Difference between revisions of "2009 AMC 10B Problems/Problem 2"
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== Solution 2 (Full Solution) == | == Solution 2 (Full Solution) == | ||
| − | We write both the numerator and denominator with a denominator of <math>12</math> first, then simplify. <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{\dfrac{1}{2}}</math>. | + | We write both the numerator and denominator with a denominator of <math>12</math> first, then simplify. <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 | -sosiaops | ||
Revision as of 21:05, 27 December 2020
Problem
Which of the following is equal to 
?
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 
, getting 
.
Alternately, we can directly compute that the numerator is 
, the denominator is 
, and hence their ratio is 
.
Solution 2 (Full Solution)
We write both the numerator and denominator with a denominator of first, then simplify. 
.
-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. 
