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

(Video Solution by Hawk Math)
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--abhinavg0627
 
--abhinavg0627
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==Solution 2 (Answer Choices)==
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<math>\textbf{(A) }6</math>
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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.
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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.
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<math>\textbf{(B) }9</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.
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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) }9</math> is incorrect.
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<math>\textbf{(C) }12</math>
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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.
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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) }15</math> and <math>\textbf{(E) }18.</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.
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<math>\textbf{(D) }15</math>
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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.
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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.
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<math>\textbf{(E) }18</math>
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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) }18</math> is incorrect.
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~MRENTHUSIASM
  
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==

Revision as of 06:15, 12 February 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

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 (Answer Choices)

$\textbf{(A) }6$

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, $\textbf{(A) }6$ is incorrect.


$\textbf{(B) }9$

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, $\textbf{(B) }9$ is incorrect.


$\textbf{(C) }12$

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 $\textbf{(D) }15$ and $\textbf{(E) }18.$ 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.


$\textbf{(D) }15$

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, $\textbf{(D) }15$ is incorrect.


$\textbf{(E) }18$

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, $\textbf{(E) }18$ is incorrect.


~MRENTHUSIASM

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

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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