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: 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: 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: Patterns in Unit Digits, Perfect Squares

d) It must be an even number.
[Solution Description]
From the properties of perfect squares:
– Numbers ending with 4 may or may not be perfect squares (e.g., $2^2 = 4$ and $8^2 = 64$, but 14 is not a perfect square).
– The tens digit does not provide conclusive information.
– The number could be even (as squares of even numbers end with even digits like 0, 4, 6).
– The only certainty is that the number is even.

Key Concept: 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: 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]
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, 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

c) Yes, because its prime factors can be paired identically.
[Solution Description]
The prime factorisation of 1156 is $2 \times 2 \times 17 \times 17$. Since these factors can be grouped into two identical groups ($2 \times 17$ and $2 \times 17$), 1156 is a perfect square. Specifically, $(2 \times 17)^2 = 34^2 = 1156$.

Key Concept: Perfect Squares and Units Digit

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

Key Concept: Estimating Square Roots

b) 44
[Solution Description]
To find $\sqrt{1936}$, we note that $40^2 = 1600$ and $50^2 = 2500$, so $\sqrt{1936}$ lies between 40 and 50. The last digit being 6 suggests the root could be 44 or 46. Calculating $44^2 = 1936$ confirms the answer.

Key Concept: 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: 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

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: 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: 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: 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: 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: 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: Assertion and Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that 144 is a perfect square. We know that $12^2 = 144$, so 144 is indeed a perfect square. The reason states that a perfect square is an integer that is the square of another integer, which is the correct definition. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Square Numbers Ending with 6, Sum of Odd Numbers

a) 36
[Solution Description]
To determine which number is a perfect square ending with 6:
1) Check each option to see if it ends with 6.
2) Verify if the number can be expressed as the sum of consecutive odd numbers starting from 1.
– For 36: $1 + 3 + 5 + 7 + 9 + 11 = 36$ (valid).
– For 26: Ends with 6 but cannot be expressed as the sum of consecutive odd numbers starting from 1.
– For 46: Ends with 6 but fails the sum test.
– For 56: Ends with 6 but fails the sum test.
Thus, only 36 satisfies both conditions.

Key Concept: Units Place Property

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

Key Concept: 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 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 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: Sum of Odd Numbers

d) 16
[Solution Description]
The sum of the first $n$ odd numbers is equal to $n^2$. Here, $n = 4$.
Step 1: Calculate $n^2 = 4^2 = 16$.
Therefore, the sum of the first 4 odd numbers is $16$.

Key Concept: 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: 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 $5$-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 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: Triangular Numbers and Perfect Squares

c) 21
[Solution Description]
The sum of two consecutive triangular numbers gives a perfect square. Here, $36 = 6^2$. The two consecutive triangular numbers whose sum is 36 are 15 and 21 ($15 + 21 = 36$). Hence, the larger triangular number is 21.

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: 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: Square Root Calculation

c) 12
[Solution Description]
We need to find a number $x$ such that $x^2 = 144$. From the given syllabus, we know that $12 \times 12 = 144$. Therefore, the positive square root of 144 is 12.

Key Concept: Square Roots of Perfect Squares

b) 34
[Solution Description]
To find the positive square root of 1156, we can use prime factorisation. First, factorise 1156 into its prime factors:
$1156 = 2 \times 2 \times 17 \times 17$
Now, group the prime factors into two identical groups:
$(2 \times 17) \times (2 \times 17)$
Hence, $\sqrt{1156} = 2 \times 17 = 34$.

Key Concept: 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 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 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: 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: 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: 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: 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

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: 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

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
64 can be factorized as $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$. Since all the exponents are multiples of 3, they can be grouped into triplets ($2^3 \times 2^3$). Hence, 64 is a perfect cube and the Reason correctly explains the Assertion.

Key Concept: 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: 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: 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: Consecutive Odd Numbers and Cubes

b) 6
[Solution Description]
From the given pattern in the syllabus, the sum of the first n consecutive odd numbers equals $n^3$.
Step 1: For $6^3 = 216$, the pattern shows that the sum of 6 consecutive odd numbers gives $6^3$.
Step 2: Example from the syllabus: $31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3$. There are 6 numbers here.
Therefore, 6 consecutive odd numbers must be added.

Key Concept: 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: Prime Factorisation of Cubes

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

Key Concept: 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: 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 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: Cube Root

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The cube root of a number $n$ is the value that, when multiplied by itself three times, gives $n$. Here, $3 \times 3 \times 3 = 27$, so the cube root of 27 is indeed 3. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Cube 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: Taxicab Number Identification

b) $(2, 16)$
[Solution Description]
The number $4104$ is a taxicab number and satisfies:
$4104 = 2^3 + 16^3 = 9^3 + 15^3.$
Among the options, only $(2, 16)$ corresponds to one valid pair.

Key Concept: Cube of a Number

c) 343
[Solution Description]
The cube of 7 is calculated as:
$7^3 = 7 \times 7 \times 7 = 343$
Therefore, $7^3 = 343$.

Key Concept: 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, 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: Square Pair Arrangement, Combinatorics

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

Key Concept: Taxicab Numbers

c) 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, 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: 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: 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: Cube Roots, Prime Factorisation

b) The cube root of $1728$ is $12$.
[Solution Description]
To determine the cube root of $1728$, we perform prime factorisation:
$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
Grouping into triplets:
$1728 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) = 2^3 \times 2^3 \times 3^3$
Simplifying:
$1728 = (2 \times 2 \times 3)^3 = 12^3$
Therefore, $\sqrt[3]{1728} = 12$.

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

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: 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: Successive Differences for Cubic Numbers

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

Key Concept: 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: 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: Prime Factorisation of a Number and Its Cube

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

Key Concept: 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: Aryabhata’s Definition

c) Both the square figure/its area and the square of a number
[Solution Description]
Aryabhata stated that the term *varga* refers both to a square figure with four equal sides and the area of such a figure. Additionally, it denotes the product of two equal quantities, i.e., the square of a number.

Key Concept: Sanskrit Terminology

b) Both the square figure and the square power
[Solution Description]
As per Aryabhata’s description, *varga* refers to both the square figure (or its area) and the square power (product of two equal quantities).

Key Concept: Historical 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: 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: 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: 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, 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 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: 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: 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: Mathematical etymology, Root operations

b) Sanskrit through Arabic and Latin
[Solution Description]
The concept originated from Sanskrit ‘mula’ meaning plant root, which was adopted into Arabic as ‘jidhr’ and then Latin as ‘radix’. These terms all referred both to plant roots and mathematical roots. The English term ‘root’ comes from this lineage via Latin influence on European mathematics during the Middle Ages.

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: 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: 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: 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: 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: 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: 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: Famous Mathematicians

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

Key Concept: Cube Roots

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: 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: 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: Varga, Ghana, Mula

b) The square of a square, representing the fourth power
[Solution Description]
The term varga-varga is derived from combining two varga terms. Since varga stands for square or second power, varga-varga represents the fourth power. This is because squaring a square gives the fourth power (e.g., $(x^2)^2 = x^4$).

Key Concept: Assertion-Reason

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