key Concept: Cube Roots, Patterns in Squares

a) 6
[Solution Description]
We need $n^3 \equiv 216 \ (\text{mod} \ 1000)$.
First, solve $n^3 \equiv 216 \ (\text{mod} \ 8)$:
$216 \equiv 0 \ (\text{mod} \ 8)$, so $n$ must be even.
Next, solve $n^3 \equiv 216 \ (\text{mod} \ 125)$:
$216 = 6^3$, so $n \equiv 6 \ (\text{mod} \ 5)$ (using Fermat's Little Theorem).
Combining, the smallest $n$ satisfying both is $6$ (since $6^3 = 216$, which ends with $216$).
However, checking higher candidates:
$16^3 = 4096$ (does not end with $216$),
$26^3 = 17576$ (ends with $576$),
$36^3 = 46656$ (ends with $656$),
$46^3 = 97336$ (ends with $336$),
$56^3 = 175616$ (ends with $616$),
$66^3 = 287496$ (ends with $496$),
$76^3 = 438976$ (ends with $976$),
$86^3 = 636056$ (ends with $056$),
$96^3 = 884736$ (ends with $736$),
$106^3 = 1191016$ (ends with $016$),
$116^3 = 1560896$ (ends with $896$),
$126^3 = 2000376$ (ends with $376$),
$136^3 = 2515456$ (ends with $456$),
$146^3 = 3112136$ (ends with $136$),
$156^3 = 3796416$ (ends with $416$),
$166^3 = 4574296$ (ends with $296$),
$176^3 = 5451776$ (ends with $776$),
$186^3 = 6434856$ (ends with $856$),
$196^3 = 7529536$ (ends with $536$).
The pattern repeats every $100$ units. The smallest such $n$ is $6 + 100k$ where $k$ is an integer. Here, $6$ is the smallest.

Key Concept: Successive Differences for Cubic Numbers

c) 7
[Solution Description]
The first level difference is calculated by subtracting the previous term from the next term. Here, subtract $1$ from $8$: $8 - 1 = 7$.

Key Concept: First Differences in Cubic Numbers

c) 30
[Solution Description]
First, calculate the first differences for consecutive cubes:
$8 - 1 = 7, \quad 27 - 8 = 19, \quad 64 - 27 = 37, \quad 125 - 64 = 61, \quad 216 - 125 = 91.$
The question asks for the difference between the first differences of $216$ and $125$, which is $91 - 61 = 30$.

Key Concept: Historical Origin of Mathematical Terms

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]

The term $\textit{varga-mula}$ indeed refers to the square root in Sanskrit, as it combines $\textit{varga}$ (square) and $\textit{mula}$ (root). The reason correctly explains that $\textit{mula}$ was used to denote the origin or basis of a square, and this concept was later adopted in Arabic $(\textit{jidhr})$ and Latin $(\textit{radix}),$ eventually leading to the modern term "root." Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Babylonian Contributions

a) Quickly finding square roots and cube roots for geometric problems
[Solution Description]
The Babylonian clay tablets contained lists of perfect squares and cubes, which were used for quick calculations in land measurement and architectural design.

Key Concept: Historical terminology, Sanskrit mathematical terms

d) varga-varga-mula
[Solution Description]
In ancient India:
- Square root was called varga-mula
- Cube root was called ghana-mula
- Fourth power was called varga-varga
Therefore, the root of a fourth power (varga-varga) would logically follow the same pattern and be called varga-varga-mula.
The calculation: $\sqrt[4]{81} = 3$ since $3^4 = 81$

Key Concept: Mathematical History, Perfect Cubes

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that *varga-mula* refers to the cube root of a number. However, according to the historical context provided, *varga-mula* specifically refers to the square root (*varga* means square, *mula* means root). For cube roots, the term *ghana-mula* was used. Therefore, the Assertion is false. The Reason correctly explains that *mula* was indeed used metaphorically for roots in mathematical operations, but it does not justify the Assertion because *varga-mula* is not related to cube roots.

Key Concept: Sanskrit Terminology

b) Both the square figure and the square power
[Solution Description]
As per Aryabhata's description, *varga* refers to both the square figure (or its area) and the square power (product of two equal quantities).

Key Concept: Cube root calculation

c) 30
[Solution Description]
To find the cube root of 27000, we can factorize it:
$27000 = 27 \times 1000 = 3^3 \times 10^3 = (3 \times 10)^3 = 30^3$
Therefore, the cube root of 27000 is 30.

Key Concept: Use of *mula*

b) Because it represents the origin or basis of the square or cube
[Solution Description]
In Sanskrit, *mula* means the root of a plant, basis, cause, or origin. This term was adopted for mathematical roots (*varga-mula* for square root and *ghana-mula* for cube root) because it signifies the "origin" or "basis" of the square or cube. This terminology later influenced Arabic and Latin terms for roots.

Key Concept: Historical Mathematical Terminology, Babylonian Lists

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that $\textit{varga}$ was used exclusively for the square of a number, which is correct as per the given historical context. The Reason explains that the term $\textit{varga}$ originated from the graphical representation of a square figure, connecting both the shape and its area. This explanation aligns with Aryabhata's statement and supports the Assertion. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Babylonian Lists

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the Babylonians compiled the first known list of perfect squares and cubes around 1700 BCE, which is true as historical records confirm this. The reason given is that these lists were used for land measurement and architectural design, which is also true and correctly explains why the Babylonians needed such lists. Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Perfect Cubes

b) 7
[Solution Description] First, factorize 1323:
$1323 = 3^3 \times 7^2$
To make it a perfect cube, the exponents of all prime factors must be multiples of 3. Here, 7 has an exponent of 2. We need to multiply by 7 to get $3^3 \times 7^3 = (3 \times 7)^3 = 21^3$. Hence, the required number is 7.

Key Concept: Mathematical etymology, Root operations

b) Sanskrit through Arabic and Latin
[Solution Description]
The concept originated from Sanskrit 'mula' meaning plant root, which was adopted into Arabic as 'jidhr' and then Latin as 'radix'. These terms all referred both to plant roots and mathematical roots. The English term 'root' comes from this lineage via Latin influence on European mathematics during the Middle Ages.

Key Concept: Ancient Mathematical Terminology

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the term $varga$ was used in ancient India to refer to both the square figure and its area as well as the square power. This is true as per the given syllabus text which mentions, "In ancient Sanskrit works the term $varga$ was used both for the square figure or its area, as well as the square power."
The reason states that this usage originates from the graphical representation of a square figure. This is also true as Aryabhata's statement explains, "Thus, the term $varga$ for square power has its origin in the graphical representation of a square figure."
Moreover, the reason correctly explains why $varga$ was used for square power, linking it to the visual basis of the concept. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Concept: Early compilation of perfect squares and cubes

d) Babylonians
[Solution Description]
The Babylonians compiled the earliest known lists of perfect squares and cubes on clay tablets around 1700 BCE, aiding in land measurement and architectural design.

Key Concept: Indian Contributions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that *varga-varga* was used for the fourth power, which is correct as per the syllabus. The reason explains that *varga* means square, and *varga-varga* is derived by applying the square operation twice (square of square), which correctly explains why it represents the fourth power. Thus, both are true, and the reason correctly explains the assertion.

Key Concept: Historical Mathematical Terms, Sanskrit Origins

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that $\textit{varga-mula}$ denotes the square root as it signifies the origin/basis $(mula)$ of a square $(varga),$ which is true based on historical Sanskrit usage. The reason claims that the concept of roots originated from the graphical representation of squares/cubes in ancient Indian texts, which is also true but does not directly explain why $\textit{mula}$ (root) was specifically chosen for the mathematical operation. While the terms were indeed derived from geometric representations, the immediate explanation lies in the meaning of $\textit{mula}$ itself (basis/origin), not just the graphical connection. Hence, both statements are true, but the reason does not correctly explain the assertion.

Key Concept: Perfect Cubes and Their Properties

c) 64
[Solution Description]
A perfect cube is a number that can be expressed as $n^3$ where $n$ is an integer. Checking each option:
$64 = 4^3$
Hence, 64 is a perfect cube.

Key Concept: Perfect squares, Units digit pattern

c) 12347
[Solution Description]
A number cannot be a perfect square if it ends with 2, 3, 7, or 8. From the options:
12347 ends with 7, which means it cannot be a perfect square. The other options end with digits that may or may not be perfect squares.

Key Concept: Cube roots, Perfect cubes

c) 17
[Solution Description]
To find the cube root of 4913 without factorisation, observe the last digit. The number ends with 3. The only single-digit cube ending with 3 is $7^3 = 343$. Checking $17^3$ gives $17 \times 17 = 289$, then $289 \times 17 = 4913$. Thus, the cube root is 17.

Key Concept: Cube Roots

c) 5
[Solution Description]
The cube root of a number $x$ is a number $y$ such that $y^3 = x$. Here,
$5^3 = 125$
So, the cube root of 125 is 5.

Key Concept: Perfect Cubes

b) 27000
[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.
- 27000 can be written as $30^3$, making it a perfect cube.
- Other options do not satisfy this condition.

Key Concept: Varga, Ghana, Mula

b) The square of a square, representing the fourth power
[Solution Description]
The term varga-varga is derived from combining two varga terms. Since varga stands for square or second power, varga-varga represents the fourth power. This is because squaring a square gives the fourth power (e.g., $(x^2)^2 = x^4$).

Key Concept: Basic Concept: Cube Root Application

b) 11
[Solution Description]
Since 1331 is known to be a perfect cube, and observing that $11^3 = 1331$, the cube root must be 11.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]

Key Concept: Mula, Square Root

c) It symbolized the origin or basis of the number
[Solution Description]
The Sanskrit word 'mula' means the root of a plant, basis, cause, or origin. It was metaphorically extended to describe the mathematical operation of finding roots because it signifies the 'origin' or 'basis' of a number (e.g., the number that, when squared or cubed, gives the original number). For example, $\sqrt{x}$ finds the 'origin' of $x$ (the number which squared gives $x$).

Key Concept: Concept: Origin of mathematical terms varga and mula

d) Assertion is false, but Reason is true.
[Solution Description]
Assertion (A) is false because the term 'varga' was not exclusively used for the square power of a number. It was also used for the square figure or its area, as mentioned in the syllabus. Reason (R) is true because Aryabhata indeed stated that the term 'varga' for square power has its origin in the graphical representation of a square figure. Therefore, Assertion is false, but Reason is true.