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: 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 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: Perfect Squares

c) 64
[Solution Description]
To find the square of 8, we multiply 8 by itself: $8 \times 8 = 64$. Thus, the square of 8 is 64.

Key Concept: 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, Prime Factorisation

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

Key Concept: Perfect Squares and Units 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: Perfect Squares and Units Place

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

Key Concept: Estimating Square Roots

b) 44
[Solution Description]
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: 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: 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: 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: 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: 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: 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: 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: 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: 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: Sum of First n Odd Numbers

c) 13
[Solution Description]
The sum of the first $n$ odd numbers is $n^2$. Given that the sum is 169:
$n^2 = 169 \Rightarrow n = \sqrt{169} = 13$.
Therefore, the value of $k$ is 13.

Key Concept: Difference Between Consecutive Squares

c) 11
[Solution Description]
The difference between consecutive squares is an odd number. Specifically, the difference between $(n+1)^2$ and $n^2$ is $2n + 1$.
Step 1: Calculate $6^2 = 36$ and $5^2 = 25$.
Step 2: Compute the difference: $36 - 25 = 11$.
Therefore, the difference is $11$.

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

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

Key Concept: Perfect Squares and 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: 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: 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: 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 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: 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 Numbers, Square Roots

c) Assertion is true, but Reason is false.
[Solution Description]
To determine if 1125 is a perfect square, we first analyze its prime factorization:
$1125 = 3^2 \times 5^3$. For a number to be a perfect square, all exponents in its prime factorization must be even. Here, the exponent of 5 is 3 (odd), so 1125 is not a perfect square. However, the Reason is incorrect because a perfect square ending with 5 does not necessarily require an even number of zeros in its prime factorization. Instead, the correct condition is that all exponents in its prime factorization must be even. Thus, Assertion is true, but Reason is false.

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: 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: 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: Prime Factorisation, Perfect Cubes

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

Key Concept: Cubic Numbers

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

Key Concept: 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: 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: Identifying Perfect Cubes

b) 27
[Solution Description]
A perfect cube is a number obtained by multiplying an integer by itself three times. Checking the options:
- 9 is not a perfect cube ($2^3 = 8$ and $3^3 = 27$).
- 27 is a perfect cube ($3 \times 3 \times 3 = 27$).
- 36 is not a perfect cube.
- 49 is not a perfect cube.
Thus, 27 is the perfect cube among the given options.

Key Concept: 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: 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: 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: Prime Factorisation

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

Key Concept: Perfect Cubes, 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: 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: Prime Factorisation of Cubes, Cube Roots

b) 90
[Solution Description]
To find the cube root of $N^3$, we first express $N$ in terms of its prime factors:
$N = 2 \times 3^2 \times 5$.
The cube of $N$ is:
$N^3 = (2 \times 3^2 \times 5)^3 = 2^3 \times 3^6 \times 5^3.$
Taking the cube root of $N^3$ gives:
$\sqrt[3]{N^3} = N = 2 \times 3^2 \times 5 = 90.$

Key Concept: Taxicab 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: Taxicab Number

c) 1729
[Solution Description]
A taxicab number is expressible as the sum of two cubes in two different ways. The smallest known example is 1729:
$1729 = 1^3 + 12^3 = 9^3 + 10^3$
Among the options, 1729 fits this property.

Key Concept: 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: 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 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: 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

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

Key Concept: Taxicab Numbers

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: 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, 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

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: Cube Roots

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

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: 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: 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: 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: 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: 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: 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: Cubic Numbers: Successive Differences

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for perfect cubes, the third differences are always constant. This is true as demonstrated in the syllabus example where the third differences were all 6 for the sequence $1, 8, 27, 64, 125, 216, \ldots$.
The reason claims that this property is unique to cubic sequences and helps identify them. This is also true because, unlike squares or higher powers, only cubic numbers exhibit this specific pattern of constant third differences.
Since both the assertion and reason are true, and the reason correctly explains why perfect cubes have constant third differences, the correct option is the one where both are true and the reason is the correct explanation.

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: Origin of the Term 'Root'

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

Key Concept: 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: 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: 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: 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: 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: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the Babylonians compiled lists of perfect squares and cubes, which is historically accurate as evidenced by ancient clay tablets dating back to 1700 BCE. The reason provided explains the practical application of these lists in solving problems related to land measurement and architectural design, which aligns with historical records. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

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

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

Key Concept: 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: 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 and Their Properties

c) 64
[Solution Description]
A perfect cube is a number that can be expressed as $n^3$ where $n$ is an integer. Checking each option:
$64 = 4^3$
Hence, 64 is a perfect cube.

Key Concept: Cube roots, Perfect cubes

c) 17
[Solution Description]
To find the cube root of 4913 without factorisation, observe the last digit. The number ends with 3. The only single-digit cube ending with 3 is $7^3 = 343$. Checking $17^3$ gives $17 \times 17 = 289$, then $289 \times 17 = 4913$. Thus, the cube root is 17.

Key Concept: Taxicab Numbers

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: 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: 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: Assertion-Reason

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

Key Concept: Basic Concept: Mula Meaning

d) Because it means the root of a plant, basis, cause, or origin
[Solution Description]
The word *mula*, meaning the root of a plant, was used in ancient India to represent the basis or origin of square and cube roots, as it signifies the foundation or source.

Key Concept: Mula, Square Root

c) It symbolized the origin or basis of the number
[Solution Description]
The Sanskrit word 'mula' means the root of a plant, basis, cause, or origin. It was metaphorically extended to describe the mathematical operation of finding roots because it signifies the 'origin' or 'basis' of a number (e.g., the number that, when squared or cubed, gives the original number). For example, $\sqrt{x}$ finds the 'origin' of $x$ (the number which squared gives $x$).

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