SOLUTION: A train travels the distance between the station A and B in 45 minutes. if the train reduces its speed by 5km per hour it takes 48 minutes to cover this distance. what is the dist

Question 1044434: A train travels the distance between the station A and B in 45 minutes. if the train reduces its speed by 5km per hour it takes 48 minutes to cover this distance. what is the distance between the two stations?

Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39615)   (Show Source): You can put this solution on YOUR website!
Travel Rate Rule to relate rate, time, distance, RT=D
Forty-five minutes is and 48 minutes is .

RATE TIME DISTANCE REGULAR r t d SLOWER r-k t+h d

UNKNOWN VARIABLES: r,d
(unit for r is kilometers per hour)
(unit for d is kilometers)
GIVEN VARIABLE ASSIGNMENTS:

You would make some algebra steps to solve this system:

Answer by ikleyn(52772)   (Show Source): You can put this solution on YOUR website!
.
A train travels the distance between the station A and B in 45 minutes.
If the train reduces its speed by 5 km per hour it takes 48 minutes to cover this distance,
what is the distance between the two stations?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

45 minutes = of an hour. 48 minutes = of an hour. Let "u" be the speed of the train in the first scenario. in . Then the speed in the second scenario is (u-5) . Then you have this equation = , which says that the distance is the same. Simplify it: = , = 4, = 4, = 4, u = 80. Thus the train speed is 80 km/h. Hence, the distance is = 60 km. Answer. The distance is 60 km.

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

A train travels the distance between the station A and B in 45 minutes. if the train reduces its speed by 5km per hour it takes 48 minutes to cover this distance. what is the distance between the two stations?

Let the distance be D
Then speed of train =
Reduced speed =
We then get the following SPEED equation:
16D - 60 = 15D ------- Multiplying by LCD, 12
16D - 15D = 60
D, or distance =
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