Question

A train travels a distance of 480 k.m at a uniform speed, if the speed had been. 8 km/hr less, then it would have taken 3 hours more to cover the same distance find the speed?​

Answer 1

Answer:

Let the speed of the train be x km/hr. Then

Time taken to travel a distance of 480km=

x

480

hr

Time taken by the train to travel a distance of 480km with the speed (x−8)km/hr=

x−8

480

hr

It is given that if the speed had been 8km/hr less, then the train would have taken 3 hours more to cover the same distance

∴

x−8

480

=

x

480

+3

⇒

x−8

480

−

x

480

=3⇒

x(x−8)

480(x−x+8)

=3⇒

x(x−8)

480×8

=3

⇒3x(x−8)=480×8⇒x(x−8)=160×8⇒x

2

−8x−1280=0

This is the required quadratic equation.

Step-by-step explanation:

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Answer 2

40km/h

Step-by-step explanation:

Given -

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

To find -

• Speed of train.

Solution -

Let the speed of the train be 'x' km/hr .

Time taken by train to cover a distance of 480km

$\longmapsto \large \dfrac{480}{x} \: \: \: \: \: \: \: \mathrm{ \normalsize (time = \dfrac{distance}{speed}) }$

Now , the new speed is decreased by 8km/h , so it can be written as (x - 8) km/hr.

$\mathrm{ \normalsize{Time \: \: taken }= \large\dfrac{480}{x - 8} }$

Now , according to the provided conditions we can infer that,

$\longmapsto\dfrac{480}{x - 8} = \dfrac{480}{x } + \small3$

$\longmapsto\dfrac{480}{x - 8} - ( \dfrac{480}{x} ) = \small3$

$\longmapsto \dfrac{480}{1}( \dfrac{1}{x - 8} - \dfrac{1}{x} ) = \small3$

$\longmapsto \dfrac{480}{1} [ \dfrac{x - (x - 8)}{(x - 8)(x)}] = \small3$

$\longmapsto \dfrac{ \cancel{480}}{ \cancel3}( \dfrac{ \cancel x \cancel {- x } + 8}{{x}^{2} - 8x } ) = \small1$

$\longmapsto \dfrac{ 160}{ 1}( \dfrac{ 8}{{x}^{2} - 8x } ) = \small1$

$\longmapsto 160 \times 8 = {x}^{2} - 8x$

$\longmapsto1280= {x}^{2} - 8x$

$\longmapsto{x}^{2} - 8x - 1280 = 0$

$\longmapsto{x}^{2} - 40x + 32x- 1280 = 0$

$\longmapsto x(x - 40) + 32(x- 40 )= 0$

$\longmapsto (x - 40) (x + 32 )= 0$

$\longmapsto x = 40 \mathrm{km/h}\;\;\;\;\;\;\;, \;\;\;\;\;x = - 32\mathrm{km/h}$

Since speed cannot be negative (x ≠ -32)

Therefore , the speed of train is 40km/h.