Solving the Card Suit Distribution Problem

This problem involves determining the number of cards of each suit a player holds, based on the total number of cards, the count of black and red cards, and specific relationships between the number of cards in certain suits.

We are given the following information:

• Total number of cards held by the player: 13
• Number of black cards: 7
• Number of red cards: 6
• Suits: Spades (black), Clubs (black), Hearts (red), Diamonds (red)
• The number of diamonds is twice the number of spades.
• The number of hearts is twice the number of diamonds.

Our goal is to find the number of clubs the player holds.

Setting Up the Problem with Variables

Let's use variables to represent the number of cards in each suit:

• Let $S$ be the number of spades.
• Let $C$ be the number of clubs.
• Let $H$ be the number of hearts.
• Let $D$ be the number of diamonds.

Formulating Equations from the Given Information

Based on the problem statement, we can write down the following equations:

1. Total cards: $S + C + H + D = 13$
2. Black cards: $S + C = 7$ (Since Spades and Clubs are black)
3. Red cards: $H + D = 6$ (Since Hearts and Diamonds are red)
4. Relationship between Diamonds and Spades: $D = 2S$
5. Relationship between Hearts and Diamonds: $H = 2D$

Solving for the Number of Cards in Each Suit

We can use the relationships between suits and the total counts of red and black cards to solve for the number of cards in each suit.

Finding the Number of Red Cards per Suit

We know that $H + D = 6$. We also have the relationships $D = 2S$ and $H = 2D$. Let's substitute the relationship $H = 2D$ into the red cards equation:

Substituting $H = 2D$ into $H + D = 6$:

$\qquad 2D + D = 6$

$\qquad 3D = 6$

Now, we can solve for $D$:

$\qquad D = \frac{6}{3}$

$\qquad D = 2$

So, the player holds 2 diamonds.

Finding the Number of Spades

We know the relationship $D = 2S$. We have found that $D = 2$. Let's substitute the value of $D$ to find $S$:

Substituting $D = 2$ into $D = 2S$:

$\qquad 2 = 2S$

Now, we can solve for $S$:

$\qquad S = \frac{2}{2}$

$\qquad S = 1$

So, the player holds 1 spade.

Finding the Number of Hearts

We know the relationship $H = 2D$. We have found that $D = 2$. Let's substitute the value of $D$ to find $H$:

Substituting $D = 2$ into $H = 2D$:

$\qquad H = 2 \times 2$

$\qquad H = 4$

So, the player holds 4 hearts.

Let's verify the number of red cards: $H + D = 4 + 2 = 6$. This matches the given information.

Finding the Number of Clubs

We know that the total number of black cards is 7, and black suits are Spades and Clubs. So, $S + C = 7$. We have found that $S = 1$. Let's substitute the value of $S$ to find $C$:

Substituting $S = 1$ into $S + C = 7$:

$\qquad 1 + C = 7$

Now, we can solve for $C$:

$\qquad C = 7 - 1$

$\qquad C = 6$

So, the player holds 6 clubs.

Let's verify the total number of cards: $S + C + H + D = 1 + 6 + 4 + 2 = 13$. This also matches the given information.

Summary of Cards Held

Based on our calculations, the player holds the following number of cards for each suit:

Total Black Cards: $1 + 6 = 7$

Total Red Cards: $4 + 2 = 6$

Total Cards: $7 + 6 = 13$

The relationships hold true:

• Diamonds ($D=2$) is twice Spades ($S=1$) (2 = 2 × 1).
• Hearts ($H=4$) is twice Diamonds ($D=2$) (4 = 2 × 2).

The number of clubs the player holds is 6.

This corresponds to option 3.

Revision Table: Card Suit Problem

Steps Followed:

1. Used $H = 2D$ in $H + D = 6$ to find $D$.
2. Used $D = 2S$ to find $S$.
3. Used $H = 2D$ to find $H$.
4. Used $S + C = 7$ to find $C$.

Additional Information: Playing Card Suits

A standard deck of playing cards contains 52 cards divided into four suits:

• Spades (♠): One of the two black suits.
• Clubs (♣): One of the two black suits.
• Hearts (♥): One of the two red suits.
• Diamonds (♦): One of the two red suits.

Each suit typically has 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King.

This problem simplifies the concept by only focusing on the total count of cards in each suit within a player's hand, rather than dealing with a full deck or specific card ranks. It's a basic application of algebraic problem-solving using simultaneous equations derived from the given conditions.