speed distance

[sairakarim07]

Plz solve the math with explanation
A train which travels at a uniform speed due to some fault after traveling for an hour goes at 3/5 th of the original speed and reached the destination 2 hours late.If the fault had occured after traveling another 50 miles,the train would have reached 40 minutes earlier.What is the distance between two stations?
A)200
B)280
C)300
D)310
E)320

[Atekihcan]

Let the distance between the two stations is d miles and the actual speed of the train is v miles per hour.

Now, only difference in the two scenario is in the first case, the train covered 50 miles at speed 3v/5 and in the second case, the train covered those 50 miles at speed v.

Time taken to cover 50 miles at speed v = 50/v hours
Time taken to cover 50 miles at speed 3v/5 = 50/(3v/5) = 250/(3v) hours

So, 250/3v - 50/v = 40/60 = 2/3
--> 250 - 150 = 2v
--> v = 50

Now, the train should take d/50 hours to cover the distance.
And by traveling at speed 50 mph for 1 hour and the rest distance (d - 50) at speed 3*50/5 = 30 mph, the train took 2 extra hours.

So, 1 + (d - 50)/30 = d/50 + 2
--> 150 + 5(d - 50) = 3d + 300
--> 5d - 3d = 300 + 5*50 - 150 = 300 + 250 - 150
--> 2d = 300 + 100 = 400
--> d = 200

Answer : A

[GMATGuruNY]

Plz solve the math with explanation
A train which travels at a uniform speed due to some fault after traveling for an hour goes at 3/5 th of the original speed and reached the destination 2 hours late. If the fault had occurred after traveling another 50 miles, the train would have reached 40 minutes earlier. What is the distance between two stations?
A)200
B)280
C)300
D)310
E)320

Rate and time are RECIPROCALS.
When the train travels at 3/5 its regular rate, it takes 5/3 its regular time.

TIME:
Let t = the regular time.
When the train is 2 hours late, it travels 1 hour at its regular rate.
Thus, the regular time remaining = t-1.
Since the train is now traveling at 3/5 its regular rate, it now takes 5/3 its regular time:
(5/3)(t-1).
Since the train is 2 hours late -- implying a total time of t+2 hours -- we get:
1 + (5/3)(t-1) = t+2
3 + 5t - 5 = 3t + 6
2t = 8
t = 4.

RATE:
Let r = the regular rate and (3/5)r = the slower rate.

Case 1: 2 hours late
After the 1st hour, the train travels the next 50 miles at the SLOWER RATE.
Time to travel these 50 miles at the slower rate = d/r = 50 / [(3/5)r]

Case 2: 4/3 hours late
After the 1st hour, the train travels the next 50 miles at the REGULAR RATE.
Time to travel these 50 miles at the regular rate = d/r = 50/r.

Since the difference between the two times = 2 - 4/3 = 2/3 of an hour, we get:
50 / [(3/5)r] - 50/r = 2/3
250/(3r) - 150/(3r) = 2/3
100/(3r) = 2/3
100/r = 2
r = 50.

Thus:
Distance = r*t = 50*4 = 200.

The correct answer is A.

[sairakarim07]

Plz solve the math with explanation
A train which travels at a uniform speed due to some fault after traveling for an hour goes at 3/5 th of the original speed and reached the destination 2 hours late. If the fault had occurred after traveling another 50 miles, the train would have reached 40 minutes earlier. What is the distance between two stations?
A)200
B)280
C)300
D)310
E)320

Rate and time are RECIPROCALS.
When the train travels at 3/5 its regular rate, it takes 5/3 its regular time.

TIME:
Let t = the regular time.
When the train is 2 hours late, it travels 1 hour at its regular rate.
Thus, the regular time remaining = t-1.
Since the train is now traveling at 3/5 its regular rate, it now takes 5/3 its regular time:
(5/3)(t-1).
Since the train is 2 hours late -- implying a total time of t+2 hours -- we get:
1 + (5/3)(t-1) = t+2
3 + 5t - 5 = 3t + 6
2t = 8
t = 4.

RATE:
Let r = the regular rate and (3/5)r = the slower rate.

Case 1: 2 hours late
After the 1st hour, the train travels the next 50 miles at the SLOWER RATE.
Time to travel these 50 miles at the slower rate = d/r = 50 / [(3/5)r]

Case 2: 4/3 hours late
After the 1st hour, the train travels the next 50 miles at the REGULAR RATE.
Time to travel these 50 miles at the regular rate = d/r = 50/r.

Since the difference between the two times = 2 - 4/3 = 2/3 of an hour, we get:
50 / [(3/5)r] - 50/r = 2/3
250/(3r) - 150/(3r) = 2/3
100/(3r) = 2/3
100/r = 2
r = 50.

Thus:
Distance = r*t = 50*4 = 200.

The correct answer is A.

Thank you very much for such explanation