Difference between revisions of "2009 AMC 10B Problems/Problem 2"
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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 | + | 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 \boxed{(C)\frac {1} {2}}. |
== See Also == | == See Also == | ||
Revision as of 15:49, 27 December 2020
Problem
Which of the following is equal to 
?
Solution
Multiplying the numerator and the denominator by the same value does not change the value of the fraction.
We can multiply both by 
, getting 
.
Alternately, we can directly compute that the numerator is 
, the denominator is 
, and hence their ratio is \boxed{(C)\frac {1} {2}}.
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. 
