Difference between revisions of "2021 AMC 12A Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | + | 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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| + | After <math>4</math> black cards are added, there are <math>2x+4</math> black cards. This time, the probability of choosing a red card is <math>\frac{1}{4}</math> so the ratio of red to black cards is <math>1:3</math>. This means in the new deck the number of black cards is also <math>3x</math> for the same <math>x</math> red 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. | ||
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| + | So the answer is <math>8+4 = 12 = \boxed{\textbf{(C)}}</math>. | ||
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==Note== | ==Note== | ||
See [[2021 AMC 12A Problems/Problem 1|problem 1]]. | See [[2021 AMC 12A Problems/Problem 1|problem 1]]. | ||
Revision as of 15:05, 11 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 
.
Note
See problem 1.
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. 
