Difference between revisions of "2021 AMC 12A Problems/Problem 6"
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==Solution 4 (Ratios and Proportions)== | ==Solution 4 (Ratios and Proportions)== | ||
By looking at the ratio of black cards to the total number of cards, we can say that there are <math>2x</math> black cards and <math>3x</math> cards in total. We can write the probability of getting a black card as <math>\frac{2x}{3x}</math>. But when we increase the number of black cards by 4 the probability becomes <math>\frac{3}{4}</math>. So the equation is: | By looking at the ratio of black cards to the total number of cards, we can say that there are <math>2x</math> black cards and <math>3x</math> cards in total. We can write the probability of getting a black card as <math>\frac{2x}{3x}</math>. But when we increase the number of black cards by 4 the probability becomes <math>\frac{3}{4}</math>. So the equation is: | ||
| − | <math>\frac{2x+4}{3x+4}=\frac{3}{4}</math> | + | <math>\frac{2x+4}{3x+4}=\frac{3}{4}</math>, |
| − | + | <math>8x+16=9x+12</math>, | |
| − | <math>8x+16=9x+12</math> | + | <math>x=4</math>. |
| − | <math>x=4</math> | + | The answer is <math>3x=\boxed{\textbf{(C) }12}.</math> |
| − | |||
| − | ~Param Gor, | + | ~Param Gor |
| + | |||
| + | ==Solution 5 (Simpler Ratios)== | ||
| + | This could in a way be considered a more efficient/alternate version of Solution 4. Since the problem mentions that there is initially a <math>\frac{1}{3}</math> chance of getting a red card, we can represent the number of red cards as <math>\frac{x}{3}</math>, where <math>x</math> is the number of cards in the deck originally. Since the modification to the deck makes no changes to the number of red cards, you can see that if there is a <math>\frac{1}{4}</math> probability of having a red card in a deck of <math>x+4</math> cards, <math>\frac{x}{3} = \frac{x+4}{4}</math>. Solving this for <math>x</math> gives <math>x=\boxed{\textbf{(C) }12}.</math> | ||
| + | |||
| + | ~Almond_Oil | ||
==Video Solution (Quick and Easy)== | ==Video Solution (Quick and Easy)== | ||
Latest revision as of 22:51, 19 July 2024
Contents
- 1 Problem
- 2 Solution 1 (Algebra)
- 3 Solution 2 (Arithmetic)
- 4 Solution 3 (Observations)
- 5 Solution 4 (Ratios and Proportions)
- 6 Solution 5 (Simpler Ratios)
- 7 Video Solution (Quick and Easy)
- 8 Video Solution by Aaron He
- 9 Video Solution by Hawk Math
- 10 Video Solution by OmegaLearn (Using Probability and System of Equations)
- 11 Video Solution by TheBeautyofMath
- 12 See also
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 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 
.
--abhinavg0627
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.
~MRENTHUSIASM
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 
~MRENTHUSIASM
Solution 4 (Ratios and Proportions)
By looking at the ratio of black cards to the total number of cards, we can say that there are black cards and cards in total. We can write the probability of getting a black card as 
. But when we increase the number of black cards by 4 the probability becomes 
. So the equation is:
,
,
.
The answer is 
~Param Gor
Solution 5 (Simpler Ratios)
This could in a way be considered a more efficient/alternate version of Solution 4. Since the problem mentions that there is initially a chance of getting a red card, we can represent the number of red cards as 
, where is the number of cards in the deck originally. Since the modification to the deck makes no changes to the number of red cards, you can see that if there is a probability of having a red card in a deck of cards, 
. Solving this for gives
~Almond_Oil
Video Solution (Quick and Easy)
~Education, the Study of Everything
Video Solution by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=5m51s
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn (Using Probability and System of Equations)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
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. 
