# Difference between revisions of "2021 AMC 12A Problems/Problem 6"

## Problem

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?

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

## Solution

The probability of choosing a red card is 1/3. That means that there are x red cards and 2x black cards. Then, we have 4 black cards added to the deck. Now, the number of black cards is 2x+4. We have 2x+4 is 3x. Solving for x, we get x = 4. We want the original number of cards, and the original deck is x+2x = 3x. Therefore, the answer is $\boxed{(C)12}$.

## Solution 1

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 $8+4 = 12 = \boxed{\textbf{(C)}}$.

--abhinavg0627

## Solution 2 (Arithmetics)

For the number of cards, the final deck is $\frac43$ times the original deck. In other words, adding $4$ cards to the original deck is the same as increasing the original deck by $\frac13$ of itself. Since $4$ cards are equal to $\frac13$ of the original deck, the original deck has $4\cdot3=\boxed{\textbf{(C) }12}$ cards.

~MRENTHUSIASM

### Solution 3.1 (Observations)

Suppose there were $x$ cards in the deck originally. Now, the deck has $x+4$ cards, which must be a multiple of $4.$

Only $12+4=16$ is a multiple of $4.$ So, the answer is $x=\boxed{\textbf{(C) }12}.$

~MRENTHUSIASM

### Solution 3.2 (Plug in the Answer Choices)

• If there were $6$ cards in the deck originally, then there were $6\cdot\frac13=2$ red cards in the deck originally. Now, the deck has $6+4=10$ cards, and $\frac{2}{10}\neq\frac{1}{4}.$ So, $\textbf{(A) }$ is incorrect.
• If there were $9$ cards in the deck originally, then there were $9\cdot\frac13=3$ red cards in the deck originally. Now, the deck has $9+4=13$ cards, and $\frac{3}{13}\neq\frac{1}{4}.$ So, $\textbf{(B) }$ is incorrect.
• If there were $12$ cards in the deck originally, then there were $12\cdot\frac13=4$ red cards in the deck originally. Now, the deck has $12+4=16$ cards, and $\frac{4}{16}=\frac{1}{4}.$ So, $\boxed{\textbf{(C) }12}$ is correct. WOOHOO!!!

For completeness, we will check $\textbf{(D) }$ and $\textbf{(E)}$ too. If you decide to use this solution on the real test, then you will not need to do that, as you want to save more time.

• If there were $15$ cards in the deck originally, then there were $15\cdot\frac13=5$ red cards in the deck originally. Now, the deck has $15+4=19$ cards, and $\frac{5}{19}\neq\frac{1}{4}.$ So, $\textbf{(D) }$ is incorrect.
• If there were $18$ cards in the deck originally, then there were $18\cdot\frac13=6$ red cards in the deck originally. Now, the deck has $18+4=22$ cards, and $\frac{6}{22}\neq\frac{1}{4}.$ So, $\textbf{(E) }$ is incorrect.

~MRENTHUSIASM

~ pi_is_3.14

~IceMatrix