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: 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: Units Digit Patterns

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

Key Concept: Square Numbers - Properties of Perfect Squares

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

Key Concept: 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, Prime Factorisation

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

Key Concept: Perfect Squares and Units Place

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

Key Concept: Estimating Square Roots

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

Key Concept: Patterns in Perfect Squares, Odd Number Summation

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

Key Concept: Perfect Squares 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: 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: 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: 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: 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: Patterns in Units Place

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

Key Concept: Square Numbers, 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: 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: Difference Between Consecutive Squares

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

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

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

Key Concept: 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: 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: 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: 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: 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: 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: Perfect Squares and Odd Numbers

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

Key Concept: Square 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 Squares

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

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

Key Concept: 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 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: Estimating Square Roots

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

Key Concept: 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) 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: 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: 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: Cube Roots

b) 2
[Solution Description]
The cube root of a number is a value that, when multiplied by itself three times, gives the original number. So, to find $\sqrt[3]{8}$, we look for a number such that $x \times x \times x = 8$. Since $2 \times 2 \times 2 = 8$, the cube root of 8 is 2.

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: 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: 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, 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: 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: 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: 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: Last Digit Pattern in Cubes

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

Key Concept: Properties of Cube Numbers

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

Key Concept: 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: Cube of a Number

c) 343
[Solution Description]
The cube of 7 is calculated as:
$7^3 = 7 \times 7 \times 7 = 343$
Therefore, $7^3 = 343$.

Key Concept: Cube Root Calculation

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

Key Concept: 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 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 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, 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

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, 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

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

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: 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: Prime Factorisation

c) Yes, and its cube root is 7.
[Solution Description]
Let us check the prime factorisation of 343:
$343 = 7 \times 7 \times 7 = 7^3$.
Since the factors can be grouped into three identical groups, 343 is a perfect cube, and its cube root is $7$.

Key Concept: Perfect Cubes and Consecutive Odd Numbers

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

Key Concept: 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: 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: Third Differences in Cubic Numbers

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

Key Concept: Second Differences in Cubic Numbers

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

Key Concept: Value of Stabilized Difference

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

Key Concept: 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: 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: 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: 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: 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: 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: 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: 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: Historical terms in mathematics, Square roots, Varga-mula

b) Basis of the square power
[Solution Description]
The term 'varga-mula' was used for square root in ancient India. It literally means "the basis/origin of the square" ('varga' = square, 'mula' = root/basis). This relates to finding the original number that was squared to obtain a given value. Aryabhata and Brahmagupta explained this concept where mula represents the root operation.

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: 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 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: Concept: Origin of the term 'root' in mathematics

d) It originated from the Sanskrit word $mula$, meaning root
[Solution Description]
In ancient India, the Sanskrit word $mula$, meaning root of a plant, basis, cause, or origin, was used for taking roots mathematically. Terms like $varga-mula$ (square-root) and $ghana-mula$ (cube-root) originated from this.

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: 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: Cube Roots

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

Key Concept: 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: 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: 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: 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: 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: 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: 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.