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

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<math>\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18</math>
 
<math>\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18</math>
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==Solution==
  
 
==Solution 1==
 
==Solution 1==

Revision as of 19:14, 24 March 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

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 (Answer Choices)

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

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

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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