So my answer is within the rules, when they need to answer which color hat is on their head, the can say it loudly if the person in front of them is wearing red, or answer their answer quietly to indicate that the person in front of them is wearing blue...
Sorry, I didn't understand your answer first time around.....
Clever solution, I must say! I'm assuming the warden must've not allowed them to answer quietly, since the actual solution is different.....
Sorry, I didn't understand your answer first time around.....
Clever solution, I must say! I'm assuming the warden must've not allowed them to answer quietly, since the actual solution is different.....
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I figured they may only say Exactly blue or red and that is why I chose My solution over the other one. (That is why I originally asked if they can add words)
Sorry, I didn't understand your answer first time around.....
Clever solution, I must say! I'm assuming the warden must've not allowed them to answer quietly, since the actual solution is different.....
The prisoners come up with the following plan: If the first prisoner to speak—the one in the back of the line—sees an even number of blue hats, he'll say BLUE, if he sees even number of red hats he'll say RED. He has a 50% chance of survival. Next prisoner counts if there's an even amt of the color first prisoner said in front of HIM, that will mean he has the other color, if there's odd amount, then he must have that color to make it even. Every prisoner now has to keep track of the one's behind them too one by one, plus counting the ones in front of them, always bearing in mind that the total amt of the color first prisoner said is even.
The prisoners come up with the following plan: If the first prisoner to speak—the one in the back of the line—sees an even number of blue hats, he'll say BLUE, if he sees even number of red hats he'll say RED. He has a 50% chance of survival. Next prisoner counts if there's an even amt of the color first prisoner said in front of HIM, that will mean he has the other color, if there's odd amount, then he must have that color to make it even. Every prisoner now has to keep track of the one's behind them too one by one, plus counting the ones in front of them, always bearing in mind that the total amt of the color first prisoner said is even.
Alice and Barbara together can finish a project in 3 hours.
Barbara and Cindy together can finish the same project in 4 hours.
Alice and Cindy together can finish it in 6 hours.
How long will it take them to finish this project if they all three work on it together? (Considering that everyone consistently works at her own pace)
Alice and Barbara together can finish a project in 3 hours.
Barbara and Cindy together can finish the same project in 4 hours.
Alice and Cindy together can finish it in 6 hours.
How long will it take them to finish this project if they all three work on it together? (Considering that everyone consistently works at her own pace)
I took the time it takes each of them. Added it up and divided in 3 for an average.
You would need to state the time for the person whom it takes longest since they're all working at the same time. I'm lazy now to think, but you're on the right track, I think.
You would need to state the time for the person whom it takes longest since they're all working at the same time. I'm lazy now to think, but you're on the right track, I think.
That doesn’t make sense because the quicker workers should make up for the slower workers.
And it should take quicker than the first scenario if a third person is helping.
You would need to state the time for the person whom it takes longest since they're all working at the same time. I'm lazy now to think, but you're on the right track, I think.
Given the info, You tell me who it takes the longest.
