Difference between revisions of "2021 AMC 12A Problems/Problem 6"
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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. 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>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. WOOHOO! For completeness, we will check <math>\textbf{(D) }</math> and <math>\textbf{(E)} | + | * 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. <b>WOOHOO!!!</b> |
+ | |||
+ | 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 do not have 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>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. |
Revision as of 12:55, 23 March 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 (Arithmetics)
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 do not have 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.