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.
- The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
- 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.
Solution
Clues given to us :-
- Out of 9 intersections, only 6 have ATMs, (which means 3 intersections are empty).
- ATMs with highest capacity (15 Lakhs) and lowest capacity (7 Lakhs) is on the same road.
- 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.
According to clue 1, The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
=> 15L and 7L are on the same road.
Also, this clue gives us the maximum and minimum cash in six ATMs.
CASE - 1 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 22 | ||
| R-B | 20 | |||
| R-C | 20 | |||
| Total | 15 | 21 | 26 |
CASE - 2 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 22 | ||
| R-B | 20 | |||
| R-C | 20 | |||
| Total | 15 | 21 | 26 |
Finding place for 15L ATM
15L ATM Cannot come at (RB, V3) or (RC, V3)
since the R-B and R-C capacity is 20 each , and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2).
Therefore, we can say that,
15L ATM has to either be at (RA, V1) or (RA, V3).
Hence two case will be formed.
CASE - 1 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 22 | ||
| R-B | 20 | |||
| R-C | 20 | |||
| Total | 15 | 21 | 26 |
Taking case 1 - 15L ATM is on the intersection (RA, V3).
We see that, V3 has a total of 26.
Which means, there has to be an ATM with cash of 11L (26L - 15L).
We can place the 11L ATM at (RB, V3) or (RC, V3).
if we place them at either of these intersections, the remaining ATM has to have a capacity of 20L - 11L = 9L for the same reason.
CASE - 1 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 22 | ||
| R-B | 0 | 20 | ||
| R-C | 0 | 9 | 11 | 20 |
| Total | 15 | 21 | 26 |
Finding place for 9L :-
- 9L cannot be at (RC, V1) or (RB, V1).
since the total capacity of V1 is 15L and there cannot be an ATM with 6L.
Also,
- 9L cannot be at (RB, V2).
Since, if we put it there, (RC,V2) will be 12L (21L - 9L)
which implies 12L will be second highest capacity ATM
But we know, 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 will be contradicted.
Therefore, 9L has to be at (RC, V2).
CASE - 1 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 7 | 0 | 15 | 22 |
| R-B | 0 | 20 | ||
| R-C | 0 | 9 | 11 | 20 |
| Total | 15 | 21 | 26 |
NoW, from Clue 1 we know, 15L and 7L will be on the same road.
=> 7L will be at (RA, V1) or (RA, V2)
But if it is on (RA, V2) then 21L - 9L - 7L = 5L
But minimum cash is 7L.
Therefore, 7L will be at (RA, V1)
CASE - 1 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 7 | 0 | 15 | 22 |
| R-B | 8 | 12 | 0 | 20 |
| R-C | 0 | 9 | 11 | 20 |
| Total | 15 | 21 | 26 |
Final case 1.
CASE - 2 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 22 | ||
| R-B | 0 | 20 | ||
| R-C | 0 | 20 | ||
| Total | 15 | 21 | 26 |
When 15L is at (RA, V1).
We see that, V3 has a total of 26.
Which means, therefore there will be no other ATM on V1.
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 0 | 7 | 22 |
| R-B | 0 | 20 | ||
| R-C | 0 | 20 | ||
| Total | 15 | 21 | 26 |
Finding place for 7L
here can be no other ATM on V1 in this scenario.
7L ATM which is on RA cannot be on (RA, V2) .
Since, considering the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and RC is 20.
So, 7L has to be on V3.
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 0 | 7 | 22 |
| R-B | 0 | 12 | 8 | 20 |
| R-C | 0 | 9 | 11 | 20 |
| Total | 15 | 21 | 26 |
since there cannot be a single ATM of 19L on V3, there has to be two other ATMs on V3 adding up to 19.
Which with the help of previous case we know that will be 11L and 8L.
Rearranging the numbers, we get the scenario for the second case.
Final two case we get are as follows :-
CASE - 1 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 7 | 0 | 15 | 22 |
| R-B | 8 | 12 | 0 | 20 |
| R-C | 0 | 9 | 11 | 20 |
| Total | 15 | 21 | 26 |
CASE - 2 :-
| V1 | V2 | V3 | Total | |
|---|---|---|---|---|
| R-A | 15 | 0 | 7 | 22 |
| R-B | 0 | 12 | 8 | 20 |
| R-C | 0 | 9 | 11 | 20 |
| Total | 15 | 21 | 26 |
The only statement that is correct considering both the cases is Option A.
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs."