A motarboat traveling with the current went 54 mi in 3 hours. Against the current, it took 3.6 hours to travel the same distance.  Find the rate of the boat in calm water and the rate of the current.
126849
54 mi in 3 hours is a speed of 18mph (54/3=18),\n54 mi in 3.6 hours is a speed of 15mph (54/3.6=15),\nTHe current is half the difference between the two - it helps as much as it stops you....\nTherefore current is 1.5mph ((18-15)/2=1.5) and the boatspeed in calm water is 16.5mph (15+1.5 or 18-1.5).\n\nThe equations are, with boat rate=x and current=c;\nx+c=18      (1)\nx-c=15      (2)\n(1) ==> c=18-x\ninsert c into (2) ==> x-18+x=15\n==>2x=33 ==>x=33/2=16.5\nInsert x into (1) or (2) to get c=1.5   \n\nHope it helps,\nFredrik, MSc
523844
Total distance to be covered:690miles\nJulia+Maria:60+55\n=115miles\nTotal time taken:690/115\n=6hours
219042
distance = 160miles\nave. speed = 50mph\n\n160/50 = 3.2 hours or 3hours and 12 minutes\n\nfor the guy to arrive at exactly 11:30am\n\n(11:30) - (3hours12minutes) = 8:18am\n\nANSWER: 8:18am
514801
Something's wrong with the writeup for the first question.  I'm guessing it should have said THR = RHR + 0.65(MHR-RHR), because otherwise you're gonna get a THR of, like, 10000 beats per minute.\n\nProceeding from that assumption, the THR is RHR + 0.65(MHR-RHR).  You're told that the RHR is 54 bpm.  The MHR is 220 - the age(43), so it's 177.  Therefore the THR is 54 + 0.65(177-54) = 54 + 0.65(123) = 54 + 79.95 = 133.95 bpm.\n\nIn the second question, you didn't provide the "following equations," but think of it this way.  The airplane flew for a total of 15 hours at an average speed of 350 miles per hour.  That means it went 5250 miles.  In the final 7 hours it went 2275 miles.  So in the first 8 hours it had to go 2975 miles.  2975 miles/8 hours = 371.875 mph.  I don't exactly know how to write that in an equation, but now that you understand the reasoning, hopefully you'll be able to recognize the correct equation.  If all else fails, plug the data into each equation and see which one returns 371.875.
400545
28 hours..6.5 mph\n\nnot difficult..10 miles the first hour\n                   3 miles the next hour\n\n13 miles in 2 hours therefore 6.5mph\n\n182/6.5=28
289739
Everyone is wrong, it is 48MPH, think of it this way.  Lets say the distance is 60 miles, it takes 1.5 hours at 40, and 1 hour at 60, so a total of 2.5 hours, divide 120 by 2.5 and it is 48mph.  This will always be the answer reguardless of distance, whether 1 mile in each direction or a million miles in each direction.
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) )
508536
Draw a diagram to represent the positions of the two ships at time t.  This will be a right triangle.  Side A will be the position of ship A and side B will be the position of ship B.  The distance between the ships will be the hypotenuse.  Let's call this S.  By the Pythagorean theorem,\n     S^2 = A^2 + B^2\nDifferentiate with respect to time, t.:\n\n2 S dS/dt = 2 A dA/dt + 2 B dB/dt\n Simplify by dividing the entire equation by 2\n\nS dS/dt = A dA/dt + B dB/dt\nSolve for dS/dt:\n\ndS/dt = 1/S (A dA/dt + B dB/dt)\n\ndA/dt is constant at 24 knots\ndB/dt is constant at 18 knots\n\nAt 5 PM,    A = 30 + 24(5) = 150\n                  B = 18(5) = 90\n                  S = SQRT( A^2 + B^2) = SQRT(150^2 + 90^2)\n                     = 174.9\nTherefore dS/dt = (1/174.9)(150(24) + 90(18)) = 29.8 knots.\n\nIt's just a coincidence that spoonish's answer is about the same as mine.  Only differential calculus can solve this type of problem.
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.
540172
d=r*t\nhighway= 60*highwaytime\ncity = 30*citytime\ncity = highway -  100\ncitytime =highway time - 5/6  50 min = 5/6 hour.\nsub in\n\nhighway - 100 = 30 (highwaytime- 5/6)\n60 * highwaytime - 100 = 30 * highwaytime - 30*5/6\nsimplify\n30* highwaytime = 75\nhighwaytime = 2 1/2 hours\nh  =2.5*60 = 150 miles\ncity =30* 1 2/3 = 50 miles\ntotal 200 miles
609282
a)7:30 + 3:42 =  11:12    Just add minutes it gives 72 but sixty minutes is an hour so 72 = 1 hour and 12 minutes keep the 12 in minutes section and carry 1 hour to hour section.7+3 =10\nAdd carry 1 you get 11.\n\nb) 18:30-3:42 =   14:48  ...Proceed similarly this time you need to borrow 1 from hours section that becomes sixty minutes subtract 42 from 60+30\n\nc)3+42/60 = 3.7 hrs\n\nd)370/3.7= 100 km/hr
113940
The way I draw the picture is the port is at the origin.  Then there is a line in the 2nd quadrant 53 degrees up from the x axis, and another in the 3rd quadrant 67 degrees down from the x axis.  This gives the total angle between the 2 as 120 degrees.  If it's supposed to go the other way, 53 degrees left of the y-axis and 67 degrees left of the y-axis, then the total angle between the 2 would be 60 degrees.\n\nEither way, you use law of cosines to solve.  You have the north side of the triangle at 36 mi (3hr * 12 mph) and the south side at 48 mi (3hr * 16 mph) with I'm going to use the 120 degree angle between.\n\nLaw of cosines if you have 2 sides of a triangle, and the angle between them is\n\nc^2 = a^2 + b^2 - 2ab*cos(C) - we'll let a = the 36 side, b = the 48 side, and c is the distance between the ships at noon.\n\nc^2 = 36^2 + 48^2 - 2(36)(48)(cos 120)\nc^2 = 1296 + 2304 - (3456)(-1/2)\nc^2 = 5,328\nc = approx 73 miles.\n\nIf you think it should be a different angle, just re-solve the problem with that angle instead, but this is how you would do it.
551314
Call the speed the kayaker would make in still water x (mph).\n\nWhen he paddled with the current, the actual speed of the kayak was x+6.\n\nWhen he paddled against the current, the actual speed of the kayak was x-6.\n\nSince the lengths of both trips are the same, we have:\n2*(x+6) = 3*(x-6) <-> 2x + 12 = 3x - 18 <-> x = 30.\n\nHe would manage 30mph in still water (wow, what a beast :)))
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
184248
A car is traveling at a rate of 100 kilometers per hour to a city that is 350 kilometers away. Write an equation that represents the remaining distance to the city at any given time of travel.\n\ny = remaining distance\nx = time\n350 is the distance away\n-100 is the distance closed per hour\n\ny = -100x + 350
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).
406462
start here -- After the first two hours, the horse and roadster are already 16 miles apart. The roadster did not travel any distance in this time, while the horse moved north at 8mph. \n\ndistance = rate x time = 8mph x 2 hrs = 16 miles\n\nNext, set up your equation to take into account only the remaining distance that the two object need to move to achieve the desired 118 miles apart (this would be 118 miles - 16 miles = 102 miles)\n\n(rate of the horse * t) + (rate of the roadster * t) = 102 miles\n\n(8mph * t) + (60mph * t) = 102 miles\n\nremove the mph and miles for clarity and get\n\n8t + 60t = 102\n68t = 102\nt = 1.5 hours\n\nBe careful....add this to the initial two hours that the buggy moved while the roadster did not and get \n\n1.5 hours + 2.0 hours = 3.5 hours