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. | ||
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| + | ==Solution 3 == | ||
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| + | If there are initially <math>B</math> blue marbles in the bag, after red marbles are added, then the total number of marbles in the bag must be <math>3B</math>. Then after the yellow marbles are added, the number of marbles in the bag must be <math>5B</math>. Finally, adding <math>B</math> blue marbles to the bag gives <math>2B</math> blue marbles out of <math>6B</math> total marbles. Thus <math>\boxed{1/3}</math> of the marbles are blue. | ||
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| + | ==Solution 4== | ||
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| + | Just before the number of blue marbles is doubled, the ratio of blue marbles to non-blue marbles is 1 to 4. Doubling the number of blue marbles makes the ratio 2 to 4, so <math>\dfrac{2}{2+4} = \boxed{1/3}</math> of the marbles are blue. | ||
== See also == | == See also == | ||
Latest revision as of 16:25, 2 November 2023
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.
Solution 3
If there are initially 
blue marbles in the bag, after red marbles are added, then the total number of marbles in the bag must be 
. Then after the yellow marbles are added, the number of marbles in the bag must be 
. Finally, adding blue marbles to the bag gives 
blue marbles out of 
total marbles. Thus 
of the marbles are blue.
Solution 4
Just before the number of blue marbles is doubled, the ratio of blue marbles to non-blue marbles is 1 to 4. Doubling the number of blue marbles makes the ratio 2 to 4, so 
of the marbles are blue.
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 | ||
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