I am working through "Thinking Mathematically" by Mason, Burton and Stacey. One of the questions goes as follows:
Three slices of bread are to be toasted under a grill. The grill can hold two slices at once but only one side is toasted at a time. It takes 30 seconds to toast one side of a piece of bread, 5 seconds to put a piece in or take a piece out and 3 seconds to turn a piece over. What is the shortest time in which the three slices can be toasted?
I am wondering if I am overlooking something in my solution method, so I have come here. Some sources online say 130 seconds and others say 139. My solution, if non-fallacious, should be able to get it done in 118 seconds. The process goes as follows. Allow $T_n$ to represent the first side of the $n$th piece of toast and $T_n'$ to represent the opposite side of the $n$th piece of toast. Then, assuming a piece of toast is cooking as soon as it is placed, then the suggested sequence of events
Sorry for the convoluted timeline in the picture, but essentially I place the first piece of toast and then the second. $T_1$ and $T_2$'s cooking time will overlap, but there is necessarily a five-second window where they do not. You flip $T_1 \to T_1'$ where $T_1'$ officially begins toasting at 38 seconds. This gives another thirty-second window (until $t = 68$) to get a lot of the time-wasting flipping and removing done. For example, $T_2$ finished at $t=40$ and the most original part of this solution is to remove $T_2$ rather than flip it over and begin cooking the other side immediately. As soon as we are done removing $T_2$ the running total will be $t=45$ at which point we immediately begin placing $T_3$ bringing us to a running total of $t=50$. Note, that $T_3$ will be done at $t = 80$. At this point, we wait until $t=68$ to remove $T_1'$ which brings us to a running total of $t = 73$, at which point we immediately begin placing $T_2'$ which we put aside before. When we are done placing $T_2'$ we will be at $t = 78$ and at $t = 80$ it will be time to flip $T_3$, which brings us to $t = 83$ and which point $T_3'$ will begin toasting. $T_3'$ will thus be done at $t = 113$ and in that time we will have to remove $T_2'$ which began cooking at $t = 78$. Finally, we remove $T_3'$ at $t = 113$ and ending at $t = 118$.
Are there any problems with this solution that I am not detecting? I think it works well, indeed if it works at all, because it uses the long 30-second windows of cooking to do all the time-consuming activities of flipping etc.