CAT 2023 Slot 2 · DILR · Set 1: Coins in Sacks in a 3×3 Array of Boxes

The full set and all 5 questions (Q1–Q5) from CAT 2023 Slot 2, with answers and solutions.

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.

Question 1 · Caselet DI

What is the total number of coins in all the boxes in the 3rd row?
  1. A
    36
  2. B
    30
  3. C
    15
  4. D
    45
Pick an option, then check.
Answer & solution

Answer: Option D

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

From Step 1, each row has 45 coins. (Row 3: 7 + 8 + 9 + 1 + 8 + 9 + 1 + 1 + 1 = 45.)

Hence, the answer is option 4.

Question 2 · Caselet DI

How many boxes have at least one sack containing 9 coins?
  1. A
    3
  2. B
    4
  3. C
    5
  4. D
    8
Pick an option, then check.
Answer & solution

Answer: Option C

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

Five boxes contain a sack with 9 coins.

Hence, the answer is option 3.

Question 3 · Caselet DI

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.

Question 4 · Caselet DI

How many sacks have exactly one coin?
Answer & solution

Answer: 9

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

Total sacks with exactly one coin = 2 + 1 + 1 + 1 + 1 + 3 = 9.

Hence, the answer is 9.

Question 5 · Caselet DI

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

Answer: 5

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

All three sacks differ in five boxes: (1,3), (2,1), (2,2), (3,1) and (3,2).

Hence, the answer is 5.

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