In a metre-bridge experiment the unknown resistance X is in the left gap and a resistance box R is in the right gap. The box is set to R=100 Ω with a…

First note that $X$ is in the left gap with balance length $\ell$ measured from the left end, so $\dfrac{X}{R}=\dfrac{\ell}{100-\ell}$, i.e. $X=R\,\dfrac{\ell}{100-\ell}$. Take logarithms and differentiate for the maximum error: $\dfrac{\Delta X}{X}=\dfrac{\Delta R}{R}+\dfrac{\Delta \ell}{\ell}+\dfrac{\Delta\ell}{100-\ell}$, because an error $\Delta\ell$ in $\ell$ produces an error of the same magnitude in $100-\ell$. With $\ell=50$ and $\Delta\ell=0.5$: $\dfrac{0.5}{50}+\dfrac{0.5}{50}=0.02=2\%$. Adding the resistance-box error $1\%$ gives $3\%$. Hence the maximum percentage error in $X$ is $3$. Remember that for a product or quotient the maximum fractional errors add, whatever the sign of the exponent, and that the error in the box setting and the error in the length are independent so they are simply summed in the worst case. Numerically $X=100\times\dfrac{50}{50}=100\ \Omega$ with $\Delta X=3\ \Omega$, i.e. $97\ \Omega$ to $103\ \Omega$ at the extremes.