1. Problem: A ship has a load draft of $7\text{ m}$ and currently reads $T_F=7.1\text{ m}$ and $T_A=7.5\text{ m}$. 2. Compute the current trim: the trim (stern minus bow) is $T_A-T_F=7.5-7.1=0.4\text{ m}=40\text{ cm}$. 3. Required trimming moment to correct the 40\text{ cm} trim is $M=(\text{trim in cm})\times MCT_{1\text{cm}}=40\times 15.5=620\text{ t-m}$. 4. Choose a reference for placing the added weight: $L_{BP}=150\text{ m}$ so amidships to forward perpendicular (FP) is $L_{BP}/2=75\text{ m}$. 5. Given $LCF$ is $3\text{ m}$ aft of amidships, the distance from $LCF$ to FP is $75+3=78\text{ m}$. 6. Place the weight at FP so that its moment about $LCF$ is $W\times 78$ and set this equal to the required trimming moment $620\text{ t-m}$. 7. Solve for the required weight: $W=620/78=7.9487179487\text{ t}\approx 7.95\text{ t}$. 8. Check change in mean draft from adding $W$: change in mean draft in cm is $\Delta\bar{T}_{\text{cm}}=W/\text{TPC}=7.9487179487/15=0.5299\text{ cm}$, so $\Delta\bar{T}=0.005299\text{ m}$, which is negligible. 9. Initial mean draft was $(7.1+7.5)/2=7.3\text{ m}$, so final mean draft becomes $7.3+0.005299=7.305299\text{ m}\approx 7.305\text{ m}$. 10. Final drafts (even keel) are $T_F'=T_A'=7.305\text{ m}$. 11. Summary answer: add approximately $7.95\text{ t}$ at the forward perpendicular (which is $78\text{ m}$ forward of the LCF or $75\text{ m}$ forward of amidships) to bring the vessel to even keel with negligible change in mean draft.