A student has an unknown resistance of 12 Ω in the left gap of a metre bridge…

Step 1. The balance points. $l = \dfrac{100R}{R + S}$. For $S = 3\ \Omega$: $\dfrac{100 \times 12}{15} = 80.0$ cm. For $S = 12\ \Omega$: $50.0$ cm. For $S = 48\ \Omega$: $\dfrac{100 \times 12}{60} = 20.0$ cm.

Step 2. The errors. $\dfrac{\Delta R}{R} = \dfrac{0.1}{l} + \dfrac{0.1}{100 - l}$. At $80.0$ cm: $\dfrac{0.1}{80.0} + \dfrac{0.1}{20.0}$ $= 0.00125 + 0.005$ $= 0.625\%$. At $50.0$ cm: $0.002 + 0.002 = 0.40\%$. At $20.0$ cm: $0.625\%$.

Rule. The best result comes when the known resistance is nearly equal to the unknown one, so that the balance point is near $50$ cm.