GRE Question of the Day (October 24)

Math

If $r > 0$ , s ≠ $\frac{1}{2}$ and $r = \frac{3 s + 1}{1 - 2 s}$ , then $s =$

A) $\frac{3 r}{12 + 6 r}$

B) $\frac{- 2 r}{3}$

C) $\frac{r - 1}{3 + 2 r}$

D) $\frac{2}{3 (1 - r)}$

E) $\frac{- 1 + r}{3}$

Correct Answer - C - (click and drag your mouse to see the answer)

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