Okay, Here's a hint that might help: The fuel dumps won't be separated by equal distances. If no one gets it by this evening, I'll post the answer.
yes patty-rick, both of your solutions work. there is however one solution requiring only nine minutes, but, me not being specific gives you the correct answer.
Ok, got down to 42 drums Starting Dist - Gas Needed 0- 418 47- 225 95- 124 147- 66 193- 43 251- 25 300- 20 500- 0 I got a formula for how much gas it takes to move X gas, and then calculated an overall mpg (which includes dropped gas). Then I just started at 300, and moved back the most efficient distance. 42 is a little high, as I rounded up.
turn over both hourglasses. when the 4 minutes run out, turn it back over. after 7 minutes, there is 1 minute of sand left in the 4 min glass. restart the 7 minute timer, after 1 minute elapses (the 4 minute runs out again) 8 minutes have passed, 1 minute has passed through the 7 minute hourglass. turn the 7 minute hourglass back over and 1 minute of sand will pass through arriving you at exactly 9 minutes provided you compensated for the time it took you to flip them.
that's it Good Job On Radiactive Waves. There is the correct solution for the shortest amount of time to cook the nine minute egg. Scrambled is alright, provided the reviewer supplies the cheesePlease Register or Log in to view the hidden image!
BINGO!. Here's the solution: Jeep solution Before you read it , I think it should be fair to warn you that the solution has a problem, in terms of practicality, that I didn't notice until I wrote up the above explanation of the solution. It requires that you drop off more than 10 gals of fuel a trip most of the time. This is okay, in that you have the extra gas in the jeep's gas tank, The problem is, where are you going to put it ? If the drum holds only 10 gals, it's already full. The only way I see around this is to assume that the drums can hold more than 10 gals, but you are still limited to carrying only 20 gallons total with the jeep (A weight load factor?). Given this extra consideration, the above solution will work.
Okay, here's another one. And I guarantee this one has no hidden bugs. You have a solid sphere with a hole drilled all the way through it and passing through its center. The length of the hole, from lip to lip is 10cm. What is the volume of the remaining solid portion of the sphere?
You have a solid sphere with a hole drilled all the way through it and passing through its center. The length of the hole, from lip to lip is 10cm. When you say lip to lip... do you mean the length of the hole? (|||) where |=10cm? or (|/|) where /=10 cm?
That's a bit unfair don't you think, you should have made an amendment to the problem so that we could have tackled it from these conditions.
Yeah, you had the first completely correct answer... but the most fuel you drop off is 10.4 gallons... which seems reasonable.
Strictly you can't, it wasn't specified in the problem, but don't worry about me folks, I am just bitchin'. Good problem though, stumped everyone for a while.Please Register or Log in to view the hidden image!
Missing Janus58, Unless I'm missing something here, I believe you must give us the diameter of the hole. I haven't bothered yet to compute the zones of one base but just assuming two different hole diameters I get different sphere diameters and net volumes that seem to indicate hole size will affect the answer. Just in case not, I'm still calculating the two zones. BACK: Assuming a hole diameter of 1 cm the solid volume equals 124.9945 (with some rounding and including the two spherical zones at the ends. Assuming a hole diameter of 2 cm I get 107.103677 What is the hole diameter?
10 cm r Persol, A calculation of 4,188 cm^3 = a sphere with a radius of 9.99937 without any hole? 10 cm lip to lip should be length of the hole. The radius is therefore something greater than 5 cm as a function of the hole diameter.
