...at the same time from towns 288 miles apart and met in three hours. The rate of one train was six miles per hour slower than that of the other train. Find the rate of each train.
2007-02-05 15:46:52 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Be v1 and v2 the speeds, being v2=v1-6 (according to problem)
We know that the total time for each was 3 hours.
Calculating distances:
v1*3 for first train
(v1-6)*3 for the second train
The total distance is 288 miles and is also the sum of both distances:
3v1+3v1-18=288 | + 18 on both sides
6v1=306
v1=51 miles/hour
v2=45 miles/hour
Double check:
51*3 = 153
45*3 = 135
Adding both together:
153+135=288
2007-02-05 15:54:01 · answer #1 · answered by F B 3 · 0⤊ 0⤋
One train was going 45 miles per hour and the other 51 miles per hour. 3 hours @ 45 mph = 135 miles; 3 hours @ 51 mph = 153 miles. 135 miles + 153 miles = 288 miles in 3 hours. So the average speed is 96 mph. One train is going at speed X mph and the other is going at speed X - 6 mph. Together they have to be going 96 mph. So, X + (X - 6) = 96 mph. Or, 2X - 6 = 96. Then 2x = 96 + 6 or 2X = 102. Then, X = 102 divide by 2, or X = 51. So one train is going 51 mph (X) and the other 6 mph slower, or 45 mph
(X - 6). And 45 mph plus 51 mph = 96 mph.
2007-02-05 16:02:17 · answer #2 · answered by spartanuni 1 · 0⤊ 0⤋
Ok, obviously the sum of their complete elapsed distances after the three hours is 288miles, if we let the speed of the faster train be x mi/hr then,
=>3(x-6)+3x=288
=>6x-18=288<=>6x=306
=>x=51mi/hr is the speed of the slightly faster train and 45mi/hr is the speed of the slower train
2007-02-05 15:55:29 · answer #3 · answered by RobLough 3 · 0⤊ 1⤋
Let r = rate of slower train r + 6 = rate of faster train 2r + 6 = combined rate of approach of trains rate = distance/time 2r + 6 = 450/5 = 90 2r = 90 - 6 = 84 r = 42 r + 6 = 42 + 6 = 48 The speed of the two trains is 42 mph and 48 mph, respectively.
2016-05-23 22:29:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋
train x: x miles/hour 3 hours
train y: y miles/hour = x-6 miles/ houir 3 hours
3 hours (x + y) = 288 mph
x + (x-6) = 96
2x -6 = 96
2x = 102
x = 51
train x = 51 mph
train y = 46 mph
x=
2007-02-05 15:58:14 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Distance = 288 x 2 = 576
576 = 3x + 3y
y = x + 6
576 = 3x + 3x + 18
= 6x + 18
6x = 558
x = 93
y = 99
So one trains goes 93 miles per hour, another goes 99 miles per hour.
2007-02-05 15:52:22 · answer #6 · answered by Who_am_i 1 · 0⤊ 1⤋
Combined rate is 288/3 MPH which is 96 MPH, so one is 51 and one is 45.
2007-02-05 15:51:15 · answer #7 · answered by Anonymous · 0⤊ 0⤋
288 / 2
144 / 3
48 + 3 = 51
48 - 3 =45
One train was going 51 mph, the other was going 45 mph
2007-02-05 15:52:26 · answer #8 · answered by Aaron 4 · 0⤊ 0⤋
A 228 miles; met 3 hrs later B
|| ------------------------------------------- || -------------------------------- ||
>>>>>>> x miles/h (x-6)miles/h <<<<<<<<
thus : total distance travel by A train + B train = 228
Distance = speed x time
which led to >> x*3 + (x-6)*3 = 228
x = 41 miles/hour
A train travels at 41mph and B train travels at 35mph
AHHH.. pls change your distance to 228 miles
2007-02-05 16:06:00 · answer #9 · answered by Anonymous · 0⤊ 1⤋
3(x+[x+6])=288
3(2x+6)=288
6x+18=288
6x=270
x=45
One train's traveling at 45mph; the other's traveling at 51mph.
2007-02-05 15:50:08 · answer #10 · answered by PsychoCola 3 · 0⤊ 1⤋