Quantitative Aptitude and Reasoning

The circumscribed circle passes through all four vertices of the square. Its diameter is equal to the length of the diagonal of the square.
Diagonal of the square = s * √2 = 2 * √2 m.
Radius of the circumscribed circle (r_circum) = (diagonal)/2 = (2√2)/2 = √2 m.
Area of the circumscribed circle = π * (r_circum)² = π * (√2)² = π * 2 = 2π m².

The difference of the areas = Area of circumscribed circle – Area of inscribed circle
Difference = 2π m² – π m² = π m².

According to the second condition:
25% profit on A + 20% profit on B = ₹ 105
0.25 * CA + 0.20 * CB = 105 (Equation 2)

We want to find the sum of the cost prices, which is CA + CB.
Adding Equation 1 and Equation 2:
(0.20 * CA + 0.25 * CA) + (0.25 * CB + 0.20 * CB) = 120 + 105
0.45 * CA + 0.45 * CB = 225
0.45 * (CA + CB) = 225

Now, solve for CA + CB:
CA + CB = 225 / 0.45
CA + CB = 225 / (45/100)
CA + CB = 225 * (100/45)
CA + CB = (225/45) * 100
CA + CB = 5 * 100
CA + CB = 500

The sum of the cost price of items ‘A’ and ‘B’ is ₹ 500.

Since the HCF is 3, both $x$ and $y$ must be multiples of 3. We can write $x = 3a$ and $y = 3b$, where $a$ and $b$ are positive integers.
Substituting these into the product equation:
$(3a)(3b) = 54$
$9ab = 54$
$ab = 6$.

Furthermore, the HCF of $x$ and $y$ is 3, which means HCF$(3a, 3b) = 3 \times \text{HCF}(a, b) = 3$. This implies HCF$(a, b) = 1$, i.e., $a$ and $b$ must be coprime.

We need to find pairs of positive integers $(a, b)$ such that $ab=6$ and HCF$(a, b)=1$.
Possible pairs $(a,b)$ for $ab=6$:
1. (1, 6): HCF(1, 6) = 1. This pair is valid.
If $a=1, b=6$, then $x = 3 \times 1 = 3$ and $y = 3 \times 6 = 18$.
Check: HCF(3, 18) = 3, LCM(3, 18) = 18. Correct.
Sum $x+y = 3 + 18 = 21$.
2. (6, 1): HCF(6, 1) = 1. This pair is valid.
If $a=6, b=1$, then $x = 3 \times 6 = 18$ and $y = 3 \times 1 = 3$.
Check: HCF(18, 3) = 3, LCM(18, 3) = 18. Correct.
Sum $x+y = 18 + 3 = 21$.
3. (2, 3): HCF(2, 3) = 1. This pair is valid.
If $a=2, b=3$, then $x = 3 \times 2 = 6$ and $y = 3 \times 3 = 9$.
Check: HCF(6, 9) = 3, LCM(6, 9) = 18. Correct.
Sum $x+y = 6 + 9 = 15$.
4. (3, 2): HCF(3, 2) = 1. This pair is valid.
If $a=3, b=2$, then $x = 3 \times 3 = 9$ and $y = 3 \times 2 = 6$.
Check: HCF(9, 6) = 3, LCM(9, 6) = 18. Correct.
Sum $x+y = 9 + 6 = 15$.

The possible sums of the two integers are 21 and 15.
The minimum possible value of their sum is 15.

After removing the highest mark:
Number of remaining students = 39
Average marks of 39 students = 58
Total marks of 39 students = Average $\times$ Number of students
Total marks (39 students) = $58 \times 39$.
$58 \times 39 = 58 \times (40 – 1) = 58 \times 40 – 58 \times 1 = 2320 – 58 = 2262$.

The highest mark is the difference between the total marks of 40 students and the total marks of the remaining 39 students.
Highest mark = Total marks (40 students) – Total marks (39 students)
Highest mark = $2360 – 2262$
Highest mark = 98.

Now we need to find the value of $a-b$. We know the identity $(a-b)^2 = a^2 – 2ab + b^2$.
We can rewrite this as $(a-b)^2 = (a^2 + b^2) – 2ab$.
Substitute the values of $a^2+b^2$ and $ab$ that we found:
$(a-b)^2 = 52 – 2(24)$
$(a-b)^2 = 52 – 48$
$(a-b)^2 = 4$.
Taking the square root of both sides:
$a-b = \pm \sqrt{4} = \pm 2$.
The problem states that $a > b$, which means $a-b$ must be a positive value.
Therefore, $a-b = 2$.