Question

A train travels 180 km at a uniform speed. If the speed had been 9 km/hour more, it would have

taken 1 hour less for the same journey. Find the speed of the train​

Answer 1

Answer:

Let the speed of the train be xkmph

The time taken by the train to travel 180km is

x

180

h

The increased speed is x+9

The time taken is

x+9

180

According to the question,

The time taken is

x

180

−1

⟹

x

180

−1=

x+9

180

⟹

x

180−x

=

x+9

180

⟹180x−x

2

+1680−9x=180x

⟹x

2

+9x−1680=0

⟹x

2

+45x−36x−1680=0

⟹x(x+45)−36(x+45)=0

x=36,−45

Speed cannot be negative (In this case,)

So The speed of the train is 36kmph

Step-by-step explanation:

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Answer 2

Answer:

${ \huge{ \pmb{ \sf{Required \: Answer... }}}}$

From Question,

Distance = 180km

Time = 180/x

In question they given that,

Speed had been 9 km/hour more

So, Time taken = 180/x + 9

Let's find the speed of train:

${ \implies{ \sf{ \frac{180}{x} - \frac{180}{x + 9} = 1}}}$

By taking LCM,

$\: { \implies{ \sf{ \frac{180(x + 9) - 180x}{x(x + 9)} = 1 }}}$

$\: { \implies{ \sf{ \frac{180x + 1620 - 180x}{ {x}^{2} + 9x } = 1 }}}$

Denominator going to RHS side,

$\: { \implies{ \sf{180x - 180x + 1620 = {x}^{2} + 9x }}}$

By solving,

$\: { \implies{ \sf{1620 = {x}^{2} + 9x}}}$

$\: { \implies{ \sf{ {x}^{2} + 9x - 1620 = 0}}}$

Now, By solving the equation,

$\: { \implies{ \sf{ {x}^{2} + 45x - 36x - 1620 = 0}}}$

$\: { \implies{ \sf{x(x + 45) - 36(x + 45) = 0}}}$

$\: { \implies{ \sf{(x - 36)(x + 45) = 0}}}$

$\: { \implies{ \sf{x = 36 \: \: or \: \: - 45}}}$

" Speed Won't be negative "

${ \therefore{ \sf \red{speed \: of \: the \: train = 36km/h}}}$

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