Question

There are nine boxes arranged in a 3×3 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.
For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same? [This Question was asked as TITA]

• A. 3 boxes
• B. 2 boxes
• C. 4 boxes
• D. 1 box

Answer 1

The Correct Option is C

(Note: Median is termed as middle here, bag and pack are interchangeably used, Normal=Average)) We are given that each container contains three sacks. Each sack has a specific number of coins, somewhere in the range of 1 and 9, both comprehensive. The typical number of coins per sack in the crates are unmistakable numbers. The complete number of coins in each line is something similar. The complete number of coins in every segment is additionally something similar.
 => The complete number of coins in a case range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9) Since, it is given that the normal number of coins per sack in the crates are particular numbers
 => The all out number of coins in a container would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => midpoints of 1, 2, 3, 4,....,9 => Total = 45. 
=> Amount of midpoints coins in a container in succession or segment = 45/3 = 15 [The all out number of coins in each column is something similar. The complete number of coins in every segment is likewise the same.] 
==> First, let us use the following boxes to represent the final arrangement of the sacks:

		Table
	C-1	C-2	C-3
R-1
R-2
R-3

We are given that each container contains three sacks. Each sack has a specific number of coins, somewhere in the range of 1 and 9, both comprehensive. The typical number of coins per sack in the crates are unmistakable numbers. The complete number of coins in each line is something similar. The complete number of coins in every segment is additionally something similar. => The complete number of coins in a case range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9) Since, it is given that the normal number of coins per sack in the crates are particular numbers => The all out number of coins in a container would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => midpoints of 1, 2, 3, 4,....,9 => Total = 45. => Amount of midpoints coins in a container in succession or segment = 45/3 = 15 [The all out number of coins in each column is something similar. The complete number of coins in every segment is likewise the same.] ==> First, let us use the following boxes to represent the final arrangement of the sacks:

		Table
	C-1	C-2	C-3
R-1		3,9,9(7)
R-2	1,2,9(4)
R-3	7,8,9(8)

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3. 
From (1), The average in bag (1,3) is 15-3-7 = 5. 
 

		Table
	C-1	C-2	C-3
R-1		3,9,9(7)	Avg=5
R-2	1,2,9(4)
R-3	7,8,9(8)

Think about pack (1,1) 
Avg = 3, 1 sack has more than 5 and ** => 2 circumstances are being fulfilled. => (can't be condition-3 => 9 coins as the absolute amount of coins is it self 3*3 = 9) 
=> pack (1,1) has 1, 1, 7 coins with normal = 3. 
Think about sack (1,3) 
Avg. 5> Sum = 15. 
Middle = 6 and 2 sacks have more than 5 and * => (1 condition is fulfilled) Not condition ii as the middle is 6 and Not condition iii as the amount of 2 sacks itself will become 6 + 9 = 15 
=> 1, 6, c are the coins => For aggregate = 15 => c = 15-1-6=8 
=> pack (1,3) has 1, 6, 8 coins with average = 5.
 

		Table
	C-1	C-2	C-3
R-1	1,1,7(3)	3,9,9(7)	1,6,8(5)
R-2	1,2,9(4)
R-3	7,8,9(8)

Take into consideration bag (3,3):
 0 sacks contain more than 5 coins, and ** indicates that conditions i and ii are met, 
1,1,c are the coins. Presently c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a different of 3. Be that as it may, c = 1 as no other sack has the likelihood to get avg. = 1 as sack (2,2) ought to have 1, b, c coins and b and c ought to be more than 1 as just 1* => pack (3,3) has 1, 1, 1 coins with normal = 1. Presently, we can fill the midpoints in every one of the sacks.

		Tables
	C-1	C-2	C-3
R-1	1,1,7(3)	3,9,9(7)	1,6,8(5)
R-2	1,2,9(4)	avg=2	avg=9
R-3	7,8,9(8)	avg=6	1,1,1(1)

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins. 
In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens-t should be 1.
 => 1, b, c are the coins where 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.
 

		Table
	C-1	C-2	C-3
R-1	1,1,7(3)	3,9,9(7)	1,6,8(5)
R-2	1,2,9(4)	1,2,3(2)	9,9,9(9)
R-3	7,8,9(8)	Avg=6	1,1,1(1)

Considering bag (3,2) 
Avg. = 6=> Sum = 18. 2 sacks more than 5 coins and ** 
=> 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. 
==> Final required table, bracket number => average coins per sack in the bag.
 

		Table
	C-1	C-2	C-3
R-1	1,1,7(3)	3,9,9(7)	1,6,8(5)
R-2	1,2,9(4)	1,2,3(2)	9,9,9(9)
R-3	7,8,9(8)	1,8,9(6)	1,1,1(1)

Average = Median in boxes (3,1), (2,2), (2,3) and (3,3): = 4 boxes.
So, The correct answer is 4 Boxes.