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 "2009 AMC 10B Problems/Problem 2"

m (→‎Solution)
 
(4 intermediate revisions by 3 users not shown)
Line 15: Line 15:
 
</math>
 
</math>
  
−
== Solution ==
+
== Solution 1 ==
  
 
Multiplying the numerator and the denominator by the same value does not change the value of the fraction.
 
Multiplying the numerator and the denominator by the same value does not change the value of the fraction.
 
We can multiply both by <math>12</math>, getting <math>\dfrac{4-3}{6-4} = \boxed{\dfrac 12}</math>.
 
We can multiply both by <math>12</math>, getting <math>\dfrac{4-3}{6-4} = \boxed{\dfrac 12}</math>.
  
−
Alternately, we can directly compute that the numerator is <math>\dfrac 1{12}</math>, the denominator is <math>\dfrac 16</math>, and hence their ratio is <math>\boxed{(C)\frac {1} {2}}</math>.
+
Alternately, we can directly compute that the numerator is <math>\dfrac 1{12}</math>, the denominator is <math>\dfrac 16</math>, and hence their ratio is <math>\boxed{(C)\frac{1}{2}}</math>.
 +
 
 +
== Solution 2 (Full Solution) ==
 +
We write both the numerator and denominator with a denominator of <math>12</math> first, since the LCM of <math>2,3,</math> and <math>4</math> is <math>3\cdot4=12</math>.
 +
Next, we multiply both the numerator and denominator by <math>12</math>. <math>\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}=\dfrac{\frac{4}{12}-\frac{3}{12}}{\frac{6}{12}-\frac{4}{12}}=\dfrac{\frac{1}{12}}{\frac{2}{12}}=\boxed{(C)\dfrac{1}{2}}</math>.
 +
-sosiaops
  
 
== See Also ==
 
== See Also ==

Latest revision as of 21:08, 27 December 2020

Problem

Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$?

$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$

Solution 1

Multiplying the numerator and the denominator by the same value does not change the value of the fraction. We can multiply both by $12$, getting $\dfrac{4-3}{6-4} = \boxed{\dfrac 12}$.

Alternately, we can directly compute that the numerator is $\dfrac 1{12}$, the denominator is $\dfrac 16$, and hence their ratio is $\boxed{(C)\frac{1}{2}}$.

Solution 2 (Full Solution)

We write both the numerator and denominator with a denominator of $12$ first, since the LCM of $2,3,$ and $4$ is $3\cdot4=12$. Next, we multiply both the numerator and denominator by $12$. $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}=\dfrac{\frac{4}{12}-\frac{3}{12}}{\frac{6}{12}-\frac{4}{12}}=\dfrac{\frac{1}{12}}{\frac{2}{12}}=\boxed{(C)\dfrac{1}{2}}$. -sosiaops

See Also

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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

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=2009_AMC_10B_Problems/Problem_2&oldid=140801"