Difference between revisions of "2019 AMC 10B Problems/Problem 1"
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Cross-multiplying allows us to get <math>\frac{x}{y} = \frac{3}{4} \cdot \frac{6}{5} = \frac{18}{20} = \frac{9}{10}</math>. Thus the ratio of the volume of the first container to the second container is <math>\boxed{\textbf{(D) } \frac{9}{10}}</math> | Cross-multiplying allows us to get <math>\frac{x}{y} = \frac{3}{4} \cdot \frac{6}{5} = \frac{18}{20} = \frac{9}{10}</math>. Thus the ratio of the volume of the first container to the second container is <math>\boxed{\textbf{(D) } \frac{9}{10}}</math> | ||
| + | |||
| + | iron (can you people stop taking credit away, looking at you @Xmidnightfirex) | ||
==Solution 3== | ==Solution 3== | ||
An alternate solution is to plug in some maximum volume for the first container - let's say <math>72</math>, so there was a volume of <math>60</math> in the first container, and then the second container also has a volume of <math>60</math>, so you get <math>60 \cdot \frac{4}{3} = 80</math>. Thus the answer is <math>\frac{72}{80} = \frac{9}{10}</math>. | An alternate solution is to plug in some maximum volume for the first container - let's say <math>72</math>, so there was a volume of <math>60</math> in the first container, and then the second container also has a volume of <math>60</math>, so you get <math>60 \cdot \frac{4}{3} = 80</math>. Thus the answer is <math>\frac{72}{80} = \frac{9}{10}</math>. | ||
| + | |||
| + | iron | ||
==See Also== | ==See Also== | ||
Revision as of 19:27, 17 February 2019
- The following problem is from both the 2019 AMC 10B #1 and 2019 AMC 12B #1, so both problems redirect to this page.
Problem
Alicia had two containers. The first was 
full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was 
full of water. What is the ratio of the volume of the first container to the volume of the second container?
Solution
Let the first jar's volume be 
and the second's be 
. It is given that 
. We find that 
We already know that this is the ratio of smaller to larger volume because it is less than 
Solution 2
We can set up a ratio to solve this problem. If 
is the volume of the first container, and 
is the volume of the second container, then:
![\[\frac{5}{6}x = \frac{3}{4}y\]](http://latex.artofproblemsolving.com/4/f/a/4fab360603aed7aedc30f35b9971df74b33bfcdf.png)
Cross-multiplying allows us to get 
. Thus the ratio of the volume of the first container to the second container is 
iron (can you people stop taking credit away, looking at you @Xmidnightfirex)
Solution 3
An alternate solution is to plug in some maximum volume for the first container - let's say 
, so there was a volume of 
in the first container, and then the second container also has a volume of , so you get 
. Thus the answer is 
.
iron
See Also
| 2019 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 | ||
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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. 
