i like such stuff 
if u solve it, post the next one 
brainteasers
- denne
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There are 148 replies in this Thread. The last Post () by (IOC)Wanderer.
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ok, maybe u are too high educated. here is the result (if u want):
next one:
well, this is simple.
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69
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I would have thought 68
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Heres another one
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huor had his thoughts somewhere else
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hmm, i think 79 is right
we miss 116=68 here so 117=79 -
really like OP's comment, he he
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For Dented
Figuaratively speaking? This message serves to prove. How our minds can do amazing things! Impressive things! In the beginning, it was hard but now, on this line your mind is readding automatically with out even thinking about it, be proud. Only certain people can read this. Please forward if you can read this
after seeing picture i found :
0=O
1=i
3=e
4=a
5=s
7=t
as message said : i read this meassage automatically -
this is one of my favourites. it looks simple, but the result is amazing
u drive a complete lap with 60 km/h. how fast u have to drive the second lap (same distance) to get an average speed of 120 km/h?
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Normally it is 180 but let me think more
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do you have to get the average speed in the 2nd lap only or the average speed in two laps? makes a difference in calculation...
180 would be my guess too: for 2 round the average is 120, so 2 x 120 = 240, 240 - 60 = 180.
Not sure if there is something hidden i dont see right now... -
@denne I'm not sure if this is right but here goes:
A drives one lap at 60km/h, while B drives 2 laps at 120 km/h, they finish at the same time. A is one lap behind B. He would need to drive his second lap in exactly 0 seconds to catch up and make his average the same as B.
answer is infinite180 would work if there were 4 laps and he would drive 180 as long as he drove 60: 1*60(1 lap)+1*180(3 laps) = 2*120(2*2 laps)
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@Apache, nice
here is the comlete solution in formulars.
1. velocity for 1 lap: v1 = s/t1 (where s is the length of the lap and t1 is the time needed for the first lap)
2. average velocity for 2 laps: v2 = 2*s/(t1+t2) (where s is the length of the lap, t1 is the time needed for the first lap and t2 is needed for the second lap)
3. since v2 = 2*v1 u can also write: 2*v1 = 2*s/(t1+t2)
4. this u can devide with 2 on both sides and u get: v1 = s/(t1+t2)
5. v1 was already defined in point 1 (v1 = s/t1) --> s/t1 = s/(t1+t2)
6. devide it with s on both sides and multiply the denominators of both sides: t1+t2 = t1
--> only t2 = 0 can solve this.means the velocity in second lap must be light speed. it is the only way to finish this lap in zero seconds
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damn
that makes sense - proves me right. dont do math after wake up -
i disagree with the solution
lightspeed is not infinte... therefore the result for t2 can not be 0... eventhough its a value extremely close to 0at this moment I however have problems how to interprete the expression "average speed"
to hit exactly 120km/h you would need to travel an extremely tiny bit faster than light speed..... which equals a travel back in time
question is if the original 60km/h of the first lap were accurate considering the expansion of the universe (you have to negate that, as there is less or none time involved in the 2nd round)its truely hard to tell the correct answer.
the smallest measurable time interval is 5,39116 x 10^-44 seconds so we would never reach the t2=0 -
Vote for OP. That question like this : 1+1=? there are no exactly answer here. BTW we are fun, thats all
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And I'm just sitting here looking at those answers thinking WTF!!
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oh, nothing special here, just a crowd of nerds
