GATE 2018 EE GA Q.10: solution and answer

Official answer: CSource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Count the rowing turns. There are 3 forward trips with 2 people each and 2 return trips with 1 person each, so 5 trips in all. Exactly one person rows each trip (v), and each of the 4 people rows at least once (iv). So exactly one person rows twice.

2. What P can row. By (ii), P rows only when alone in the boat. Forward trips always carry two people, so P can row only a return trip. Everyone starts on the near side and ends on the far side, so each person crosses forward exactly once more than they cross back.

3. P rows exactly one return. Suppose P rowed both returns. Then P crosses forward 3 times, so P is on every forward trip. Q cannot row a return because both are taken, so Q must row a forward trip, and by (iii) R must be the other person on it. But every forward trip already holds P and one other person, so there is no room for both Q and R. Contradiction. So P rows exactly one return and crosses forward twice. The other return is made, and rowed, by someone else.

4. Q never makes a return trip. Suppose Q made the other return. Then Q also crosses forward twice, so the six forward seats are $P,P,Q,Q,R,S$. A trip $\{P,Q\}$ has nobody who can row it: P cannot row with a passenger (ii), and Q rows only with R (iii). So P's two trips must be $\{P,R\}$ and $\{P,S\}$, which leaves $\{Q,Q\}$ for the third trip. That is impossible. So Q never goes back and crosses exactly once, forward. Q must row at least once (iv), so Q rows that one trip, and by (iii) it is the trip $\{Q,R\}$ with Q rowing.

5. Who makes the second return? It is R or S, and that person crosses forward twice. The other of R and S never goes back, so they cross forward only once. Suppose S made the second return. Then R crosses forward only once, and that crossing is the trip $\{Q,R\}$, which Q rows. R makes no return either, because P and S have them. So R rows no trip at all, which breaks (iv). Therefore R makes and rows the second return, and R crosses forward twice.

6. P's two crossings. The six forward seats are $P,P,Q,R,R,S$. The trip $\{Q,R\}$ uses one Q seat and one R seat, which leaves $P,P,R,S$, so P's two crossings are $\{P,R\}$ and $\{P,S\}$. P cannot row either one (ii), so R rows $\{P,R\}$ and S rows $\{P,S\}$.

7. Tally and check. P rows 1 trip (a return), Q rows 1 ($\{Q,R\}$), S rows 1 ($\{P,S\}$) and R rows 2 ($\{P,R\}$ and a return). That makes $1+1+1+2=5$ trips. One order that works: Q rows Q and R across; R rows back; S rows S and P across; P rows back; R rows R and P across. Every condition (i) to (v) holds. So R rowed twice, which is option (C).