Question 581527
The train from C to D breaks down after 1 hour.
 The train has stopped because of 1/2 hour repairs, then, it continues on at 3/4 of its beginning rate, arriving at B 2 hours late.
 If the train had covered 75 miles more before the break down, it could be only 1 hour late.
 What is the beginning rate of the train? Explain how you would get the answer.
:
Let s = original speed of the train, also distance traveled in 1 hr
then
.75s = speed after the breakdown
:
let d = distance from C to D
then
{{{d/s}}} the normal time for the trip
Distance traveled to the breakdown: 1 * s = s miles
(d-s) = distance traveled at .75s
and
{{{(d-s)/(.75s)}}} = time at slower rate
write a time equation
1 + .5 + {{{(d-s)/(.75s)}}} = {{{d/s}}} + 2
1.5 + {{{(d-s)/(.75s)}}} = {{{d/s}}} + 2
{{{(d-s)/(.75s)}}} = {{{d/s}}} + 2 - 1.5
{{{(d-s)/(.75s)}}} = {{{d/s}}} + .5
multiply by .75s to clear the denominators
d - s = .75d + .5(.75s)
d - s = .75d + .375s
d - .75d = .375s + s
.25d = 1.375s
multiply both sides by 4
d = 5.5s
:
"If the train had covered 75 miles more before the break down, it could be only 1 hour late."
I made a big mistake here, instead "75 mi more" I used 75 mi, try to correct this now
:
write a time equation,
{{{((s+75))/s}}} + .5 + {{{(d-(s+75))/(.75s)}}} = {{{d/s}}} + 1
{{{((s+75))/s}}} + {{{(d-(s+75))/(.75s)}}} = {{{d/s}}} + 1 - .5
{{{((s+75))/s}}} + {{{(d-(s+75))/(.75s)}}} = {{{d/s}}} + .5
Multiply by .75s to clear the denominators
.75(s+75) + (d-s-75) = .75d + .5(.75s)
.75s + 56.25 + d - s - 75 = .75d = .375s
Combine like terms
.75s - s - .375s + d - .75d = 75 - 56.25
-.625s + .25d = 18.75
Replace d with 5.5s
-.625s + .25(5.5s) = 18.75
-.625s + 1.375s = 18.75
.75s = 18.75
s = {{{18.75/.75}}}
s = 25 mph is the initial speed of the train
:
:
See if this checks out, d = 5.5(25) = 137.5 mi and .75(25) = 18.75 mph
{{{((25+75))/25}}} + .5 + {{{(137.5-100))/(18.75)}}} = {{{137.5/25}}} + 1
{{{100/25}}} + {{{37.5/18.75}}} = {{{137.5/25}}} + .5
4 + 2 = 5.5 + .5; checks out