there are 4 guys A, B, C, and D.these thre guys are standing on the bank of a river. they have one boat with them, on which maximum of two guys can travel at a time. when A alone moves on the boat, it takes 10 minutes to cross the river. similarly B takes 5 minutes, C takes 2 minutes, and D takes 1 minute. and if two guys travel at a time, it takes maximum of the two times, associated with the two guys.\nnow the problem is that you have to find out what can be the minimum time , in which they can cross the river.
125726
C+D cross first 2 mins\nD returns for 1 min\nA+B cross for 10 mins\nC returns for 2 mins\nC+D cross for 2 mins\nTotal 17 minutes
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
257741
If I'm reading your description right, I think your first boat is heading up and to the right ("direction N.68 degrees E") making an angle of 68 degrees with north, or 22 degrees with east.  And I think your second boat is headed directly east ("S.90 degrees E").  So I'm assuming that they form a triangle, with one angle being 22 degrees and another one being a 90 degree angle, since we're looking for the point where one boat will be directly south of the other.\n\nThe first boat traveled 12t miles so far.  The second boat left an hour later and traveled 16(t-1) miles.\n\nThe first boat's path turns out to be the hypotenuse of the triangle, and the second boat's path will be one of the adjacent legs to the 22 degree angle.  So we can set up an equation with trig identities:\n\ncos 22 = 16(t-1)/12t\n0.9272 = 4(t-1)/3t\n2.7816t = 4t - 4\n4 = 1.2184t\nt = 3.3 hours\n\nSo after 3.3 hours, one boat will be directly south of the other.\n\nFinding that distance can be done one of several ways: using the Pythagorean theorem, sine relationships, law of cosines, etc.  I'll use sine, since we're already doing trig.  After 3.3 hours, the northernmost boat will have traveled 39.6 miles.  The distance between the boats becomes our side opposite the 22 degree angle.  So,\n\nsin 22 = distance between/39.6\ndistance = 39.6sin 22\n= 14.83 miles\n\nNow if I read the angle wrong, then simply go back and change all the "22" angles to the correct one and recalculate.
144302
if the question is: If it takes 3 hours to travel 24 miles downstream and 6 hours to travel UPstream, how fast is the current? then,\nlets call the travelling speed Vt and the current speed Vc\ndownstream we have Vt+Vc=24miles/3hours=8mph\nupstream we have Vt-Vc=24/6=4mph\nto solve the system of two equations we can add the equations and get 2Vt=12, Vt=6mph\nVc=8-6=6-4=2mph\nThe current is 2mph fast.
424373
See picture: http://mudandmuck.com/str2/BOAT.JPG\nShe must travel √(x^2 +4) miles by boat, and 6-x miles walking.\nTherefore, the time t is the sum of the distances divided by the rates, or\nt = √(x^2 +4) / 2 + (6-x)/5\n\nTake the derivitive of t with respect to x and set = 0 to find the minimum time.\n\nt = (1/2)(x^2 + 4)^(1/2) - x/5 + 6/5\ndt/dx = (1/2)(1/2)[(x^2+4)^(-1/2)](2x) - 1/5\n= x/[2√(x^2+4)] - 1/5 = 0\n5x = 2√(x^2+4)\n25x^2 = 4x^2 + 16\n21x^2 = 16\nx^2 = 16/21\nx = √(16/21) = 4√21/21 ≈ .87287156 miles from the point on shore nearest the boat.\n\n{Substituting this value back into the equation to find time, we see that the quickest time is ≈2.116515139 hours.}
297281
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
415867
They travelled a total of 476 miles in 4 hours.  So together they are travelling (476/4) mph or 119 mph.\n\nSo you have two equations:\nJ + S = 119\nJ - S = 5\n\n2J = 124\nJ = 62 mph\nS = 57 mph
673313
Tim is right.  i just want to explain it further so you will understand.\n\nso the problem is we are looking for the time spent in both trips. lets call it "X". one is 1 hour longer than the other so we can call it "X + 1". \n\nif trip1 with a distance of 180 miles is taken in X with an additional 1 hour we can call this equation as 180 per X+1 or in mathematical form its 180/(X+1)\n\nif trip 2 with a distance of 135 miles is taken in X we can call it 135 per X or 135/X\n\nwe have now 2 equations 180/X+1 and 135/X\n\nsince the rates are the same we can say that the rate for 180/(X+1) is equal to rate for 135/X.  or in equational form it is\n180/(X+1)=135/X.\n\nWe now have a working equation.  If you notice, we are dealing with fractions and to equate it we need to transpose it using cross multiplication.  Cross multiplication is done by pairing the numerator of one side to the denominator of the other side and then proceed to multiplication.\n\nlooking at the working equation, we will pair 180 and X or 180X and another pair which is 135 and (X+1) or 135(X+1)\n\nnow we have a new working equation 135(X+1) = 180X\n\nto simplify 135(X+1) we multiply 135 with X and with 1. 135 and X is equal to 135X and 135 and 1 is equal to 135\n\nnow we have simplified the equation to 135X + 135 = 180X\n\nto get the value of X, we combine similar terms.  they are 135X and 180X.  We combine this by transposition.  Remember that in order not to change the value of the equation we must change the sign of what is moved to another side.  In this case i will move 180X to pair it with 135X and move 135 to the other side. so now we get 135X - 180X = -135\n\nwe simplify the like terms and we get -45X.\n\nnow we have -45X = -135\n\nbut we are only looking for X so we must eliminate -45 from the X.  identify the numerical coefficient of X which is -45 and devide it from the original. what you do to one side, you should also do it to the other.\n\n-45X/-45 = -135/-45 with this equation we can simplify it into\nX=3\n\nnow we have the value for X\n\nso one trip will be 3 and since another is 1 hour longer then it must 3 + 1 = 4\n\nheres the solution\n\n180/(X+1)    =      135/X\n135(X+1)     =      180X\n135X + 135   =    180X\n135X - 180X   =    -135  \n            -45X   =    -135\n         -45X/-45=    -135/-45\n                   X =  3\n          \n             X + 1 =  trip 2\n             3+1 = 4
501796
After trying to solve this multiple times, I've come to a simple conclusion.  The dog is constantly moving at a velocity of 4.5(m/s).  Distance is time*velocity.  The dog is in motion until the lady reaches the end.  If we use an inversion of that formula, we get time=distance/velocity.  \n\n4000(m)/2.5(m/s) = 1600 seconds.  \n       (time it takes her to reach the river)\n1600(s)*4.5(m/s)=7200 meters or 7.2km \n       (that time multiplied by the dogs velocity to figure distance)\n\n     So, after failed attempts using calculus and other math that is unnecessary, we've come to the answer of 7.2km of distance covered by the dog.  Hope this helps you out!
4658
you left out the number of seconds.  I'll give you the formulas you need though:\n\n(a)  a=v/t  so 6 divided by the time you forgot to put.\n(b)  F=ma  so .2 times the acceleration you just solved for in (a)\n(c)  same as (b), the puck exerts an equal force on the stick, the puck just goes flying farther because it's smaller.
194133
For every kilometer of the distance from Rome, Sam spent a total of 36 minutes in travel:  30 minutes riding out on the elephant (it takes 30 minutes to go one kilometer at 2 kilometers per hour) and 6 minutes riding back on the chariot (the time it takes to go 1 kilometer at 10 kilometers per hour).  \n\nThere is 1080 minutes in 18 hours (18*60=1080).  Divide 1080 by 36 and you get 30 kilometers.\n\nCheck your answer:\n\nTo go 30 kilometers out by elephant at 2 kilometers per hour takes 15 hours.\n\nTo go 30 kilometers back by chariot at 10 kilometers per hour takes 3 hours.\n\n15 hours + 3 hours = 18 hours.
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
668152
10  seconds.\n\nfirst  you find the relative speed.\nsince the two are headed toward each other you add their respective speeds  \n                   vr = v1+v2 = 10+5= 15 meters/second \n\nnow  you want to find the time it takes to cross 150 meters of the trains length  at this speed \n                          speed =distance/time\n                          time = distance /speed =  150/15= 10seconds
394357
Let us call x the minutes hand on the clock.\nIf 0<x<=5, then it is train B. (5 minutes)\nIf 5<x<=12, then it is train A. (7 minutes)\nIf 12<x<= 24, then it is train A. (12 minutes)\nIf 24<x<= 25, then it is train B. (1 minute)\nIf 25<x<= 45, then it is train B. (20 minutes)\nIf 45<x<=48, then it is train A. (3 minutes)\nIf 48<x<=60, then it is train A. (12 minutes)\n\nWe add it up to get that for 34 of the minutes, he gets on A, and that for of the 26 minutes he gets on B. There are 60 minutes in an hour, so the probability is 34/60, which reduces to 17/30.
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.
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! :-)
253730
Use the pythagorean theorum.  Train One is the left leg, train 2 is the right leg, and the distance between them is the hypotenuse.  Since they only traveled an hour, their speed is irrelevant.  That triangle gives the formula of (x+10)^2 + x^2 = 71^2\n\nWork it out from there.  X will equal the distance that the second train traveled, and since they traveled for an hour, that is the magnitude of its speed as well.