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2021 AMC 12A Problems/Problem 6: Difference between revisions

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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 <imath>\frac13</imath>. When <imath>4</imath> black cards are added to the deck, the probability of choosing red becomes <imath>\frac14</imath>. How many cards were in the deck originally?


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


<math>\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18</math>
==Solution 1 (Algebra)==
If the probability of choosing a red card is <imath>\frac{1}{3}</imath>, the red and black cards are in ratio <imath>1:2</imath>. This means at the beginning there are <imath>x</imath> red cards and <imath>2x</imath> black cards.


==Solution==
After <imath>4</imath> black cards are added, there are <imath>2x+4</imath> black cards. This time, the probability of choosing a red card is <imath>\frac{1}{4}</imath> so the ratio of red to black cards is <imath>1:3</imath>. This means in the new deck the number of black cards is also <imath>3x</imath> for the same <imath>x</imath> red 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.


After <math>4</math> black cards are added, there are <math>2x+4</math> black cards. This time, the probability of choosing a red card is <math>\frac{1}{4}</math> so the ratio of red to black cards is <math>1:3</math>. This means in the new deck the number of black cards is also <math>3x</math> for the same <math>x</math> red cards.
So, <imath>3x = 2x + 4</imath> and <imath>x=4</imath> meaning there are <imath>4</imath> red cards in the deck at the start and <imath>2(4) = 8</imath> 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 <imath>8+4 = 12 = \boxed{\textbf{(C) }12}</imath>.


So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>.
--abhinavg0627
 
==Solution 2 (Arithmetic)==
In terms of the number of cards, the original deck is <imath>3</imath> times the red cards, and the final deck is <imath>4</imath> times the red cards. So, the final deck is <imath>\frac43</imath> times the original deck. We are given that adding <imath>4</imath> cards to the original deck is the same as increasing the original deck by <imath>\frac13</imath> of itself. Since <imath>4</imath> cards are equal to <imath>\frac13</imath> of the original deck, the original deck has <imath>4\cdot3=\boxed{\textbf{(C) }12}</imath> cards.
 
~MRENTHUSIASM
 
==Solution 3 (Observations)==
Suppose there were <imath>x</imath> cards in the deck originally. Now, the deck has <imath>x+4</imath> cards, which must be a multiple of <imath>4.</imath>
 
Only <imath>12+4=16</imath> is a multiple of <imath>4,</imath> so the answer is <imath>x=\boxed{\textbf{(C) }12}.</imath>
 
~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 <imath>2x</imath> black cards and <imath>3x</imath> cards in total. We can write the probability of getting a black card as <imath>\frac{2x}{3x}</imath>. But when we increase the number of black cards by 4 the probability becomes <imath>\frac{3}{4}</imath>. So the equation is:
<imath>\frac{2x+4}{3x+4}=\frac{3}{4}</imath>,
<imath>8x+16=9x+12</imath>,
<imath>x=4</imath>.
The answer is <imath>3x=\boxed{\textbf{(C) }12}.</imath>
 
~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 <imath>\frac{1}{3}</imath> chance of getting a red card, we can represent the number of red cards as <imath>\frac{x}{3}</imath>, where <imath>x</imath> 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 <imath>\frac{1}{4}</imath> probability of having a red card in a deck of <imath>x+4</imath> cards, <imath>\frac{x}{3} = \frac{x+4}{4}</imath>. Solving this for <imath>x</imath> gives <imath>x=\boxed{\textbf{(C) }12}.</imath>
 
~Almond_Oil


==Solution 6 (Observation)==
Notice that in a deck with 1 red card and 2 black cards, the probability of a red ball is <imath>\frac{1}{3}</imath>


--abhinavg0627
Adding a single black card would make the chance <imath>\frac{1}{4}</imath>
 
If we scale this deck by 4, making so that there are 4 red cards and 8 black cards, the original chance is still <imath>\frac{1}{3}</imath>
 
Adding 4 black cards will make the chance <imath>\frac{1}{4}</imath>
 
The answer is <imath>\boxed{\textbf{(C) }12}</imath>.
 
 
==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


==Note==
See [[2021 AMC 12A Problems/Problem 1|problem 1]].
==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 12:09, 11 November 2025

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

Solution 6 (Observation)

Notice that in a deck with 1 red card and 2 black cards, the probability of a red ball is $\frac{1}{3}$

Adding a single black card would make the chance $\frac{1}{4}$

If we scale this deck by 4, making so that there are 4 red cards and 8 black cards, the original chance is still $\frac{1}{3}$

Adding 4 black cards will make the chance $\frac{1}{4}$

The answer is $\boxed{\textbf{(C) }12}$.


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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