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.

How many ATMshave cash requirements of Rs. 10 Lakhs or more?

To solve this puzzle, let us first analyze the structure of the road network and the given sums of the ATM cash requirements on each road:
Horizontal roads:
R-A = 22 Lakhs
R-B = 20 Lakhs
R-C = 20 Lakhs
Vertical roads:
V1 = 15 Lakhs
V2 = 21 Lakhs
V3 = 26 Lakhs

The total cash requirement across all ATMs is:
$22+20+20=62$
Since there are 6 ATMs placed at 6 of the 9 intersections, 3 intersections will have no ATMs (value of 0). Each of the 6 ATMs has a distinct integer cash requirement.

Step 1: Finding the location and values of the Minimum (7) and Maximum (15) ATMs
From Statement 1, the minimum cash requirement is 7 Lakhs and the maximum is 15 Lakhs, and they are placed on the same road.
Let's consider vertical road V1, which has a sum of 15 Lakhs. Since the minimum value of any ATM is 7 Lakhs, V1 cannot contain more than one ATM (as two ATMs would sum to at least 7 + 8 = 15, but then they would have to be exactly 7 and 8, which means the maximum ATM of 15 cannot be on V1). Therefore, V1 must contain exactly one ATM, which must have a value of 15 Lakhs.
Since 15 is on V1, and the minimum ATM (7) and maximum ATM (15) are on the same road, this shared road must be horizontal. Let us check the horizontal roads:
- If the shared road is R-B (sum = 20): The sum of 7 and 15 is 22, which exceeds 20. Thus, they cannot be on R-B.
- If the shared road is R-C (sum = 20): Similarly, 7 + 15 = 22 > 20, so they cannot be on R-C.
- If the shared road is R-A (sum = 22): Since 7 + 15 = 22, R-A must contain exactly these two ATMs (7 and 15).
Thus, the ATM at the intersection of R-A and V1 must be 15, i.e., A1 = 15. The other ATM on R-A has a value of 7, which can be at either A2 or A3.

Step 2: Identifying the ATM with the second highest cash requirement
From Statement 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.
Let us calculate the road distances from the intersection of R-C and V3 (C3) to other intersections using the segment lengths shown in the image (horizontal: V1-V2 = 4 km, V2-V3 = 7 km; vertical: R-A to R-B = 3 km, R-B to R-C = 5 km):
- Distance to B2 (via C3 → B3 → B2 or C3 → C2 → B2) = 5 km + 7 km = 12 km.
Thus, the intersection B2 is located exactly 12 km from C3. Therefore, the ATM with the second highest cash requirement is at B2. Let us denote its value as $X$.

Step 3: Determining the active ATM positions
Since V1 contains only one ATM of value 15 (at A1), there are no ATMs at B1 and C1.
This means:
- Row R-B must have ATMs at B2 and B3 (since B1 is empty, and R-B needs a sum of 20).
- Row R-C must have ATMs at C2 and C3 (since C1 is empty, and R-C needs a sum of 20).
- Row R-A has ATMs at A1 (15) and either A2 or A3 (7).

Step 4: Finding the ATM values
Let's analyze the two possible cases for the position of 7 on R-A:
Case 1: A2 = 7, A3 is empty (0)
The sum of V2 is:
$\mathrm{A2}+\mathrm{B2}+\mathrm{C2}=21\Rightarrow 7+\mathrm{B2}+\mathrm{C2}=21\Rightarrow \mathrm{B2}+\mathrm{C2}=14$
Since all ATM values must be distinct integers ≥ 7, B2 and C2 would have to be at least 7 and 8, which sum to 15. Since 14 is less than 15, this case is impossible.
Case 2: A3 = 7, A2 is empty (0)
Using the vertical sums:
For V2:
$\mathrm{B2}+\mathrm{C2}=21$
For V3:
$\mathrm{A3}+\mathrm{B3}+\mathrm{C3}=26\Rightarrow 7+\mathrm{B3}+\mathrm{C3}=26\Rightarrow \mathrm{B3}+\mathrm{C3}=19$
We also have:
$\mathrm{B2}+\mathrm{B3}=20$
$\mathrm{C2}+\mathrm{C3}=20$
From these equations, we find:
$\mathrm{C2}=\mathrm{B3}+1$
$\mathrm{B2}=20-\mathrm{B3}$
$\mathrm{C3}=19-\mathrm{B3}$
Testing integer values for B3 such that all ATMs are distinct integers in the range [7, 15] with B2 being the second highest:
If B3 = 8:
- A3 = 7
- B3 = 8
- C2 = 9
- C3 = 11
- B2 = 12
- A1 = 15
This yields the set of ATM values: {7, 8, 9, 11, 12, 15}. All numbers are distinct, and 12 (at B2) is indeed the second highest cash requirement.

Conclusion:
The cash requirements of the six ATMs are 7, 8, 9, 11, 12, and 15 Lakhs.
The ATMs with cash requirements of Rs. 10 Lakhs or more are:
- 11 Lakhs (at C3)
- 12 Lakhs (at B2)
- 15 Lakhs (at A1)
There are exactly 3 ATMs that meet this condition.

Therefore, the correct answer is 3.