CAT 2023 Slot 2 · DILR · Question 3

CAT 2023 Slot 2 previous year question on Caselet DI, with the answer and a step-by-step solution.

Caselet DIData InterpretationType in the answer (TITA)Hard

Set 1: Coins in Sacks in a 3×3 Array of Boxes — this question belongs to a set. Read the full set with all 5 questions →

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?

Answer & solution

Answer: 4

Solving the set

Step 1 – The nine box averages are distinct integers, and each lies between 1 and 9 (every sack has 1 to 9 coins). So the averages are exactly 1, 2, …, 9 and the total is 3 × 45 = 135 coins. Equal row totals and equal column totals mean every row and every column has 45 coins, i.e., the averages in each row and column add up to 15.

Step 2 – Read each clue: the number in Table 2 = sacks with more than 5 coins; * = exactly one of (min 1, median 1, max 9); ** = at least two of them.

Step 3 – Fix the boxes one by one (the order below is the order of deduction):

Box (row, column)	Clues	Deduction	Sacks	Average
(2,1)	median 2, 1**	One sack > 5; median 2 ≠ 1, so min 1 and max 9	1, 2, 9	4
(3,1)	median 8, 3*	All > 5, so min ≠ 1; the one condition is max 9; (a + 17) divisible by 3 with 6 ≤ a ≤ 8	7, 8, 9	8
(1,2)	median 9, 2*	Two sacks are 9, third ≤ 5; max 9 is the one condition, so min ≠ 1; (a + 18) divisible by 3	3, 9, 9	7
(3,2)	2**	Two sacks > 5 so median > 5; needs min 1 and max 9: 1, m, 9 with (10 + m) divisible by 3	1, 8, 9	6
(3,3)	0**	No sack > 5, so max ≠ 9; needs min 1 and median 1; average from magic sums = 1	1, 1, 1	1
(1,1)	1**	Average 3 (column 1: 3 + 4 + 8 = 15); sum 9 with one sack > 5 and two conditions	1, 1, 7	3
(2,2)	0*	Average 2 (column 2); sum 6, all ≤ 5, min 1 but median ≠ 1	1, 2, 3	2
(1,3)	median 6, 2*	Average 5 (row 1); sum 15 = a + 6 + c with c ≥ 6; only (1, 6, 8) meets exactly one condition	1, 6, 8	5
(2,3)	3*	Average 9 (row 2)	9, 9, 9	9

Step 4 – Box averages (each row and column adds up to 15):

Averages	Column 1	Column 2	Column 3	Row total (coins)
Row 1	3	7	5	15 × 3 = 45
Row 2	4	2	9	45
Row 3	8	6	1	45

Step 5 – Final contents of every box:

Sacks	Column 1	Column 2	Column 3
Row 1	1, 1, 7	3, 9, 9	1, 6, 8
Row 2	1, 2, 9	1, 2, 3	9, 9, 9
Row 3	7, 8, 9	1, 8, 9	1, 1, 1

This question

Box	Sacks	Has a 9?	Average = median?	Sacks with 1 coin	All different?
(1,1)	1, 1, 7	No	3 vs 1 – No	2	No
(1,2)	3, 9, 9	Yes	7 vs 9 – No	0	No
(1,3)	1, 6, 8	No	5 vs 6 – No	1	Yes
(2,1)	1, 2, 9	Yes	4 vs 2 – No	1	Yes
(2,2)	1, 2, 3	No	2 = 2 – Yes	1	Yes
(2,3)	9, 9, 9	Yes	9 = 9 – Yes	0	No
(3,1)	7, 8, 9	Yes	8 = 8 – Yes	0	Yes
(3,2)	1, 8, 9	Yes	6 vs 8 – No	1	Yes
(3,3)	1, 1, 1	No	1 = 1 – Yes	3	No
Total		5	4	9	5

The average equals the median in four boxes: (2,2), (2,3), (3,1) and (3,3).

Hence, the answer is 4.

CAT 2023 Slot 2 · DILR · Question 3

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