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: 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: 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: 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 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: Estimating Square Roots

b) 44
[Solution Description]
From the syllabus, we know that 1936 is a perfect square. To estimate its square root:
1. It lies between $1600$ ($40^2$) and $2500$ ($50^2$).
2. The last digit is 6, meaning the square root must end with 4 or 6 (i.e., 44 or 46).
3. Calculating $45^2 = 2025$, which is greater than 1936.
4. Now calculate $44^2 = 1936$.
Hence, the correct answer is 44.

Key Concept: Perfect Squares, Prime Factorisation

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

Key Concept: 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 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 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: Sum of Consecutive Odd Numbers, Finding the Next Square

a) 1296
[Solution Description]
The sum of the first $n$ odd numbers is given by $n^2 = 1225$. Therefore, $n = \sqrt{1225} = 35$. The next square is $(n + 1)^2 = 36^2$. To find $36^2$, we add the $(n+1)^{th}$ odd number to the previous sum. The $(n+1)^{th}$ odd number is $2(n + 1) - 1 = 2(36) - 1 = 71$. Thus, $36^2 = 35^2 + 71 = 1225 + 71 = 1296$.

Key Concept: Patterns in Units Place

c) 537
[Solution Description]
The units digit of a perfect square can only be $0, 1, 4, 5, 6,$ or $9$. Numbers ending with $2, 3, 7,$ or $8$ are not perfect squares. Among the options, only the number ending with $7$ does not fit this pattern.

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: Perfect Squares

c) 49
[Solution Description]
A perfect square is the square of a natural number. From the options, only 49 can be written as $7^2 = 49$.

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

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

Key Concept: 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: Sum of Odd Numbers

c) 25
[Solution Description]
Sum of first $n$ odd natural numbers equals $n^2$. Here, $n = 5$, so sum $= 5^2 = 25$.

Key Concept: Difference Between Consecutive Squares

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

Key Concept: Perfect Squares and Odd Numbers

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

Key Concept: Perfect Squares and Odd Numbers

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

Key Concept: Difference Between Consecutive Squares

b) 25
[Solution Description]
The difference between consecutive squares is an odd number. The general pattern shows that the difference between $n^2$ and $(n+1)^2$ is given by $(2n + 1)$.
Here, $n = 12$.
So, the difference is $2(12) + 1 = 25$.

Key Concept: Perfect 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: Square Numbers and Triangular Numbers, Patterns in Perfect Squares

b) 25
[Solution Description]
The sequence follows the pattern where the sum of two consecutive triangular numbers gives a square number. The triangular numbers are $1, 3, 6, 10, 15, 21, \ldots$. The next pair is $10 + 15 = 25$, which is $5^2$.

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: 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: 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

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

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: 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: Perfect Square Identification

a) 100
[Solution Description]
A perfect square is obtained by multiplying an integer by itself. Checking each option:
- 100: $10 \times 10 = 100$ (perfect square).
- 110: No integer multiplied by itself gives 110.
- 120: No integer multiplied by itself gives 120.
- 130: No integer multiplied by itself gives 130.
Thus, 100 is the only perfect square in the options.

Key Concept: 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: 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: 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: 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: 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

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: Cubic Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
27 can be expressed as $3^3$, which means it is a perfect cube. The reason correctly defines a perfect cube as a number that can be written in the form $n^3$ where $n$ is an integer. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Perfect Cubes

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 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: 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 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: Taxicab numbers, Sum of two cubes

b) 4104
[Solution Description]
A taxicab number can be expressed as the sum of two cubes in two distinct ways. Among the options, $4104$ is known to satisfy this property: $4104 = 2^3 + 16^3 = 9^3 + 15^3$. We verify: $2^3 = 8$, $16^3 = 4096$, $8 + 4096 = 4104$; similarly, $9^3 = 729$, $15^3 = 3375$, $729 + 3375 = 4104$.

Key Concept: 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: 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: Properties of Cube Numbers, 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 and known as the Hardy-Ramanujan number or taxicab number. The Reason explains that no smaller number satisfies this condition, which is also true. Furthermore, the Reason correctly justifies why 1729 holds this unique property. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

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: 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: Perfect Cube Identification

c) 64
[Solution Description]
A perfect cube is a number that can be expressed as the cube of an integer. Checking each option:
$64 = 4 \times 4 \times 4 = 4^3$
64 is a perfect cube.

Key Concept: Cube 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 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: 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: Taxicab Numbers

c) 4104
[Solution Description]
The next two taxicab numbers after 1729 are 4104 and 13832. Here, 4104 can be expressed as:
$4104 = 2^3 + 16^3$ and $4104 = 9^3 + 15^3$.

Key Concept: 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: 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

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

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: Perfect Cubes and Consecutive Odd Numbers

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

Key Concept: Perfect Cubes and Consecutive Odd Numbers

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion is false because the sum of the first $n$ consecutive odd numbers is equal to $n^2$, not $n^3$. For example, $1 + 3 = 4 = 2^2$. The Reason is also false because the sum of any 10 consecutive odd numbers depends on the starting number and is not necessarily a perfect cube. For instance, $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100 = 10^2$, which is not a perfect cube. Thus, both Assertion and Reason are false.

Key Concept: Cube Roots

c) 8
[Solution Description]
To find the cube root of 512, we can use prime factorization.
$512 = 8 \times 64 = (2^3) \times (4^3) = (2 \times 4)^3 = 8^3$
Therefore, $\sqrt[3]{512} = 8$.

Key Concept: 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: 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: 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: 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: Third Differences in Cubic Numbers

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

Key Concept: 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: 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 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: Historical terminology, Sanskrit mathematical terms

d) varga-varga-mula
[Solution Description]
In ancient India:
- Square root was called varga-mula
- Cube root was called ghana-mula
- Fourth power was called varga-varga
Therefore, the root of a fourth power (varga-varga) would logically follow the same pattern and be called varga-varga-mula.
The calculation: $\sqrt[4]{81} = 3$ since $3^4 = 81$

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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: Perfect Cubes

b) 27000
[Solution Description]
To determine if a number is a perfect cube, we check if it can be expressed as $n^3$ where $n$ is an integer.
- 27000 can be written as $30^3$, making it a perfect cube.
- Other options do not satisfy this condition.

Key Concept: Perfect squares, Units digit pattern

c) 12347
[Solution Description]
A number cannot be a perfect square if it ends with 2, 3, 7, or 8. From the options:
12347 ends with 7, which means it cannot be a perfect square. The other options end with digits that may or may not be perfect squares.

Key Concept: Famous Mathematicians

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that 1729 is the smallest number expressible as the sum of two cubes in two different ways, which is true ($1729 = 1^3 + 12^3 = 9^3 + 10^3$). The reason explains Ramanujan's talent for recognizing numerical patterns, which is also true and directly explains why he identified this property of 1729. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Perfect Cubes, 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: Assertion-Reason

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

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: Perfect Cubes

a) The cube of any odd number is even.
[Solution Description]
Let us analyze each statement:
(i) The cube of an odd number is odd, not even. Hence this statement is false.
(ii) There are perfect cubes that end with 8, such as $\sqrt[3]{512} = 8$. So this is false.
(iii) True, e.g., $4^3 = 64$ is a 2-digit cube of a 1-digit number; $5^3 = 125$ is a 3-digit number.
(iv) False, since the cube of a 2-digit number (maximum 99) would be $99^3 = 970299$, which has 6 digits.
(v) True, because perfect squares have an odd number of factors, but cubes may not necessarily have an odd number unless they are also perfect squares.
Therefore, the false statements are (i), (ii), and (iv). Among the options provided, only one is given as the main answer.

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, $\sqrt[3]{27000} = 30$.