Understanding the set :- 1) We are given nine boxes arranged in a 3×3 array - in the same way as shown in tables, in the question. 2) There are 3 sacks in each box as well , which implies we need to find three numbers for a particular box. 3) 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 4) The total number of coins in each row is the same. The total number of coins in each column is also the same. 5) 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) 1) Median = 2 2) 1 sack has more than 5 coins 3) 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) 1) Median = 9 2) 2 elements are more than 5. 3) 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) 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 2) 1 sack has more than 5 3) ** => 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) 1) 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 2) Median = 6 and 2 sacks have more than 5 3) * => (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) 1) 0 sacks have more than 5 coins 2) ** => 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) 1) Avg. = 2 2) Sum = 6 3) 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) 2 sacks more than 5 coins 3) ** => 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

CAT (DILR) PYQs

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