1. **State the problem:** We have a cuboid water tank where volume $V$ varies directly as height $h$ and base area $A$. Given $V=90$ when $h=2$ and $A=30$, find the new height if volume increases by 20%. Also, if base area is insufficient but volume unchanged, find the best solution. 2. **Formula and explanation:** Since $V$ varies directly as $h$ and $A$, we write: $$V = k h A$$ where $k$ is the constant of proportionality. 3. **Find $k$ using given values:** $$90 = k \times 2 \times 30$$ $$90 = 60k$$ $$k = \frac{90}{60} = \frac{3}{2} = 1.5$$ 4. **Find new volume after 20% increase:** $$V_{new} = 90 + 0.20 \times 90 = 90 + 18 = 108$$ 5. **Find new height $h_{new}$ with $A=30$ and $V_{new}=108$:** $$108 = 1.5 \times h_{new} \times 30$$ $$108 = 45 h_{new}$$ $$h_{new} = \frac{108}{45} = \frac{\cancel{108}}{\cancel{45}} = 2.4$$ 6. **Answer for (a):** The new height is $2.4$ meters. 7. **For (b), volume unchanged but base area insufficient:** Since $V = k h A$, if $V$ is constant and $A$ decreases, then $h$ must increase to compensate. 8. **Best solution:** Increase the height of the tank to maintain the same volume with smaller base area. **Final answers:** (a) New height = $2.4$ m (b) Increase the height of the tank to compensate for smaller base area.