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 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: 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: 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 of Perfect Squares

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that 144 is a perfect square because its units digit is 4. The reason states that a number is a perfect square only if its units digit is 0, 1, 4, 5, 6, or 9.
Checking the assertion: $12^2 = 144$, so 144 is indeed a perfect square.
Evaluating the reason: According to the syllabus, perfect squares can only have 0, 1, 4, 5, 6, or 9 as their units digit. However, having such a unit digit does not guarantee that the number is a perfect square — it is a necessary but not sufficient condition. For example, 24 has a units digit of 4 but is not a perfect square.
Thus, both the assertion and reason are true, but the reason does not fully explain why 144 is a perfect square (it just provides a necessary condition).

Key Concept: Perfect Squares and Units Digit

c) 338
[Solution Description]
Perfect squares can only end with 0, 1, 4, 5, 6, or 9. Numbers ending with 2, 3, 7, or 8 cannot be perfect squares.

Key Concept: Prime Factorisation, Estimating Square Roots

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

Key Concept: Perfect Squares 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: 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: 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: 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: Square Numbers Ending with 6, Patterns in Units Place

d) 2
[Solution Description]

If a perfect square ends in 6, then the number being squared must end in 4 or 6. The possible last two digits of such perfect squares are:

$16,\ 36,\ 56,\ 76,\ 96$

So, the possible tens digits are:

$1,\ 3,\ 5,\ 7,\ 9$

Therefore, the digit 2 cannot be the tens digit of a perfect square ending in 6.

Correct Option: D. 2

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, 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: Perfect Squares and Odd Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The sum of the first 5 odd numbers (1, 3, 5, 7, 9) is calculated as $1 + 3 + 5 + 7 + 9 = 25$. This matches the assertion since $5^2 = 25$. The reason states that every square can be expressed as the sum of consecutive odd numbers starting from 1, which is a property of perfect squares and explains why the sum is equal to the square of the count of odd numbers. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Difference Between Consecutive Squares

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

Key Concept: 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: 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: Squares Ending with 1

c) 29
[Solution Description]
According to the properties of square numbers, any number ending with 1 or 9 will have a square ending with 1. Checking the options:
- Option 1: 29 (units digit is 9, so $29^2$ ends with 1)
- Option 2: 14 (units digit is 4, so $14^2$ ends with 6)
- Option 3: 33 (units digit is 3, so $33^2$ cannot be a perfect square as per the property)
- Option 4: 47 (units digit is 7, so $47^2$ cannot be a perfect square as per the property)
Therefore, only 29 satisfies the condition.

Key Concept: Zeros in Perfect Squares

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

Key Concept: 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: 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 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

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

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

b) $T_4$ and $T_5$
[Solution Description]
Triangular numbers follow the pattern $T_n = \frac{n(n+1)}{2}$. The sum of consecutive triangular numbers $T_{n-1} + T_n = n^2$. For example, $T_2 + T_3 = 3 + 6 = 9 = 3^2$. Thus, the correct pair must satisfy this relationship.

Key Concept: Perfect Squares and Units Digit

c) 87654
[Solution Description]
A perfect square can only end with digits 0, 1, 4, 5, 6, or 9. Therefore, any number ending with 2, 3, 7, or 8 cannot be a perfect square.

Key Concept: Properties of Perfect Squares

d) 2
[Solution Description]
Perfect squares can only end with 0, 1, 4, 5, 6, or 9. Among the given options, 2 is not one of these digits. Hence, 2 cannot be the units digit of a perfect square.

Key Concept: 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 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: 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: 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 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: 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: 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: 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, 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: 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: 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: Perfect cubes, Prime factorization

c) Yes, the cube root is 1500
[Solution Description]
To check if the number is a perfect cube, we verify if all exponents in its prime factorization are divisible by 3. Here, $6$ is divisible by 3 ($6/3 = 2$), $3$ is divisible by 3 ($3/3 = 1$), and $9$ is divisible by 3 ($9/3 = 3$). Thus, the number is a perfect cube. The cube root is obtained by taking each exponent divided by 3: $2^{6/3} \times 3^{3/3} \times 5^{9/3} = 2^2 \times 3^1 \times 5^3 = 4 \times 3 \times 125 = 1500$.

Key Concept: Sum of consecutive odd numbers, Perfect cubes

c) 7
[Solution Description]
The sum of consecutive odd numbers forming a perfect cube follows the pattern where the sum of the first $n$ such numbers is $n^3$. Given that $343 = 7^3$, the number of terms being added must be $7$.

Key Concept: Prime Factorization of Cubes

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

Key Concept: 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: Taxicab Numbers

c) 1729
[Solution Description]
The smallest such number is called a taxicab number. From the syllabus, the first taxicab number is 1729, which can be written as $1^3 + 12^3$ and $9^3 + 10^3$.

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: Cubes of Fractions/Decimals, Numerical Accuracy

c) 0.421875
[Solution Description]
To compute $\left(\frac{3}{4}\right)^3$, multiply the numerator and denominator three times:
$\left(\frac{3}{4}\right)^3 = \frac{3^3}{4^3} = \frac{27}{64}.$
Converting to decimal form gives $0.421875$.

Key Concept: Properties of Cube Numbers

d) Assertion is false, but Reason is true.
[Solution Description]
To check if a perfect cube can end with exactly two zeros (00), we need to analyze the prime factorization of such a number. A number ending with two zeros must be divisible by 100. The prime factorization of 100 is $2^2 \times 5^2$. For a number to be a perfect cube, each prime factor must appear in multiples of three. However, 100 has $2^2$ and $5^2$, which are not multiples of three. Therefore, a perfect cube cannot end with exactly two zeros. The Reason states that a number is a perfect cube if its prime factors appear in triplets, which is true. However, this does not explain why a perfect cube cannot end with two zeros, as the assertion is false.

Key Concept: Cube Roots, Taxicab Numbers

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

Key Concept: Cube Roots, Estimation

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

Key Concept: 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 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: 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: 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: Taxicab Numbers, Prime Factorization

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that 4104 is a taxicab number, which is correct as shown in the syllabus ($4104 = 2^3 + 16^3 = 9^3 + 15^3$).
The reason (R) claims that for any taxicab number, the prime factorization of its cube must have each prime factor appearing exactly three times. While this is true for any number's cube (as per the syllabus example: $12^3 = 2^3 \times 2^3 \times 3^3$), this property is not unique to taxicab numbers—it applies to all cubes. Therefore, while both statements are true, the reason does not explain why 4104 is a taxicab number; it merely states a general property of cubes.
Thus, the correct option is (b).

Key Concept: Taxicab Numbers, Sum of Two Cubes

a) 1729
[Solution Description]
The smallest such $k$ is 1729, which has two pairs: $(1, 12)$ and $(9, 10)$. This makes 1729 the smallest taxicab number.

Key Concept: Taxicab Numbers, 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

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 as proven by Ramanujan. The reason provides the two valid cube pairs $(1^3 + 12^3)$ and $(9^3 + 10^3)$ that sum to 1729, correctly explaining why the assertion is true.

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: 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: 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: Cube Roots

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

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

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: Second Differences in Cubic Numbers

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

Key Concept: Cubic Numbers: Successive Differences, 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: 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: 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: 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: Historical Origin of Mathematical Terms

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]

The term $\textit{varga-mula}$ indeed refers to the square root in Sanskrit, as it combines $\textit{varga}$ (square) and $\textit{mula}$ (root). The reason correctly explains that $\textit{mula}$ was used to denote the origin or basis of a square, and this concept was later adopted in Arabic $(\textit{jidhr})$ and Latin $(\textit{radix}),$ eventually leading to the modern term "root." Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Ancient Mathematical Terminology

b) Cube root
[Solution Description]
In ancient Sanskrit mathematics, the term *ghana* was used for the cube of a number or the solid cube figure, and *mula* means root (origin). Hence, *ghana-mula* refers to the cube root.

Key Concept: Historical 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 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: 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: Babylonian Lists

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

Key Concept: Properties of cubes, Historical context

a) Varga-varga of any number has an even number of factors
[Solution Description]
We need to identify which statement contradicts known properties of cubes from historical context.
Analyzing each concept:
1) Cubes of odd numbers: Always odd (3³=27, 5³=125)
2) Perfect cubes ending with 8: Exist (2³=8, 12³=1728)
3) Two-digit cubes: Yes (4³=64, 5³=125)
4) Two-digit cubes becoming 7+ digits: Impossible (max 99³=970299 is 6 digits)
5) Factors of cubes: All have odd number of factors (as perfect squares)
Ancient texts would recognize all except the false claim about seven digit cubes from two-digit numbers.

Key Concept: 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: 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: 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: 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: 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: Historical applications, Perfect squares

c) Land measurement and architecture
[Solution Description]
The Babylonian clay tablets containing lists of perfect squares were primarily used for land measurement calculations and architectural designs. Having pre-computed square values allowed quick determination of areas and dimensions needed in construction projects and land surveys, which were crucial for their advanced civilization.

Key Concept: 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: Perfect Cubes, Taxicab Numbers

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

Key Concept: 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: 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: 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

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: 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: 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: 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: 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: 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.