Understanding the Michaelis-Menten Equation, Enzyme Kinetics

For an enzyme which obeys Henri-Michaelis-Menten kinetics,
(1) At what substrate concentration will an enzyme characterized by a kcat of 30 s-1 and Km of 0.005 M show one quarter of its maximal rate?

(2) Calculate the fraction of Vmax that would be found when [S]=0.5 Km, [S]=3Km, and [S]=10Km

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You must also understand turnover number, or kcat. kcat = Vmax/[E]T, where [E]T=the total amount of enzyme present. The form of the Henri-Michaelis-Menten (HMM) equation shown above can be rewritten substituting kcat*[E] instead of Vmax (not shown).

(1) When kcat = 30 s-1 and Km = 0005 M, what is [S] when v=one quarter of Vmax ...