A train made a trip at a uniform speed. Had the rate been 5 miles per hour faster, the trip would have taken?

Question

24 minutes less time, but had the been 4 miles per hour less, the trip would have taken 24 minutes longer. Find the distance and rate.

Answer 1

Equation basis: Distance = speed x time.
Let v be the speed.
Then the two statements become (distance being the same) vt = (v+5)(t-.4)
vt = (v-4)(t+.4)
Then: vt = vt + 5t - .4 v -2
vt = vt -4t +.4 v -1.6
Rearranging: 2 = 5t - .4 v
1.6 = -4t + .4 v
Adding 3.6 =t in hours.
From 1st eqtn: 2 = 18 - .4 v
v=40 mph
D= v*t = 144 miles

Answer 2

Let train moved at x mph.
the time taken for the trip t= d/x
where d is the distance of the trip.

Now speed = x+5

time taken t1=d/(x+5)
but t1=t-24/60
d/(x+5)=d/x -24/60
multiply by 60x(x+5)
60dx=60d(x+5)-24x(x+5)
60dx=60dx+300d-24x^2-120x
24x^2+120x-300d=0
2x^2+10x-25d=0.........(1)

In second case,
speed=x-4
time taken t2=d/(x-4)
t2=t+24/60
d/(x-4)=d/x +24/60
multiply by 60x(x-4)
60dx=60d(x-4)+24x(x-4)
60dx=60dx-240d+24x^2-96x
24x^2-96x-240d=0
x^2-4x-10d=0-----(2)

Multiply (1) by 2
4x^2+20x-50d=0....(3)
Multiply (2) by 5
5x^2-20x-50d=0....(4)
(4) - (3) will give you
x^2-40x=0
x^2=40x
x=40mph.....since x=0 is not possible.

Answer 3

If you have to figure nonsense like this out, the trip isn't any fun anymore. Just stay home..