Difference between revisions of "2021 AMC 12A Problems/Problem 6"

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==Solution==
 
==Solution==
The solutions will be posted once the problems are posted.
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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 16:05, 11 February 2021

Problem

A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$. When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$. How many cards were in the deck originally?

Solution

If the probability of choosing a red card is $\frac{1}{3}$, the red and black cards are in ratio $1:2$. This means at the beginning there are $x$ red cards and $2x$ black cards.

After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the ratio of red to black cards is $1:3$. This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.

So, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.

So the answer is $8+4 = 12 = \boxed{\textbf{(C)}}$.

Note

See problem 1.

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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