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

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==Problem==
 
==Problem==
 
A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is <math>\frac13</math>. When <math>4</math> black cards are added to the deck, the probability of choosing red becomes <math>\frac14</math>. How many cards were in the deck originally?
 
A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is <math>\frac13</math>. When <math>4</math> black cards are added to the deck, the probability of choosing red becomes <math>\frac14</math>. How many cards were in the deck originally?
 
  
 
<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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So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>.
 
So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>.
 
  
 
--abhinavg0627
 
--abhinavg0627
  
==Solution 2 (Arithmetics)==
+
==Solution 2 (Arithmetic)==
For the number of cards, the final deck is <math>\frac43</math> of the original deck. Adding <math>4</math> cards to the original deck is the same as increasing the original deck by <math>\frac13.</math> So, the original deck has <math>\boxed{\textbf{(C) }12}</math> cards.
+
For the number of cards, the final deck is <math>\frac43</math> times the original deck. In other words, adding <math>4</math> cards to the original deck is the same as increasing the original deck by <math>\frac13</math> of itself. Since <math>4</math> cards are equal to <math>\frac13</math> of the original deck, the original deck has <math>4\cdot3=\boxed{\textbf{(C) }12}</math> cards.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
==Solution 3 (Observations)==
+
==Solution 3 (Answer Choices)==
 +
===Solution 3.1 (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>
 
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>
  
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
==Solution 4 (Answer Choices)==
+
===Solution 3.2 (Plug in the 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.
+
* 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. Now, the deck has <math>6+4=10</math> cards, and <math>\frac{2}{10}\neq\frac{1}{4}.</math> So, <math>\textbf{(A) }</math> is incorrect.
 
 
Now, the deck has <math>6+4=10</math> cards, and <math>\frac{2}{10}\neq\frac{1}{4}.</math> So, <math>\textbf{(A) }6</math> is incorrect.
 
  
 +
* If there were <math>9</math> cards in the deck originally, then there were <math>9\cdot\frac13=3</math> red cards in the deck originally. Now, the deck has <math>9+4=13</math> cards, and <math>\frac{3}{13}\neq\frac{1}{4}.</math> So, <math>\textbf{(B) }</math> is incorrect.
  
If there were <math>9</math> cards in the deck originally, then there were <math>9\cdot\frac13=3</math> red cards in the deck originally.
+
* If there were <math>12</math> cards in the deck originally, then there were <math>12\cdot\frac13=4</math> red cards in the deck originally. Now, the deck has <math>12+4=16</math> cards, and <math>\frac{4}{16}=\frac{1}{4}.</math> So, <math>\boxed{\textbf{(C) }12}</math> is correct. <i><b>WOOHOO!!!</b></i>
  
Now, the deck has <math>9+4=13</math> cards, and <math>\frac{3}{13}\neq\frac{1}{4}.</math> So, <math>\textbf{(B) }9</math> is incorrect.
+
For completeness, we will check <math>\textbf{(D) }</math> and <math>\textbf{(E)}</math> 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 <math>15</math> cards in the deck originally, then there were <math>15\cdot\frac13=5</math> red cards in the deck originally. Now, the deck has <math>15+4=19</math> cards, and <math>\frac{5}{19}\neq\frac{1}{4}.</math> So, <math>\textbf{(D) }</math> is incorrect.
  
If there were <math>12</math> cards in the deck originally, then there were <math>12\cdot\frac13=4</math> red cards in the deck originally.
+
* If there were <math>18</math> cards in the deck originally, then there were <math>18\cdot\frac13=6</math> red cards in the deck originally. Now, the deck has <math>18+4=22</math> cards, and <math>\frac{6}{22}\neq\frac{1}{4}.</math> So, <math>\textbf{(E) }</math> is incorrect.
 
 
Now, the deck has <math>12+4=16</math> cards, and <math>\frac{4}{16}=\frac{1}{4}.</math> So, <math>\boxed{\textbf{(C) }12}</math> is correct. WOOHOO! For completeness, we will check <math>\textbf{(D) }</math> and <math>\textbf{(E)}.</math> 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 <math>15</math> cards in the deck originally, then there were <math>15\cdot\frac13=5</math> red cards in the deck originally.
 
 
 
Now, the deck has <math>15+4=19</math> cards, and <math>\frac{5}{19}\neq\frac{1}{4}.</math> So, <math>\textbf{(D) }15</math> is incorrect.
 
 
 
 
 
If there were <math>18</math> cards in the deck originally, then there were <math>18\cdot\frac13=6</math> red cards in the deck originally.
 
 
 
Now, the deck has <math>18+4=22</math> cards, and <math>\frac{6}{22}\neq\frac{1}{4}.</math> So, <math>\textbf{(E) }18</math> is incorrect.
 
 
 
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
 +
==Video Solution by Aaron He==
 +
https://www.youtube.com/watch?v=xTGDKBthWsw&t=5m51s
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
 
https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
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~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/cckGBU2x1zg
 +
 +
~IceMatrix
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}
 
{{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:12, 22 April 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 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 (Arithmetic)

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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