Key Concept: Sum of Successive Odd Numbers

c) 8
[Solution Description]
Perfect squares can be expressed as the sum of successive odd numbers starting from 1. Since $64 = 8^2$, we need to add the first 8 odd numbers: $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64$.

Key Concept: Square Pairs, Adjacent Sums

c) 6 and 10
[Solution Description]
We test each possible pair:
- $1 + 3 = 4$ (perfect square).
- $1 + 6 = 7$ (not a perfect square).
- $1 + 10 = 11$ (not a perfect square).
- $3 + 6 = 9$ (perfect square).
- $3 + 10 = 13$ (not a perfect square).
- $6 + 10 = 16$ (perfect square).
Thus, multiple pairs work, but the question asks for one such pair.

Key Concept: Patterns in Unit Digits, Perfect Squares

d) It must be an even number.
[Solution Description]
From the properties of perfect squares:
- Numbers ending with 4 may or may not be perfect squares (e.g., $2^2 = 4$ and $8^2 = 64$, but 14 is not a perfect square).
- The tens digit does not provide conclusive information.
- The number could be even (as squares of even numbers end with even digits like 0, 4, 6).
- The only certainty is that the number is even.

Key Concept: Square Numbers - Properties of Perfect Squares

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that 144 is a perfect square. Calculating $12 \times 12 = 144$, we confirm this is true. The reason states that a number ending with digit 4 can be a perfect square. Examples include $2^2 = 4$ and $8^2 = 64$. While both are true, the reason does not explain why 144 is a perfect square; it merely supports the possibility. Hence, they are independently true but not causally related.

Key Concept: Units Digit of Perfect Squares

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that 144 is a perfect square because its units digit is 4. The reason states that a number is a perfect square only if its units digit is 0, 1, 4, 5, 6, or 9.
Checking the assertion: $12^2 = 144$, so 144 is indeed a perfect square.
Evaluating the reason: According to the syllabus, perfect squares can only have 0, 1, 4, 5, 6, or 9 as their units digit. However, having such a unit digit does not guarantee that the number is a perfect square — it is a necessary but not sufficient condition. For example, 24 has a units digit of 4 but is not a perfect square.
Thus, both the assertion and reason are true, but the reason does not fully explain why 144 is a perfect square (it just provides a necessary condition).

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: Prime Factorisation and Perfect Squares

a) 144
[Solution Description]
A perfect square has its prime factors grouped into two identical sets. Let's check each option:
1. 144: $12^2 = 144$ and its prime factors are $2 \times 2 \times 2 \times 2 \times 3 \times 3$, which can be grouped as $(2 \times 2 \times 3) \times (2 \times 2 \times 3)$, so it is a perfect square.
2. 99: Its prime factors are $3 \times 3 \times 11$, which cannot be grouped equally.
3. 50: Its prime factors are $2 \times 5 \times 5$, which cannot be grouped equally.
4. 72: Its prime factors are $2 \times 2 \times 2 \times 3 \times 3$, which cannot be grouped equally.
Therefore, only 144 is a perfect square.

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: Estimating Square Roots

b) 44
[Solution Description]
From the syllabus, we know that 1936 is a perfect square. To estimate its square root:
1. It lies between $1600$ ($40^2$) and $2500$ ($50^2$).
2. The last digit is 6, meaning the square root must end with 4 or 6 (i.e., 44 or 46).
3. Calculating $45^2 = 2025$, which is greater than 1936.
4. Now calculate $44^2 = 1936$.
Hence, the correct answer is 44.

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 Place

d) 273
[Solution Description]
According to the syllabus, perfect squares can only end with 0, 1, 4, 5, 6, or 9 in their units place. We need to find the number that does not end with any of these digits. Looking at the options:
1. 324 ends with 4.
2. 841 ends with 1.
3. 576 ends with 6.
4. 273 ends with 3, which is not in the list of possible unit digits for perfect squares. Thus, 273 cannot be a perfect square.

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: 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: 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: Patterns in Units Place

d) 7
[Solution Description]
According to the syllabus, perfect squares end with digits 0, 1, 4, 5, 6, or 9. Therefore, 7 cannot be the units place of a perfect square.

Key Concept: Square Numbers: Observations on Square Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To check if the assertion is true, we need to verify if 1444 is a perfect square. Calculating the square root of 1444 gives $38^2 = 1444$, so it is indeed a perfect square. The reason states that a number ending with 4 in the units place can be a perfect square, which is correct as seen from examples like $2^2 = 4$ and $12^2 = 144$. Moreover, the reason correctly explains why the assertion is true because the property of ending with 4 does not contradict the possibility of being a perfect square. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Difference Between Consecutive Squares

b) 25
[Solution Description]
Difference between $(k+1)^2$ and $k^2$ is $2k + 1$. Given $2k + 1 = 11$, solving gives $k = 5$. Thus, smaller square is $5^2 = 25$.

Key Concept: Properties of Square Numbers - Sum of Odd Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify if 25 is a perfect square, we can express it as the sum of consecutive odd numbers starting from 1. The sum of the first $n$ odd numbers is equal to $n^2$. Let's check:
1 + 3 + 5 + 7 + 9 = 25
Here, the sum of the first 5 odd numbers is 25, which equals $5^2$. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Squares Ending with 1

c) 29
[Solution Description]
According to the properties of square numbers, any number ending with 1 or 9 will have a square ending with 1. Checking the options:
- Option 1: 29 (units digit is 9, so $29^2$ ends with 1)
- Option 2: 14 (units digit is 4, so $14^2$ ends with 6)
- Option 3: 33 (units digit is 3, so $33^2$ cannot be a perfect square as per the property)
- Option 4: 47 (units digit is 7, so $47^2$ cannot be a perfect square as per the property)
Therefore, only 29 satisfies the condition.

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: 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: Assertion and Reason

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. We know that $12^2 = 144$, so 144 is indeed a perfect square. The reason states that a perfect square is an integer that is the square of another integer, which is the correct definition. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Difference Between Consecutive Squares

b) 25
[Solution Description]
The difference between consecutive squares is an odd number. The general pattern shows that the difference between $n^2$ and $(n+1)^2$ is given by $(2n + 1)$.
Here, $n = 12$.
So, the difference is $2(12) + 1 = 25$.

Key Concept: Difference Between Consecutive Squares

c) 11
[Solution Description]
The difference between consecutive squares is an odd number. Specifically, the difference between $(n+1)^2$ and $n^2$ is $2n + 1$.
Step 1: Calculate $6^2 = 36$ and $5^2 = 25$.
Step 2: Compute the difference: $36 - 25 = 11$.
Therefore, the difference is $11$.

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: Perfect Squares and Odd Numbers, Differences Between Consecutive Squares

a) The difference between consecutive squares is always an odd number.
[Solution Description]
To solve this, we analyze each option based on the properties of perfect squares:
1) The difference between consecutive squares is always an odd number: This is true as shown by examples like $4 - 1 = 3$, $9 - 4 = 5$.
2) A perfect square can end with the digit 8: This is false since perfect squares can only end with $0, 1, 4, 5, 6,$ or $9$.
3) The sum of the first $n$ odd numbers is $(n+1)^2$: This is false; the correct formula is $n^2$.
4) A number with 5 zeros at the end cannot be a perfect square: This is true because squares must have an even number of zeros at the end.
Thus, options 1 and 4 are correct, but since only one option is allowed, the most encompassing truth is option 1.

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 Square Units Digit

c) 7
[Solution Description]
A perfect square can only end with 0, 1, 4, 5, 6, or 9 in its units place. Thus, 7 is not a possible units digit for a perfect square.

Key Concept: Perfect Squares

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

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

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: Square Roots and Differences

c) 11
[Solution Description]
Let the two consecutive odd numbers be $n$ and $n+2$. The difference of their squares is $(n+2)^2 - n^2 = 4n + 4 = 40$. Solving for $n$, we get $4n = 36 \implies n = 9$. Thus, the larger number is $n + 2 = 11$.

Key Concept: Triangular Numbers and Perfect Squares

c) 21
[Solution Description]
The sum of two consecutive triangular numbers gives a perfect square. Here, $36 = 6^2$. The two consecutive triangular numbers whose sum is 36 are 15 and 21 ($15 + 21 = 36$). Hence, the larger triangular number is 21.

Key Concept: Square Roots and Perfect Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if 324 is a perfect square, we analyze its prime factorization: $324 = 2 \times 2 \times 3 \times 3 \times 3 \times 3$. These factors can be grouped as $(2 \times 3 \times 3) \times (2 \times 3 \times 3) = 18^2$, which shows that 324 is a perfect square. The Reason correctly explains the Assertion because it states the condition for a number to be a perfect square based on its prime factorization.

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: Estimating Square Roots

d) 47
[Solution Description]
To estimate $\sqrt{2209}$, note that it lies between $46^2 = 2116$ and $47^2 = 2209$. Calculate $47^2$:
$47^2 = (40 + 7)^2 = 40^2 + 2 \times 40 \times 7 + 7^2 = 1600 + 560 + 49 = 2209$
Thus, $\sqrt{2209} = 47$.

Key Concept: Square Roots and Perfect Squares

a) 441
[Solution Description]
To find if a number is a perfect square using prime factorisation, we need to ensure all prime factors can be paired into two identical groups.
Let's check each option:
Option a) 441: Prime factorisation is $3 \times 3 \times 7 \times 7 = (3 \times 7)^2 = 21^2$. Hence, it is a perfect square.
Option b) 200: Prime factorisation is $2 \times 2 \times 2 \times 5 \times 5$. The prime factor 2 does not have a pair (three 2s). Hence, it is not a perfect square.
Option c) 500: Prime factorisation is $2 \times 2 \times 5 \times 5 \times 5$. The prime factor 5 does not have a pair (three 5s). Hence, it is not a perfect square.
Option d) 300: Prime factorisation is $2 \times 2 \times 3 \times 5 \times 5$. The prime factors 3 and 5 do not form pairs. Hence, it is not a perfect square.
Therefore, the correct answer is option a) 441.

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: Cube Roots

c) 7
[Solution Description]
The cube root of a number $y$ is a number $x$ such that $x^3 = y$. Here, we need to find $\sqrt[3]{343}$.
We know that $7 \times 7 \times 7 = 343$, so:
$\sqrt[3]{343} = 7.$
Therefore, the correct answer is 7.

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: 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: Taxicab Numbers, Sum of Cubes

c) 4104
[Solution Description]
Taxicab numbers can be expressed as two distinct pairs of cubes. From the syllabus, the next taxicab number after 1729 is 4104:
$4104 = 2^3 + 16^3 = 9^3 + 15^3$
Verify:
$2^3 + 16^3 = 8 + 4096 = 4104$.
$9^3 + 15^3 = 729 + 3375 = 4104$.

Key Concept: Cubic Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
27 can be expressed as $3^3$, which means it is a perfect cube. The reason correctly defines a perfect cube as a number that can be written in the form $n^3$ where $n$ is an integer. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Definition of Cubic Numbers

c) 64
[Solution Description]
To find the cube of 4, multiply 4 by itself three times: $4 \times 4 \times 4 = 64$. Thus, the cube of 4 is 64.

Key Concept: Taxicab numbers, Sum of two cubes

b) 4104
[Solution Description]
A taxicab number can be expressed as the sum of two cubes in two distinct ways. Among the options, $4104$ is known to satisfy this property: $4104 = 2^3 + 16^3 = 9^3 + 15^3$. We verify: $2^3 = 8$, $16^3 = 4096$, $8 + 4096 = 4104$; similarly, $9^3 = 729$, $15^3 = 3375$, $729 + 3375 = 4104$.

Key Concept: Perfect Cubes, Prime Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if 5832 is a perfect cube, we first find its prime factorisation:
$5832 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^3 \times 3^6$
Since both prime factors ($2$ and $3$) appear in multiples of three, 5832 is indeed a perfect cube. Further, $(2 \times 3^2)^3 = 2^3 \times 3^6 = 5832$, so $\sqrt[3]{5832} = 18$.
Therefore, both Assertion (A) and Reason (R) are true, and Reason correctly explains why 5832 is a perfect cube.

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: Sum of consecutive odd numbers, Perfect cubes

c) 7
[Solution Description]
The sum of consecutive odd numbers forming a perfect cube follows the pattern where the sum of the first $n$ such numbers is $n^3$. Given that $343 = 7^3$, the number of terms being added must be $7$.

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: 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: 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: Properties of Cube Numbers

d) Assertion is false, but Reason is true.
[Solution Description]
To check if a perfect cube can end with exactly two zeros (00), we need to analyze the prime factorization of such a number. A number ending with two zeros must be divisible by 100. The prime factorization of 100 is $2^2 \times 5^2$. For a number to be a perfect cube, each prime factor must appear in multiples of three. However, 100 has $2^2$ and $5^2$, which are not multiples of three. Therefore, a perfect cube cannot end with exactly two zeros. The Reason states that a number is a perfect cube if its prime factors appear in triplets, which is true. However, this does not explain why a perfect cube cannot end with two zeros, as the assertion is false.

Key Concept: Last Digit Pattern in Cubes

c) 3
[Solution Description]
From the pattern observed in cubes:
- Numbers ending with 7 have cubes ending with 3.
For example:
$7^3 = 343 \quad (\text{last digit is } 3)$
$17^3 = 4913 \quad (\text{last digit is } 3)$
Therefore, if a number ends with 7, its cube ends with 3.

Key Concept: Properties of Cube Numbers

b) 3
[Solution Description]
Observing the last digit pattern for cubes, numbers ending with 7 have cubes ending with 3 because $7^3 = 343$. Thus, the cube of any number ending with 7 will also end with 3.

Key Concept: Perfect Cube Identification

c) 64
[Solution Description]
A perfect cube is a number that can be expressed as the cube of an integer. Checking each option:
$64 = 4 \times 4 \times 4 = 4^3$
64 is a perfect cube.

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: Cube Roots, Estimation

b) 23
[Solution Description]
Observe the last digit of 12167, which is 7. The cube of a number ending with 3 is the only cube ending with 7 (since $3^3 = 27$). Now, check the nearest cubes of two-digit numbers around 20-30:
- $20^3 = 8000$
- $25^3 = 15625$
- $23^3 = 12167$
Thus, $\sqrt[3]{12167} = 23$.

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: Taxicab Number Identification

b) $(2, 16)$
[Solution Description]
The number $4104$ is a taxicab number and satisfies:
$4104 = 2^3 + 16^3 = 9^3 + 15^3.$
Among the options, only $(2, 16)$ corresponds to one valid pair.

Key Concept: Perfect Cube Verification

b) 125
[Solution Description]
A number is a perfect cube if its prime factors can be split into three identical groups. Let's check each option:
- $125$: $5 \times 5 \times 5 = 5^3$, so it is a perfect cube.
- $100$: $2 \times 2 \times 5 \times 5$, not a perfect cube.
- $144$: $2 \times 2 \times 2 \times 2 \times 3 \times 3$, not a perfect cube.
- $200$: $2 \times 2 \times 2 \times 5 \times 5$, not a perfect cube.
Hence, only $125$ is a perfect cube.

Key Concept: Taxicab Numbers, Sum of Cubes

d) 87539319
[Solution Description]
The smallest taxicab number expressible as a sum of two cubes in three distinct ways is 87539319. It can be written as:
$87539319 = 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3$.

Key Concept: Square Pair Arrangement, Combinatorics

a) 1
[Solution Description]
Only one unique sequence exists for numbers 1 to 17 where adjacent pairs sum to a perfect square. For example, [8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16] forms such a sequence (sums: 9, 16, 25, 16, 9, 16, 25, 16, 9, 16, 25, 16, 9, 16, 25).

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) 1729
[Solution Description]
The smallest number expressible as the sum of two positive cubes in two distinct ways is 1729, also known as the Hardy-Ramanujan Number. The two expressions are $1^3 + 12^3 = 1729$ and $9^3 + 10^3 = 1729$.

Key Concept: Taxicab Numbers

b) 2
[Solution Description]
There are exactly two pairs of positive integers $(a, b)$ that satisfy $a^3 + b^3 = 1729$. These are $(1, 12)$ and $(9, 10)$, since $1^3 + 12^3 = 1729$ and $9^3 + 10^3 = 1729$.

Key Concept: Perfect Cubes and Consecutive Odd Numbers

c) 1000
[Solution Description]
The given series is a set of 10 consecutive odd numbers starting from 91. Following the pattern, the sum of the first $n$ consecutive odd numbers equals $n^3$. Here, $n = 10$. Thus, the sum is $10^3 = 1000$.

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: Cube Roots, Prime Factorisation

b) The cube root of $1728$ is $12$.
[Solution Description]
To determine the cube root of $1728$, we perform prime factorisation:
$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
Grouping into triplets:
$1728 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) = 2^3 \times 2^3 \times 3^3$
Simplifying:
$1728 = (2 \times 2 \times 3)^3 = 12^3$
Therefore, $\sqrt[3]{1728} = 12$.

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 and Consecutive Odd Numbers, Taxicab Numbers

c) Assertion is true, but Reason is false.
[Solution Description]
Let's verify the assertion first:
The given sequence is:
$91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109$.
Step 1: Count the number of terms. There are $10$ consecutive odd numbers here.
Step 2: Find the average of the first and last term to determine the median value or central tendency.
$\text{Average} = \frac{91 + 109}{2} = 100$.
Step 3: Multiply the average by the number of terms to find the sum.
$\text{Sum} = 100 \times 10 = 1000$.
Now check if $1000$ is a perfect cube:
$1000 = 10^3$, so it is indeed a perfect cube. Therefore, the Assertion (A) is true.
Now let's evaluate the Reason (R):
The reason states that the sum of any set of consecutive odd numbers will always result in a perfect cube. However, this is not necessarily true.
For example, consider the sum of two consecutive odd numbers: $1 + 3 = 4$, which is $2^2$ (a perfect square, not a cube).
Thus, Reason (R) is false.
Hence, Assertion (A) is true, but Reason (R) is false.

Key Concept: Cubic Numbers: Successive Differences, Prime Factorisation

b) 48
[Solution Description]
First, calculate the cubes:
$7^3 = 343$, $8^3 = 512$, $9^3 = 729$, $10^3 = 1000$.
First differences:
$512 - 343 = 169$,
$729 - 512 = 217$,
$1000 - 729 = 271$.
Second differences:
$217 - 169 = 48$,
$271 - 217 = 54$.
The second-level differences are $48$ and $54$.

Key Concept: Value of Stabilized Difference

b) 6
[Solution Description]
According to the syllabus, after three levels of successive differences, the differences stabilize to a constant value of $6$.

Key Concept: Cubic Numbers: Successive Differences

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 compute the successive differences for perfect cubes (a cubic polynomial sequence):
1. **First-level differences**:
$8 - 1 = 7, \quad 27 - 8 = 19, \quad 64 - 27 = 37, \quad 125 - 64 = 61, \ldots$
2. **Second-level differences**:
$19 - 7 = 12, \quad 37 - 19 = 18, \quad 61 - 37 = 24, \ldots$
3. **Third-level differences**:
$18 - 12 = 6, \quad 24 - 18 = 6, \ldots$
The differences stabilize at the third level. The reason states a general property of polynomial sequences, where a degree $n$ polynomial stabilizes after $n$ levels of differences. Since perfect cubes are degree 3 polynomials, the assertion aligns with the reason.

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: Prime Factorisation of a Number and Its Cube

a) $2^6 \times 5^3 \times 7^3$
[Solution Description]
Given $N = 2^2 \times 5 \times 7$, then:
$N^3 = (2^2 \times 5 \times 7)^3 = 2^{6} \times 5^3 \times 7^3$.
Each exponent in the prime factorisation is multiplied by $3$.

Key Concept: Ancient Mathematical Terminology

b) Cube root
[Solution Description]
In ancient Sanskrit mathematics, the term *ghana* was used for the cube of a number or the solid cube figure, and *mula* means root (origin). Hence, *ghana-mula* refers to the cube root.

Key Concept: Origin of the Term 'Root'

c) Because it represented the basis or origin of a square/cube
[Solution Description]
The term *mula* was used because it signified the basis or origin of a square or cube, similar to how roots are the basis of plants. This terminology influenced Arabic (*jidhr*) and Latin (*radix*).

Key Concept: Historical Context

b) Due to its symbolic meaning of origin or base
[Solution Description]
The word *mula* in Sanskrit means "root of a plant," which symbolically represents the origin or basis. This metaphorical usage was extended to mathematical roots, as they represent the base value from which powers (like squares or cubes) are derived.

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: 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: Properties of cubes, Historical context

a) Varga-varga of any number has an even number of factors
[Solution Description]
We need to identify which statement contradicts known properties of cubes from historical context.
Analyzing each concept:
1) Cubes of odd numbers: Always odd (3³=27, 5³=125)
2) Perfect cubes ending with 8: Exist (2³=8, 12³=1728)
3) Two-digit cubes: Yes (4³=64, 5³=125)
4) Two-digit cubes becoming 7+ digits: Impossible (max 99³=970299 is 6 digits)
5) Factors of cubes: All have odd number of factors (as perfect squares)
Ancient texts would recognize all except the false claim about seven digit cubes from two-digit numbers.

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: 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: 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: Cube Roots and Perfect Cubes

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

Key Concept: Terminology

c) $varga-mula$
[Solution Description]
The Sanskrit term $varga-mula$ was used for square-root, where $mula$ means the root of a plant or origin. This term was later adapted into Arabic and Latin.

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: Mathematical Term Origin

d) Originated from the Sanskrit word $mula$.
[Solution Description]
The term ‘root’ comes from the Sanskrit word $mula$, meaning the root of a plant, basis, or origin, which was used for operations involving roots in ancient India.

Key Concept: Conceptual Understanding

c) Babylonians
[Solution Description]
The Babylonians were the first to compile lists of perfect squares and cubes on clay tablets, which were used for geometric calculations in land measurement and architecture.

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: Perfect cubes, Taxicab numbers

d) 1729
[Solution Description]
The smallest number that can be expressed as the sum of two positive cubes in two different ways is 1729. This was famously stated by Ramanujan to Hardy. The two representations are:
$1729 = 1^3 + 12^3$ and $1729 = 9^3 + 10^3$.

Key Concept: Cube Roots

c) 17
[Solution Description]
Observing the pattern of perfect cubes:
- $17^3 = 4913$.
This can be deduced by checking nearby integers since $16^3 = 4096$ and $18^3 = 5832$.

Key Concept: Famous Mathematicians

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 ($1729 = 1^3 + 12^3 = 9^3 + 10^3$). The reason explains Ramanujan's talent for recognizing numerical patterns, which is also true and directly explains why he identified this property of 1729. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

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: 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: Historical Terminology, Babylonians

c) To aid in land measurement and architectural design
[Solution Description]
The Babylonians created these lists to simplify calculations in practical applications like land measurement and architectural design. Knowing perfect squares and cubes allowed them to quickly find square roots or cube roots without solving complex problems repeatedly.

Key Concept: Perfect Cubes

a) The cube of any odd number is even.
[Solution Description]
Let us analyze each statement:
(i) The cube of an odd number is odd, not even. Hence this statement is false.
(ii) There are perfect cubes that end with 8, such as $\sqrt[3]{512} = 8$. So this is false.
(iii) True, e.g., $4^3 = 64$ is a 2-digit cube of a 1-digit number; $5^3 = 125$ is a 3-digit number.
(iv) False, since the cube of a 2-digit number (maximum 99) would be $99^3 = 970299$, which has 6 digits.
(v) True, because perfect squares have an odd number of factors, but cubes may not necessarily have an odd number unless they are also perfect squares.
Therefore, the false statements are (i), (ii), and (iv). Among the options provided, only one is given as the main answer.

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: Varga and Ghana Definitions

b) The square figure or its area and the square power of a number
[Solution Description]
According to Aryabhata, the term *varga* refers to both the square figure or its area and the square power of a number.

Key Concept: Assertion-Reason

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