Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Further fixed punctuation and deleted irrelevant message)
m (→‎Solution 1)
 
(25 intermediate revisions by 11 users not shown)
Line 8: Line 8:
  
 
==Solution==
 
==Solution==
 +
===Solution 1===
  
−
Let the first jar's volume be <math>A</math> and the second's be <math>B</math>. It is given that <math>\frac{3}{4}A=\frac{5}{6}B</math>. We find that <math>\frac{B}{A}=\frac{3/4}{5/6}=\boxed{\frac{9}{10}}.</math>  
+
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>  
  
−
We already know that this is the ratio of smaller to larger volume because it is less than <math>1.</math>
+
We already know that this is the ratio of the smaller to the larger volume because it is less than <math>1.</math>
  
−
==Solution 2==
+
===Solution 2===
−
We can set up a ratio to solve this problem. If <math>x</math> is the volume of the first container, and <math>y</math> is the volume of the second container, then:
 
−
<cmath>\frac{5}{6}x = \frac{3}{4}y</cmath>
 
  
−
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>.
+
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>.
  
−
==Solution 3==
+
~IronicNinja
  
−
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>.
+
==Video Solution==
 +
https://youtu.be/fGpUnH4sUc4
 +
 
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

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 $\tfrac{5}{6}$ 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 $\frac{5}{6}A=\frac{3}{4}B$. 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

https://youtu.be/fGpUnH4sUc4

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_1&oldid=198110"