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

Revision as of 05:26, 12 February 2021

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

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

~MRENTHUSIASM

Solution 3 (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, $\textbf{(A) }6$ 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, $\textbf{(B) }9$ 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 $\textbf{(E)}.$ If you decide to use this approach on the real test, you don't have 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, $\textbf{(D) }15$ 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, $\textbf{(E) }18$ is incorrect.

~MRENTHUSIASM

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

@@ Line 17: / Line 17: @@
 --abhinavg0627
+==Solution 2 (Observations)==
+Suppose there were <math>x</math> cards in the deck originally. Now, the deck has <math>x+4</math> cards, which must be a multiple of <math>4.</math>
-==Solution 2 (Answer Choices)==
+Only <math>12+4</math> is a multiple of <math>4.</math> So, the answer is <math>\boxed{\textbf{(C) }12}.</math>
+~MRENTHUSIASM
+==Solution 3 (Answer Choices)==
 If there were <math>6</math> cards in the deck originally, then there were <math>6\cdot\frac13=2</math> red cards in the deck originally.

2021 AMC 12A (Problems • Answer Key • Resources)
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