A logic problem by Aaditya Pitke

17min is the answer

First A and B will cross ( 2min ) and then A will return with lamp ( 1min )

then C with D will cross (10min) and then B will return with lamp ( 2min )

then A with B again ( 2min )

Hence Total Time = 2+1+10+2+2 = 17 Min ;)

First, I tried to add all of the friends' total time: $1+2+7+10 = 20$ minutes. It means if the friends want to shorten the time, $A$ or $B$ must be the ones who carries the lamp back. If $C$ and $D$ carries the lamp, the total time must be $> 7+10$ minutes.

As usual, let two of the fastest, $A$ and $B$ , cross the bridge. It takes $2$ minutes, plus $A$ carries the lamp back, which the total is $\color{#D61F06}{3}$ minutes.

Then, the two slower one, $C$ and $D$ cross the river. It takes $10$ minutes to cross, plus $B$ carries the lamp back, total is $\color{#3D99F6}{12}$ minutes. (Why not $B$ and $C$ or $D$ ? Because the total time for crossing and return for $C$ is $7+2$ minutes, for $D$ is $10+2$ minutes, and they must travel in different times, which is $21$ minutes!)

To the last trip, $A$ and $B$ crosses the bridge, which takes $3$ minutes.

Therefore, I think the shortest time for $4$ friends to cross is $\boxed{17}$ minutes.

Answer
= ceiling[1,2] + 1 + ceiling[10,7] + 2 + ceiling[1,2]
= 2 + 1 + 10 + 2 + 2
= 17 minutes