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>.
 
~IronicNinja
 
  
 
==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>.
 
~IronicNinja
 
  
 
==See Also==
 
==See Also==

Revision as of 00:17, 18 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 $\tfrac{5}{6}$ 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 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 $\frac{B}{A}=\frac{\left(\frac{3}{4}\right)}{\left(\frac{5}{6}\right)}=\boxed{\frac{9}{10}}.$

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

Solution 2

We can set up a ratio to solve this problem. If $x$ is the volume of the first container, and $y$ is the volume of the second container, then: \[\frac{5}{6}x = \frac{3}{4}y\]

Cross-multiplying allows us to get $\frac{x}{y} = \frac{3}{4} \cdot \frac{6}{5} = \frac{18}{20} = \frac{9}{10}$. Thus the ratio of the volume of the first container to the second container is $\boxed{\textbf{(D) } \frac{9}{10}}$.

Solution 3

An alternate solution is to plug in some 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 $\frac{72}{80} = \frac{9}{10}$.

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

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