Question

solve it plzzzz its urgent

Answer 1

Let usual speed = x
Total distance = 300km

◆ time taken= 300/x
New speed of train is x+5

time taken= 300/x+5

A/q

300/x-300/x+5=2 hrs

300x+1500-300x/x(x+5). = 2

1500=2x2+10x

2x2+10x-1500=0

x2+5x-750=0

(x+30)(x+25)=0

but 30 is rejected answer

the correct answer is 25kmph

okkk

phle wrong tha


now
you can mark me as brainiest

Answer 2

Hi there!
Here's the answer:

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Let the Normal Speed be x kmph
Distance = 300 km

Time t1 = $\frac{Distance}{Speed}$
=> t1 = $\frac{300}{x}$

If the speed is increased,
Speed = x+5 kmph
distance = 300 km

Time t1 = $\frac{Distance}{Speed}$
=> t2 = $\frac{300}{x+5}$

As per the given data,

If speed is changed from x to (x+5) ; then
$t2 = t1 - 2$

Substitute t2 & t1

$\frac{300}{x+5} = \frac{300}{x} - 2$

$\frac{300}{x+5} - \frac{300}{x} = 2$

=> $300(\frac{1}{x+5}) - 300(\frac{1}{x}) = 2$

=> $(\frac{1}{x+5} - \frac{1}{x}) = \frac{2}{300}$

=> $(\frac{1}{x+5} - \frac{1}{x}) = \frac{1}{150}$

=> $\frac{x+5-x}{x(x+5)} = \frac{1}{150}$

=> $\frac{5}{x(x+5)} = \frac{1}{150}$

=> x(x+5) = 750

=> x² + 5x + 750 = 0

=> x² + 30x - 25x + 750 = 0

=> x(x+30) - 25(x+30) = 0

=> (x+30)(x-25) = 0

•°• x = 25 (As it can't be -Ve)

•°• Speed = 25 kmph

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Hope it helps