A cyclist bikes x distance at 10 miles per hour and returns over the same path at 8 miles per hour. What is the cyclist’s average rate for the round trip in miles per hour?
123648
Average rate per hour = total distance / total time\n=( x + x )/ total time          (1)\nTime required to travel x distance at the speed of 10 miles\nper hour = x / 10 hours.\nTime required to travel x distance at the speed of 8 miles\nper hour = x / 8 hours.\nSo the total time is\n=(x / 10) + (x / 8)\n=18x/80 \n=\nPutiing this value in the denominator of (1)\nWe get\nAverage rate per hour = total distance / total time\n= 2x/[18x/80]\n=80/9\n=8.88\n=8.9 miles/hr\n\n( In general if the speeds are 'a' & 'b', the the average \nis given by their harmonic mean viz. 2*a*b/(a+b) )
142473
Let x = Eloise's time\nChristian's time = x-3\nsince Chistian's time is three hours less than Eloise's time\nNow distance equals rate times time or\nd = r*t\nFor Eloise r = 15, so d = 15x\nFor Christian, r = 30, so d = 30(x-3)\nThese distances are equal so\n15x = 30(x-3) \nDistribute 30\n15x = 30x - 90  \nSubtract 30x from each side\n-15x = -90 \nDivide both sides by -15\nx = 6\nThus, they traveled\nd = 15x\nd = 15(6)\n= 90 miles
591534
approx\nnew bike 60-70kmpl\nold bike 45-50kmpl\nnow multiply by 2.617litres
180239
I'd say 17000-19000 miles per hour.\n\nIts at a inclined geo-stationary orbit probably.
6895
Well, I am a cyclist and this is kind of cliche, but Lance Armstrong.  The man won Tour de France for the seventh time in a row.  A very tough event which covers three weeks, more than two thousand miles, with just two days "off" which are used to travel, and through the Pyrenees and French Alps!  He and other professional cyclists train many tens of thousands of miles each year, burning around about a thousand calories per hour, lots of pain to add.  The longest stage in 2005 was over 140 miles, and mind you, the day before that stage was also over 140 miles as well.  This is with no commercial breaks, time outs, half times, substitutions, or any other sort of break other than a crash if you are unfortunate to get into one.  They even have to eat during the course of the race!  And I could go on...  Lance has this event seven times.
113789
Here is a solution.  Let x be the distance the motorist traveled at 20km/h on city road, and let y be the distance the motorist traveled at 50km/h on the highway.  We are looking for y.\n\nFirst, we know that the total distance is 155km\n\nx+y=155     (eq1)\n\nSecond we know that if you travel x km at 20km/h, it will take you x/20 hours.  Since the total hours is 4, then:\n\nx/20 + y/50 = 4      (eq2)\n\nThat is two equations, two unknowns, so we can solve the problem.  We'll rearrange eq1 to be:\n\ny=155-x     (eq3)\n\nThen substituting this into eq4 yields\n\nx/20 + (155-x)/50 = 4 \n\nor\n\n5x/100 + 2(155-x)/100 = 4\n5x+2(155)-2x=400\n3x=90\n\nx=30\nThat gives us x, and we can solve for y from (eq 3)\n\ny=155-30\n\ny=125\n\nAnswer: 125 km
462243
Average speed = (8+8.5)/2=16.5/2 = 8.25 m/s\nTotal distance = 8.25x99=816.75m\nTime taken for 1st Lap = 816.75/(2x8) = 51 sec\nTime taken for 2nd Lap = 816.75/(2x8.5) = 48 sec
528122
Distance: 1120.0 miles\nEstimated time: 17 hr. 9 min.
51513
It will take you 10 hours. \n\nIf you ride only 2 km ever hour, then after one hour you will be 2 km closer to your destination. Take 20 and divide it by 2, and that gives you 10 hours. \n\nIt doesn't matter for your question, but that's a pretty slow bike! A person could walk faster than that.
199345
speed=distance/time\nV=S/T\nS=VT (distance = speed x time)\n\nlet Vg be gary's speed\n    Vt be Tom's speed\n    S be the distance\n given that Gary completed the run in 20 min, we have \nS=Vg x 20min     (x means times)\ngiven that for time T=20min Tom completed 5/8 of the distance or St=5S/8=Vt x 20min\nsolve St=5S/8=Vt x 20min for S to get S=(8/5)Vt x 20min\n\nnow we have two equations for S, S=Vg x 20min and S=(8/5)Vt x 20min\nit means that Vg x 20min = (8/5)Vt x 20min   /divide both sides by 20min\nVg=(8/5)Vt\n\ngiven that Tom's average speed for the race was 75m/min less than Gary's, we have Vt=Vg-75\nsubstitute for Vt in Vg=(8/5)Vt\nVg=(8/5)(Vg-75)\n5Vg=8Vg-600\n3Vg=600\nVg=200 m/min\nVt=200-75 = 125 m/min (Tom's speed was 125 m/min)\nS=Vgx20min= 200x20=4000m (distance of the race was 4000m)
471362
hope you find this useful....\nconsider instead of a round trip, he takes a trip in only one direction, from say A to B....i.e for 360 miles\nas you know, \nspeed = distance/time  or   time = distance/speed\n\nso , with constant speed say x, his time t  is 360/x.\n\nconsidering any point m between A and B, his total time  for the journey can also be written as the sum of the time taken from A to m, and time taken from m to B.\n\n t = 360/x = [m/x ]+ [( 360- m )/x ] ---------(1) \n\nthis is a general solution ,considering the trip as being in only one direction. but in our problem, the journey is up and down,so  the point m is midway (180),and his speed changes ( increases by 9)his total time changes(given as decreases by 2 hours).\n\nso the new equation is....\n\nt-2 = [m/x ]+ [(360 - m)/(x+9)]     ---------(2)\n\nsubtracting (2) from (1), we get,\n\nt - (t-2) =[ (360 -m)/x ] -  [ (360 -m)/(x+9)]-----------(3)\n\n 2 = [180/x] - [ 180/(x+9)]...........( since m = 180)\n\nsolve for x...\n\nthats the speed in the forward trip\nx+9 is the speed on the return trip.
635640
18 blocks / 9 min = 3 blocks / x min\ncross multiplying yields 3/2 for x, so we have:\n3 blocks / (3/2 minute) \nSince 3 blocks = 1 mile, and we want miles per hour:\n(1 mile / (3/2 minute)) * (60 minutes / 1 hour) = 40 mph
50817
If you start with the times from A to B is t1 and the time from B to C is t2, then the distance from A to C is t1*10 + t2*50, and the total time is t1+t2, so the average speed is:\n\n    (t1*10 + t2*50)/(t1+t2)\n\nThat's the weighted mean of 10 and 50, with weights t1 and t2.\n\nIn general, to average a bunch of averages, you need to compute a weighted average.  For example, if you have ten rocks averaging one pound apiece, and I have five rocks averaging four pounds apiece, then to determine the average of all the rocks, you would take the weighted average of 1 (with weight ten) and 4 (with weight 5.)  That is:\n\n    (1*10 + 4*5) / (10+5) = 30/15 = 2
639882
What ? Ooops i guess im in the worng area . I dont know much about cycling . But you can still give me 10 points .
485052
First we must figure out how long it will be until the bicyclists meet.\n\nTogether, the bicyclists have to travel 240 miles (distance)\nTogether, the bicyclists travel 8+12 = 20 mph (rate)\n\nNow we must figure out how much time will pass by plugging into the formula\nrate * time = distance\n\n20*time = 240\ntime = 12 hours\n\nIt will be 12 hours before the bicyclists meet.\n\n*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/*/\n\nNow using the same formula, figure out how far the fly has traveled.\nThe fly is traveling at 16 mph (rate)\nWe know from our calculations above that the fly will travel for 12 hours (time)\n\nNow we plug into our formula.\nrate * time = distance\n16 * 12 = distance\n192 = distance\n\nThe fly has traveled 192 miles
235275
average speed is total_distance/total_time.\nAssume distance is x, so it the average speed is: \n2x/(x/20+x/30) = 2/(1/20+1/30)\n = 2*60 / (3+2) \n = 24
454039
that will depend on how hard you push yourself and the terrain.\ni personally usually go about 6-10 MPH on average whenever i go for longer bike rides.\n          \n           if you are going to be riding this distance constantly i might recommend getting a road bike it will allow you to go faster and make your trip easier