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

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