A, B, C, D and E play a game of cards. A tells B that if B gives A 3 cards, B will then have as many cards as A has now. A also says that if D takes 5 cards from B, D will then have as many cards as E has. A and C together have twice as many cards as E. B and D together also have the same number of cards as A and C together. If the total number of cards with all five players is 150, how many cards does C have?

Introduction / Context:
This is a challenging logical and algebraic reasoning problem involving five players A, B, C, D and E sharing a total of 150 cards. Several conditions describe how the number of cards with one player changes when cards are given or taken, and how these counts compare to others. The goal is to determine how many cards C has. Problems like this are common in higher level aptitude tests and require careful equation setup and simultaneous solution.

Given Data / Assumptions:

• Let the cards with A, B, C, D and E be a, b, c, d, and e respectively.
• When B gives A 3 cards, B will have as many cards as A has now: b - 3 = a.
• When D takes 5 cards from B, D will have as many cards as E: d + 5 = e.
• A and C together have twice as many cards as E: a + c = 2e.
• B and D together have the same number of cards as A and C together: b + d = a + c.
• Total cards: a + b + c + d + e = 150.

Concept / Approach:
We convert each statement into an algebraic equation in the variables a, b, c, d and e. This gives a system of linear equations. The strategy is to use substitution and elimination to reduce the system step by step until we find c, the number of cards with C. Because there are five variables and five independent equations, the system has a unique solution. Once we find values for all variables, we verify them against the original statements.

Step-by-Step Solution:
Step 1: From “B gives A 3 cards and then B has as many as A has now”, write b - 3 = a.Step 2: From “D takes 5 cards from B and then has as many as E”, write d + 5 = e.Step 3: From “A and C together have twice as many as E”, write a + c = 2e.Step 4: From “B and D together have the same as A and C together”, write b + d = a + c.Step 5: Total cards: a + b + c + d + e = 150.Step 6: Solve this system (by substitution or elimination). The consistent solution is a = 32, b = 35, c = 28, d = 25, e = 30.Step 7: Therefore C has c = 28 cards.

Verification / Alternative check:
Verify each condition using the solution a = 32, b = 35, c = 28, d = 25, e = 30. First, b - 3 = 35 - 3 = 32 which equals a, so the first statement is satisfied. Second, d + 5 = 25 + 5 = 30 which equals e, so the second statement is satisfied. Third, a + c = 32 + 28 = 60 and 2e = 2 * 30 = 60, so the third condition holds. Fourth, b + d = 35 + 25 = 60, which equals a + c = 60, so the fourth condition holds. Finally, total cards a + b + c + d + e = 32 + 35 + 28 + 25 + 30 = 150, matching the given total. All conditions are met, so the solution is fully consistent.

Why Other Options Are Wrong:
Option B (29), Option C (31) and Option D (35): Any choice other than 28 for C would break at least one of the relationships between A, B, C, D and E when you attempt to find consistent values for the other variables. No other value for c satisfies all the equations simultaneously.

Common Pitfalls:
This problem is error prone because of the number of variables and relationships. A common mistake is to misinterpret the phrase “B will have as many as A has at this moment” and write the equation incorrectly. Another pitfall is to forget to use the total number of cards to close the system. To handle such questions efficiently, it is helpful to label each equation clearly, avoid arithmetic slips, and consider solving them systematically, even with scratch algebra or a matrix approach if you are comfortable with it.

Final Answer:
C has 28 cards.