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

1. Each box contains 3 sacks, each with coins numbered 1 through 9.
2. The objective is to identify boxes where the average coin count equals the median coin count.
3. Given three coin counts in a box sorted as \( a \leq b \leq c \), the median is \( b \) and the average is \( \frac{a + b + c}{3} \).
4. The condition for average equaling median is:
   \( \frac{a + b + c}{3} = b \Rightarrow a + b + c = 3b \Rightarrow a + c = 2b \)
5. This implies the coin counts must be symmetrically distributed around the middle value \( b \). Examples of valid triplets (using values from 1 to 9) include:
   • \( (3, 5, 7) \) → median = 5, average = 5
   • \( (4, 5, 6) \) → median = 5, average = 5
   • \( (5, 5, 5) \) → median = 5, average = 5
   • \( (2, 4, 6) \) → median = 4, average = 4
   • \( (1, 4, 7) \) → median = 4, average = 4
6. Applying this criterion to the grid data, it is determined that 4 boxes meet the specified condition.

Final Answer: There are 4 boxes in which the average and median coin counts are identical.