No machine hands back every joule: friction, heat and noise all take their cut. Efficiency is the honest ratio $\eta = \dfrac{W_{out}}{W_{in}}$ (read aloud: eta equals W-out over W-in — $\eta$ is the Greek letter eta). The subscripts say which side of the machine you are standing on: $W_{in}$ is everything paid in, $W_{out}$ is the useful work that comes back out, both in joules, and $\eta$ is what you are solving for. Joules over joules: the units cancel, leaving a pure number, usually dressed as a percent.

Two rails to check every answer against. An efficiency over 100% means more energy out than in — the first law of thermodynamics files an immediate complaint. And 100% minus your answer is the share LOST — a real number, answering a different question. Know which one the paper asked for.