There are nine boxes arranged in a array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a, if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by. if two or more of these conditions are satisfied.
i) The minimum among the numbers of coins in the three sacks in the box is 1.
ii) The median of the numbers of coins in the three sacks is 1.
iii) The maximum among the numbers of coins in the three sacks in the box is 9.
Solution
Understanding the set :-
- We are given nine boxes arranged in a 3×3 array - in the same way as shown in tables, in the question.
- There are 3 sacks in each box as well , which implies we need to find three numbers for a particular box.
- The average number of coins per sack in the boxes are all distinct integers, which means average of each box will be a distinct integer from 1 to 9
- The total number of coins in each row is the same. The total number of coins in each column is also the same.
- Also, question gives 2 tables where,
Table 1 - gives median of number of coins in a box
Table 2 - gives a number which represents the number of sacks in that box having more than 5 coins.
In this question we will try to find the number on each sack of each box.
Let us represent the final configuration of the sacks in boxes as follows:
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 45 | |||
| 2nd row | 45 | |||
| 3rd row | 45 | |||
| Total | 45 | 45 | 45 | 135 |
We know the average of all the boxes is a distinct integer from 1 to 9
Therefore, the total of all averages = 1+2+3+4+5+6+7+8+9 = 45
Also, we know The total number of coins in each row is the same. The total number of coins in each column is also the same
=> Sum of averages coins in a box in a row or column = 45/3 = 15
Since there are 3 rows and 3 columns the total of each one will come out to be 15 * 3 = 45
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 45 | |||
| 2nd row | 45 | |||
| 3rd row | 7, 8, 9 (8) | 45 | ||
| Total | 45 | 45 | 45 | 135 |
Consider bag (3,1) {3dr row and 1st column}
=> From Table-1 => Median = 8
From Table-2 - all 3 sacks have more than 5 coins
Also * - which implies the sacks in that box satisfy exactly one among the three conditions given in the question.
But we know minimum can't be 1 (as all 3 sacks have more then 3 coins)
also, Median of 3 sacks is not equal to 1
=> it follows 3rd condition
=> There is a 9 in one of the sacks.
=> c, 8, 9 are the coins in bag (3,1)
now c > 5 & c + 8 + 9 should be a multiple of 3
=> c = 7 is the only possibility. with average = 8
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 45 | |||
| 2nd row | 1, 2, 9 (4) | 45 | ||
| 3rd row | 7, 8, 9 (8) | 45 | ||
| Total | 45 | 45 | 45 | 135 |
Consider bag (2,1)
-
Median = 2
-
1 sack has more than 5 coins
-
Also ** => conditions i & iii should be satisfied.
=> 1, 2, 9 are the coins in bag (2,1). with average = 4
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 3, 9, 9 (7) | 45 | ||
| 2nd row | 1, 2, 9 (4) | 45 | ||
| 3rd row | 7, 8, 9 (8) | 45 | ||
| Total | 45 | 45 | 45 | 135 |
Consider bag (1,2)
-
Median = 9
-
2 elements are more than 5.
-
Also * => (9 is present & 1 is not present)
=> c, 9, 9 are the coins in bag (1,2)
and c is not equal to 1 and less than 5
=> c = 3 for c + 18 to be a multiple of 3.
=> 3, 9, 9 are the coins in bag (1,2) with average =7
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 1, 1, 7 (3) | 3, 9, 9 (7) | 45 | |
| 2nd row | 1, 2, 9 (4) | 45 | ||
| 3rd row | 7, 8, 9 (8) | 45 | ||
| Total | 45 | 45 | 45 | 135 |
Consider bag (1,1)
- Avg = 3
sum of average of each row = 15, and we already have (2, 1) average = 4 and (3, 1) average = 8
=> average of (1,1) = 3
- 1 sack has more than 5
- ** => 2 conditions are being satisfied.
condition 3 won't be fulfilled as average = 3
therefore, total in itself = 9
1, 1, 7 coins with average = 3
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 1, 1, 7 (3) | 3, 9, 9 (7) | 1, 6, 8 (5) | 45 |
| 2nd row | 1, 2, 9 (4) | 45 | ||
| 3rd row | 7, 8, 9 (8) | 45 | ||
| Total | 45 | 45 | 45 | 135 |
Consider bag (1,3)
- Avg. = 5 => Sum = 15.
As we know average of total= 15 and we already found average of (1,1) = 3 and (1,2) = 7
therefore, average of (1,3) = 15 - 3 - 7 = 5
- Median = 6 and 2 sacks have more than 5
-
- => (1 condition is satisfied)
Not condition ii as the median is 6
Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15
=> 1, 6, c are the coins
=> For sum = 15 => c = 15 - 1 - 6 = 8
=> bag (1,3) has 1, 6, 8 coins with average = 5
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 1, 1, 7 (3) | 3, 9, 9 (7) | 1, 6, 8 (5) | 45 |
| 2nd row | 1, 2, 9 (4) | 45 | ||
| 3rd row | 7, 8, 9 (8) | 1, 1, 1 (1) | 45 | |
| Total | 45 | 45 | 45 | 135 |
Consider bag (3,3)
- 0 sacks have more than 5 coins
- ** => conditions i & ii are being satisfied.
=> 1,1,c are the coins.
Now c = 1 or 2 or 3 or 4
=> c = 1 or 4 for number of coins to be a multiple of 3
But c will be 1
as no other bag has the possibility to get avg. = 1
=> bag (3,3) has 1, 1, 1 coins with average = 1
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 1, 1, 7 (3) | 3, 9, 9 (7) | 1, 6, 8 (5) | 45 |
| 2nd row | 1, 2, 9 (4) | 9, 9, 9 (9) | 45 | |
| 3rd row | 7, 8, 9 (8) | 1, 1, 1 (1) | 45 | |
| Total | 45 | 45 | 45 | 135 |
In bag (2,3)
Avg. = 9
=> 9, 9, 9 are the coins.
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 1, 1, 7 (3) | 3, 9, 9 (7) | 1, 6, 8 (5) | 45 |
| 2nd row | 1, 2, 9 (4) | 1, 2, 3 (2) | 9, 9, 9 (9) | 45 |
| 3rd row | 7, 8, 9 (8) | 1, 1, 1 (1) | 45 | |
| Total | 45 | 45 | 45 | 135 |
In bag (2,2)
- Avg. = 2
- Sum = 6
- 1* => smallest element hould be 1.
=> 1, b, c are the coins
=> b + c = 5 and b,c can't be equal to 1 and less than 5
=> 2 + 3 = 5 is the only possibility
=> 1, 2, 3 are the coins with average = 2
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 1, 1, 7 (3) | 3, 9, 9 (7) | 1, 6, 8 (5) | 45 |
| 2nd row | 1, 2, 9 (4) | 1, 2, 3 (2) | 9, 9, 9 (9) | 45 |
| 3rd row | 7, 8, 9 (8) | 1, 8, 9 (6) | 1, 1, 1 (1) | 45 |
| Total | 45 | 45 | 45 | 135 |
Considering bag (3,2)
1)Avg. = 6 => Sum = 18.
- 2 sacks more than 5 coins
- ** => 2 sacks have 1 and 9 coins.
=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8
=> bag (3,2) has 1, 8, 9 coins with average = 6 coins
| 1st column | 2nd column | 3rd column | Total | |
|---|---|---|---|---|
| 1st row | 1, 1, 7 (3) | 3, 9, 9 (7) | 1, 6, 8 (5) | 45 |
| 2nd row | 1, 2, 9 (4) | 1, 2, 3 (2) | 9, 9, 9 (9) | 45 |
| 3rd row | 7, 8, 9 (8) | 1, 8, 9 (6) | 1, 1, 1 (1) | 45 |
| Total | 45 | 45 | 45 | 135 |
for 4 boxes median and average will be same