Answer: A
Let \(a = 2k\), \(b = 3k\), and \(c = 5k\) ∴ \(\frac{2a + 3b - 3c}{c}\) \(= \frac{2 \times 2k + 3 \times 3k - 3 \times 5k}{5k}\) \(= \frac{4k + 9k - 15k}{5k}\) \(= \frac{-2k}{5k}\) \(= -\frac{2}{5}\)
Let \(a = 2k\), \(b = 3k\), and \(c = 5k\) ∴ \(\frac{2a + 3b - 3c}{c}\) \(= \frac{2 \times 2k + 3 \times 3k - 3 \times 5k}{5k}\) \(= \frac{4k + 9k - 15k}{5k}\) \(= \frac{-2k}{5k}\) \(= -\frac{2}{5}\)