Double

$2^{60}=\text{full}\\\\ 2^t=\dfrac{1}{2}\text{full}\\\\ 2^t=\dfrac{1}{2}\cdot2^{60}\\\\ 2^t=2^{59}\\\\ t=59$

Since it is full at 60 minutes, and we know it doubled in the last minute, then at 59 minutes it must have been half full.

it doubles itself every minute.

so, if after 59 minutes it covers half of the area , the next or after 60 minutes it covers the full area

again, if after 60 minutes it covers full area ,so, 1 minute before it covered half area.

2^60 = 1.15 X 10^18 = number of bacteria that just fill the plate.

Half that number will fill half the plate = 5.76 x 10^17 bacteria.

Antilog base 2 of 5.76 x 10^17 = 59.

This is the same as Matt Doe's solution, arranged a little differently. But the most elegant solutions are those of Geoff Pilling and Mohammad Khaza -- no math required, just logic.

once the surface is 100%, it stays, no information of death. the last minute had to be 59 or fewer minutes. maybe 10.

Let initial area covered by bacteria be x so at t=0 it has x then t=2 it has 2x similarly at t=60 it has 2^60x so its half is 2^59 hence it will corespond to 59 sec