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: 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: 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 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: 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: 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: 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: Prime Factorisation and Perfect Squares

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

Key Concept: 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: Difference Between Consecutive Squares

c) 9
[Solution Description]
The next perfect square after $16$ ($4^2$) is $25$ ($5^2$). The difference is $25 - 16 = 9$, which matches one of the options.

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: 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: 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: 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: 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: 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: 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: 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: Perfect Square Check

d) 25
[Solution Description]
A number is a perfect square if it can be expressed as the sum of consecutive odd numbers starting from 1.
For option 25:
25 - 1 = 24,
24 - 3 = 21,
21 - 5 = 16,
16 - 7 = 9,
9 - 9 = 0.
Since subtraction results in zero after five steps, $25$ is a perfect square ($5^2$).

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: 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

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

Key Concept: Perfect Squares and Odd Numbers

c) 36
[Solution Description]
According to the concept, a number is a perfect square if it can be expressed as the sum of consecutive odd numbers starting from 1. Let’s verify each option:
1. For 49: Subtract successive odd numbers until zero is obtained.
$49 - 1 = 48$, $48 - 3 = 45$, $45 - 5 = 40$, $40 - 7 = 33$, $33 - 9 = 24$, $24 - 11 = 13$, $13 - 13 = 0$. Since it results in zero, 49 is a perfect square ($7^2$).
2. For 30: $30 - 1 = 29$, $29 - 3 = 26$, $26 - 5 = 21$, $21 - 7 = 14$, $14 - 9 = 5$, $5 - 11 = -6$. Doesn’t result in zero, so not a perfect square.
3. For 36: $36 - 1 = 35$, $35 - 3 = 32$, $32 - 5 = 27$, $27 - 7 = 20$, $20 - 9 = 11$, $11 - 11 = 0$. It results in zero, so 36 is a perfect square ($6^2$).
4. For 50: $50 - 1 = 49$, $49 - 3 = 46$, $46 - 5 = 41$, $41 - 7 = 34$, $34 - 9 = 25$, $25 - 11 = 14$, $14 - 13 = 1$, $1 - 15 = -14$. Doesn’t result in zero, so not a perfect square.
Thus, the correct answer is 36.

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 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: 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: 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: Perfect Squares and Triangular Numbers

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

Key Concept: 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 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: 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: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the square root of 144 is 12, which is correct since $12^2 = 144$. The reason given is $12 \times 12 = 144$, which is also true and correctly explains why 12 is the square root of 144.

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: 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: 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: 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: 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 and Prime Factorization

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

Key Concept: Prime Factorisation of Cubes

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

Key Concept: 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: 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: 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: 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: 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: Cube Roots

c) 6
[Solution Description]
The cube root of 216 is a number that, when multiplied by itself three times, gives 216. Since $6 \times 6 \times 6 = 216$, $\sqrt[3]{216} = 6$.

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: Cubic Numbers: Properties of Cube Numbers

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

Key Concept: Properties of Cube Numbers

b) 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: Taxicab Numbers, Perfect Cubes

b) 1729
[Solution Description]
The smallest taxicab number is 1729, which can be written as:
$1729 = 1^3 + 12^3 = 9^3 + 10^3.$
The next taxicab numbers are 4104 and 13832, but 1729 is the most well-known example.

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 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: 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: 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 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, Taxicab Numbers

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

Key Concept: Cube 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 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

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

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: 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, 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) 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

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 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the Babylonians compiled lists of perfect squares and cubes, which is historically accurate as evidenced by ancient clay tablets dating back to 1700 BCE. The reason provided explains the practical application of these lists in solving problems related to land measurement and architectural design, which aligns with historical records. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Babylonian Lists

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

Key Concept: 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 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: 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: 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: 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: 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: 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: 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: 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, 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: 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 $x$).