A student measures a resistor of 50 Ω with a voltmeter and an ammeter…

Step 1. The calculated resistance. The ammeter reads the current of the resistor and the voltmeter together. So $R' = \dfrac{V}{I} = \dfrac{RR_V}{R + R_V}$ $= \dfrac{50 \times 500}{50 + 500}\ \Omega = 45.5\ \Omega$.

Step 2. The error. $\dfrac{50 - 45.5}{50} \times 100\% = 9.1\%$. The result is too low.

Step 3. A better voltmeter. We want $\dfrac{RR_V}{R + R_V} = 0.99R$. Then $R_V = 0.99(R + R_V)$, so $0.01R_V = 0.99R$ and $R_V = 99R = 4950\ \Omega$.