maths based riddles... See if you can answer them.
Discussion
So many maths based riddles that I thought I would start a new thread for them all as a little game were the person who answers with the correct answer gets to ask the next one...
I had some fun confusing my friends last night with this little one that they couldnt get their head round.
If you have a one mile race track and you must complete two laps, for a total of two miles.
no more, no less.
If you average 30mph on the first one mile lap, how fast would you have to go on the second lap to average 60 MPH?
I await the first smart arsed comment.
I had some fun confusing my friends last night with this little one that they couldnt get their head round.
If you have a one mile race track and you must complete two laps, for a total of two miles.
no more, no less.
If you average 30mph on the first one mile lap, how fast would you have to go on the second lap to average 60 MPH?
I await the first smart arsed comment.
illmonkey said:
Stupid wording:
To average 60MPH on the second lap, its 60mph.
To average 60MPH over both laps, its 90MPH on the second lap.
How do you work that?To average 60MPH on the second lap, its 60mph.
To average 60MPH over both laps, its 90MPH on the second lap.
Since many have it I will explain why its not possible by answering 90mph.
Since it has already taken 2 minutes to cover 1st mile and you cant cover the 2nd in 0 time, doing it at 90mph would take 40 seconds, which would mean it just took you 2mins 40s to cover 2 laps or an average of 45 mph.
Edited by Du1point8 on Monday 15th August 12:15
illmonkey said:
Einion Yrth said:
HTH. If not I'll have another stab - but that seems clear to me.
I was talking to the OP. His post does not mention it being done in 2 minutes. I'm probably being a thicky thick thick.Just because something is not said, it can be an assumption in your hypothesis or educated guess and then come to the conclusion that it is impossible.
dvance said:
Einion Yrth said:
samdale said:
Einion Yrth said:
I'll try and keep this one alive for a bit - again with a logic puzzle rather than mathematics.
You are trapped in a room with two guarded doors; one door leads to certain death, the other to escape. One of the guards always tells the truth, the guard of the other door always lies. You are permitted to ask one question of either, but not both, of the guards. What question will ensure that you exit through the safe door?
Been watching Labyrinth?You are trapped in a room with two guarded doors; one door leads to certain death, the other to escape. One of the guards always tells the truth, the guard of the other door always lies. You are permitted to ask one question of either, but not both, of the guards. What question will ensure that you exit through the safe door?
Did you just change the wording from the monte card game and 2 black cards and 1 red, to 2 goats and an Mx-5?
RizzoTheRat said:
Du1point8 said:
dvance said:
A similar one is: You're at a TV show where you're faced with three doors. There are two goats and an MX-5 behind them. Your task is to choose the door behind which the ultimate driving machine is placed. You choose a door, but then the TV host opens up one of the remaining doors to reveal a goat. The question now is what is the best strategy (the one that maximizes your chances of winning) -- to keep your choice or switch your chosen door?
The short answer is yes. Did you just change the wording from the monte card game and 2 black cards and 1 red, to 2 goats and an Mx-5?
You're playing three-card monte. Two cards are red, one is black. (Note: In three-card monte, the three cards are face down and you try to pick the black card in order to win.) You pick the middle card. After you pick, the dealer shows that one of the cards you have not chosen is red. You are given the chance to switch your selection. Should you?
Yes (below is text book answer I have for it, I recognised the format straight away)
By switching, you are betting that the card you initially chose was red. By not switching, you are betting that the card you initially chose was black. And because two out of three cards are red, of course, betting on red is the way to go.
Let's break it down, starting with the not switching case. Say the first card you chose was the black one. This happens one-third of the time. If you do not switch your choice, you win. Needless to say, the other two-thirds of the time, having picked a red card, and deciding not to switch, you lose. In other words, if you do not switch, you win a third of the time.
Now let's examine what happens when you switch cards. Say the first card you chose was the black one. Again, this would happen one-third of the time. If, after being shown a red card, you switch, you lose. The other two-thirds of the time, if you switch, you win because the dealer has already shown you that one of the cards you did not pick is red. Given the premise that your original pick was a red card, the card you are switching to must be the black one. You will win two-thirds of the time.
17 mins
First have the 1 minute and 2 minute person cross together with the flashlight. You have used 2 minutes.
Now have the 2 minute person go back with the flashlight. You have used 4 minutes.
Now have the 5 minute and 10 minutes person cross with the flashlight. You have used 14 minutes.
Now have the 1 minute person go back with the flashlight. You have used 15 minutes.
Now have the 1 minute person and 2 minute person cross with the flashlight. You have used 17 minutes and everyone is across.
First have the 1 minute and 2 minute person cross together with the flashlight. You have used 2 minutes.
Now have the 2 minute person go back with the flashlight. You have used 4 minutes.
Now have the 5 minute and 10 minutes person cross with the flashlight. You have used 14 minutes.
Now have the 1 minute person go back with the flashlight. You have used 15 minutes.
Now have the 1 minute person and 2 minute person cross with the flashlight. You have used 17 minutes and everyone is across.
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