CAT 2023 Slot 2 DILR Set 1 Q1–Q5

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2023 Slot 2, with 5 questions (Grids & Matrices). Answers are CAT's official key.

The set

This set relies on a figure in the original paper, which is not reproduced here. The questions and answers are below.

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.
[Figure in the original paper]

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

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) 45 Grids & Matrices MCQ

Solution

  1. Averages are distinct integers from 1 to 9, so they are exactly 1..9 (box totals 3, 6, ..., 27, grand total 135); every row and column of averages sums to 15.
  2. Every row's averages sum to 15, so each row holds \(3 \times 15 = 45\) coins.
  3. Row 3 (averages 8, 6, 1): \(24 + 18 + 3 = 45\).

Shortcut. Grand total \(3(1+2+\dots+9) = 135\), split equally over 3 rows gives 45.

Question 2

How many boxes have at least one sack containing 9 coins?
  1. A
    3
  2. B
    5
  3. C
    4
  4. D
    8

Answer: (B) 5 Grids & Matrices MCQ

Solution

  1. Averages are distinct integers from 1 to 9, so they are exactly 1..9 (box totals 3, 6, ..., 27, grand total 135); every row and column of averages sums to 15.
  2. Clues: (3,3) is 0** so its sacks are \((1,1,x)\), \(x\in\{1,4\}\); (2,1) is 1** with median 2, forcing \((1,2,9)\), avg 4; (3,1) is 3* with median 8, forcing \((7,8,9)\), avg 8; the all-9 box (avg 9) has 3 sacks above 5 and only condition iii, so it is (2,3).
  3. Column 1 gives (1,1) = 3; row 2 gives (2,2) = 2, so (3,3) = 1 and (3,2) = 6; then (1,2) = 7 and (1,3) = 5. Averages: row 1: 3, 7, 5; row 2: 4, 2, 9; row 3: 8, 6, 1.
  4. Sacks from the medians and stars: (1,1) \((1,1,7)\); (1,2) \((3,9,9)\); (1,3) \((1,6,8)\); (2,1) \((1,2,9)\); (2,2) \((1,2,3)\); (2,3) \((9,9,9)\); (3,1) \((7,8,9)\); (3,2) \((1,8,9)\); (3,3) \((1,1,1)\).
  5. Boxes containing a 9: (1,2), (2,1), (2,3), (3,1), (3,2), so 5.

Question 3

For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same?

TITA: type in the answer. There are no options and no negative marking.

Answer: 4 Grids & Matrices TITA

Solution

  1. Averages are distinct integers from 1 to 9, so they are exactly 1..9 (box totals 3, 6, ..., 27, grand total 135); every row and column of averages sums to 15.
  2. Clues: (3,3) is 0** so its sacks are \((1,1,x)\), \(x\in\{1,4\}\); (2,1) is 1** with median 2, forcing \((1,2,9)\), avg 4; (3,1) is 3* with median 8, forcing \((7,8,9)\), avg 8; the all-9 box (avg 9) has 3 sacks above 5 and only condition iii, so it is (2,3).
  3. Column 1 gives (1,1) = 3; row 2 gives (2,2) = 2, so (3,3) = 1 and (3,2) = 6; then (1,2) = 7 and (1,3) = 5. Averages: row 1: 3, 7, 5; row 2: 4, 2, 9; row 3: 8, 6, 1.
  4. Sacks from the medians and stars: (1,1) \((1,1,7)\); (1,2) \((3,9,9)\); (1,3) \((1,6,8)\); (2,1) \((1,2,9)\); (2,2) \((1,2,3)\); (2,3) \((9,9,9)\); (3,1) \((7,8,9)\); (3,2) \((1,8,9)\); (3,3) \((1,1,1)\).
  5. Average = median holds for (2,2) \((1,2,3)\), (2,3) \((9,9,9)\), (3,1) \((7,8,9)\) and (3,3) \((1,1,1)\), so 4 boxes.

Question 4

How many sacks have exactly one coin?

TITA: type in the answer. There are no options and no negative marking.

Answer: 9 Grids & Matrices TITA

Solution

  1. Averages are distinct integers from 1 to 9, so they are exactly 1..9 (box totals 3, 6, ..., 27, grand total 135); every row and column of averages sums to 15.
  2. Clues: (3,3) is 0** so its sacks are \((1,1,x)\), \(x\in\{1,4\}\); (2,1) is 1** with median 2, forcing \((1,2,9)\), avg 4; (3,1) is 3* with median 8, forcing \((7,8,9)\), avg 8; the all-9 box (avg 9) has 3 sacks above 5 and only condition iii, so it is (2,3).
  3. Column 1 gives (1,1) = 3; row 2 gives (2,2) = 2, so (3,3) = 1 and (3,2) = 6; then (1,2) = 7 and (1,3) = 5. Averages: row 1: 3, 7, 5; row 2: 4, 2, 9; row 3: 8, 6, 1.
  4. Sacks from the medians and stars: (1,1) \((1,1,7)\); (1,2) \((3,9,9)\); (1,3) \((1,6,8)\); (2,1) \((1,2,9)\); (2,2) \((1,2,3)\); (2,3) \((9,9,9)\); (3,1) \((7,8,9)\); (3,2) \((1,8,9)\); (3,3) \((1,1,1)\).
  5. Sacks with 1 coin: (1,1) 2, (1,3) 1, (2,1) 1, (2,2) 1, (3,2) 1, (3,3) 3, giving 9.

Question 5

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

TITA: type in the answer. There are no options and no negative marking.

Answer: 5 Grids & Matrices TITA

Solution

  1. Averages are distinct integers from 1 to 9, so they are exactly 1..9 (box totals 3, 6, ..., 27, grand total 135); every row and column of averages sums to 15.
  2. Clues: (3,3) is 0** so its sacks are \((1,1,x)\), \(x\in\{1,4\}\); (2,1) is 1** with median 2, forcing \((1,2,9)\), avg 4; (3,1) is 3* with median 8, forcing \((7,8,9)\), avg 8; the all-9 box (avg 9) has 3 sacks above 5 and only condition iii, so it is (2,3).
  3. Column 1 gives (1,1) = 3; row 2 gives (2,2) = 2, so (3,3) = 1 and (3,2) = 6; then (1,2) = 7 and (1,3) = 5. Averages: row 1: 3, 7, 5; row 2: 4, 2, 9; row 3: 8, 6, 1.
  4. Sacks from the medians and stars: (1,1) \((1,1,7)\); (1,2) \((3,9,9)\); (1,3) \((1,6,8)\); (2,1) \((1,2,9)\); (2,2) \((1,2,3)\); (2,3) \((9,9,9)\); (3,1) \((7,8,9)\); (3,2) \((1,8,9)\); (3,3) \((1,1,1)\).
  5. All three different: (1,3), (2,1), (2,2), (3,1), (3,2), so 5 boxes.

Same concept in other years

See every question on 3x3 grid of boxes each holding three sacks/pouches of coins, equal row/column totals.

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