Question

A train averages a speed of 90 miles per hour across the plains and 37.5 miles per hour through the mountains. if a trip of 300 miles took 3 hours and 48 minutes, how much of it was through the mountains?

Answer 1

This is a tough one. What we are looking for is how long (how much time) was spent in the mountains. So we are looking for t, time. 3 hours 48 minutes is 3.8 hours. Both have to be in terms of one or the other, you cant have unknown mountain time and plain time, so let's say that mountain time is t. That means that plain time is 3.8-t. Multiply the rates times the times for each and add them together and set them equal to the total distance. 90(3.8-t)+37.5t=300. Solve for t. The time spent in the mountains was 48 minutes

Answer 2

Answer:

The distance traveled by train in mountains is:

                     30 miles.

Time taken by train in mountains is:

                 0.8 hours( 48 minutes)

Step-by-step explanation:

Let the distance traveled by a train in plains be: x miles.

Time taken in plains to travel x mile= y hours.

Hence, distance traveled by train in mountains is: 300-x miles.

( Since a train travels  a total of 300 miles)

Time taken in mountains is: 3.8-y hours

( Since train total takes a time of 3 hours and 48 minutes

3 hours and 48 minutes=3+48/60 hours

                                       =3+0.8 hours

                                       = 3.8 hours )

Now, average speed in plains is: 90 miles per hour.

i.e.

$\dfrac{x}{y}=90------(1)$

Also,

Average speed of train in mountains is:37.5 miles per hour

i.e.

$\dfrac{300-x}{3.8-y}=37.5------(2)$

From equation (1) we have:

$x=90y$

on putting the value of x in equation (2) we get:

$\dfrac{300-90y}{3.8-y}=37.5$

Now on solving we get:

y=3

Similarly x=270.

Hence, the distance traveled by the train in mountains is:

30 miles.

Time taken by train in mountains is: 48 minutes

( Since, 300-x=300-270=30 miles.

and 3.8-y=3.8-3=0.8 hours

which is equal to 48 minutes)