Difference between revisions of "2019 AMC 10B Problems/Problem 1"

(Video Solution 1)
Line 20: Line 20:
 
~IronicNinja
 
~IronicNinja
  
==Video Solution 1==
+
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 
https://youtu.be/fGpUnH4sUc4
 
https://youtu.be/fGpUnH4sUc4
  

Revision as of 12:23, 29 May 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 $\tfrac{5}{6}$ 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 $\tfrac{3}{4}$ full of water. What is the ratio of the volume of the first container to the volume of the second container?

$\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}$

Solution

Solution 1

Let the first jar's volume be $A$ and the second's be $B$. It is given that $\frac{3}{4}A=\frac{5}{6}B$. We find that $\frac{A}{B}=\frac{\left(\frac{3}{4}\right)}{\left(\frac{5}{6}\right)}=\boxed{\textbf{(D) }\frac{9}{10}}.$

We already know that this is the ratio of the smaller to the larger volume because it is less than $1.$

Solution 2

An alternate solution is to substitute an arbitrary maximum volume for the first container - let's say $72$, so there was a volume of $60$ in the first container, and then the second container also has a volume of $60$, so you get $60 \cdot \frac{4}{3} = 80$. Thus the answer is $\frac{72}{80} = \boxed{\textbf{(D) }\frac{9}{10}}$.

~IronicNinja

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/fGpUnH4sUc4

~Education, the Study of Everything

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png