Four people want to cross a bridge at night with just one torch (the torch is needed to cross)

Why

Rohan's method feels natural — the fastest person seems like the obvious one to handle every back-and-forth trip — but it hides a costly waste: every time the two slowest people (5 and 10 minutes) cross SEPARATELY, their slow speeds get paid for twice instead of once. Rohan's plan: 1&2 cross (2 min), 1 returns (1), 1&5 cross (5), 1 returns (1), 1&10 crosses (10) — total 2+1+5+1+10=19 minutes, with the 10-minute person's slowness paid for on its own trip. The better plan sends the two SLOWEST together, so their slowness is only paid for once: 1&2 cross (2 min), 1 returns (1 min), 5&10 cross together (10 min, both slow people in one trip), 2 returns (2 min), 1&2 cross again (2 min) — total 2+1+10+2+2=17 minutes. Test it: the saving comes from bundling the two slow crossers into a single 10-minute trip instead of two separate slow trips (5+10=15 minutes' worth, done as one 10-minute trip) — the 'obvious' fastest-person-escorts-everyone method actually wastes the slow crossers' time by not pairing them up. His sister is right: sending the two slowest people together, even though it looks odd since they are not naturally paired, saves 2 full minutes over the seemingly obvious approach.