Difference between revisions of "2012 AMC 10A Problems/Problem 7"
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− | == Problem | + | {{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #4]] and [[2012 AMC 10A Problems|2012 AMC 10A #7]]}} |
+ | |||
+ | == Problem == | ||
In a bag of marbles, <math>\frac{3}{5}</math> of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red? | In a bag of marbles, <math>\frac{3}{5}</math> of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red? | ||
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<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{4}{7}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{4}{5} </math> | <math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{4}{7}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{4}{5} </math> | ||
− | == Solution == | + | == Solution 1 == |
Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling. | Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling. | ||
− | There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is <math>\ | + | There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is <math>\boxed{\textbf{(C)}\ \frac{4}{7}}</math>. |
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Let us say that there are <math>x</math> marbles in the bag. Therefore, <math>\frac{3x}{5x}</math> are blue, and <math>\frac{2x}{5x}</math> are red. When the red marbles are doubled, we now have <math>\frac{2*2x}{5x+2x} = \frac{4x}{7x} = \frac{4}{7} \Rightarrow \boxed{\textbf{(C)}}</math> | ||
+ | |||
+ | == Solution 3 (Fakesolve) == | ||
+ | |||
+ | Say there are <math>5</math> marbles in the bag. So, we have <math>3</math> blue marbles and <math>2</math> red marbles. Doubling the number of red, marbles, we end up with <math>3</math> blue marbles and <math>4</math> red marbles. So, we have <math>\frac{4}{4+3}=\boxed{\textbf{(C)}\ \frac{4}{7}}</math> | ||
+ | ~Sosiaops | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/SjMfmDWhWX0 | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2012|ab=A|num-b=6|num-a=8}} | ||
+ | {{AMC12 box|year=2012|ab=A|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:17, 15 January 2021
- The following problem is from both the 2012 AMC 12A #4 and 2012 AMC 10A #7, so both problems redirect to this page.
Problem
In a bag of marbles, of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?
Solution 1
Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.
There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is .
Solution 2
Let us say that there are marbles in the bag. Therefore, are blue, and are red. When the red marbles are doubled, we now have
Solution 3 (Fakesolve)
Say there are marbles in the bag. So, we have blue marbles and red marbles. Doubling the number of red, marbles, we end up with blue marbles and red marbles. So, we have ~Sosiaops
Video Solution
~savannahsolver
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 10 Problems and Solutions |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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.