In order to save time we need to send the 2 slowest together or get the fast guy to return always
The route when A(fastest) returns every time is B+A+D+A+C
The route when C and D(slowest two) are sent together is B+A+D+B+B
Clearly the difference is on account of A+C in first and B+B in second route.
Which ever of the 2 expressions is smaller is the right way. Therefore the answer in with 3,8,11,17 should be route 1(2*8>3+11)
2*3+11+8+17=42 seconds
CAT-holics 
A and B cross > 8 min
A returns > 3 min
A and D cross > 17 min
A returns > 3 min
A and C cross > 11 min
———–
Total Time > 42 min ?
Looks good, Catatonic…
regards
J
In order to save time we need to send the 2 slowest together or get the fast guy to return always
The route when A(fastest) returns every time is B+A+D+A+C
The route when C and D(slowest two) are sent together is B+A+D+B+B
Clearly the difference is on account of A+C in first and B+B in second route.
Which ever of the 2 expressions is smaller is the right way. Therefore the answer in with 3,8,11,17 should be route 1(2*8>3+11)
2*3+11+8+17=42 seconds
Well reasoned, Arvind…
regards
J