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 from 1 to 9.
2. We are to find boxes where the average equals the median.
3. Let the three values in a box be \( a \leq b \leq c \). The median is \( b \), and the average is \( \frac{a + b + c}{3} \).
4. For average = median: 
   \( \frac{a + b + c}{3} = b \Rightarrow a + b + c = 3b \Rightarrow a + c = 2b \)
5. This means the values must be symmetric around the middle value \( b \). Some valid triplets (with 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. Using this and checking across the grid, we find 4 boxes satisfy the condition.

Final Answer: There are 4 boxes where the average and median of the number of coins are the same.