Source: High school physics (Chinese)
Problem
Two resistors $R_1 = 2 k\Omega$ and $R_2 = 3 k\Omega$ are connected in series across a constant supply of $10 V$. A real (non-ideal) voltmeter is used to measure the voltage across $R_2$. On its $6 V$ range, the meter reads $5 V$.
On the $12 V$ range, the voltmeter reads $U \approx 5.45 V$ (exactly $60/11 V$).
A non-ideal voltmeter has finite internal resistance and loads the circuit by drawing current. Let $R_{v,6}$ denote the voltmeter's resistance on the $6 V$ range. With it in parallel across $R_2$, define $X = R_2 \parallel R_{v,6}$.
The reading equals the voltage across $X$:
$$5 = 10 \cdot \dfrac{X}{R_1 + X} \implies X = R_1 = 2 k\Omega.$$Then $\dfrac{1}{R_{v,6}} = \dfrac{1}{X} - \dfrac{1}{R_2} = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$, so $R_{v,6} = 6 k\Omega$.
For an analog voltmeter, $R_v = V_{\text{full}} / I_g$ where $I_g$ is the galvanometer's full-scale current (independent of range). Hence $R_v \propto V_{\text{full}}$, giving
$$R_{v,12} = R_{v,6} \cdot \dfrac{12}{6} = 12 k\Omega.$$New parallel load: $R_2 \parallel R_{v,12} = \dfrac{3 \times 12}{15} = 2.4 k\Omega$.
Reading on the $12 V$ range:
$$U_{12} = 10 \cdot \dfrac{2.4}{2 + 2.4} = \dfrac{24}{4.4} = \dfrac{60}{11} V \approx 5.45 V.$$