2021 AMC 12A Problems/Problem 6

Revision as of 19:14, 11 February 2021 by Pi is 3.14 (talk | contribs) (Video Solution by Hawk Math)


A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$. When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$. How many cards were in the deck originally?

$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$


If the probability of choosing a red card is $\frac{1}{3}$, the red and black cards are in ratio $1:2$. This means at the beginning there are $x$ red cards and $2x$ black cards.

After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the ratio of red to black cards is $1:3$. This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.

So, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.

So the answer is $8+4 = 12 = \boxed{\textbf{(C)}}$.


Video Solution by Hawk Math


Video Solution (Using Probability and System of Equations)


~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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