Key Concept: Square Pairs, Adjacent Sums

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

Key Concept: 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 - 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: 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: 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: 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: 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: Prime Factorisation and Perfect Squares

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

Key Concept: Perfect Squares, 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: 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: 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: Difference Between Consecutive Squares, Sum of Consecutive Odd Numbers

c) 81
[Solution Description]
Let the two consecutive integers be $n$ and $n+1$. The difference between their squares is $(n+1)^2 - n^2 = 81$. Expanding $(n+1)^2$, we get $n^2 + 2n + 1 - n^2 = 81$, which simplifies to $2n + 1 = 81$. Solving for $n$, we get $2n = 80$ and $n = 40$. The two consecutive integers are 40 and 41, so their sum is $40 + 41 = 81$.

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: 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: Sum of Consecutive Odd Numbers

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

Key Concept: Sum of Consecutive Odd Numbers

c) 1, 3, 5, 7, 9
[Solution Description]
A perfect square can be expressed as the sum of consecutive odd numbers starting from 1. Let's evaluate the options:
- Option 1: 5 + 7 + 9 = 21 (Not a perfect square)
- Option 2: 11 + 13 + 15 = 39 (Not a perfect square)
- Option 3: 1 + 3 + 5 + 7 + 9 = 25 (Which is $5^2$, a perfect square)
- Option 4: 3 + 5 + 7 + 9 = 24 (Not a perfect square)
Hence, the correct set is 1, 3, 5, 7, 9.

Key Concept: Zeros in Perfect Squares

d) 6
[Solution Description]
According to the properties, if a number has 3 zeros at the end, its square will have 6 zeros. Since 5000 has 3 zeros ($5000 = 5 \times 10^3$), squaring it gives $(5 \times 10^3)^2 = 25 \times 10^6 = 25,000,000$. Thus, the square of 5000 has 6 zeros at the end.

Key Concept: 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: 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: Units Place Property

c) 198
[Solution Description]
Perfect squares can only end with digits 0, 1, 4, 5, 6, or 9. The number ending with 8 is not a perfect square.

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 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: 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, 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: 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: Perfect Squares

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

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 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: 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 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: Square Roots and Perfect Squares

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

Key Concept: Square 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 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 Roots and Perfect Squares

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

Key Concept: Square 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: 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: Prime Factorisation, Perfect Cubes

a) 216
[Solution Description]
To determine if a number is a perfect cube, its prime factors must form identical triplets. Let's analyze the options:
Option A: 216 = $2^3 \times 3^3$ → Perfect cube ($6^3$).
Option B: 500 = $2^2 \times 5^3$ → Not a perfect cube (exponent of 2 is not a multiple of 3).
Option C: 1000 = $2^3 \times 5^3$ → Perfect cube ($10^3$).
Option D: 800 = $2^5 \times 5^2$ → Not a perfect cube (exponents are not multiples of 3).
Thus, both A and C are correct, but since only one option can be chosen, we select the smallest valid choice.

Key Concept: 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: 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 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: 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: 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: Perfect Cubes, Prime Factorisation

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

Key Concept: Prime Factorisation

c) 8
[Solution Description]
A number is a perfect cube if its prime factors can be grouped into triplets. For example, 8 can be written as $2 \times 2 \times 2 = 2^3$, making it a perfect cube.

Key Concept: 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: Prime Factorisation of Cubes

c) $2^3 \times 3^3$
[Solution Description]
The cube of a number has each prime factor raised to three times its original power. For $6 = 2 \times 3$, its cube is:
$6^3 = (2 \times 3)^3 = 2^3 \times 3^3$

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: 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: 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: 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: 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: Cube Roots, Estimation

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

Key Concept: Cube 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: 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: Perfect Cube Verification

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

Key Concept: 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, 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: Square Pair Arrangement, Combinatorics

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

Key Concept: Taxicab Numbers

c) 3
[Solution Description]
There are three known taxicab numbers below 20000: 1729, 4104, and 13832.

Key Concept: Taxicab Numbers

c) 4104
[Solution Description]
After 1729, the next taxicab numbers are 4104 and 13832. Here, 4104 is one such taxicab number. It can be expressed as $2^3 + 16^3 = 4104$ and $9^3 + 15^3 = 4104$.

Key Concept: Taxicab Numbers

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

Key Concept: Cube Roots

c) 9
[Solution Description]
We know that $9 \times 9 \times 9 = 729$. Therefore, $\sqrt[3]{729} = 9$.

Key Concept: Perfect Cubes via Prime Factorisation

b) Yes, because it can be written as $6^3$.
[Solution Description]
Prime factorize the number 216:
$216 = 6 \times 6 \times 6 = (2 \times 3) \times (2 \times 3) \times (2 \times 3) = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$
Since the factors can be grouped into three identical groups, 216 is a perfect cube.

Key Concept: Perfect Cubes and Consecutive Odd Numbers

b) 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

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

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: 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: Successive Differences for Cubic Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To solve this, let's compute the successive differences for a simple sequence of perfect cubes: $1, 8, 27, 64$.
Level 1 differences: $8 - 1 = 7$, $27 - 8 = 19$, $64 - 27 = 37$.
Level 2 differences: $19 - 7 = 12$, $37 - 19 = 18$.
Level 3 differences: $18 - 12 = 6$.
The assertion states that the stabilized difference is 6, which matches our calculation. The reason explains why this stabilization occurs due to the cubic nature of the sequence. Since the reason correctly explains the assertion, both are true and connected logically.

Key Concept: Successive Differences for Cubic Numbers

c) 7
[Solution Description]
The first level difference is calculated by subtracting the previous term from the next term. Here, subtract $1$ from $8$: $8 - 1 = 7$.

Key Concept: 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: 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: 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 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: Origin of the Term 'Root'

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

Key Concept: 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: 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: Perfect cubes, Multiplication factors

c) 36, ghana
[Solution Description]
To find the smallest multiplier to make 8232 a perfect cube and identify its correct Sanskrit terminology.
Step 1: Factorize 8232:
$8232 = 2^3 \times 3 \times 7^3 \times 2 = 2^4 \times 3^1 \times 7^3$
Step 2: For perfect cube, all exponents must be multiples of 3:
- Need 2 more 2's (to make exponent 6)
- Need 2 more 3's (to make exponent 3)
- 7's exponent is already 3
Step 3: Multiplier needed:
$2^2 \times 3^2 = 4 \times 9 = 36$
Step 4: Brahmagupta would call the perfect cube "krti" (square) or "ghana" (cube), but since it's a perfect cube, "ghana" is correct.

Key Concept: Perfect Cubes

b) 7
[Solution Description] First, factorize 1323:
$1323 = 3^3 \times 7^2$
To make it a perfect cube, the exponents of all prime factors must be multiples of 3. Here, 7 has an exponent of 2. We need to multiply by 7 to get $3^3 \times 7^3 = (3 \times 7)^3 = 21^3$. Hence, the required number is 7.

Key Concept: 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: 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: Indian Contributions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that *varga-varga* was used for the fourth power, which is correct as per the syllabus. The reason explains that *varga* means square, and *varga-varga* is derived by applying the square operation twice (square of square), which correctly explains why it represents the fourth power. Thus, both are true, and the reason correctly explains the assertion.

Key Concept: 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: Concept: Early compilation of perfect squares and cubes

d) Babylonians
[Solution Description]
The Babylonians compiled the earliest known lists of perfect squares and cubes on clay tablets around 1700 BCE, aiding in land measurement and architectural design.

Key Concept: Conceptual Understanding

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

Key Concept: Historical Context

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

Key Concept: Perfect Cubes 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: 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: 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: 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: 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: 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: 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.

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: 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., $(x^2)^2 = x^4$).