6 votes

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?

- $7$
- $9$
- $11$
- $13$
- $15$

8 votes

2 votes

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: $C$

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: $C$

1

In step 4, how did the boat come back?

So, 3 M's on the other side and an R came rowing back.

Step 5: it will take 2 other R's one by one. That makes total 8+3 = 11

11 is the correct answer.

So, 3 M's on the other side and an R came rowing back.

Step 5: it will take 2 other R's one by one. That makes total 8+3 = 11

11 is the correct answer.

0

answer is clearly wrong it should be 11 steps. @Arjun sir, kindly remove the best tag from this answer.

0

**The answer seems right till step 4. The modification there after is given below.**

At the end of step 4 the boat is at the right bank with 3M and 1R. The boat has to return to the left bank to pick the rest of the Rakshasas. So the boat goes back to left bank with 1R and the count becomes 8 meanwhile at the right bank we've 3M.

Now we've 3R in the left bank. We need to row with 2R in the boat to the right bank, leaving 1R with other 3M. The count now is 9. The last Rakshasa can be taken in 2 crosses with any R/M accompanying the last R.

**Thus the final count is 9 + 2 = 11.**

@Arjun sir please note!

2 votes