Question

A, B, C and D play a game of cards. A says to B, "If I give you 8 cards, you will have as many as C has and I shall have 3 less than what C has. Also, if I take 6 cards from C, I shall have twice as many as D has". If B and D together have 50 cards, how many cards has A got?

• A. 40
• B. 37
• C. 27
• D. 23

Answer 1

Answer: (A)

40

Answer 2

The problem can be solved by setting up a system of linear equations based on the given statements. Let A, B, C, and D represent the number of cards each person has.

From the first statement: "If I (A) give you (B) 8 cards, you will have as many as C has and I shall have 3 less than what C has."

1. B's new cards = B + 8. This equals C's cards: $B + 8 = C \quad \cdots (1)$
2. A's new cards = A - 8. This is 3 less than C's cards: $A - 8 = C - 3 \quad \cdots (2)$

From the second statement: "Also, if I (A) take 6 cards from C, I shall have twice as many as D has."

3. A's new cards = A + 6. This is twice D's cards: $A + 6 = 2D \quad \cdots (3)$

From the third statement: "If B and D together have 50 cards"

4. $B + D = 50 \quad \cdots (4)$

Now, we solve this system of equations:

From equation (2), express A in terms of C: $A = C - 3 + 8$ $A = C + 5 \quad \cdots (5)$

Substitute C from equation (1) into equation (5): $A = (B + 8) + 5$ $A = B + 13 \quad \cdots (6)$

From equation (3), express D in terms of A: $D = \frac{A + 6}{2} \quad \cdots (7)$

Substitute A from equation (6) into equation (7): $D = \frac{(B + 13) + 6}{2}$ $D = \frac{B + 19}{2} \quad \cdots (8)$

Now substitute D from equation (8) into equation (4): $B + \frac{B + 19}{2} = 50$

Multiply the entire equation by 2 to eliminate the fraction: $2B + (B + 19) = 100$ $3B + 19 = 100$ $3B = 100 - 19$ $3B = 81$ $B = \frac{81}{3}$ $B = 27$

Finally, substitute the value of B back into equation (6) to find A: $A = B + 13$ $A = 27 + 13$ $A = 40$

Thus, A has 40 cards.