Difference between revisions of "2021 AMC 12A Problems/Problem 6"
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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. | 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) } | + | 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) }.</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. |
Revision as of 05:18, 12 February 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
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 (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 
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.
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 Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution (Using Probability and System of Equations)
~ pi_is_3.14
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. 
