SSC JE Electrical Engineering 2017 Set 1 Paper

Ampere hour efficiency = $\frac{\mathrm{Ampere}\left(\mathrm{Discharge}\right)\times \mathrm{Time}\left(\mathrm{hours}\right)}{\mathrm{Ampere}\left(\mathrm{Charge}\right)\times \mathrm{Time}\left(\mathrm{hours}\right)}\times 100$
Watt-hour efficiency = $\frac{\mathrm{Ampere}\mathrm{Voltage}\left(\mathrm{Discharge}\right)\times \mathrm{Time}\left(\mathrm{hours}\right)}{\mathrm{Ampere}\mathrm{Voltage}\left(\mathrm{Charge}\right)\times \mathrm{Time}\left(\mathrm{hours}\right)}\times 100$
Voltage at discharge is lower then charge. So Amperehour efficiency is higher 90% then watt-hour efficiency is 75% approx, and
over all efficiency < 75%.