2021 AMC 12A Problems/Problem 6
- 1 Problem
- 2 Solution 1 (Algebra)
- 3 Solution 2 (Arithmetic)
- 4 Solution 3 (Observations)
- 5 Video Solution (Quick and Easy)
- 6 Video Solution by Aaron He
- 7 Video Solution by Hawk Math
- 8 Video Solution by OmegaLearn (Using Probability and System of Equations)
- 9 Video Solution by TheBeautyofMath
- 10 See also
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 (Algebra)
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 .
Solution 2 (Arithmetic)
In terms of the number of cards, the original deck is times the red cards, and the final deck is times the red cards. So, the final deck is times the original deck. We are given that 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.
Solution 3 (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
Video Solution (Quick and Easy)
~Education, the Study of Everything
Video Solution by Aaron He
Video Solution by Hawk Math
Video Solution by OmegaLearn (Using Probability and System of Equations)
Video Solution by TheBeautyofMath
|2021 AMC 12A (Problems • Answer Key • Resources)|
|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|