Difference between revisions of "2004 AMC 10B Problems/Problem 14"
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We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed <math>\frac{1}{5}</math> of the marbles in the bag. This means that there were <math>x</math> blue and <math>4x</math> other marbles, for some <math>x</math>. When we double the number of blue marbles, there will be <math>2x</math> blue and <math>4x</math> other marbles, hence blue marbles now form <math>\boxed{\mathrm{(C)\ }\frac{1}{3}}</math> of all marbles in the bag. | We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed <math>\frac{1}{5}</math> of the marbles in the bag. This means that there were <math>x</math> blue and <math>4x</math> other marbles, for some <math>x</math>. When we double the number of blue marbles, there will be <math>2x</math> blue and <math>4x</math> other marbles, hence blue marbles now form <math>\boxed{\mathrm{(C)\ }\frac{1}{3}}</math> of all marbles in the bag. | ||
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| + | ==Solution 2 == | ||
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| + | Assume that there are <math>2</math> blue marbles and <math>1</math> red marble. You don't have to add any red marbles to make the quantity <math>\frac{1}{3}</math>. Then, to make the blue marbles <math>\frac{1}{5}</math>, add <math>7</math> yellow marbles. Doubling the blue marbles makes <math>4</math> blue marbles out of <math>12</math> total marbles, or <math>\boxed{\mathrm{(C)\ }\frac{1}{3}}</math> of all marbles in the bag. | ||
== See also == | == See also == | ||
Revision as of 20:09, 26 July 2020
Contents
Problem
A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only 
of the marbles in the bag are blue. Then yellow marbles are added to the bag until only 
of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?
Solution
We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed of the marbles in the bag. This means that there were 
blue and 
other marbles, for some . When we double the number of blue marbles, there will be 
blue and other marbles, hence blue marbles now form 
of all marbles in the bag.
Solution 2
Assume that there are 
blue marbles and 
red marble. You don't have to add any red marbles to make the quantity . Then, to make the blue marbles , add 
yellow marbles. Doubling the blue marbles makes 
blue marbles out of 
total marbles, or of all marbles in the bag.
See also
| 2004 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
