Quadratic Equation Question-urgent?

Question

A journey of 192 km from A to B takes two hours less by a fast train than by a slow train. If the average speed of the slow train is 18 km/hr less than that of the fast train, find the average speed of each train.

Answer 1

let x be the speed of the slow train and t be the time of slow train. x + 18 is the speed of the fast train and t - 2 is the time of the fast train.

both travel the same distances.
192 = xt
192 = (x + 18) (t - 2)

solve for t
t = 192/x

192 = (x + 18) (192/x - 2)

192 = 192 - 2x + 3456/x - 36

multiply x for boht sides
192x = 192x - 2x^2 + 3456 - 36x
0 = -2x^2 - 36x + 3456

use quadratic formula and you'll get x = 33.53

speed of slow train is 33.53 km/hr
speed of fast train is 51.53 km/hr

Answer 2

ok so let's make x the rate of the slow train
and y the rate of the fast train
h = hours for slow train
h - 2= hours for fast train

so you can make the equations
x + 18 = y
xh = 192
y(h-2) = 192

so x + 18 = y
is the same thing as
y - 18 = x
replace x in this equation : xh = 192 with (y - 18)
and you get
yh - 18h = 192
y(h-2) = 192 = yh - 2y = 192
now you can subsitute to solve for y or h
yh - 18h = 192 is also the same as
(192 + 18h)/h = y
now input that for the other equation
h((192 + 18h)/h) - 2 ((192 + 18h)/h) = 192
so you get:
192 + 18h - 2 ((192 + 18h)/h) = 192
now multiply the entire equation by h
192h + 18h^2 - 384 -36h = 192h
so now you get 18h^2 - 36h - 384 = 0
6h^2 - 12h - 128 = 0
3h^2 - 6h - 64 = 0

once you use the quadratic you get h = 5.725815626
so now solve for the rest with the original equations
you should get x = 33.53234064 (rate of slow train)
and y = 51.53234064 (rate of fast train)

Cheers!

Answer 3

Use the equation:

x = v*t

Let x1,v1, and t1 refer to the fast train.
Let x2,v2, and t2 refer to the slow train.

v2 = (v1-18) the velocity of the slow train is 18 less than the
fast train.

t1 = (t2-2) the time of fast train is two hours less than the
slow train.

x1 = v1*t1 or 192 = v1*(t2-2)
x2=v2*t2 or 192 = (v1-18)*t2

Solve the second eq. for t2 and obtain.

t2 = 192/(v1-18)

Plug this t2 into the first eq. to get

192 = v1[192/(v1-18) - 2]

Solve for v1, which is 51.5323 km/hr

The average is (v1+v2)/2 = (51.5323 + 33.5323)/2 = 42.5323

answer is 42.5 km/hr

Check answer for time

train 1 192/51.5323 = 3.72 hours

train 2 192/33.5323 = 5.72 hours

Which is 2 hours quicker.

Answer 4

slow=v1=v-18
time=t1=t+2
fast=v2=v
time=t2=t
v2*t=192
v1=?, v2=?
v1*t1=v2*t2
(v-18)(t+2)=vt
vt+2v -18t-36=vt

2v-18t=36
vt=192

18t^2+36t-384=0
t=3.7
fast train velocity=v=51.9

slow speed=v1=51.9-18=33.9