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

(→‎Solution 3 (Observations))
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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:
 
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>
 
<math>\frac{2x+4}{3x+4}=\frac{3}{4}</math>
−
Solving for x:
 
 
<math>8x+16=9x+12</math>
 
<math>8x+16=9x+12</math>
 
<math>x=4</math>
 
<math>x=4</math>
−
Now we know what the ratio is scaled by so taking our previous expression <math>3x</math> and substituting our value for <math>x</math> we get that the answer is <math>x=\boxed{\textbf{(C) }12}.</math>
+
Substituting to <math>3x</math>.
 +
The answer is <math>3x=\boxed{\textbf{(C) }12}.</math>
  
 
~Param Gor, Terrill Middle School
 
~Param Gor, Terrill Middle School

Revision as of 15:55, 5 May 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: $\frac{2x+4}{3x+4}=\frac{3}{4}$ $8x+16=9x+12$ $x=4$ Substituting to $3x$. The answer is $3x=\boxed{\textbf{(C) }12}.$

~Param Gor, Terrill Middle School

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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