# CAT Quant Practice Problems

Question: Let f(x) = max (2x + 1, 3 - 4x), where x is any real number. Then the minimum possible value of f(x) is:

 $\frac{1}{3}$ $\frac{1}{2}$ $\frac{2}{3}$ $\frac{4}{3}$ $\frac{5}{3}$

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CAT Quant Practice Problems
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