A player holds $13$ cards of four suits, of which seven are black and six are red. There are twice as many diamonds as spades and twice as many hearts as diamonds. How many clubs does he hold?

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Solution

The correct option is B $6$
Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts.
Let the number of spades be $x$ . Then, number of clubs $= (7 - x)$ .
Number of diamonds $= 2 \times$ number of spades $= 2 x$ ;
Number of hearts $= 2 \times$ number of diamonds $= 4 x$ .
Total number of cards $= x + 2 x + 4 x + 7 - x = 6 x + 7$ .
Therefore $6 x + 7 = 13 \Leftrightarrow 6 x = 6 \Leftrightarrow x - 1$ .
Hence, number of clubs $= (7 - x) = 6$ .

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A player holds 13 cards of four suits, of which seven are black and six are red. There are twice as many diamonds as spades and twice as many hearts as diamonds. How many clubs does he hold?

A player holds $13$ cards of four suits, of which seven are black and six are red. There are twice as many diamonds as spades and twice as many hearts as diamonds. How many clubs does he hold?