1. Problem statement: A ship has a target mean (load) draft of $9$ m, current forward draft $8.9$ m and aft draft $8.5$ m, and the only spaces left for cargo are no.2 at $+50$ m (forward) and no.4 at $-40$ m (aft) relative to amidships; given $\Delta=20000$ t, $TPC=20$ t/cm and $MCT1cm=21$ t-m with LCF amidships, find the masses to load in each hold to achieve even keel. 2. Compute the current mean draft by averaging forward and aft drafts. Current mean draft: $$d_{mean}=\frac{8.9+8.5}{2}=8.7\ \text{m}$$ 3. Find the required increase in mean draft to reach $9$ m. Required increase: $9-8.7=0.3$ m $=30$ cm. Weight needed using TPC: $$W= TPC\times 30\ \text{cm}=20\times 30=600\ \text{t}$$ 4. Determine the trim correction needed to achieve even keel. Current trim (forward minus aft) is $$\text{trim}=8.9-8.5=0.4\ \text{m}=40\ \text{cm}$$ We must change trim by $-40$ cm (remove bow-down trim). The trimming moment required (about amidships, since LCF is amidships) is $$M_{req}=-MCT1cm\times 40=-21\times 40=-840\ \text{t-m}$$ 5. Set unknowns for cargo masses and form equations. Let $x$ be mass in no.2 (at $+50$ m) and $y$ be mass in no.4 (at $-40$ m). The weight (mean draft) condition gives: $$x+y=600$$ The trimming moment condition (moments about amidships) gives: $$50x-40y=-840$$ 6. Solve the two simultaneous equations. From $x+y=600$ we have $$y=600-x$$ Substitute into the moment equation: $$50x-40(600-x)=-840$$ Simplify the left-hand side: $$50x-24000+40x=-840$$ Combine like terms: $$90x-24000=-840$$ Solve for $x$: $$90x=23160$$ Thus $$x=\frac{23160}{90}=\frac{772}{3}\approx257.333\ \text{t}$$ Now find $y$: $$y=600-x=600-\frac{772}{3}=\frac{1028}{3}\approx342.667\ \text{t}$$ 7. Check and final answer. Check sum: $\frac{772}{3}+\frac{1028}{3}=600$ t, as required. Check moment: $50\times257.333-40\times342.667\approx12866.667-13706.667=-840$ t-m, as required. Answer: Load no.2 (at $+50$ m) with $\dfrac{772}{3}\approx257.333$ t and load no.4 (at $-40$ m) with $\dfrac{1028}{3}\approx342.667$ t to achieve even keel.