2021 AMC 12A Problems/Problem 6

1 Problem
2 Solution 1 (Algebra)
3 Solution 2 (Arithmetic)
4 Solution 3 (Observations)
5 Solution 4 (Ratios and Proportions)
6 Solution 5 (Simpler Ratios)
7 Video Solution (Quick and Easy)
8 Video Solution by Aaron He
9 Video Solution by Hawk Math
10 Video Solution by OmegaLearn (Using Probability and System of Equations)
11 Video Solution by TheBeautyofMath
12 See also

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 1 (Algebra)

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

--abhinavg0627

Solution 2 (Arithmetic)

In terms of the number of cards, the original deck is $3$ times the red cards, and the final deck is $4$ times the red cards. So, the final deck is $\frac43$ times the original deck. We are given that 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 (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 4 (Ratios and Proportions)

By looking at the ratio of black cards to the total number of cards, we can say that there are $2x$ black cards and $3x$ cards in total. We can write the probability of getting a black card as $\frac{2x}{3x}$ . But when we increase the number of black cards by 4 the probability becomes $\frac{3}{4}$ . So the equation is: $\frac{2x+4}{3x+4}=\frac{3}{4}$ , $8x+16=9x+12$ , $x=4$ . The answer is $3x=\boxed{\textbf{(C) }12}.$

~Param Gor

Solution 5 (Simpler Ratios)

This could in a way be considered a more efficient/alternate version of Solution 4. Since the problem mentions that there is initially a $\frac{1}{3}$ chance of getting a red card, we can represent the number of red cards as $\frac{x}{3}$ , where $x$ is the number of cards in the deck originally. Since the modification to the deck makes no changes to the number of red cards, you can see that if there is a $\frac{1}{4}$ probability of having a red card in a deck of $x+4$ cards, $\frac{x}{3} = \frac{x+4}{4}$ . Solving this for $x$ gives $x=\boxed{\textbf{(C) }12}.$

~Almond_Oil

Video Solution (Quick and Easy)

https://youtu.be/oYGeTN2-82s

~Education, the Study of Everything

Video Solution by Aaron He

https://www.youtube.com/watch?v=xTGDKBthWsw&t=5m51s

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn (Using Probability and System of Equations)

https://youtu.be/C6x361JPLzU

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/cckGBU2x1zg

~IceMatrix

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 12 Problems and Solutions

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