Difference between revisions of "2008 AMC 12A Problems/Problem 2"
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What is the [[reciprocal]] of <math>\frac{1}{2}+\frac{2}{3}</math>? | What is the [[reciprocal]] of <math>\frac{1}{2}+\frac{2}{3}</math>? | ||
| − | <math>\ | + | <math>\mathrm{(A)}\ \frac{6}{7}\qquad\mathrm{(B)}\ \frac{7}{6}\qquad\mathrm{(C)}\ \frac{5}{3}\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{7}{2}</math> |
| − | ==Solution== | + | ==Solution 1== |
| − | <math>\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\frac{6}{7} | + | |
| + | Here's a cheapshot: | ||
| + | Obviously, <math>\frac{1}{2}+\frac{2}{3}</math> is greater than <math>1</math>. Therefore, its reciprocal is less than <math>1</math>, and the answer must be <math>\boxed{\frac{6}{7}}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | <math>\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\boxed{\mathrm{(A)}\ \frac{6}{7}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2008|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:22, 7 September 2021
Contents
Problem
What is the reciprocal of 
?
Solution 1
Here's a cheapshot:
Obviously, is greater than 
. Therefore, its reciprocal is less than , and the answer must be 
.
Solution 2
.
See Also
| 2008 AMC 12A (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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
