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Question

A, B, C, D and E play a game of cards. A says to B, "If you give me three cards, you will have as many as E has and if I give you three cards, you will have as many as D has". A and B together have 10 cards more than what D and E together have. If B has two cards more than what C has and the total number of cards be 133, how many cards does B have?

The correct answer is

25

Solving a Card Game Word Problem Using Algebra

This problem involves determining the number of cards each player has based on several given conditions. We will use algebraic equations to represent these conditions and solve for the unknown quantities.

Understanding the Card Game Conditions

We are given information about the relative number of cards held by five players: A, B, C, D, and E. Let's denote the number of cards each player has by $n_A, n_B, n_C, n_D,$ and $n_E$ respectively.

  • Condition 1: If B gives 3 cards to A, B will have as many cards as E.
  • Condition 2: If A gives 3 cards to B, B will have as many cards as D.
  • Condition 3: A and B together have 10 cards more than D and E together.
  • Condition 4: B has two cards more than C.
  • Condition 5: The total number of cards held by all five players is 133.

Setting up the Algebraic Equations

Let's translate each condition into a mathematical equation:

  1. "If you give me three cards (B to A), you will have as many as E has": B's cards become $n_B - 3$. A's cards become $n_A + 3$. The condition states $n_B - 3 = n_E$. Equation 1: $n_B - n_E = 3$
  2. "...and if I give you three cards (A to B), you will have as many as D has": A's cards become $n_A - 3$. B's cards become $n_B + 3$. The condition states $n_B + 3 = n_D$. Equation 2: $n_D - n_B = 3$
  3. "A and B together have 10 cards more than what D and E together have": $(n_A + n_B) = (n_D + n_E) + 10$. Equation 3: $n_A + n_B - n_D - n_E = 10$
  4. "If B has two cards more than what C has": $n_B = n_C + 2$. Equation 4: $n_B - n_C = 2$
  5. "and the total number of cards be 133": $n_A + n_B + n_C + n_D + n_E = 133$. Equation 5: $n_A + n_B + n_C + n_D + n_E = 133$

We now have a system of 5 linear equations with 5 variables.

Solving the System of Equations to Find B's Cards

Our goal is to find the value of $n_B$. We can express the other variables in terms of $n_B$ and substitute them into the total sum equation (Equation 5).

From Equation 1, we get: $n_E = n_B - 3$.

From Equation 2, we get: $n_D = n_B + 3$.

From Equation 4, we get: $n_C = n_B - 2$.

Now, let's use Equation 3: $n_A + n_B - n_D - n_E = 10$. Substitute the expressions for $n_D$ and $n_E$:

$\qquad n_A + n_B - (n_B + 3) - (n_B - 3) = 10$

$\qquad n_A + n_B - n_B - 3 - n_B + 3 = 10$

$\qquad n_A - n_B = 10$

This gives us an expression for $n_A$ in terms of $n_B$: $n_A = n_B + 10$.

Now we have expressions for $n_A, n_C, n_D,$ and $n_E$ all in terms of $n_B$:

  • $n_A = n_B + 10$
  • $n_C = n_B - 2$
  • $n_D = n_B + 3$
  • $n_E = n_B - 3$

Substitute these expressions into Equation 5 (the total number of cards):

$\qquad (n_B + 10) + n_B + (n_B - 2) + (n_B + 3) + (n_B - 3) = 133$

Combine the terms with $n_B$ and the constant terms:

$\qquad (n_B + n_B + n_B + n_B + n_B) + (10 - 2 + 3 - 3) = 133$

$\qquad 5n_B + 8 = 133$

Now, solve for $n_B$:

$\qquad 5n_B = 133 - 8$

$\qquad 5n_B = 125$

$\qquad n_B = \frac{125}{5}$

$\qquad n_B = 25$

So, B has 25 cards.

Verifying the Solution

Let's find the number of cards for all players based on $n_B = 25$ and check if they satisfy all conditions:

  • $n_B = 25$
  • $n_A = n_B + 10 = 25 + 10 = 35$
  • $n_C = n_B - 2 = 25 - 2 = 23$
  • $n_D = n_B + 3 = 25 + 3 = 28$
  • $n_E = n_B - 3 = 25 - 3 = 22$

Check conditions:

  • Condition 1: $n_B - 3 = n_E \implies 25 - 3 = 22$. (Correct)
  • Condition 2: $n_B + 3 = n_D \implies 25 + 3 = 28$. (Correct)
  • Condition 3: $n_A + n_B = n_D + n_E + 10 \implies 35 + 25 = 28 + 22 + 10 \implies 60 = 50 + 10 \implies 60 = 60$. (Correct)
  • Condition 4: $n_B = n_C + 2 \implies 25 = 23 + 2$. (Correct)
  • Condition 5: $n_A + n_B + n_C + n_D + n_E = 133 \implies 35 + 25 + 23 + 28 + 22 = 133$. $35 + 25 = 60$ $23 + 28 = 51$ $60 + 51 + 22 = 111 + 22 = 133$. (Correct)

All conditions are satisfied, confirming our calculated value for $n_B$ is correct.

The number of cards B has is 25.

Revision Table: Key Equations

Relationship Equation
B gives 3 to A, B has E's cards $n_B - 3 = n_E$
A gives 3 to B, B has D's cards $n_B + 3 = n_D$
A & B vs D & E difference $n_A + n_B = n_D + n_E + 10$
B vs C difference $n_B = n_C + 2$
Total cards $n_A + n_B + n_C + n_D + n_E = 133$

Additional Information: Solving Word Problems with Linear Equations

Word problems often describe relationships between unknown quantities. To solve them using algebra, follow these general steps:

  1. Read Carefully: Understand the problem and identify what is being asked.
  2. Define Variables: Assign variables (like $n_A, n_B$, etc.) to the unknown quantities you need to find.
  3. Translate to Equations: Convert each statement or condition in the problem into a mathematical equation using the defined variables.
  4. Solve the System: Use methods like substitution or elimination to solve the system of equations. The goal is to find the value of the variable(s) you are looking for.
  5. Check Your Answer: Substitute the values you found back into the original equations or the problem description to ensure they satisfy all conditions.
  6. State the Answer: Clearly state the answer to the specific question asked in the problem.

This card game problem is a good example of setting up multiple linear equations and solving them simultaneously, often using substitution to reduce the number of variables.

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Important Questions from Quant Based Puzzle

  1. A number is subtracted from 4 times of it and then, the number obtained is added to its (the resultant’s) next number. If this gives the answer as 91, what was the original number?

  2. When twice of a number added to 3 is multiplied by 5 and added to the number itself, it gives 158. What is the square of that number?

  3. In a class of 72 students, the number of boys is twice the number of girls. Find the number of boys.

  4. Two years ago, T was twice as old as P. P is thrice as old as R. In five years, P will be 29. What is the present age of T?

  5. When a number is added to its multiple of 5 and its square, the sum of these three numbers is 91. Find the number.

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