If it takes an hour to dig a hole, how long does it take to dig half a hole?
104137
I can dig half a hole.  But not half a whole.
186250
There's no such thing as "half a hole." Funny question. Thanks for the smile.
496519
12 minutes (1/5 of an hour)\n\nDistance (d) = 10 miles\nRate (r) = 50 miles per hour\nTime (t) = Time in hours\nd = r * t\n\nSubstituting:\n10 = 50 * t\n\nt = 10/50\nt = 1/5 hours\n\n1/5 hour * 60 minutes/hour = 12 minutes\n\nThe answer is 12 minutes
982506
what hole?\n\neither way diameter is length across, straight through the middle point of the 'hole'\n\nthis line should then make two perfect (or close to perfect) semi-holes (or semicircles)
747324
3 hours. \n\nIf it takes 12 hours to go from low tide to high tide and back to low tide, that means it takes 12 hours to go from 1.0m, to 3.0m and back to 1.0m.\n\nSo in half of 12 hours= 6 hours it goes from 1 to 3\nAnd in half of that= 3 hours it goes from 1 to 1.5.\n\nTry drawing a wave starting from the lowest point (1.0m) going to the highest point (3.0m) and coming back to the lowest point (1.0m). \nDraw a line through where it is 1.5m, which is half way between 1m and 3m. If the whole wave is 12 hours you will see that at a quarter of twelve hours the wave crosses the line you drew for 1.5m.
651108
We know that the Earth is 360 degrees(as the Earth is a sphere). Now the Earth takes about 24 hrs to complete a rotation or 360 degrees. Therefore in 1 hour, it will travel 15 degrees. But some times countries are so small that they don't even occupy 15 degree area. So its half - 7.5 degrees is taken into consideration. The main purpose of taking all this trouble is to make the calculation of time easier.
109042
13 and a third hours is the number i came up with...\n\nJust plug some fake numbers in, say, for example, the tank that these two pipes are filling is 500 gallans.  then you can figure out how much each pipe put into it, take that five hundred, divide by 8, then multiply by 3, that's how many gallens the smaller pipe put into the tank in 5 hours...   now, take that number and divide by the 5 hours, that's how much that smaller pipe can pour out into the tank in one hour...  to take that 500 gallens and divide by the amount that can be poured in one hour and it'll give you how many hours it'd take to fill at that slower rate...\n\nThis is not a standard way of figuring out this problem, but it seemed like it worked to me, it should have taken more then double the 5 hours, but less then 3 times...   does this help at all??
696825
Distance = 100\nRate = 180\nTime for the first half = t\n\n100 = 180t\nt = 100/180\nt = 5/9 hours\n\nSecond half:\n100 = 110T\nT = 100/110\nT = 10/11 hours\n\nAdding the time together:\n5/9 + 10/11 = 55/99 + 90/99 = 145/99 hours\n\n145/99 hours ≈ 1 hour 27 minutes 52.72 seconds\n\nAnd if you want the average speed it is:\n200 miles = R * 145/99\nR = 200 * 99 / 145\nR ≈ 136.6 mph\n\nTime = 1 hour 27 minutes 52.72 seconds (*assuming you could instantaneously drop your speed from 180 to 110 at the halfway point)
307426
Let e = a standard unit of 1 man-hour per weekday effort\n3.5 e = effort made during 1 man-hour during the weekend\n\nSunday 2am +19 hours = Sunday 9pm\neffort = 3 men, 19 hours, weekend effort only = 3*19*3.5e = 199.5e\nto dig 12 m^3 = 12*(100/2.54)^3 cu.in. (about 700k)\n\ne (using shovel) in 1 man-hour (week-day) = (12*(100/2.54)^3 / 199.5) cu.in.\nwhich is roughly 4000 cu.in. (I'll do the exact math shortly)\n\nStarting at Saturday 5pm to dig 17 in. cubed with a tablespoon (1/5 the size)\n\n17 in. cubed is 17^3 cu.in. (somewhere less than 8000 cu.in.)\n\nNow, a standard tablespoon is maybe 9" long (so that puts the shovel just under 4 feet... so reality check on the shovel \n\nchecks out at least)... the question is, is the tablespoon (1/5)^3 the volume lifting capacity of the shovel?  let's assume \n\nyes... not as long, not as wide, not as deep... so let's use (1/5)^3 for the shoveling volume.\n\nthat means it takes 5^3 as much effort to dig the same cubic volume with a tablespoon as with the shovel.\n\nso this should put us somewhere at (8000/4000)*5^3 times the effort made to dig the first hole, in order to dig the second \n\nhole...or roughly 2*125 = man-hours... coronary time for the man\n\nIf the first 24 hours is at 3.5e efficiency, that's 84 man-hours done...\nthen he drops to half-efficiency on Sunday 5pm. 1.75e from 5pm-12am (7hours) = 12.25 man-hours done\nnow he's still at half-efficiency for the week... 24*5/2=60 per set of 5 weekdays\n\nuntil he's set for a coronary at ((17^3)/(12*(100/2.54)^3)/199.5))*(5^3) = 167.30911 man-hours\n\n84+12.25+60=156.25, so he needs another 11 or so man-hours at 3.5/2 efficiency on next Saturday, or 6.3194914 hours or\n\n1 week later Saturday 6:19:10am, just in time for the morning shift of the ER at the local hospital.
1001541
260 miles.\n\nFrom the half-hour difference we can find the length of the hill:\n\nt = d/v, t2-t1 = d (1/v2 - 1/v1)\nd= .5/(1/72-1/90) = 180\n\nThis takes 2 hours at 90 mph, so the other hour covers 80 miles.\n180+80=260
687277
1 hour or 60 mins or 3600 seconds :-) is the time it takes but it will still be 12:30 :-p
213241
Probably a half hour ask bush
979265
Sand falls onto a conical pile at rate of 10 ft^3/min. The radius of the base of the pile is always equal to 1/2 of its altitude. How fast is the altitude of the pile increasing when it is 5 ft deep?\n\nLet radius = r and altitude = h\nThen r = ½h\n\nV = ⅓πr²h\n= ⅓π(½h)²h\n= πh³/12\n\nNow dV/dt = dV/dh * dh/dt\nSo 10 = πh²/4 * dh/dt\nWhen h = 5\ndh/dt = 10/(π*5²/4)\n= 8/5π ft/min\n≈ 0.51 ft/min\n(≈ 6.1 inches/min)
996796
20 min.\n\njim goes 1 1/3 miles in 20 min\njohn goes 1 2/3 miles in 20 min\ntogether = 3 miles
338411
You can do it in x/30 + x/20 time, where x is the total hours.  Since each hour, you accomplish 1/30 of the task.\n\nx/30 + x/20 = 1.  Solve and get 12.\n\nIf you think about it, this answer checks.  You won't both get the half done.  After 12 hours, you would have done 12/30 of the work and your friend would have done 12/20 of the work, assuming the pace is constant.  12/30 + 12/20 adds to all of the work being done.
1062962
so Joe is taking 20 hrs and Fred is taking 32 hours. If these two are to be given work, then 20/52% should be given to Fred and 32/52% should be given to Joe to get things done in time ( I mean both finishes at the same time)\n\nSo Joe's portion of work would be 64% and Fred wil do 36% work...\n\nsimple !
414725
t=time it takes in hours to do the job\nSusan does half the job in an hour\nJaime does a third\nThey have one job to do\nso 1/2t+1/3t=1\nfinding like denominators, we get\n3/6t+2/6t=1\n5/6t=1\n5t=6\nt=6/5\n\nThat's roughly 1 hour, 12 min