I was busy doing my 7th maths assignment for my tutor, when i came up with these 2 questions. If you people find it too easy than good for you. Ok so 10 points for best answerer, who shows their working.\n\nQ1: A girl drives the 251 miles from southampton to york in 4 hours and 15 minutes. What is her average speed in MPH?\n\nQ2: A guy called Martin drives 38 KM at an average speed of 63 km/hour.\nCalculate how long his journey takes, and give your answer in minutes.\n\nThen without using a calculator(if you can), show how to check your answer is the right order of magnitude. You must show all your working.\n\nFYI: This is NOT my homework! I just wanna see how smart u people are at maths.
130417
Q1:  4 hours and 15 minutes is 4.25 hours.  The girl's speed is 251/4.25 miles/hour.  Without a calculator, I would estimate that this is 60 miles per hour (60 = 240/4).  With a calculator, it is 59 miles per hour. \n\nQ2.  Martin drives 38 km at 63 km/hr.  He was driving for 38/63 hours.  Without a calculator, I would estimate that this is a little less than 38 minutes (38/60 hours would be 38 minutes).  With a calculator, this is 0.60 hours, which is 36 minutes (0.60 * 60 = 36).
387156
Equate distances and solve for time.\n\ngraces distance: 5*10 + 5*t    \nwill's distance: 15*t\n\nNote that I've started the clock when will starts.  That's because the question asks how long will must ride.\n\nNow, just equate the distances and solve for t.\n\nI get 5 minutes.\n\ngrace = 50+25 = 75\nwill = 5*15 = 75\n\nBtw, I graphed these on my TI-83 and looked for the intersection.  It's helpful to draw the lines.
196125
Let's think about how to solve these and maybe that will help you with others,\n\n1. Christian makes the trip twice as fast as Eloise.  Therefore his trip would take half the time.  If his trip was three hours shorter, then 3 hours is half the time.  His trip was 3 hours, Eloises was 6 hours.  So if Christian traveled at 30 mph for 3 hours, and Eloise traveled at 15mph for 6 hours .. how far is that?\n\n[modified below with details for additional answers]\n2. Basically \n4mph x SomeTime = SomeDistance\n5mph x (Sometime-1 hour) = SomeDistance\n\nor, \nif 4x=y and 5(x-1)=y\nthen,\n4x=5x-5\n4x-5x=-5\n-x=-5\nx=5  (this is the SomeTime)\n\ngoing back to the first equation\n4mph x SomeTime = SomeDistance\n4mph x 5hours = 20 Miles\n\ndoublecheck with the second equation,\n5mph x (Sometime-1hour) = SomeDistance\n5mph x (5hours-1hour) = 5mph x 4hours = 20 miles\n\n\n3. If the fast freight was going 60mph for 6 hours (noon to 6pm), then the whole trip is how far? .. 360 miles.  If the slow freight takes 8 hours to go 360 miles, then what is the speed of the slow freight?  360mph/8m = 45 mph
540146
(a) too trivial to answer\n(b) Take the average speed times the number of seconds over which it had that average speed.
63705
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
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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
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)
500410
1.) Ms. Jordan 275 / 2500 = 0.11 \n     0.11  x  100 = 11%\n\n2.) 900 students x 1.07 = 963 students .\n\n3.) Virus Scan  300 sec / 0.40 = 750 sec for 100%\n\n4.) Gasoline... 8882 / 0.85 = 10,449 BBLs per day.\n\n5.) Jerry's new salary is\n     18,500 x 1.105 = $ 20,442.50\n\n6.) Chen's Investment wound up being \n     6400 x 0.918 = $ 5875.20\n\nHave a nice day,\nZahbudar
576902
Point A and Point B are 600 miles apart. A train leaves Point A travelling toward Point B at 5:30AM at a constant speed of 65 kph. Another train leaves Point B heading toward Point A at 6:20AM at a constant speed of 40 kph. What time do the two trains meet?1 kilometer = 0.6214 miles\n\nx=kilometers     5/6(50 minutes/60 minutes)*65= 54.166...km\nx=.6214 (600 miles)\nx=372.84 kilometers apart\n5/6(50 minutes/60 minutes)*65= 54.166...km\n372.84-54.17= 318.67km between the 2 trains at 6:20\nthats as far as I can get (I don't remember how to do this... hope this helps!)
441184
Answer: 4PM\n\nSouth<=======Start======>North\n4PM===3PM===1PM==3PM===4PM\n120 M==60 M===0 M==110 M==165 M\n\n120 Miles +165 Miles = 285 Miles
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At 1 p.m the 1st bus starts moving west with a velocity of 44 mph  that means at 2 pm the bus will be 44 miles away from the station because  D = V x t = 44 mph x 1 h = 44 miles .\nAt 2 pm the 2nd bus starts moving east i.e in the opposite direction with a velocity of 48 mph \n              \nB1  ____________B2      ( B1 B2 = 44 miles )\n                      \n             \n\nAfter 2 pm the distance between the 2 buses at any time is equal \nto : \nD = 44 +( V1 x t ) + (V2 x t ) = 44 + (44 x t ) + (48 x t ) = 44 + 92 t \nwhere D >> miles , t >> hrs \n                                         \n                                          B1    _________S___________ B2\nWhere B1 S = 44 miles + 44 t \n            B2 S = 48 t \n                                             \nPut D = 274 miles to get  t \n274 = 44 + 92 t     >>  t = 2.5 h\nTherefore the two buses will be a distance of 274 miles apart at : \n2 pm + 2.5 h =  4 : 30 pm
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Lets call the distance of the first part S1 and the total distance S2, and the corresponding times T1 and T2.  We know:\nS1 = 40\nS2 = 100\nT2 = 9.85\n\nIf V is the maximum velocity we know S1 = 0.5VT1 (const acc.)\nFor the second part we know S2-S1 = V(T2-T1) = VT2 - VT1\n\nVT1 = 2S1 so S2-S1 = VT2 - 2S1 or V = (S2+S1)/T2 = 140/9.85\n\nSo a) - V = 14.21 m/s\n\nWe know V^2 = 2AS1 where A is acceleration so A = V^2./2S1\nor A = 14.21^2 /(2x40)\n\nSo b) A = 2.525 m/s^2\n\nWe know V = AT1 so T1 = V/A = 14.21/2.525 = 5.62 s\n\nSo c) the two parts are 5.62s and 4.23s
1802
As I am travelling with a velocity 90 kph & the other car with 120 kph, I will appear as going back to the other driver but with a slower speed ie\na) 120-90 =30 kph\n\nThe other car wil appear to move ahead of me but with a slower speed as I am also chasing it\nb)120-90 =30 kph\n\nBut as we are talking about velocity we have to consider direction also. Hence my velocity will be -ve i.e. (-30 kph) & other car will be +30 kph
428707
Distance = Speed * Time.\nWe don't know the "distance" variable for either. We need to find the "time" at which the "distance" will be the same. Let distance = d\n\nWe want to find Thomas's time, so let Thomas's time = t.\nAlexis left 3 hours earlier, so her time is t + 3.\n\nThomas speed is 30, Alexis is 20. So lets fill this in.\nThomas:\nD = 30t\nAlexis:\nD = 20(t+3)\n\nThe distance will be the same, so the D in both equations is the same variable. A basic principle is that if a = b and b = c, then a = c.  Therefore, we can combine the equations for the two people as follows:\n30t = 20(t+3)\n30t = 20t + 60\n10t = 60\nt = 6 hrs\n\n\nThe same applies to the second.\nTimothy:\nD = 58t\nSavanah:\nD = 47(t+2)\n\n58t = 47(t+2)\n58t = 47t + 94\n11t = 94\nt = 8 6/11 OR 8.5454545454545\nt = 8 hours, 32 minutes, 43 seconds\n\n\n\nThe last one is slightly different.\nJet 1 Distance = 331t\nJet 2 Distance = 333t\nThe two distances aren't equal this time, but the two distances added up will equal 1,705, SO:\n331t + 333t = 1705\n664t = 1705\nt = 1705/664 = 2 377/664 = 2.56777108\nt = 2 hrs, 34 minutes, 4 seconds
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
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let V1= 4 mph, V2= 5 mph\n\nV1*t + V2*t = 3\nt= 20 min\n\n19. a 20 min\n\n\n14.16/3 = 4.72\n\n4.72*14 = 66.08\n\n20 c 66.08
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***************\nQUESTION 1\n***************\nOn the first question, we first have to find the equation that we will be solving for.  \n\nAssume that x is one integer.\nWe know that x+1 is the other integer because they are consecutive.\n\nSo we now put the wording into an equation\n1/3(x + (x+1)) = x - 5\n\nNow we solve the equation\n1/3(2x + 1) = x - 5\n2/3x + 1/3 = x - 5\n16/3 = 1/3x\n16 = x\n17 = x + 1\n\nOur two integers are 16 and 17.\n\nNow we check our answer.\n1/3 of 33 = 11 & 16 - 5 = 11 We are good! :)\n\n\n***************\nQUESTION 2\n***************\nEvery hour, Amy and Randy together walked 3.6 + 4.2 km apart from each other.\n\n3.6 + 4.2 = 7.8\n\nSo, every hour they will be 7.8 more km apart from each other.  We only want 10 minutes though, and there are 60 minutes in an hour. To get how far apart they will be in 10 minutes, we take:\n7.8 * 10/60 = 1.3\n\nRandy and Amy will be 1.3 km apart after 10 minutes.\n\nWe are done! :)