# LR - Board Games - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic LR - Board Games. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**Answer the next 4 questions based on the information given below.**

Twenty-five coloured beads are to be arranged in a grid comprising of five rows and five columns. Each cell in the grid must contain exactly one bead. Each bead is coloured either Red, Blue or Green.

While arranging the beads along any of the five rows or along any of the five columns, the rules given below are to be followed:

- Two adjacent beads along the same row or column are always of different colors.
- There is at least one Green bead between any two Blue beads along the same row or column.
- There is at least one Blue and at least one Green bead between any two Red beads along the same row or column.

Every unique, complete arrangement of twenty-five beads is called a configuration.

**CAT 2020 LRDI Slot 2 | LR - Board Games**

The total number of possible configurations using beads of only two colors is:

Answer: 2

**Explanation** :

We can use beads of only two colors.

Since there must be a blue and a green bead between two red beads, we cannot choose one the two colors as red.

Hence, the two colors that can be used are green and blue.

Also, two adjacent beads along the same row or column are always of different colors. Hence, blue and green beads can only be placed alternately.

∴ The following 2 configurations are possible.

**Configuration 1:**

**Configuration 2:**

Hence, 2.

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**CAT 2020 LRDI Slot 2 | LR - Board Games**

What is the maximum possible number of Red beads that can appear in any configuration?

Answer: 9

**Explanation** :

There must be a blue and a green bead between two red beads, we cannot choose one the two colors as red, and

There is at least one Green bead between any two Blue beads along the same row or column.

∴ We can have maximum two red beads in any row or column.

Consider the following configuration

∴ The maximum number of red beads that can be used is 9.

Hence, 9.

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**CAT 2020 LRDI Slot 2 | LR - Board Games**

What is the minimum number of Blue beads in any configuration?

Answer: 6

**Explanation** :

To minimize blue beads, we have to use maximum number of red and green beads.

There must be a blue and a green bead between two red beads, we cannot choose one the two colors as red, and

There is at least one green bead between any two blue beads along the same row or column.

Consider the configuration in the previous questions. We can replace blue beads, which are not in between two red beads, with green beads.

Hence, 6.

Workspace:

**CAT 2020 LRDI Slot 2 | LR - Board Games**

Two Red beads have been placed in ‘second row, third column’ and ‘third row, second column’. How many more Red beads can be placed so as to maximize the number of Red beads used in the configuration?

Answer: 6

**Explanation** :

According to the information given in the questions:

To maximum the number of red beads we can place the remaining beads in the following manner.

∴ Maximum 6 more red beads can be placed.

Hence, 6.

Workspace:

**Answer the following questions based on the information given below.**

Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.

There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.

**CAT 2019 LRDI Slot 2 | LR - Board Games**

What is the total amount of money (in rupees) in the three pouches kept in the first column of the second row?

Answer: 13

**Explanation** :

Consider C1R1: Maximum coins in a pouch = 4 and minimum coins = 2. The third bag could have 2, 3 or 4 coins. Therefore, sum = 8, 9 or 10

Consider C1R2: Maximum coins in a pouch = 5 and minimum coins = 3. The third bag could have 3, 4 or 5 coins. Therefore, sum = 11, 12 or 13

Consider C1 R3: Maximum coins in a pouch = 2 and minimum coins = 1. As the sum = 4, third pouch has 1 coin.

As the sum of coins in the nine pouches in the column are divisible by 9, the coins in C1R1 and C1R2 has to be 10 (i.e., 2, 4, 4) and 13 (i.e., 3, 5, 5) respectively.

Consider column 2:

C2R1: Maximum coins in a pouch = 8 and minimum coins = 6. The third bag could have 6, 7 or 8 coins. Therefore, sum = 20, 21 or 22

C2R2: All the three pouches have one coin each. Sum = 3

C2R3: Maximum coins in a pouch = 2 and minimum coins = 1. The third bag could have 1 or 2 coins. Therefore, sum = 1 or 4

In order to have number of coins in the cells of the column divisible by 9, sum of the coins in C2R1 = 20(i.e., 6, 6, 8) and in C2R3 = 4(i.e., 1,1,2)

Now consider R1

First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.

Consider R2: First two cells together have 16 coins. Coins in the third cell are in the range 6 + 6 + 20 = 32 to 6 + 20 + 20 = 46. Therefore the third cell has to have 38 (i.e., 6, 12, 20) coins.

Consider R3: First two cells together have 8 coins. So the third cell has to have 10 (i.e., 2, 3, 5) coins.

Thus, we have

First column of the second row has 13 coins. i.e., Rs. 13

Answer: 13

Workspace:

**CAT 2019 LRDI Slot 2 | LR - Board Games**

How many pouches contain exactly one coin?

Answer: 8

**Explanation** :

Consider C1R1: Maximum coins in a pouch = 4 and minimum coins = 2. The third bag could have 2, 3 or 4 coins. Therefore, sum = 8, 9 or 10

Consider C1R2: Maximum coins in a pouch = 5 and minimum coins = 3. The third bag could have 3, 4 or 5 coins. Therefore, sum = 11, 12 or 13

Consider C1 R3: Maximum coins in a pouch = 2 and minimum coins = 1. As the sum = 4, third pouch has 1 coin.

As the sum of coins in the nine pouches in the column are divisible by 9, the coins in C1R1 and C1R2 has to be 10 (i.e., 2, 4, 4) and 13 (i.e., 3, 5, 5) respectively.

Consider column 2:

C2R1: Maximum coins in a pouch = 8 and minimum coins = 6. The third bag could have 6, 7 or 8 coins. Therefore, sum = 20, 21 or 22

C2R2: All the three pouches have one coin each. Sum = 3

C2R3: Maximum coins in a pouch = 2 and minimum coins = 1. The third bag could have 1 or 2 coins. Therefore, sum = 1 or 4

In order to have number of coins in the cells of the column divisible by 9, sum of the coins in C2R1 = 20(i.e., 6, 6, 8) and in C2R3 = 4(i.e., 1,1,2)

Now consider R1

First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.

Consider R2: First two cells together have 16 coins. Coins in the third cell are in the range 6 + 6 + 20 = 32 to 6 + 20 + 20 = 46. Therefore the third cell has to have 38 (i.e., 6, 12, 20) coins.

Consider R3: First two cells together have 8 coins. So the third cell has to have 10 (i.e., 2, 3, 5) coins.

Thus, we have

Two pouches in Row 3 Column 1 slot have 1 coin each.

Three pouches in Row 2 Column 2 have 1 coin each.

Two pouches in Row 3 Column 3 slot have 1 coin each.

One pouch in Row 1 Column 3 slot has 1 coin.

Total number of pouches = 2 + 3 + 2 + 1 = 8

Answer: 8

Workspace:

**CAT 2019 LRDI Slot 2 | LR - Board Games**

What is the number of slots for which the average amount (in rupees) of its three pouches is an integer?

Answer: 2

**Explanation** :

Consider C1R1: Maximum coins in a pouch = 4 and minimum coins = 2. The third bag could have 2, 3 or 4 coins. Therefore, sum = 8, 9 or 10

Consider C1R2: Maximum coins in a pouch = 5 and minimum coins = 3. The third bag could have 3, 4 or 5 coins. Therefore, sum = 11, 12 or 13

Consider C1 R3: Maximum coins in a pouch = 2 and minimum coins = 1. As the sum = 4, third pouch has 1 coin.

As the sum of coins in the nine pouches in the column are divisible by 9, the coins in C1R1 and C1R2 has to be 10 (i.e., 2, 4, 4) and 13 (i.e., 3, 5, 5) respectively.

Consider column 2:

C2R1: Maximum coins in a pouch = 8 and minimum coins = 6. The third bag could have 6, 7 or 8 coins. Therefore, sum = 20, 21 or 22

C2R2: All the three pouches have one coin each. Sum = 3

C2R3: Maximum coins in a pouch = 2 and minimum coins = 1. The third bag could have 1 or 2 coins. Therefore, sum = 1 or 4

In order to have number of coins in the cells of the column divisible by 9, sum of the coins in C2R1 = 20(i.e., 6, 6, 8) and in C2R3 = 4(i.e., 1,1,2)

Now consider R1

First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.

Consider R2: First two cells together have 16 coins. Coins in the third cell are in the range 6 + 6 + 20 = 32 to 6 + 20 + 20 = 46. Therefore the third cell has to have 38 (i.e., 6, 12, 20) coins.

Consider R3: First two cells together have 8 coins. So the third cell has to have 10 (i.e., 2, 3, 5) coins.

Thus, we have

For average amount to be an integer, we need to consider slots having total number of coins in the pouches divisible by 3.

There are two such slot i.e., (row 1 column 3) and (row 2 column 2).

Answer: 2.

Workspace:

**CAT 2019 LRDI Slot 2 | LR - Board Games**

The number of slots for which the total amount in its three pouches strictly exceeds Rs. 10 is

Answer: 3

**Explanation** :

Consider column 2:

C2R2: All the three pouches have one coin each. Sum = 3

Now consider R1

First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.

Thus, we have

The slots for which the total amount in its three pouches strictly exceeds Rs. 10 are Row 1 Column 2, Row 2 Column 1, Row 2 Column 3.

Answer: 3.

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**Answer the following question based on the information given below.**

You are given an n × n square matrix to be filled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the first or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.

**CAT 2018 LRDI Slot 1 | LR - Board Games**

What is the minimum number of different numerals needed to fill a 3×3 square matrix?

Answer: 4

**Explanation** :

A number say ‘x’ can be filled in four corner cells. Another number say ‘y’ can be used in the central cell. Now the remaining cells are middle cells of top row, bottom row, left column and right column. Two more numerals can be used to fill these cells. Assume that ‘z’ and ‘w’ are those numerals.

Thus,

Thus, four different numerals are required.

Therefore, the required answer is 4.

Answer: 4

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**CAT 2018 LRDI Slot 1 | LR - Board Games**

What is the minimum number of different numerals needed to fill a 5 × 5 square matrix?

Answer: 4

**Explanation** :

A number say ‘x’ can be filled in four corner cells. Another number say ‘y’ can be used in the central cell. Now the remaining cells are middle cells of top row, bottom row, left column and right column. Two more numerals can be used to fill these cells. Assume that ‘z’ and ‘w’ are those numerals.

Thus,

So, at least 4 different numerals are required for a 3 × 3 matrix.

Top row i.e., the first row of the 5 × 5 matrix can be (x, z, x, z, x). Similarly, other cells from the second and the third row can be filled.

For the first column i.e., the leftmost column, the fourth cell from top can be filled with the number from the second cell from top i.e. w and the bottom cell can be filled with the number from the top most cell of the column i.e., x. Similarly, remaining cells of the remaining columns can be filled.

Thus, four different numerals are sufficient to fill a 5 × 5 matrix.

Therefore, the required answer is 4.

Answer: 4

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Board Games**

Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to fill a 5×5 matrix?

- A.
9

- B.
16

- C.
25

- D.
4

Answer: Option D

**Explanation** :

A number say ‘x’ can be filled in four corner cells. Another number say ‘y’ can be used in the central cell. Now the remaining cells are middle cells of top row, bottom row, left column and right column. Two more numerals can be used to fill these cells. Assume that ‘z’ and ‘w’ are those numerals.

Thus,

4 different numerals are required to fill a 5 × 5 matrix. As one is allowed to make only one mistake, it is possible to change exactly one entry such that all other conditions are not violated.

Hence, option 4.

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Board Games**

Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a 5×5 square matrix?

- A.
25

- B.
16

- C.
9

- D.
4

Answer: Option C

**Explanation** :

Assuming that the particular cell is not a cell at one of the corner, it has 8 adjacent cells.

Hence, 8 + 1 = 9 different numerals are required to fill the particular cell and the adjacent cells.

Earlier we have seen that 4 different numerals are required to fill a 5 × 5 matrix. Hence, 9 different numerals will be definitely sufficient to fill the 5 × 5 matrix.

Hence, option 3.

Workspace:

**Answer the following question based on the information given below.**

In an 8 × 8 chessboard a queen placed anywhere can attack another piece if the piece is present in the same row, or in the same column or in any diagonal position in any possible 4 directions, provided there is no other piece in between in the path from the queen to that piece.

The columns are labelled a to h (left to right) and the rows are numbered 1 to 8 (bottom to top). The position of a piece given by the combination of column and row labels. For example, position c5 means that the piece is cth column and 5th row.

**CAT 2017 LRDI Slot 2 | LR - Board Games**

If the queen is at c5, and the other pieces at positions c2, g1, g3, g5 and a3, how many are under attack by the queen? There are no other pieces on the board.

- A.
2

- B.
3

- C.
4

- D.
5

Answer: Option C

**Explanation** :

In the diagram given above let Q represent the queen at cell C_{5}. Now let us suppose pieces P_{1}, P_{2}, P_{3}, P_{4} and P_{5} are at cells. C_{2}, G_{1}, G_{3}, G_{5} and A_{3}. As can seen that Q can attack P_{1} as they are is the same column and can attack P_{4} as both are in the same row. Further Q can attack P_{5} and P_{2} as both are in a diagonal position to Q. So it cannot be attacked by Q. So 4 pieces i.e., P_{1}, P_{2}, P_{4} and P_{5} can attacked by Q.

Hence, option 3.

Workspace:

**CAT 2017 LRDI Slot 2 | LR - Board Games**

If the other pieces are only at positions a1, a3, b4, d7, h7 and h8, then which of the following positions of the queen results in the maximum number of pieces being under attack?

- A.
f8

- B.
a7

- C.
c1

- D.
d3

Answer: Option D

**Explanation** :

Let P_{1}, P_{2}, P_{3}, P_{4}, P_{5} and P_{6} be the 6 pieces at positions A_{1}, A_{3}, B_{4}, D_{7}, H_{7} and H_{8} respectively.

1] If the queen is at position F_{8} it can attack P_{6} as it is in the same row and P_{3} as it in a diagonal position. However it cannot attack P_{2}, which is in a diagonal position as P_{3} is placed between P_{2} and Q along the same diagonal. So a total of 2 pieces can be attacked if the queen is at position F_{8}.

2] If the queen is at position A_{7} it can directly attack P_{4} as it is in the same row and P_{2} as it is in the same column. However it cannot attack P_{5} because P_{4} is in between the queen and P_{5}. Similarly it cannot attack P_{1} because P_{2} is in between the Queen and P_{1}. So a total of 2 pieces can be attacked by the queen if it is at position A_{7}.

3] If the queen is at position C_{1}, then it can only attack P_{1} which is in the same row and P_{2} which is along the same diagonal as the queen. So at positions C_{1}, the queen can attack only 2 pieces.

4] If the queen is at position D_{3} then it can attack P_{2} along the same row, P_{4} along the same column and P_{5} along the same diagonal i.e., a total of 3 pieces. So out of the given options, the queen at position D_{3} can attack maximum number of pieces.

Hence, option 4.

Workspace:

**CAT 2017 LRDI Slot 2 | LR - Board Games**

If the other pieces are only at positions a1, a3, b4, d7, h7 and h8, then from how many positions the queen cannot attack any of the pieces?

- A.
0

- B.
3

- C.
4

- D.
6

Answer: Option C

**Explanation** :

All cells in rows 1, 3, 4, 7 and 8 and columns A, B, D and H will be eliminated as one of the ways the queen can the attack us from the same row or column. If we check for cell E_{2}, F_{2}, G_{2} and G_{5}, queen will not be able to attack any of the 6 pieces. So, from a total of 4 positions, the queen cannot attack any of the 6 pieces.

Hence, option 4.

Workspace:

**CAT 2017 LRDI Slot 2 | LR - Board Games**

Suppose the queen is the only piece on the board and it is at position d5.

In how many positons can another piece be placed on the board such that it is safe from attack from the queen?

- A.
32

- B.
35

- C.
36

- D.
37

Answer: Option C

**Explanation** :

Let Q represent the position of the queen at cell D_{5}. Now none of the pieces can be along column D or along row 5 i.e., a total of 8 + 8 – 1 = 15 cells (∵ D_{5} is common to column D and row 5 it is counted twice, hence we subtract 1) Now cells along the diagonal of D5 have to be eliminated since they can be directly attacked by the Queen from that cell. Therefore cells A_{8}, B_{7}, C_{6}, E_{4}, F_{3}, G_{2}, H_{1} and G_{8}, F_{7}, E_{6}, C_{4}, B_{3}, A_{2} i.e., a total of 13 cells are not to be considered. So a total of 64 – 28 = 36 positions are considered safe from attack if the queen is at position D_{5}

Hence, option 3.

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