Difference between revisions of "2019 AMC 10B Problems/Problem 1"
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| + | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #1]] and [[2019 AMC 12B Problems|2019 AMC 12B #1]]}} | ||
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==Problem== | ==Problem== | ||
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==Solution== | ==Solution== | ||
| + | ===Solution 1=== | ||
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| + | Let the first jar's volume be <math>A</math> and the second's be <math>B</math>. It is given that <math>\frac{5}{6}A=\frac{3}{4}B</math>. We find that <math>\frac{A}{B}=\frac{\left(\frac{3}{4}\right)}{\left(\frac{5}{6}\right)}=\boxed{\textbf{(D) }\frac{9}{10}}.</math> | ||
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| + | We already know that this is the ratio of the smaller to the larger volume because it is less than <math>1.</math> | ||
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| + | ===Solution 2=== | ||
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| + | An alternate solution is to substitute an arbitrary 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} = \boxed{\textbf{(D) }\frac{9}{10}}</math>. | ||
| − | + | ~IronicNinja | |
| − | + | ==Video Solution== | |
| + | https://youtu.be/fGpUnH4sUc4 | ||
| − | + | ~Education, the Study of Everything | |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|before=First Problem|num-a=2}} | {{AMC10 box|year=2019|ab=B|before=First Problem|num-a=2}} | ||
| + | {{AMC12 box|year=2019|ab=B|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:17, 11 September 2023
- 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
Solution 1
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 the smaller to the larger volume because it is less than 
Solution 2
An alternate solution is to substitute an arbitrary 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 
.
~IronicNinja
Video Solution
~Education, the Study of Everything
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. 
