A train travels a distance of 480 km at uniform speed.If the speed had been 8km/h less,then it would have taken 3 hours more to cover the same distance. W need to find the speed of the train. Related: Using Quadratic Formula?

Question

A train travels a  distance of 480 km at uniform speed.If the speed had been 8km/h less,then it would have taken 3 hours more to cover the same distance. W need to find the speed of the train. Related:  Using Quadratic Formula?

Answer

Problem

A train travels a distance of 480 km at uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution

Let's assume the original speed of the train as x km/h.

According to the problem:

Distance travelled by the train = 480 km

Time taken at speed x = t

Time taken at speed x - 8 km/h = t + 3

Deriving the equation

Using the formula:

Distance = Speed x Time

We can write:

480 = xt

480 = (x - 8)(t + 3)

Simplifying the second equation:

480 = xt - 8t + 3x - 24

504 = xt + 3x - 8t

504 = x(t + 3) - 8t

Substituting the value of xt from the first equation:

504 = 480 + 3x - 8t

24 = 3x - 8t

Now, we have two equations:

480 = xt

24 = 3x - 8t

Using Quadratic Formula

Let's solve the second equation for t:

24 = 3x - 8t

8t = 3x - 24

t = (3x - 24)/8

Substituting this value of t in the first equation:

480 = x((3x - 24)/8)

3840 = 3x² - 24x

3x² - 24x - 3840 = 0

Using the quadratic formula:

x = (-b ± √(b² - 4ac))/2a

We get:

Answer

Answer..40km/hr.

Explanation:

Total Distance S=480km
Formula For Speed
speed=distancetime
Hence distance=speed×time
Case-I
let the uniform speed of Train =vkmh
and the time taken to complete distance =(t)hour
Distance S=speed×time=vt -(equation 1)
Case-II
If speed has been 8km/h less then train would have taken 3 hours more to cover the same distance.
Now in this situation
speed of train =(v−8)kmh
and Time taken to complete distance =(t+3)hour
Distance S=speed×time=(v−8)(t+3) - (equation 2)
Since Distance for both cases are same .
Comparing equation 1 and equation 2, we get
⇒vt=(v−8)(t+3)
⇒vt=(v)(t+3)−8(t+3)=vt+3v−8t−24
cancel vt from both side
⇒0=3v−8t−24
⇒3v−8t=24
⇒3v=24+8t
⇒v=24+8t3

but from equation 1 the value of t=Sv=480v(hour)
⇒3v=24+8(480v)
⇒3v2=24v+3840
⇒3v2−24v−3840=0
⇒v2−8v−1280=0
factorize the quadratic equation
⇒(v−40)(v+32)=0
v is either 40kmh or −32kmh
Speed is Positive hence
Speed=40km/h

HOPE YOU ARE ABLE TO UNDERSTAND

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