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 . When black cards are added to the deck, the probability of choosing red becomes . How many cards were in the deck originally?

Solution(Pretty much the same as Solution 1)

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 .

Solution 1

If the probability of choosing a red card is , the red and black cards are in ratio . This means at the beginning there are red cards and black cards.

After black cards are added, there are black cards. This time, the probability of choosing a red card is so the ratio of red to black cards is . This means in the new deck the number of black cards is also for the same red cards.

So, and meaning there are red cards in the deck at the start and black cards.

So the answer is .

--abhinavg0627

Solution 2 (Arithmetics)

For the number of cards, the final deck is times the original deck. In other words, adding cards to the original deck is the same as increasing the original deck by of itself. Since cards are equal to of the original deck, the original deck has cards.

~MRENTHUSIASM

Solution 3 (Answer Choices)

Solution 3.1 (Observations)

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

Only is a multiple of So, the answer is

~MRENTHUSIASM

Solution 3.2 (Plug in the Answer Choices)

• If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.

• If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.

• If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is correct. WOOHOO!!!

For completeness, we will check and 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 cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.

• If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.

~MRENTHUSIASM

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 (Using Probability and System of Equations)

https://youtu.be/C6x361JPLzU

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/cckGBU2x1zg

~IceMatrix

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.