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: 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: 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: 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: 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, 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) 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: Prime Factorisation, Estimating Square Roots

c) 6
[Solution Description]
We first list the squares of numbers from 21 to 29:
- $21^2 = 441$ → Prime factors: $3^2 \times 7^2$ → Sum of exponents = $2 + 2 = 4$
- $22^2 = 484$ → Prime factors: $2^2 \times 11^2$ → Sum of exponents = $2 + 2 = 4$
- $23^2 = 529$ → Prime factors: $23^2$ → Sum of exponents = $2$
- $24^2 = 576$ → Prime factors: $2^6 \times 3^2$ → Sum of exponents = $6 + 2 = 8$
- $25^2 = 625$ → Prime factors: $5^4$ → Sum of exponents = $4$
- $26^2 = 676$ → Prime factors: $2^2 \times 13^2$ → Sum of exponents = $2 + 2 = 4$
- $27^2 = 729$ → Prime factors: $3^6$ → Sum of exponents = $6$
- $28^2 = 784$ → Prime factors: $2^4 \times 7^2$ → Sum of exponents = $4 + 2 = 6$
- $29^2 = 841$ → Prime factors: $29^2$ → Sum of exponents = $2$
From above, possible sums of exponents are: 2, 4, 6, 8. The correct option is 6.

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 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]
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: 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: 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: 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: Difference Between Consecutive Squares

b) 49
[Solution Description]
The difference between consecutive squares forms a sequence of odd numbers: $3, 5, 7, 9, \dots$. Given the difference is $15$, we can find the smaller square by solving $2n + 1 = 15 \Rightarrow n = 7$. Thus, the smaller square is $7^2 = 49$.

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: 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: 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: 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: Sum of Odd Numbers

c) 25
[Solution Description]
Sum of first $n$ odd natural numbers equals $n^2$. Here, $n = 5$, so sum $= 5^2 = 25$.

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: 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: Sum of First n Odd Numbers

c) 13
[Solution Description]
The sum of the first $n$ odd numbers is $n^2$. Given that the sum is 169:
$n^2 = 169 \Rightarrow n = \sqrt{169} = 13$.
Therefore, the value of $k$ is 13.

Key Concept: Perfect Squares and Odd Numbers, Sum of Consecutive Odd Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify if 144 is a perfect square, we check if it can be expressed as the sum of consecutive odd numbers starting from 1. The sum of the first $n$ odd numbers is given by $n^2$. For 144:
$n^2 = 144 \implies n = 12$
Now, let’s verify by adding the first 12 odd numbers:
$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 144$
Thus, the Assertion is true. The Reason states that the sum of the first $n$ odd numbers equals $n^2$, which is also true. Moreover, the Reason correctly explains why 144 is a perfect square, as it follows this property.

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: 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: 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: 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 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: 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: 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: 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 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 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: Square Numbers, Square Roots

c) Assertion is true, but Reason is false.
[Solution Description]
To determine if 1125 is a perfect square, we first analyze its prime factorization:
$1125 = 3^2 \times 5^3$. For a number to be a perfect square, all exponents in its prime factorization must be even. Here, the exponent of 5 is 3 (odd), so 1125 is not a perfect square. However, the Reason is incorrect because a perfect square ending with 5 does not necessarily require an even number of zeros in its prime factorization. Instead, the correct condition is that all exponents in its prime factorization must be even. Thus, Assertion is true, but Reason is false.

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 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: 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: 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: 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: 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: 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: Perfect Cubes

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: Prime Factorization of Cubes

d) $2^3 \times 3^3 \times 5^3$
[Solution Description]
Given the prime factorization of a number as $2 \times 3 \times 5$, its cube would have each prime factor raised to the power of 3.
Step 1: Cube each prime factor individually. $(2 \times 3 \times 5)^3 = 2^3 \times 3^3 \times 5^3$.
Step 2: Thus, the prime factorization of the cube is $2^3 \times 3^3 \times 5^3$.

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: 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, 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: 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: 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

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: 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: 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: 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: 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 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: 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: 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

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 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

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that 1729 can be expressed as the sum of two cubes in exactly two different ways, which is true as per the given examples: $1729 = 1^3 + 12^3$ and $1729 = 9^3 + 10^3$. The reason states that 1729 is the smallest number satisfying this condition, which is also true as it is known as the Hardy-Ramanujan Number. Moreover, the reason correctly explains the assertion because being the smallest such number makes it significant as the first taxicab number.

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: 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: 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: 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: 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: Perfect Cubes and Consecutive Odd Numbers

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion is false because the sum of the first $n$ consecutive odd numbers is equal to $n^2$, not $n^3$. For example, $1 + 3 = 4 = 2^2$. The Reason is also false because the sum of any 10 consecutive odd numbers depends on the starting number and is not necessarily a perfect cube. For instance, $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100 = 10^2$, which is not a perfect cube. Thus, both Assertion and Reason are false.

Key Concept: Cube Roots

c) 8
[Solution Description]
To find the cube root of 512, we can use prime factorization.
$512 = 8 \times 64 = (2^3) \times (4^3) = (2 \times 4)^3 = 8^3$
Therefore, $\sqrt[3]{512} = 8$.

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: 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: 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: 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: Cubic Numbers: Successive Differences

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for perfect cubes, the third differences are always constant. This is true as demonstrated in the syllabus example where the third differences were all 6 for the sequence $1, 8, 27, 64, 125, 216, \ldots$.
The reason claims that this property is unique to cubic sequences and helps identify them. This is also true because, unlike squares or higher powers, only cubic numbers exhibit this specific pattern of constant third differences.
Since both the assertion and reason are true, and the reason correctly explains why perfect cubes have constant third differences, the correct option is the one where both are true and the reason is the correct explanation.

Key Concept: Historical Terminology in Mathematics

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 for both the square figure and its area in ancient Sanskrit works, which is true as per the historical context provided. The reason explains that this usage originated from the graphical representation of a square figure, which is also true and directly explains why $varga$ was used for square power. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Babylonian mathematics, Historical context

b) Determining right angles in construction
[Solution Description]
From the syllabus:
- Babylonians compiled lists of perfect squares/cubes
- Used these for quick calculations of roots
- Applied in land measurement and architecture
The number 16 is a perfect square ($4^2$), so it would be used as a reference for square root calculations, particularly in practical applications like determining side lengths from areas.

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 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: 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: 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: Definition of *varga*

b) The square of a number or its graphical representation
[Solution Description]
The term *varga* was used in ancient Sanskrit works to denote both the square figure or its area and the square power of a number. Aryabhata also mentions that the product of two equal quantities is called *varga*. Thus, *varga* refers to the concept of squaring a number.

Key Concept: Properties of Cubes

d) All of the above are false.
[Solution Description] Let's evaluate each statement:
(i) The cube of any odd number is even - False, as the cube of an odd number remains odd.
(ii) There is no perfect cube that ends with 8 - False, as $2^3 = 8$ and $12^3 = 1728$, which end with 8.
(iii) The cube of a 2-digit number may be a 3-digit number - True, e.g., $4^3 = 64$ and $5^3 = 125$.
(iv) The cube of a 2-digit number may have seven or more digits - False, as the largest 2-digit number is 99 and $99^3 = 970299$, a 6-digit number.
(v) Cube numbers have an odd number of factors - True, as they are perfect squares of their cube roots.
The false statements are (i) and (ii).

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: 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: 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: 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: 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: 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: 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: 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: Perfect Cubes, Taxicab Numbers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To verify:
1. Calculate $15^3 + 9^3 = 3375 + 729 = 4104$.
2. Calculate $16^3 + 2^3 = 4096 + 8 = 4104$.
However, there exists another pair: $12^3 + (-6)^3 = 1728 + (-216) = 1512 \neq 4104$. This shows that the Reason provides the correct pairs for 4104, but the explanation is incomplete because it does not consider all possible pairs, including negative integers. Nonetheless, the Assertion is true, and the Reason correctly identifies two valid pairs for 4104, although it's not a complete explanation.

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: Taxicab Numbers

b) 1729
[Solution Description]
The smallest number expressible as the sum of two positive cubes in two distinct ways is 1729, known as the Taxicab Number. It can be written as:
$1729 = 1^3 + 12^3 = 9^3 + 10^3$

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: 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: 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: Basic Concept: Mula Meaning

d) Because it means the root of a plant, basis, cause, or origin
[Solution Description]
The word *mula*, meaning the root of a plant, was used in ancient India to represent the basis or origin of square and cube roots, as it signifies the foundation or source.

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: 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.