Difference between revisions of "2021 AMC 12A Problems/Problem 6"
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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> | 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>\boxed{\textbf{(C) }12}.</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 | ~MRENTHUSIASM |
Revision as of 06:27, 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 (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 (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.