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? | ||
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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>. |
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--abhinavg0627 | --abhinavg0627 | ||
− | ==Solution 2 ( | + | ==Solution 2 (Arithmetic)== |
− | + | 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 | + | ==Solution 3 (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 | + | 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 4 (Ratios and Proportions)== | |
− | + | 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>8x+16=9x+12</math>, | |
− | + | <math>x=4</math>. | |
− | + | The answer is <math>3x=\boxed{\textbf{(C) }12}.</math> | |
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+ | ~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 <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 | |
+ | ==Video Solution (Quick and Easy)== | ||
+ | https://youtu.be/oYGeTN2-82s | ||
− | ~ | + | ~Education, the Study of Everything |
==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 | ||
~ 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}} |
Revision as of 18:14, 19 July 2024
Contents
- 1 Problem
- 2 Solution 1 (Algebra)
- 3 Solution 2 (Arithmetic)
- 4 Solution 3 (Observations)
- 5 Solution 4 (Ratios and Proportions)
- 6 Solution 5 (Simpler ratios)
- 7 Video Solution (Quick and Easy)
- 8 Video Solution by Aaron He
- 9 Video Solution by Hawk Math
- 10 Video Solution by OmegaLearn (Using Probability and System of Equations)
- 11 Video Solution by TheBeautyofMath
- 12 See also
Problem
A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is . When
black cards are added to the deck, the probability of choosing red becomes
. How many cards were in the deck originally?
Solution 1 (Algebra)
If the probability of choosing a red card is , the red and black cards are in ratio
. This means at the beginning there are
red cards and
black cards.
After black cards are added, there are
black cards. This time, the probability of choosing a red card is
so the ratio of red to black cards is
. This means in the new deck the number of black cards is also
for the same
red cards.
So, and
meaning there are
red cards in the deck at the start and
black cards.
So, the answer is .
--abhinavg0627
Solution 2 (Arithmetic)
In terms of the number of cards, the original deck is times the red cards, and the final deck is
times the red cards. So, the final deck is
times the original deck. We are given that adding
cards to the original deck is the same as increasing the original deck by
of itself. Since
cards are equal to
of the original deck, the original deck has
cards.
~MRENTHUSIASM
Solution 3 (Observations)
Suppose there were cards in the deck originally. Now, the deck has
cards, which must be a multiple of
Only is a multiple of
so the answer is
~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 black cards and
cards in total. We can write the probability of getting a black card as
. But when we increase the number of black cards by 4 the probability becomes
. So the equation is:
,
,
.
The answer is
~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 chance of getting a red card, we can represent the number of red cards as
, where
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
probability of having a red card in a deck of
cards,
. Solving this for
gives
~Almond_Oil
Video Solution (Quick and Easy)
~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)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.