2021

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, $a-b = 2$.

Hold on, let’s re-calculate the subtraction step correctly.
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
$2S – S = (1\cdot2^2 – 2\cdot2^2) + (2\cdot2^3 – 3\cdot2^3) + \dots + (9\cdot2^{10} – 10\cdot2^{10}) + 10\cdot2^{11} – 1\cdot2^1$
This is incorrect. The terms should align:
$S = \quad 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
Subtracting S from 2S:
$S = (10\cdot2^{11}) – (1\cdot2^1) – [(2-1)2^2 + (3-2)2^3 + \dots + (10-9)2^{10}]$
$S = 10\cdot2^{11} – 2 – [2^2 + 2^3 + \dots + 2^{10}]$
The geometric series is $2^2 + 2^3 + \dots + 2^{10}$. First term $a=2^2=4$, ratio $r=2$, number of terms $n=9$.
Sum $G = 4 \frac{2^9 – 1}{2-1} = 4(512 – 1) = 4 \times 511 = 2044$.
Let’s re-calculate $G$ using the formula $a(r^n-1)/(r-1)$ where $a=2$, $n=10$ for $2^1 + … + 2^{10}$, then subtract the first term $2^1$.
$G’ = 2^1 + 2^2 + \dots + 2^{10} = 2 \frac{2^{10} – 1}{2-1} = 2(1024 – 1) = 2 \times 1023 = 2046$.
The sum $2^2 + 2^3 + \dots + 2^{10} = G’ – 2^1 = 2046 – 2 = 2044$.
$S = 10\cdot2^{11} – 2 – 2044 = 10\cdot2^{11} – 2046$.
$10 \cdot 2^{11} = 10 \cdot 2048 = 20480$.
$S = 20480 – 2046 = 18434$.

Now check the options:
A) $9\times2^{11} + 2 = 9 \times 2048 + 2 = 18432 + 2 = 18434$. Matches.
B) $10\times2^{11} + 2 = 10 \times 2048 + 2 = 20480 + 2 = 20482$.
C) $10\times2^{11} – 2 = 10 \times 2048 – 2 = 20480 – 2 = 20478$.
D) $9\times2^{11} – 2 = 9 \times 2048 – 2 = 18432 – 2 = 18430$.

My original calculation of the geometric series part in the subtraction was correct:
$S = 10 \cdot 2^{11} – (2^1 + 2^2 + \dots + 2^{10})$
The geometric series $2^1 + 2^2 + \dots + 2^{10}$ has $a=2, r=2, n=10$.
Sum is $2 \frac{2^{10}-1}{2-1} = 2(1024-1) = 2(1023) = 2046$.
$S = 10 \cdot 2^{11} – 2046$.
$S = 10 \cdot 2048 – 2046 = 20480 – 2046 = 18434$.
This matches $9 \cdot 2^{11} + 2$.

Let’s re-do the AGP subtraction step structure:
$S = 1\cdot2 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
Subtracting the first from the second, aligning terms vertically:
$S = (2\cdot2^2 – 1\cdot2^2) + (3\cdot2^3 – 2\cdot2^3) + \dots + (10\cdot2^{10} – 9\cdot2^{10}) + 10\cdot2^{11} – 1\cdot2^1$ — This is wrong alignment
Correct alignment for subtraction:
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$

$2S – S = (10\cdot2^{11}) + (9\cdot2^{10} – 10\cdot2^{10}) + \dots + (2\cdot2^2 – 3\cdot2^2) + (1\cdot2^1 – 2\cdot2^1)$ — Also wrong alignment.

Correct method:
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$

$2S – S = (10\cdot2^{11}) – (1\cdot2^1) + (1\cdot2^2 – 2\cdot2^2) + (2\cdot2^3 – 3\cdot2^3) + \dots + (9\cdot2^{10} – 10\cdot2^{10})$ This is also not right.

The general formula for $\sum_{k=1}^n k r^k$ is $\frac{nr^{n+2} – (n+1)r^{n+1} + r}{(r-1)^2}$.
Here $n=10$, $r=2$.
$S = \frac{10 \cdot 2^{12} – (10+1)2^{11} + 2}{(2-1)^2} = \frac{10 \cdot 2^{12} – 11 \cdot 2^{11} + 2}{1^2}$
$S = 10 \cdot 2 \cdot 2^{11} – 11 \cdot 2^{11} + 2$
$S = 20 \cdot 2^{11} – 11 \cdot 2^{11} + 2$
$S = (20 – 11) \cdot 2^{11} + 2$
$S = 9 \cdot 2^{11} + 2$.
This matches option A and confirms the previous calculation result.

The error in manual subtraction breakdown was in the terms. It should be:
$S = 1\cdot2 + 2\cdot2^2 + 3\cdot2^3 + \dots + 10\cdot2^{10}$
$2S = \quad \quad 1\cdot2^2 + 2\cdot2^3 + \dots + 9\cdot2^{10} + 10\cdot2^{11}$
$2S – S = (10\cdot2^{11}) + (2\cdot2^2 – 1\cdot2^2) + (3\cdot2^3 – 2\cdot2^3) + \dots + (10\cdot2^{10} – 9\cdot2^{10}) – 1\cdot2^1$
This is still not quite right. The terms are offset.
$S = 1\cdot2^1 + 2\cdot2^2 + 3\cdot2^3 + \dots + (n-1)r^{n-1} + n r^n$
$rS = \quad \quad 1\cdot r^2 + 2\cdot r^3 + \dots + (n-1)r^n + n r^{n+1}$
$rS – S = n r^{n+1} – 1\cdot r^1 – [(2-1)r^2 + (3-2)r^3 + \dots + (n-(n-1))r^n]$
$(r-1)S = n r^{n+1} – r – [r^2 + r^3 + \dots + r^n]$
The geometric series in the bracket is $r^2 + \dots + r^n$. First term $r^2$, ratio $r$, number of terms $n-1$.
Sum is $r^2 \frac{r^{n-1}-1}{r-1}$.
$(r-1)S = n r^{n+1} – r – r^2 \frac{r^{n-1}-1}{r-1}$. For $r=2$:
$S = n 2^{n+1} – 2 – 4 \frac{2^{n-1}-1}{1} = n 2^{n+1} – 2 – 4(2^{n-1}-1) = n 2^{n+1} – 2 – 2^2 2^{n-1} + 4 = n 2^{n+1} – 2^{n+1} + 2$.
$S = (n-1)2^{n+1} + 2$.
With $n=10$:
$S = (10-1)2^{10+1} + 2 = 9\cdot2^{11} + 2$.
This confirms the formula derived from the difference method.

Let’s verify with the other examples:
DOG -> FQI
D(4) -> F(6) : +2
O(15) -> Q(17) : +2
G(7) -> I(9) : +2 (Consistent)

SUN -> UWP
S(19) -> U(21) : +2
U(21) -> W(23) : +2
N(14) -> P(16) : +2 (Consistent)

Applying the +2 rule to BET:
B(2) -> D(4)
E(5) -> G(7)
T(20) -> V(22)
The resulting letters are D, G, and V.
So, the code word for BET is DGV.

Alternatively, assume the first walk of 1 km is along the positive y-axis. The position is (0,1).
Turning right from the positive y-axis means moving along the positive x-axis. The walk of 2 km leads to the position (2, 1).
Turning left from the positive x-axis means moving along the positive y-axis. The walk of 3 km leads to the position (2, 1+3) = (2, 4).
Point Q is at (2, 4). The distance from P(0,0) to Q(2,4) is $\sqrt{(2-0)^2 + (4-0)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
In both common interpretations of direction and turns, the distance is $2\sqrt{5}$ km.