# PE 1 - Board Games | LR - Board Games

**Answer the next 3 questions based on the information given below.**

Minesweeper is a game played on a 8 × 8 matrix. The columns are numbered A to H from left to right and rows are numbered 1 to 8 from top to bottom.

A cell in B^{th} column and 5^{th} row is termed as B5.

There is a mine in one of cells. If a player clicks on any of the cells in the same column, row or any diagonal as the cell with mine the player will lose.

The player has to click on all the safe cells (Safe cells are those cells which are neither in the same column, row or any diagonal as the cell with mine).

The game is designed such that first two clicks by a player are always safe and mine is placed accordingly after that.

**PE 1 - Board Games | LR - Board Games**

What is the difference between the maximum and minimum number of safe cells?

- (a)
8

- (b)
6

- (c)
5

- (d)
4

- (e)
None of these

Answer: Option B

**Explanation** :

Maximum number of safe cells will be when the mine is placed at one of the corners.

If mine is placed at A8, cells marked with × denote unsafe cells.

∴ Number of unsafe cells = 22

∴ Number of safe cells = 64 – 22 = 42.

Minimum number of safe cells will be when the mine is placed in the middle.

If mine is placed at D4, cells marked with × denote unsafe cells.

∴ Number of unsafe cells = 28

∴ Number of safe cells = 64 – 28 = 36.

⇒ Difference between maximum and minimum safe cells = 42 – 36 = 6.

Hence, option (b).

Workspace:

**PE 1 - Board Games | LR - Board Games**

If the first 2 clicks are on A5 and G2, while the third click C3 is also safe. In how many cells mine can be placed now?

Answer: 14

**Explanation** :

Here × denotes safe cells.

Remaining cells = 14.

Mine can be placed in any of these 14 cells.

Hence, 14.

Workspace:

**PE 1 - Board Games | LR - Board Games**

If the first 2 clicks are on D4 and H7, in how many ways can 2 different mines be placed each in a different cell?

Answer: 420

**Explanation** :

Here × denotes safe cells.

∴ 2 mines can be placed on any of the remaining 21 cells.

⇒ Number of ways = ^{21}C_{2} × 2! = 420.

Hence, 210.

Workspace:

**PE 1 - Board Games | LR - Board Games**

**Answer the next 3 questions based on the information given below.**

The following questions relate to a game to be played by you and your friend. The game consists of a 4 × 4 board (see below) where each cell contains a positive integer. You and your friend make moves alternately. A move by any of the players consists of splitting the current board configuration into two equal halves and retaining one of them. In your moves you are allowed only to split the board vertically and decide to retain either the left half or the right half. Your friend, in his/her moves, can only split the board horizontally and can retain either the lower half or the upper half. After two moves by each player a single cell will remain which can no longer be split and the number in that cell will be treated as the gain (in rupees) of the person who has started the game. A sample game is shown below.

So, your gain is Re.1. With the same initial board configuration as above and assuming that you have to make the first move, answer the following questions.

- (a)
(retain upper) (retain lower)

- (b)
(retain lower) (retain upper)

- (c)
(retain upper) (retain upper)

- (d)
(retain lower) (retain lower)

Answer: Option D

**Explanation** :

The initial board is:

Since your first move is (retain right), the board configuration after the move is as shown below

Since your next move is (retain left), your final profit will be one from (2, 9, 4 and 1).

Since your friend wants to minimize your profit, he/she would like you to get Rs. 1.

Hence, he/she would retain lower in first attempt and again retain lower in second attempt.

Your gain now will be Rs. 1.

Hence, the moves are (retain lower), (retain lower).

Hence, option (d).

Workspace:

**PE 1 - Board Games | LR - Board Games**

With the initial board configuration as shown in the question, if both of you select your moves intelligently then at the end of the game, your gain will be?

- (a)
Rs. 4

- (b)
Rs. 3

- (c)
Rs. 2

- (d)
None of these

Answer: Option A

**Explanation** :

The initial board is:

Since both of you play intelligently, your objective is to get the maximum possible gain while your friend's objective is to give you the minimum possible gain.

**Case I:** You select (retain right).

Since your friend also plays intelligently, he will select (retain lower) so that your gain does not cross Rs. 6. If he selects (retain upper), your gain can possibly go up to Rs. 9

Now, you will select (retain left) to get a maximum possible gain of Rs. 6.

Finally, in order to minimise your gain, your friend will select (retain upper).

This will limit your gain to Rs. 4.

Thus, with intelligent moves, your maximum gain in this case is Rs. 4.

**Case II**: You select (retain left).

Now, if your friend selects (retain upper), the profit you can earn (both playing intelligently) is Rs. 4 when you again select (retain right).

On the other hand, if your friend selects (retain lower), the profit you can earn (both playing intelligently) is Rs. 2 when you again select (retain right).

Since your friend also plays intelligently, he will select (retain lower) so that your profit is only Rs. 2.

Since you want to maximize your gain, your first intelligent move will be (retain right) i.e., case I.

As per Case I, your gain will be Rs. 4.

Hence, option (a).

Workspace:

**PE 1 - Board Games | LR - Board Games**

In the previous question, if your first move is (retain left), then whatever moves your friend may select, you can always force a gain of no less than?

Answer: 2

**Explanation** :

Consider the solution to previous question of this set.

if your first move is (retain left), then whatever moves your friend may select, you can always force a gain of no less than Rs. 2.

Hence, 2.

Workspace:

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

A standard deck of 52 cards contains four suits – diamonds (♦), clubs (♣), spades (♠) and hearts (♥) – of 13 cards each. The 13 cards in each suit are 2, 3, 4, …, 10 and J, Q, K and A.

Six friends: Sumit, Nikhar, Rachit, Vishal, Achal and Romesh, are playing a game of cards sitting around a circular table. In his turn each player draws exactly one card from the deck (without replacing) and performs one of the following actions:

- Keeps it with him, if the card is a heart (♥),
- Passes the card to the player on his right, if the card is a diamond (♦),
- Passes it to the player on his left, if it is a club (♣),
- Passes it to the player opposite him, if it is a spade (♠)

A round is completed when all six friends have picked one card each.

The following table gives the information about the cards each friend had at the end of certain number of rounds.

- Akash is sitting to the right of Romesh who is sitting to the right of Vishal who is sitting to the right of Achal who is sitting to the right of Nikhar who is sitting to the right of Sumit.
- The cards ‘5’ and ‘Q’ that Achal has are both of the same suit.
- Of the cards whose suit in not known, 4 of them are spade.

**PE 1 - Board Games | LR - Board Games**

Who drew the king which Vishal had?

- (a)
Vishal

- (b)
Romesh

- (c)
Akash

- (d)
Achal

Answer: Option B

**Explanation** :

Total 24 cards are there with the six friends. Since 6 cards are picked in each round, it means that 4 rounds would have been completed.

⇒ Each friend picks 4 cards (1 in each of the 4 rounds).

Let us rearrange the table based on their seating arrangement.

If a player draws a heart, he keeps it for himself. Therefore, all the heart cards must have been drawn by the person who has it.

If a player draws a club, he passes it to the player on his left. Therefore, all the club cards must have been drawn by the person on the right of the one who has it.

If a player draws a spade, he passes it to the player opposite him. Therefore, all the spade cards must have been drawn by the person opposite to the one who has it.

If a player draws a diamond, he passes it to the player on his right. Therefore, all the diamond cards must have been drawn by the person on the left of the one who has it.

Romesh has two 3’s whose suit is unknown. The suit of these 3’s can be either spade or diamonds as club and heart are already present. 3 of spade must have come from Nikhar while 3 of diamond must have come from Akash.

From (i): Now both the 5 and Q with Achal cannot be clubs since Nikhar cannot draw any more cards. They cannot be heart (already drawn). They both cannot be diamonds as Vishal cannot draw 2 more cards. Hence, they both will be spades which were drawn by Akash.

The 4 with Sumit cannot be heart and spades (already drawn). It cannot be diamonds since Nikhar cannot draw any more cards. Hence, it has to be clubs which was drawn by Akash.

Now the 10 with Romesh cannot be hearts (10 of hearts is already present). Hence, it could have been drawn by Vishal or Akash or Nikhar. But Akash and Nikhar cannot draw any more cards, hence it must have been drawn by Vishal and it would have been a club.

From (ii): There are a total of 5 spades. We already know 6 spades and need one more spade.

Now, Achal has to draw two more cards.

**Case 1**: Achal draws A and K

For this A has to be diamonds while K should be clubs. This leaves 6 which must be spade, but Achal cannot draw another card now. Hence, this case is rejected.

**Case 4**: Achal draws 6 and K

For this 6 has to be spades while K should be clubs. This leaves A which must be drawn by Romesh which means A is spades. Now we will have total 8 spades which is not possible. Hence, this case is rejected.

**Case 3**: Achal draws A and 6

For this A has to be diamonds while 6 should be spades. This leaves K which must be picked by Romesh, hence K should be diamonds. This case is possible.

∴ Romesh drew both the kings.

Hence, option (b).

Workspace:

**PE 1 - Board Games | LR - Board Games**

Total how many clubs were drawn?

Answer: 4

**Explanation** :

Consider the solution to first question of the set.

Sumit, Achal, Romesh and Akash drew 1 club each.

Hence, 4.

Workspace:

**PE 1 - Board Games | LR - Board Games**

How many players drew at least one card of all 4 suits.

Answer: 2

**Explanation** :

Consider the solution to first question of the set.

Sumit and Achal drew 4 cards of 4 different suits.

Hence, 2.

Workspace:

**PE 1 - Board Games | LR - Board Games**

The least number of cards were drawn from which suit?

- (a)
Hearts (♥)

- (b)
Spades (♠)

- (c)
Diamonds (♦)

- (d)
Clubs (♣)

Answer: Option D

**Explanation** :

Consider the solution to first question of the set.

Number of cards drawn of each suit is:

♥ - 8

♣ - 4

♠ - 7

♦ - 5

Hence, option (d).

Workspace:

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