Three pouches (each represented by a filled circle) are kept in each of the nine slots in a grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.
There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Hence the sum of nine pouches in any row or column should be a multiple of 9. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
Solution
Understanding the set :-
- In this set, You see a 3×3 grid.
Each cell (or slot) in the grid contains 3 pouches (shown as blue dots).
Each pouch contains some ₹1 coins.
So overall:
There are 9 slots, and each slot has 3 pouches → Total 27 pouches.
Each pouch has some amount of money (in whole rupees).
- Also, For each of those 9 slots, the table gives:
The minimum and maximum money among the three pouches in that slot.
For example:
In Row 1, Column 2 → pouches contain money between ₹6 and ₹8. So maybe the pouches have ₹6, ₹7, ₹8 (or any such combination).
This is a logical reasoning + math puzzle.
You are given partial information (min & max in each slot), and you have to use that along with clues to figure out the actual values in each pouch.
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | |||
| Row 2 | |||
| Row 3 | 4 (1,1,2) |
We are given that,
total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
Now, we know, Minimum = 1
and, maximum = 2
Therefore, in the last pouch we will have =
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | |||
| Row 2 | 3 (1, 1, 1) | ||
| Row 3 | 4 (1,1,2) |
Row 2 column 2 ,
We can see that here the maximum and minimum number of coins are both 1.
=> number of coins in the last pouch will also be one.
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | |||
| Row 2 | 3 (1, 1, 1) | ||
| Row 3 | 4 (1,1,2) |
Now, the only clue have is - that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer.
So, let's try to find the maximum and minimum value possible for each slot in column 1.
For the slot, column 1 and row 1 :-
the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}
Similarly, for the slot, column 1 and row 2:-
the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | 10 (2,4,4) | ||
| Row 2 | 13 (3,5,5) | 3 (1, 1, 1) | |
| Row 3 | 4 (1,1,2) |
Since, the average amount of money in each column is an integer
=> total of every row and every column is a multiple of 9.
Therefore, in column 1;
We can see that is the only possible value for the slots in column 1.
Also, we know the value of two of the pouches in each slot (minimum, maximum)
Therefore, In R1C1 values are -
and, In R2C1 coins in each pouch =
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | 10 (2,4,4) | ||
| Row 2 | 13 (3,5,5) | 3 (1, 1, 1) | 38 (6,12,20) |
| Row 3 | 4 (1,1,2) |
We now know two elements of row 2,
whose total is
Also, we have minimum and maximum of R2C3 as 6 and 20.
Therefore, we get the total =
Now the maximum value of last pouch =
Also, the minimum value of the pouch =
Therefore, the minimum total =
and, maximum value =
Thus we can iterate from the range that is the only value which is a multiple of 9.
Therefore, sum of R2C3 =
=> Value in the last pouch of R2C3 =
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | 10 (2,4,4) | 20 (6,6,8) | |
| Row 2 | 13 (3,5,5) | 3 (1, 1, 1) | 38 (6,12,20) |
| Row 3 | 4 (1,1,2) | 4 (1,1,2) |
Similarly, we can find the amount for Column 2.
For the slot, column 2 and row 1,
the maximum value possible is - 22{6,8,8}
while the minimum value possible is 20{6,6,8}.
For the slot, column 2 and row 3,
the maximum value possible is - 5{1,2,3}
while the minimum value possible is - 4{1,1,2}.
Thus {20,3,4} is the only solution possible.
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | 10 (2,4,4) | 20 (6,6,8) | 6 (1,2,3) |
| Row 2 | 13 (3,5,5) | 3 (1, 1, 1) | 38 (6,12,20) |
| Row 3 | 4 (1,1,2) | 4 (1,1,2) | 10 (2,3,5) |
Similarly, we can find the amount for Column 3.
For the slot, column 3 and row 1,
the maximum value possible is - 7{1,3,3}
while the minimum value possible is 5{1,1,3}.
For the slot, column 3 and row 3,
the maximum value possible is 12{2,5,5}
while the minimum value possible is 9{2,2,5}.
Thus {1,2,3} is the only solution possible.
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | 10 (2,4,4) | 20 (6,6,8) | 6 (1,2,3) |
| Row 2 | 13 (3,5,5) | 3 (1, 1, 1) | 38 (6,12,20) |
| Row 3 | 4 (1,1,2) | 4 (1,1,2) | 10 (2,3,5) |
As shown the required sum is 13