Answer: D
Let the length, width, and height of the cuboid be \( a \), \( b \), and \( c \) units respectively. ∴ Original volume = \( abc \) cubic units New volume after changes: \[ \left(a + \frac{10a}{100}\right)\left(b + \frac{20b}{100}\right)\left(c - \frac{30c}{100}\right) = \left(\frac{11a}{10} \times \frac{12b}{10} \times \frac{7c}{10}\right) = \frac{924abc}{1000} \text{ cubic units} \] ∴ Percentage decrease in volume: \[ \frac{abc - \frac{924abc}{1000}}{abc} \times 100\% = \frac{\frac{76abc}{1000}}{abc} \times 100\% = \frac{76 \times 100}{1000}\% = 7.6\% \]
Let the length, width, and height of the cuboid be \( a \), \( b \), and \( c \) units respectively. ∴ Original volume = \( abc \) cubic units New volume after changes: \[ \left(a + \frac{10a}{100}\right)\left(b + \frac{20b}{100}\right)\left(c - \frac{30c}{100}\right) = \left(\frac{11a}{10} \times \frac{12b}{10} \times \frac{7c}{10}\right) = \frac{924abc}{1000} \text{ cubic units} \] ∴ Percentage decrease in volume: \[ \frac{abc - \frac{924abc}{1000}}{abc} \times 100\% = \frac{\frac{76abc}{1000}}{abc} \times 100\% = \frac{76 \times 100}{1000}\% = 7.6\% \]