Boxes and Sacks of Coins

CAT 2023 Slot 2 · DILR · Hard · Data Interpretation

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Passage / data set

There are nine boxes arranged in a 3×33 \times 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.

Table 1: Median of coins in three sacks

1st column2nd column3rd column
1st row96
2nd row2
3rd row8

Table 2: Sacks with > 5 coins and condition flags

1st column2nd column3rd column
1st row1∗∗1^{**}2∗2^*2∗2^*
2nd row1∗∗1^{**}0∗0^*3∗3^*
3rd row3∗3^*2∗∗2^{**}0∗∗0^{**}

Question 1 of 5

What is the total number of coins in all the boxes in the 3rd row?

  1. A.

    36

  2. B.

    30

  3. C.

    45

  4. D.

    15

Answer

C

Explanation

Since the average number of coins per sack in each of the 9 boxes are distinct integers, the 9 box averages must be the integers 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9.

The total sum of all box averages is 1+2+3+4+5+6+7+8+9=451 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. Since each box contains 3 sacks, the total number of coins across all 9 boxes is 45×3=13545 \times 3 = 135.

Since the total number of coins in each row is equal, each row must contain 135/3=45135 / 3 = 45 coins.

Question 2 of 5

How many boxes have at least one sack containing 9 coins?

  1. A.

    3

  2. B.

    5

  3. C.

    4

  4. D.

    8

Answer

C

Explanation

By analyzing the condition flags (∗* and ∗∗**) along with the medians and averages of each box, exactly 4 boxes satisfy condition (iii) (maximum coins in a sack is 9). Therefore, 4 boxes have at least one sack containing 9 coins.

Question 3 of 5

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

4

Explanation

By determining the exact contents or properties of each box's sacks, we find that exactly 4 boxes have an average equal to their median.

Question 4 of 5

How many sacks have exactly one coin?

Answer

6

Explanation

By detailed evaluation of the sack contents across all 9 boxes, there are 6 sacks that contain exactly 1 coin.

Question 5 of 5

In how many boxes do all three sacks contain different numbers of coins?

Answer

5

Explanation

Counting the boxes where all three sacks have distinct values gives a total of 5 boxes.

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