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 Ending with Specific Digits

d) 22
[Solution Description]
A number ending with 2, 3, 7, or 8 cannot be a perfect square. Among the options, 22 ends with 2, so it cannot be a perfect square.

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: 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: Sum of Consecutive Odd Numbers

b) 25
[Solution Description]
The sum of the first five consecutive odd numbers is calculated as follows: $1 + 3 + 5 + 7 + 9 = 25$. We know that $5^2 = 25$, so the result is the perfect square 25.

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: 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]
To verify the assertion, we check if 625 is a perfect square. Calculating $25 \times 25 = 625$, so 625 is indeed a perfect square. The reason states that all numbers ending with 5 are perfect squares, which is false. For example, 35 ends with 5 but is not a perfect square since $\sqrt{35}$ is not an integer. Therefore, the assertion is true, but the reason is false.

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: 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, Units Digit Analysis

b) Its tens digit must be odd
[Solution Description]
Perfect squares ending with 6 must have the tens digit as an odd number. Examples:
- $4^2 = 16$ (tens digit = 1, odd)
- $6^2 = 36$ (tens digit = 3, odd)
- $14^2 = 196$ (tens digit = 9, odd)
- $16^2 = 256$ (tens digit = 5, odd)
- $24^2 = 576$ (tens digit = 7, odd)
This pattern shows that the tens digit must be odd. The correct option is the one stating this condition.

Key Concept: Perfect Squares and Odd Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The sum of the first 5 odd numbers (1, 3, 5, 7, 9) is calculated as $1 + 3 + 5 + 7 + 9 = 25$. This matches the assertion since $5^2 = 25$. The reason states that every square can be expressed as the sum of consecutive odd numbers starting from 1, which is a property of perfect squares and explains why the sum is equal to the square of the count of odd numbers. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Sum of Consecutive Odd Numbers

b) 25, representing $5^2$
[Solution Description]
The sum of the first $n$ consecutive odd numbers starting from $1$ is equal to $n^2$. For $n = 5$, the sum is $1 + 3 + 5 + 7 + 9 = 25$, which represents the perfect square $5^2$.

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: 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: 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: 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: 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: 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: 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: 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 Ending with 6, Sum of Odd Numbers

a) 36
[Solution Description]
To determine which number is a perfect square ending with 6:
1) Check each option to see if it ends with 6.
2) Verify if the number can be expressed as the sum of consecutive odd numbers starting from 1.
- For 36: $1 + 3 + 5 + 7 + 9 + 11 = 36$ (valid).
- For 26: Ends with 6 but cannot be expressed as the sum of consecutive odd numbers starting from 1.
- For 46: Ends with 6 but fails the sum test.
- For 56: Ends with 6 but fails the sum test.
Thus, only 36 satisfies both conditions.

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: 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: 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: 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, 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: Sum of Consecutive Triangular Numbers

b) 49
[Solution Description]
The 6th triangular number is $T_6 = 1 + 2 + 3 + \cdots + 6 = 21$.
The 7th triangular number is $T_7 = 1 + 2 + 3 + \cdots + 7 = 28$.
Sum: $T_6 + T_7 = 21 + 28 = 49$.
Since $49 = 7^2$, it is a perfect square.

Key Concept: Triangular Numbers and Perfect Squares

b) $T_4$ and $T_5$
[Solution Description]
Triangular numbers follow the pattern $T_n = \frac{n(n+1)}{2}$. The sum of consecutive triangular numbers $T_{n-1} + T_n = n^2$. For example, $T_2 + T_3 = 3 + 6 = 9 = 3^2$. Thus, the correct pair must satisfy this relationship.

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: 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: 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: Properties of Perfect Squares

d) 2
[Solution Description]
Perfect squares can only end with 0, 1, 4, 5, 6, or 9. Among the given options, 2 is not one of these digits. Hence, 2 cannot be the units digit of 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: 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 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: 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: 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: 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: 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: Perfect Cubes

b) 216
[Solution Description]
To identify a perfect cube, we need to check if the number can be expressed as $n^3$ where $n$ is an integer. Let's evaluate each option:
- For 144: The prime factorization is $12 \times 12 = 144$, which cannot be split into three identical groups.
- For 216: $6^3 = 216$ and its prime factors are $(2 \times 3)^3 = 2^3 \times 3^3$, which forms three identical groups.
- For 99: The prime factors are $9 \times 11$, which do not form three identical groups.
- For 200: The prime factorization is $8 \times 25 = 200$, which does not split into three identical groups.
Thus, the correct answer is the perfect cube 216.

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: 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: 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: 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

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: 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: 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: 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, 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 and known as the Hardy-Ramanujan number or taxicab number. The Reason explains that no smaller number satisfies this condition, which is also true. Furthermore, the Reason correctly justifies why 1729 holds this unique property. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Cubes of Fractions/Decimals, Numerical Accuracy

c) 0.421875
[Solution Description]
To compute $\left(\frac{3}{4}\right)^3$, multiply the numerator and denominator three times:
$\left(\frac{3}{4}\right)^3 = \frac{3^3}{4^3} = \frac{27}{64}.$
Converting to decimal form gives $0.421875$.

Key Concept: Taxicab Number

c) 1729
[Solution Description]
A taxicab number is expressible as the sum of two cubes in two different ways. The smallest known example is 1729:
$1729 = 1^3 + 12^3 = 9^3 + 10^3$
Among the options, 1729 fits this property.

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: 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: 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 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: Cube Root Calculation

c) 5
[Solution Description]
To find the cube root of 125, we look for a number that, when multiplied by itself three times, gives 125. We know that $5 \times 5 \times 5 = 125$. Therefore, $\sqrt[3]{125} = 5$.

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: 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

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: 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: Taxicab Numbers, Prime Factorization

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that 4104 is a taxicab number, which is correct as shown in the syllabus ($4104 = 2^3 + 16^3 = 9^3 + 15^3$).
The reason (R) claims that for any taxicab number, the prime factorization of its cube must have each prime factor appearing exactly three times. While this is true for any number's cube (as per the syllabus example: $12^3 = 2^3 \times 2^3 \times 3^3$), this property is not unique to taxicab numbers—it applies to all cubes. Therefore, while both statements are true, the reason does not explain why 4104 is a taxicab number; it merely states a general property of cubes.
Thus, the correct option is (b).

Key Concept: Taxicab Numbers, Sum of Two Cubes

a) 1729
[Solution Description]
The smallest such $k$ is 1729, which has two pairs: $(1, 12)$ and $(9, 10)$. This makes 1729 the smallest taxicab number.

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: 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) $31 + 33 + 35 + 37 + 39 + 41$
[Solution Description]
To find the sequence of consecutive odd numbers whose sum is $216$, we refer to the pattern given in the syllabus. The sum of $n$ consecutive odd numbers starting from the appropriate term will be $n^3$. Here, $216 = 6^3$, so $n = 6$.
The first term of the sequence can be found using the formula for the first term of the $k^{th}$ group of consecutive odd numbers:
$\text{First term} = k^2 - k + 1$
For $k = 6$:
$\text{First term} = 6^2 - 6 + 1 = 36 - 6 + 1 = 31$
Thus, the sequence is $31, 33, 35, 37, 39, 41$.

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: 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: 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: 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: 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: Stabilized Differences for Cubic Numbers

c) 3
[Solution Description]
From the syllabus, we observe that for perfect cubes, the differences stabilize at the third level. At this level, all differences become $6$.

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: Aryabhata's Definition

c) Both the square figure/its area and the square of a number
[Solution Description]
Aryabhata stated that the term *varga* refers both to a square figure with four equal sides and the area of such a figure. Additionally, it denotes the product of two equal quantities, i.e., the square of a number.

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: 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: 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: 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: 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: 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: 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 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: Cube roots, Perfect cubes, Sanskrit terminology

c) ghana-mula of 10648 is 22
[Solution Description]
The problem involves finding the cube root of 10648 and understanding its representation in Sanskrit terms.
Step 1: Find prime factorization of 10648.
$10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 = 2^3 \times 11^3$
Step 2: Calculate cube root:
$\sqrt[3]{10648} = 2 \times 11 = 22$
Step 3: In Sanskrit terminology:
- Cube is called "ghana"
- Cube root is called "ghana-mula"
Thus, 22 is the ghana-mula of 10648.

Key Concept: Concept: Historical use of mathematical terms in India

c) Varga
[Solution Description]
The term $varga$ was used in ancient Sanskrit works for both the square figure or its area and the square power. Aryabhata (499 CE) explicitly mentions this usage.

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: Historical applications, Perfect squares

c) Land measurement and architecture
[Solution Description]
The Babylonian clay tablets containing lists of perfect squares were primarily used for land measurement calculations and architectural designs. Having pre-computed square values allowed quick determination of areas and dimensions needed in construction projects and land surveys, which were crucial for their advanced civilization.

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: 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: 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: 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 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: Historical Context

d) Srinivasa Ramanujan
[Solution Description]
Srinivasa Ramanujan identified 1729 as the smallest number expressible as the sum of two cubes in two different ways:
$1729 = 1^3 + 12^3 = 9^3 + 10^3$
This discovery was made during his conversation with G.H. Hardy.

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: 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: Cube Roots

c) 30
[Solution Description]
To find the cube root of 27000, we can perform prime factorization:
$27000 = 27 \times 1000 = 3^3 \times 10^3 = (3 \times 10)^3$. Therefore, $\sqrt[3]{27000} = 30$.

Key Concept: Perfect Cubes, Historical Terminology

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that $\textit{varga-mula}$ was used to denote the square root due to its connection with the origin of squares. This is correct as per historical records where $\textit{mula}$ (root) was used for the mathematical operation of taking roots. The Reason states that Aryabhata introduced $\textit{varga}$ for square numbers and the squaring operation, which is also true based on his work. However, the Reason does not explain why $\textit{varga-mula}$ denotes the square root; it only provides information about $\textit{varga}.$ Thus, both statements are true, but Reason is not the correct explanation of Assertion.

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: 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.