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: Perfect Squares, Sum of Odd Numbers

d) 50
[Solution Description]
To determine which number is not a perfect square, we check each option by verifying if it can be written as the sum of consecutive odd numbers starting from 1 or by confirming its non-membership in the list of perfect squares.
- $1 + 3 + 5 = 9$ (valid, so 9 is a perfect square).
- $1 + 3 + 5 + 7 = 16$ (valid, so 16 is a perfect square).
- The sum $1 + 3 + 5 + 7 + 9 + 11 + 13 = 49$ (valid, so 49 is a perfect square).
- The number 50 cannot be expressed as the sum of consecutive odd numbers starting from 1, nor is it listed among perfect squares.

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: 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: 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 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]
The assertion states that 49 is a perfect square, which is true since $7^2 = 49$. The reason claims that all numbers ending with the digit 9 are perfect squares. However, this is false because although some numbers ending with 9 are perfect squares (like 49), others like 19, 29, etc., are not perfect squares. Hence, the assertion is true but the reason is false.

Key Concept: Perfect Squares, Units Digit Analysis

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

Key Concept: Prime Factorisation, Estimating Square Roots

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

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

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

Key Concept: Perfect Squares and Odd Numbers, 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, Verification Method

a) 49
[Solution Description]
We apply the verification method to each option:
1) For 49: $49 - 1 = 48$, $48 - 3 = 45$, $45 - 5 = 40$, $40 - 7 = 33$, $33 - 9 = 24$, $24 - 11 = 13$, $13 - 13 = 0$. Since it reaches zero, 49 is a perfect square.
2) For 50: The process does not reach zero.
3) For 64: It also reaches zero after successive subtractions.
4) For 72: The process does not reach zero.
Both 49 and 64 are perfect squares, but since only one option is correct, we choose 49 as the primary example.

Key Concept: 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: 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 Units Digit

c) 7
[Solution Description]
A perfect square can only end with 0, 1, 4, 5, 6, or 9 in its units place. Thus, 7 is not a possible units digit for a perfect square.

Key Concept: 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: 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: 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: Perfect Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that 144 is a perfect square, which is true since $12 \times 12 = 144$. The reason states that a perfect square can be expressed as the sum of consecutive odd numbers starting from 1, which is also true. For example, $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 144$. Moreover, the reason correctly explains why 144 is a perfect square by illustrating the property of perfect squares being sums of consecutive odd numbers.

Key Concept: 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: 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: 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: Square Root Properties

d) 9
[Solution Description]
The square root of 81 can be both $+9$ and $-9$ as $9^2 = 81$ and $(-9)^2 = 81$. However, in this context, the syllabus specifies considering only the positive square root. Therefore, $\sqrt{81} = 9$.

Key Concept: Square Roots and Estimation

c) 55
[Solution Description]
To estimate $\sqrt{3025}$:
Step 1: Identify the range. We know $50^2 = 2500$ and $60^2 = 3600$, so $50 < \sqrt{3025} < 60$. Step 2: Check the last digit. 3025 ends with 5, so $\sqrt{3025}$ must end with 5 (since $5^2 = 25$). So possible candidates are 55. Step 3: Verify $55^2 = (50 + 5)^2 = 50^2 + 2 \times 50 \times 5 + 5^2 = 2500 + 500 + 25 = 3025$. Thus, $\sqrt{3025} = 55$.

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: 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 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 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: 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

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, 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

c) 5
[Solution Description]
To find the cube root of 125, we need to find a number that when multiplied by itself three times gives 125. We can break it down:
Step 1: Find the prime factorization of 125. $125 = 5 \times 5 \times 5 = 5^3$.
Step 2: The cube root of $5^3$ is $\sqrt[3]{5^3} = 5$.
Therefore, the cube root of 125 is 5.

Key Concept: Perfect Cubes and Prime Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if 216 is a perfect cube, we examine its prime factorization. Let's break it down step by step:
- Prime factorization of 216:
$216 = 6 \times 6 \times 6 = (2 \times 3)^3 = 2^3 \times 3^3$.
- Since the exponents of all prime factors (2 and 3) are multiples of 3, 216 is indeed a perfect cube.
- The assertion (A) states that 216 is a perfect cube, which is true.
- The reason (R) provides the correct prime factorization of 216, which also justifies why it is a perfect cube. Therefore, the reason correctly explains the assertion.

Key Concept: Perfect Cubes

c) 64
[Solution Description]
To find the cube of 4, multiply 4 by itself three times: $4 \times 4 \times 4 = 64$. Thus, the cube of 4 is 64.

Key Concept: Prime 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: 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: 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 Numbers, Perfect Cubes

a) 1729
[Solution Description]
The smallest taxicab number is 1729, known as the Hardy-Ramanujan Number. It can be expressed as:
$1729 = 1^3 + 12^3 = 9^3 + 10^3$.
The next taxicab numbers are larger (e.g., 4104, 13832), so 1729 is the correct answer.

Key Concept: Cube Root

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: Cube Roots, Estimation

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

Key Concept: Perfect Cube Verification

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

Key Concept: Cube Root and Prime Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if the assertion is true, let's find the prime factorization of 2744:
$2744 \div 2 = 1372$
$1372 \div 2 = 686$
$686 \div 2 = 343$
$343 \div 7 = 49$
$49 \div 7 = 7$
$7 \div 7 = 1$
So, the prime factorization is $2^3 \times 7^3$. Each prime factor appears three times, meaning they can indeed be split into three identical groups ($2^3$ and $7^3$). Therefore, 2744 is a perfect cube. The reason correctly explains the assertion.

Key Concept: Cube Root Calculation

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

Key Concept: 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 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, 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

b) 2
[Solution Description]
There are exactly two pairs of positive integers $(a, b)$ that satisfy $a^3 + b^3 = 1729$. These are $(1, 12)$ and $(9, 10)$, since $1^3 + 12^3 = 1729$ and $9^3 + 10^3 = 1729$.

Key Concept: Taxicab Numbers

b) 1729
[Solution Description]
The smallest taxicab number is 1729, as famously mentioned by Srinivasa Ramanujan. It can be expressed as:
$1729 = 1^3 + 12^3$ and $1729 = 9^3 + 10^3$.

Key Concept: Perfect Cubes and Consecutive Odd Numbers

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

Key Concept: Taxicab Numbers

b) $(2, 16)$ and $(9, 15)$
[Solution Description]
The taxicab number $4104$ can be expressed as the sum of two cubes in two different ways:
$4104 = 2^3 + 16^3 = 9^3 + 15^3$
Calculating:
$2^3 + 16^3 = 8 + 4096 = 4104$
$9^3 + 15^3 = 729 + 3375 = 4104$
Therefore, the correct pairs are $(2, 16)$ and $(9, 15)$.

Key Concept: Cube Roots

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

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: 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: 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: Prime Factorisation of a Number and Its Cube

a) $2^6 \times 5^3 \times 7^3$
[Solution Description]
Given $N = 2^2 \times 5 \times 7$, then:
$N^3 = (2^2 \times 5 \times 7)^3 = 2^{6} \times 5^3 \times 7^3$.
Each exponent in the prime factorisation is multiplied by $3$.

Key Concept: 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

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason, let's compute the successive differences for perfect cubes (a cubic polynomial sequence):
1. **First-level differences**:
$8 - 1 = 7, \quad 27 - 8 = 19, \quad 64 - 27 = 37, \quad 125 - 64 = 61, \ldots$
2. **Second-level differences**:
$19 - 7 = 12, \quad 37 - 19 = 18, \quad 61 - 37 = 24, \ldots$
3. **Third-level differences**:
$18 - 12 = 6, \quad 24 - 18 = 6, \ldots$
The differences stabilize at the third level. The reason states a general property of polynomial sequences, where a degree $n$ polynomial stabilizes after $n$ levels of differences. Since perfect cubes are degree 3 polynomials, the assertion aligns with the reason.

Key Concept: 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: 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: 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: Mathematical History, Perfect Cubes

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that *varga-mula* refers to the cube root of a number. However, according to the historical context provided, *varga-mula* specifically refers to the square root (*varga* means square, *mula* means root). For cube roots, the term *ghana-mula* was used. Therefore, the Assertion is false. The Reason correctly explains that *mula* was indeed used metaphorically for roots in mathematical operations, but it does not justify the Assertion because *varga-mula* is not related to cube roots.

Key Concept: Etymology, Mathematical terminology evolution

b) Sanskrit "mula" → Arabic "jidhr" → Latin "radix" → English "root"
[Solution Description]
The syllabus explains:
1. Sanskrit "mula" (plant root) → mathematical root operations
2. This influenced Arabic "jidhr"
3. Which then influenced Latin "radix"
4. Finally evolving into English "root"
The path must show this progression through multiple languages while maintaining the plant root → mathematical root connection.

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: 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: Use of *mula*

b) Because it represents the origin or basis of the square or cube
[Solution Description]
In Sanskrit, *mula* means the root of a plant, basis, cause, or origin. This term was adopted for mathematical roots (*varga-mula* for square root and *ghana-mula* for cube root) because it signifies the "origin" or "basis" of the square or cube. This terminology later influenced Arabic and Latin terms for roots.

Key Concept: Properties of Cubes

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

Key Concept: Cube roots, Perfect cubes, Sanskrit terminology

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

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

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

Key Concept: 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: 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: 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: 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: Taxicab Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that 1729 is the smallest number expressible as the sum of two cubes in two different ways. This is true, as verified by the equations $1729 = 1^3 + 12^3$ and $1729 = 9^3 + 10^3$. The Reason correctly explains why Ramanujan considered 1729 special, linking it directly to the Assertion. Therefore, both Assertion and Reason are true, and Reason is 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: Square Roots

b) 21 meters
[Solution Description]
The area of a square is given by $\text{Side}^2$. To find the side length, take the square root of the area:
$\sqrt{441} = 21$ meters.

Key Concept: Basic Concept: Cube Root Application

b) 11
[Solution Description]
Since 1331 is known to be a perfect cube, and observing that $11^3 = 1331$, the cube root must be 11.

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.