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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>
  
−
==Solution 1==
+
==Solution 1 (Algebra)==
 
If the probability of choosing a red card is <math>\frac{1}{3}</math>, the red and black cards are in ratio <math>1:2</math>. This means at the beginning there are <math>x</math> red cards and <math>2x</math> black cards.
 
If the probability of choosing a red card is <math>\frac{1}{3}</math>, the red and black cards are in ratio <math>1:2</math>. This means at the beginning there are <math>x</math> red cards and <math>2x</math> black cards.
  
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So, <math>3x = 2x + 4</math> and <math>x=4</math> meaning there are <math>4</math> red cards in the deck at the start and <math>2(4) = 8</math> black cards.  
 
So, <math>3x = 2x + 4</math> and <math>x=4</math> meaning there are <math>4</math> red cards in the deck at the start and <math>2(4) = 8</math> black cards.  
  
−
So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>.
+
So, the answer is <math>8+4 = 12 = \boxed{\textbf{(C) }12}</math>.
−
 
 
  
 
--abhinavg0627
 
--abhinavg0627
  
−
==Solution 2 (Arithmetics)==
+
==Solution 2 (Arithmetic)==
−
For the number of cards, the final deck is <math>\frac43</math> times the original deck. 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.
+
In terms of the number of cards, the original deck is <math>3</math> times the red cards, and the final deck is <math>4</math> times the red cards. So, the final deck is <math>\frac43</math> times the original deck. We are given that 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 (Answer Choices)==
+
==Solution 3 (Observations)==
−
===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>
  
−
Only <math>12+4=16</math> is a multiple of <math>4.</math> So, the answer is <math>x=\boxed{\textbf{(C) }12}.</math>
+
Only <math>12+4=16</math> is a multiple of <math>4,</math> so the answer is <math>x=\boxed{\textbf{(C) }12}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
===Solution 3.2 (Plug in the Answer Choices)===
+
==Solution 4 (Ratios and Proportions)==
−
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.
+
By looking at the ratio of black cards to the total number of cards, we can say that there are <math>2x</math> black cards and <math>3x</math> cards in total. We can write the probability of getting a black card as <math>\frac{2x}{3x}</math>. But when we increase the number of black cards by 4 the probability becomes <math>\frac{3}{4}</math>. So the equation is:
−
 
+
<math>\frac{2x+4}{3x+4}=\frac{3}{4}</math>,
−
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.
+
<math>8x+16=9x+12</math>,
−
 
+
<math>x=4</math>.
 +
The answer is <math>3x=\boxed{\textbf{(C) }12}.</math>
  
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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.
+
~Param Gor
  
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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.
+
==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 <math>\frac{1}{3}</math> chance of getting a red card, we can represent the number of red cards as <math>\frac{x}{3}</math>, where <math>x</math> 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 <math>\frac{1}{4}</math> probability of having a red card in a deck of <math>x+4</math> cards, <math>\frac{x}{3} = \frac{x+4}{4}</math>. Solving this for <math>x</math> gives <math>x=\boxed{\textbf{(C) }12}.</math>
  
 +
~Almond_Oil
  
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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.
+
==Video Solution (Quick and Easy)==
 +
https://youtu.be/oYGeTN2-82s
  
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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.
+
~Education, the Study of Everything
−
 
 
−
 
 
−
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>18</math> cards in the deck originally, then there were <math>18\cdot\frac13=6</math> red cards in the deck originally.
 
−
 
 
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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.
 
−
 
 
−
 
 
−
~MRENTHUSIASM
 
  
 
==Video Solution by Aaron He==
 
==Video Solution by Aaron He==
 
https://www.youtube.com/watch?v=xTGDKBthWsw&t=5m51s
 
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
  
−
== Video Solution (Using Probability and System of Equations) ==
+
== Video Solution by OmegaLearn (Using Probability and System of Equations) ==
 
https://youtu.be/C6x361JPLzU
 
https://youtu.be/C6x361JPLzU
  

Latest revision as of 22:51, 19 July 2024

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

See also

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