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.
the median of the numbers of coins in the three sacks in a box for some of the boxes
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.
How many sacks have exactly one coin?[This Question was asked as TITA]

Updated On: Sep 17, 2024
  • 11 sacks
  • 10 sacks
  • 9 sacks
  • None
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The Correct Option is C

Solution and Explanation

Each box contains three sacks, with each sack holding between 1 and 9 coins. The average number of coins per sack in each box is a distinct integer.The total number of coins is the same for each row and for each column. 
Therefore, the total number of coins in a box can range from 3 (1+1+1) to 27 (9+9+9).
 Given that the average number of coins per sack in each box is distinct, the total number of coins in a box must be 3, 6, 9, 12, 15, 18, 21, 24, or 27, corresponding to averages of 1 through 9, summing to 45. 
Thus, the sum of the average coins per box in a row or column is 15.
Each bag (x,y) is in the x-th row and y-th column. We are given two clues: Table-1 and Table-2.

For bag (3,1):

  • From Table-1: The median number of coins is 8.
  • From Table-2: All three sacks have more than 5 coins, and one sack has 9 coins.

So, the coins in bag (3,1) are c, 8, and 9. Since c > 5 and the total coins must be a multiple of 3, c = 7 is the only possibility. Thus, bag (3,1) has 7, 8, and 9 coins, with an average of 8.

For bag (2,1):

  • The median number of coins is 2.
  • One sack has more than 5 coins, and conditions (i) and (iii) must be satisfied.

Thus, the coins in bag (2,1) are 1, 2, and 9, with an average of 4.

For bag (1,2):

  • The median number of coins is 9.
  • Two sacks have more than 5 coins, and 9 is present, but 1 is not present.

So, the coins in bag (1,2) are c, 9, and 9. Since c is not 1 and is less than 5, for the total to be a multiple of 3, c = 3. Thus, bag (1,2) has 3, 9, and 9 coins, with an average of 7.
From (1), the average in bag (1,1) is 15−4−8=315−4−8=3. From (1), the average in bag (1,3) is 15−3−7=515−3−7=5.

For bag (1,1) with an average of 3:

  • One sack has more than 5 coins, and ** means two conditions are satisfied.
  • It can't be condition (iii) with 9 coins because the total is only 9 (3*3).
  • So, bag (1,1) has 1, 1, and 7 coins, averaging 3.

For bag (1,3) with an average of 5:

  • The sum of coins is 15.
  • The median is 6, with two sacks having more than 5 coins, and * means one condition is satisfied.
  • It can't be condition (ii) because the median is 6, nor condition (iii) because two sacks summing 6 + 9 = 15 leaves no room for the third sack.
  • So, bag (1,3) has 1, 6, and 8 coins, averaging 5.

For bag (3,3) with 0 sacks having more than 5 coins:

  • ** means conditions (i) and (ii) are satisfied.
  • The coins are 1, 1, and c, where c can be 1, 2, 3, or 4.
  • To be a multiple of 3, c must be 1 or 4.
  • Since no other bag can have an average of 1, bag (3,3) has 1, 1, and 1 coins, averaging 1.

For bag (2,3) with an average of 9:

  • The coins are 9, 9, and 9.

For bag (2,2) with an average of 2:

  • The sum is 6, and only one sack is greater than 5.
  • The smallest element should be 1.
  • So, bag (2,2) has 1, 2, and 3 coins, averaging 2.
  Table 
 C-1C-2C-3
R-11,1,7(3)3,9,9(7)1,6,8(5)
R-21,2,9(4)1,2,3(2)9,9,9(9)
R-37,8,9(8)Avg=61,1,1(1)

 

 

 

 

 

 

For bag (3,2) with an average of 6:

  • The sum of coins is 18.
  • Two sacks have more than 5 coins, and ** means two conditions are satisfied.
  • One sack has 1 coin, and one sack has 9 coins.
  • Therefore, bag (3,2) has 1, 8, and 9 coins, summing to 18, with an average of 6.

Final table (number in brackets is the average number of coins per sack in the bag):

  • Bag (1,1): 1, 1, 7 (average = 3)
  • Bag (1,2): 3, 9, 9 (average = 7)
  • Bag (1,3): 1, 6, 8 (average = 5)
  • Bag (2,1): 1, 2, 9 (average = 4)
  • Bag (2,2): 1, 2, 3 (average = 2)
  • Bag (2,3): 9, 9, 9 (average = 9)
  • Bag (3,1): 7, 8, 9 (average = 8)
  • Bag (3,2): 1, 8, 9 (average = 6)
  • Bag (3,3): 1, 1, 1 (average = 1)

Summary of sacks with 1 coin:

  • Bag (1,1): 2 sacks with 1 coin
  • Bag (2,1): 1 sack with 1 coin
  • Bag (2,2): 1 sack with 1 coin
  • Bag (3,2): 1 sack with 1 coin
  • Bag (1,3): 1 sack with 1 coin
  • Bag (3,3): 3 sacks with 1 coin

Total: 2 + 1 + 1 + 1 + 1 + 3 = 9 sacks with 1 coin.

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