Comprehension Passage:

The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections.

Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known:

1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.

2. The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.

Which of the following statements is correct?

The correct option is:
The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

Let us break down the step-by-step logical and mathematical derivation to understand why this option is correct:

Step 1: Understand the Grid and Cash Requirements
From the grid image, we observe three horizontal roads ($R-A$, $R-B$, and $R-C$) and three vertical roads ($\mathrm{V1}$, $\mathrm{V2}$, and $\mathrm{V3}$). The row and column sums represent the total cash requirement (in Rs. Lakhs) of the ATMs on those roads:
Row sums (horizontal roads): $R-A=22$, $R-B=20$, $R-C=20$
Column sums (vertical roads): $\mathrm{V1}=15$, $\mathrm{V2}=21$, $\mathrm{V3}=26$
The grand total cash requirement is:

$22+20+20=15+21+26=62\text{ Lakhs}$

There are $9$ possible intersections in the grid, but exactly $6$ ATMs are placed, meaning $3$ intersections have no ATM (cash requirement of $0$). The $6$ ATMs have distinct integer cash requirements.

Step 2: Determine the Set of Cash Requirements
According to Clue 1, the minimum and maximum cash requirements are $7$ Lakhs and $15$ Lakhs, respectively.
Therefore, the $6$ ATM requirements must be distinct integers selected from the set $\{7,8,9,10,11,12,13,14,15\}$, with $7$ and $15$ strictly included.
Let the sum of the remaining $4$ distinct requirements be:

$62-(7+15)=40$

We need to select $4$ distinct integers from $\{8,9,10,11,12,13,14\}$ that sum to exactly $40$.
If we do not select $8$, the minimum possible sum of the $4$ integers would be:

$9+10+11+12=42>40$

Thus, $8$ must be included. Working with the remaining set, we find two possible combinations:
Case A: The ATM cash requirements are $\{7,8,9,10,13,15\}$ (with the second highest requirement being $13$).
Case B: The ATM cash requirements are $\{7,8,9,11,12,15\}$ (with the second highest requirement being $12$).

Step 3: Analyze Distances and Locate the Second Highest ATM
Clue 2 states that the road distance between the ATM with the second highest requirement and the ATM located at the intersection of $R-C$ and $\mathrm{V3}$ is exactly $12\text{ km}$.
From the grid image, the road distances between adjacent lines are:
Horizontal distances: Between $\mathrm{V1}$ and $\mathrm{V2}$ is $4\text{ km}$; between $\mathrm{V2}$ and $\mathrm{V3}$ is $7\text{ km}$.
Vertical distances: Between $R-A$ and $R-B$ is $3\text{ km}$; between $R-B$ and $R-C$ is $5\text{ km}$.
Let us find which intersection is at a road distance of $12\text{ km}$ from $(R-C,\mathrm{V3})$:
The distance from $(R-C,\mathrm{V3})$ to $(R-B,\mathrm{V2})$ going along the grid roads is:

$\text{Vertical distance}(R-C\text{to}R-B)+\text{Horizontal distance}(\mathrm{V3}\text{to}\mathrm{V2})=5\text{ km}+7\text{ km}=12\text{ km}$

No other intersection has a road distance of exactly $12\text{ km}$ from $(R-C,\mathrm{V3})$.
Thus, the ATM with the second highest requirement is located at the intersection $(R-B,\mathrm{V2})$.

Step 4: Place the Min and Max ATMs
According to Clue 1, the minimum ($7$ Lakhs) and maximum ($15$ Lakhs) requirements are on the same road.
- They cannot be on $\mathrm{V1}$ (sum $=15$) or $\mathrm{V2}$ (sum $=21$) because $7+15=22$, which exceeds those sums.
- If they are on $\mathrm{V3}$ (sum $=26$), the third intersection on $\mathrm{V3}$ would need to have $26-22=4$ Lakhs, which is not $0$ and not in our ATM requirement sets (minimum ATM requirement is $7$).
- They cannot be on $R-B$ or $R-C$ since both have a sum of $20$, which is less than $22$.
Therefore, the $7$ and $15$ Lakhs ATMs must both be located on road $R-A$ (sum $=22$).
Since $7+15=22$, the third intersection on $R-A$ must have a requirement of $0$ (meaning no ATM is placed there).
Since the second highest ATM is at $(R-B,\mathrm{V2})$, let us analyze the vertical road $\mathrm{V2}$ (sum $=21$):

$C(R-A,\mathrm{V2})+C(R-B,\mathrm{V2})+C(R-C,\mathrm{V2})=21$

Since $R-A$'s ATM at $\mathrm{V2}$ must be either $7$, $15$, or $0$:
- If it is $7$ or $15$, then the sum on $\mathrm{V2}$ would exceed $21$ when added to the second highest value ($12$ or $13$).
- Therefore, the intersection $(R-A,\mathrm{V2})$ must have no ATM ($C(R-A,\mathrm{V2})=0$).
Thus, the sum on $\mathrm{V2}$ becomes:

$C(R-B,\mathrm{V2})+C(R-C,\mathrm{V2})=21$

Step 5: Differentiate between Case A and Case B
If Case A is correct:
The second highest is $13$, so $C(R-B,\mathrm{V2})=13$.
Then $C(R-C,\mathrm{V2})=21-13=8$.
Since $C(R-A,\mathrm{V2})=0$, the other two intersections on $R-A$ are $15$ and $7$ in some order.
- If $C(R-A,\mathrm{V1})=15$, then because the column sum of $\mathrm{V1}$ is $15$, the other two intersections on $\mathrm{V1}$ must be $0$. For row $R-B$ (sum $=20$), we get $C(R-B,\mathrm{V3})=20-13=7$. But $7$ is already used on $R-A$, violating the requirement that all ATM values must be distinct.
- If $C(R-A,\mathrm{V1})=7$, then the remaining sum on $\mathrm{V1}$ is $8$. Since we already have $8$ at $(R-C,\mathrm{V2})$, the only way to get a sum of $8$ on $\mathrm{V1}$ without duplicating the value $8$ is to have $8$ and $0$ at $(R-B,\mathrm{V1})$ and $(R-C,\mathrm{V1})$. If $C(R-B,\mathrm{V1})=8$, then row $R-B$ sum is $8+13+C(R-B,\mathrm{V3})=21>20$, which is impossible.
Hence, Case A is impossible.

If Case B is correct:
The ATM requirements are $\{7,8,9,11,12,15\}$.
The second highest value is $12$, so $C(R-B,\mathrm{V2})=12$.
Thus, the requirement at the $(R-C,\mathrm{V2})$ intersection is:

$C(R-C,\mathrm{V2})=21-12=9\text{ Lakhs}$

This configuration is fully consistent and yields the unique solution:
- ATM values: $C(R-A,\mathrm{V1})=15$, $C(R-A,\mathrm{V3})=7$, $C(R-B,\mathrm{V2})=12$, $C(R-B,\mathrm{V3})=8$, $C(R-C,\mathrm{V2})=9$, $C(R-C,\mathrm{V3})=11$.
Thus, the ATM placed at the $(R-C,\mathrm{V2})$ intersection must have a cash requirement of exactly Rs. 9 Lakhs.