The Question is,\nTom and Gary ran in a race. When Gary had completed the run in 20 minutes, Tom had only run 5/8 of the distance. Tom's average speed for the race was 75m/min less than Gary's.\n\na) Find the distance of the race.\nb) What was Tom's speed in m/min?
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)
178173
1st question: Answer is 10km/(2/3hr) = 10*3/2 km/hr, that is\n              15km/hr\n\n2nd question: Answer is 200m over 160s, that is:\n                    200m/160s = 1.25 m/s
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
607518
i am no genius but i did try. here's how i went about it.\nspeed = distance/time.\n(1/30)s = 525 lines, therefore time for 1 line = ((1/30)/525)\n=(1/15750)s\n20in = 1 line.\nresweeping the line, therefore distance = 40in.\n40in = 3.33ft (1ft = 12in)\nagain, speed = distance/time, so\nspeed = (3.33ft)/(1/15750s) = 52500 (ft/sec)\nhopefully, this is the right answer. please let me know. thanks.
537905
average speed =total distance/total time\n1.\ntotal distance=2km\ntotal time=1/2hr\naverage speed =4km/hr\n2.\ntotal distance=360km\ntotal time=5hr\naverage speed =360/5=72km/hr\nthanks & good luck
620833
Let the first car had an average x and the second y\n\nThen,\nthe first car moved 25x miles in that week\nand the second moved 20y miles\n\nNow\nx+y = 45, and\n25x+20y=975\n\nSolve the equations baby, and you are through!\n\nvisit this amazing site called www.mathguru.com for ample practice exercises all solved step-by-step... with voice and all.. its awesome
261522
This is how you solve it. Use dimmensional canellation, but you will have to do it twice and subtract the difference\n\nA. (5 miles/1)  *   (1 hr/55 mi)  * (60 min/1 hr)\n   =5.45 minutes\n \n   \nB. (5 miles/1)*  (1 hr/65 miles) * (60 min/1 hr)\n   =4.6 minutes\n    \nYou saved 1.2 minutes. Not much at all. Which is what they say, speeding only saved 2 or so minutes in a 15-20 minute trip. BUt it tears the bejesus out of your engine
216198
1= 8mph\n2= 6 hrs walked 4hrs run\n3= 60 miles to the farm
340705
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)
132140
speed=distance/timetaken\ntherefore the average speed=86/2=43miles per hour
514565
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
11497
20m/s.\nbecause the total time is 5s. and by defn of average speed it is the ratio of total distance travelled to total time taken i.e 100/5 = 20m/s
493563
Alright, so the course is ten miles and they are running at 37 miles per hour. So everyone finishes in 10/37 hours, except one man, who finishes in 10/37 + 5/60 hours. Thus, their average time is (299*10/37 + 10/37 + 1/12)/300 =  36037/133200 hours. They ran for 10 miles, so we take 10 over this to get 1332000/36037 mph as their average speed, which is about 36.96201126 mph
347635
Roman please! Don't confuse the poor girl any more than she already is.  Which is  it, 23 or 24?\n  \nTat, Roman was on the right track with this one.  He had the right concept, but the wrong constant in the end.  Guess that's why the Caesars no longer rule.\n\nIf Greg now runs 35 miles today, then six weeks ago he started running 35 - 2(6) or 23 miles per week.  Letting m be the mileage run per week, and n the number of weeks he runs, then we come up with this equation for the mileage Greg runs each week he is on the program.:\n\nm = 23 + 2n.\n\nWe defy the normal convention of putting the constant at the end of the equation this time, because that was the mileage Greg started with, and everything else is added to it.
534544
1) \nSB takes 3 hours to do the job.\n\nfraction of job done by SB in 1 hr = 1/3\n\nSimilarly Fraction of job done by SW in 1 hr = 1/2.5 = 2/5\n\nFraction of job done by both SB and SW in 1 hr = 1/3+2/5 = 11/15\n\nNow Both SB and SW can finish the job together in : 1 / (11/15) = 15/11 = 1.36 hrs.\n\n Shortcut:\n\n1 / (X and Y together) = 1/ X + 1 / Y\n\n2)\n\nNow when Mark and his agent meet they would have covered the 500 Kms together as they \n\ntravel the same road to meet each other.\n\nsuppose after x hrs they meet.\n\n\nMark has travelled : 55x km\nAgent has travelled : 65x km\n\nTotal distance : 55x + 65x = 500\n\n120x = 500 => x = 500/120 = 4.167 hrs\n\n3) Use the same approach as in 1) above.\n\nUse the shortcut too if u need.\n\n1/ ans = 1/12 + 1/9\n\nand find the ans!\n\n4)\n\nLet the cost of the pencil be x.\ncost of pen = 3x\n\n5 * x + 4 * 3x = 4.25\n17x = 4.25\nx = 0.25\n\nSo pencil costs 25 cents!\n\n\nHope the above answers help! Do let me know if they dont! :-)
385603
It's 8 meters per second.  He's gone 5000 meters in 625 seconds; 5000 meters/625 seconds = 8 meters per second.\n\n v=d/t\nv=5000 m/625 seconds\nv=8m/sec
666323
The difference in speed is 6.45-5.30 = 1.15 m/s.\n\nThe track is 198 m long - so the faster runner must catch up by this much to lap the slower.\n\nThis will take 198/1.15 seconds.\n\nThe slower one will have run 198 x 5.30/1.15 m.