Difference between revisions of "2021 AMC 12A Problems/Problem 6"
Abhinavg0627 (talk | contribs) (→Solution) |
MRENTHUSIASM (talk | contribs) m (→Solution 1: Deleted an extra space.) |
||
(34 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
==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? | ||
− | |||
<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== | + | ==Solution 1== |
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. | ||
Line 14: | Line 13: | ||
So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>. | So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>. | ||
+ | --abhinavg0627 | ||
+ | |||
+ | ==Solution 2 (Arithmetic)== | ||
+ | For the number of cards, the final deck is <math>\frac43</math> times the original deck. In other words, 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 | ||
+ | |||
+ | ==Solution 3 (Answer Choices)== | ||
+ | ===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> | ||
+ | |||
+ | 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 | |
+ | |||
+ | ===Solution 3.2 (Plug in the Answer Choices)=== | ||
+ | * 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. 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. | ||
+ | |||
+ | * 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. 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. | ||
+ | |||
+ | * 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. 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. <i><b>WOOHOO!!!</b></i> | ||
+ | |||
+ | For completeness, we will check <math>\textbf{(D) }</math> and <math>\textbf{(E)}</math> too. If you decide to use this solution on the real test, then you will not need to do that, as you want to save more time. | ||
+ | |||
+ | * 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. 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== | ||
+ | https://www.youtube.com/watch?v=xTGDKBthWsw&t=5m51s | ||
+ | ==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 | ||
+ | |||
+ | ==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}} |
Latest revision as of 21:12, 22 April 2021
Contents
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
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)
For the number of cards, the final deck is times the original deck. In other words, 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 (Answer Choices)
Solution 3.1 (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 3.2 (Plug in the Answer Choices)
- If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.
- If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.
- If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is correct. WOOHOO!!!
For completeness, we will check and too. If you decide to use this solution on the real test, then you will not need to do that, as you want to save more time.
- If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.
- If there were cards in the deck originally, then there were red cards in the deck originally. Now, the deck has cards, and So, is incorrect.
~MRENTHUSIASM
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 (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.