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Three men and three rakhsasas arrive together at a ferry crossing to find a boat with an oar, but no boatman. The boat can carry one or at the most two persons, for example, one man and one rakhsasas, and each man or rakhsasas can row. But if at any time, on any bank, (including those who maybe are in the boat as it touches the bank) rakhsasas outnumber men, the former will eat up the latter. If all have to go to the other side without any mishap, what is the minimum number of times that the boat must cross the river?

  1. $7$
  2. $9$
  3. $11$
  4. $13$
  5. $15$
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r

So, boat crosses river 11 times.. Option (C)

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I am denoting $M$ for man and $R$ for rakhsas.

Step 1 : Two rakhsasas row the boat from one side to another. So, one rakshas will reach other side. and one rakhsas will have to row back. So, first side now $(3M + 2R)$ and second side $(1R)$.

So, boat crosses $2$ times.

Step $2 : 2R$ row to other side and leaves one $R$ there and comes back to first side. So, now first side has $(3M + 1R)$ and second side has $(2R)$.

So, boat crosses $2$ times.

Step $3: 2$ $M$ row to other side and gets down at other side. So, first side has $(1R + 1M)$ and second side has $(2M+2R).$
So, boat crosses $1$ time.

Step 4: Now, $1M$ and $1R$ will have to row back from second side, and they will take $1M + 1M$ to other side. So, in the first side, we now have $2R$ and second side $(1R + 3M)$

So, boat crosses $2$ times.

Step 5: $1R$ will row back and take $1$ more $R$ with him. First side $1R$, second side $(2R, 3M).$

So, boat crosses $2$ times.

Step 6: Same step $5$ repeated.

So, boat crosses $2$ times.

Total number of times the boat crosses $=10+1 = 11$ times.

Correct Answer: $C$
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