# Problem Solving Question (Riddle)

The problem I need to solve is written as the following:

Four people want to cross a bridge on a very dark night. They all begin on the same side and want to do the crossing as quickly as possible. A maximum of two persons can cross the bridge at one time. Any group that crosses, either 1 or 2 people, must have a flashlight with them. There is only one flashlight. The flashlight must be walked back and forth, it cannot be thrown. Each person walks at a different speed. A pair must walk together at the rate of the slower person’s pace, no one can be carried.

Here is the list of persons and their speed:

• $$A$$ ~ $$1\text{ min}$$
• $$B$$ ~ $$2\text{ min}$$
• $$C$$ ~ $$5\text{ min}$$
• $$D$$ ~ $$10\text{ min}$$

What is the least amount of time needed to get all four people across the bridge?

Here's how I solved it:

Since each person inevitably has to cross the bridge, the only fact we can manipulate is that someone has to come back each time to get the next person.

$$A$$ and $$D$$ cross first, then $$A$$ goes back. $$A$$ and $$B$$ cross second, then $$A$$ goes back. $$A$$ and $$C$$ cross third. In total, that gives me $$((10+1) + (2+1) + 5)\text{ min}=(11+3+5)\text{ min}=19\text{ min}$$.

The T.A. in class, though, said $$19$$ was an incorrect answer.

Thoughts?

• This would be better on puzzling.stackexchange, a similar question with different constants is puzzling.stackexchange.com/questions/287/… The point is that you want the two slow ones to cross together, so the second slowest doesn't count. Sep 16, 2014 at 2:55

\begin{align} ABCD&|&0\\ BC&|AD&10\\ ABC&|D&11\\ B&|ACD&16\\ AB&|CD&17\\ &|ABCD&19 \end{align}
\begin{align} ABCD&|&0\\ CD&|AB&2\\ ACD&|B&3\\ A&|BCD&13\\ AB&|CD&15\\ &|ABCD&17 \end{align}