Difference between revisions of "2008 AMC 12A Problems/Problem 2"

Revision as of 13:41, 17 February 2008

What is the reciprocal of $\frac{1}{2}+\frac{2}{3}$ ?

$\textbf{(A)} \frac{6}{7} \qquad \textbf{(B)} \frac{7}{6} \qquad \textbf{(C)} \frac{5}{3} \qquad \textbf{(D)} 3 \qquad \textbf{(E)} \frac{7}{2}$

$\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\frac{6}{7}\Rightarrow A$ .

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

}

Revision as of 13:32, 17 February 2008 (view source) Xantos C. Guin (talk \| contribs) (New page: ==Problem == What is the reciprocal of <math>\frac{1}{2}+\frac{2}{3}</math>? <math>\textbf{(A)} \frac{6}{7} \qquad \textbf{(B)} \frac{7}{6} \qquad \textbf{(C)} \frac{5}{3} \qquad \textb...)		Revision as of 13:41, 17 February 2008 (view source) Xantos C. Guin (talk \| contribs) (typo) Newer edit →
Line 8:		Line 8:

	==See Also==		==See Also==
−	{{AMC12 box\|year=~~2007~~\|ab=A\|~~before~~=1\|num-a=3}}	+	{{AMC12 box\|year=2008\|ab=A\|num-b=1\|num-a=3}}}