How full was the battery when the charging began?

Question

Mark charged a battery. Each minute, the percentage of the battery's capacity that was charged increased by 2.52.52, point, 5 percent. After 171717 minutes of charging, the battery was 61.561.561, point, 5 percent full.

Answer 1

The equation of the line is y = Mx + b

 

 

 Changing the variables, 

   let y = E be the energy in the battery in terms of a percent

 

  let x = t, be the time in minutes.

 

The battery is at 61.5615% at 17 minutes, so (17, 61.5615)

The battery charges M = 2.525% every minute, which is the slope.

 

E = 2.525*t + B

 

61.5615 = 2.525*17 + B

 

B = 61.5615- 2.525*17

   = 18.6365

 

So the model is 

  E = 2.525*t + 18.6365

 

The intercept at t=0 when the charging began, there was B=18.6365%

 

The battery will be full at 100% when

  100 = 2.525*t + 18.6365

 

  t = (100 - 18.6365)/2.525 = 32.2231683

 

    32 minutes and 13.5 seconds