Physics
Grade-9
Easy

Question

An oven consumes 425 kWh of energy in order to provide 386 kWh of useful energy. What is its percent efficiency?

  1. 90%
  2. 91%
  3. 92%
  4. 93%

hintHint:

  • Efficiency (%) = (useful energy output ÷ energy input) × 100

The correct answer is: 91%


    • Efficiency (%) = (useful energy output ÷ energy input) × 100
    • E f f i c i e n c y equals 386 over 425 cross times 100 equals 90.8 space w h i c h space i s space a p p r o x i m a t e l y space e q u a l space t o space 91 space percent sign

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