Question

Solve the given problem

Answer 1

Let the usual speed of train be a km/h

Distance = 300 km

Time taken to reach the destination with usual speed = 300/a

New Speed = (a + 5) km/h

Time taken to reach the destination with new speed = 300 /(a+5)

According to question,

300/a - 300/(a+5) = 2

=> 300 [ 1/a - 1/(a+5)] = 2

=> 300 ( a + 5 - a) /(a) (a+5) = 2

=> 300 × 5 / a^2 + 5a = 2

=> 1500 = 2a^2 + 10a

=> 2a^2 +10a - 1500 = 0

=> a^2 + 5a - 750 = 0

=> a^2 +30a - 25a - 750 = 0

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

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

a = - 30 and 25

But speed can't be negative,

So, a = 25

Usual speed of train = 25 km/h

Answer 2

HLO MATE ✋✋
Simran Here !!!!

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Given That :
Distance of the train = 300km

Let the usual speed of the train = x km/hr

Increased Speed of the train = (x+5) km/hr

Total time taken = Distance / Speed = 300/x hr

Acc.To. Statement :
$= > \frac{300}{x} - \frac{300}{(x + 5) } = 2 \\ \\ = > \frac{300x + 1500 - 300x}{x(x + 5)} = 2 \\ \\ = > 1500 = 2( {x}^{2} + 5) \\ \\ = > 1500 = 2 {x}^{2} + 10x \\ \\ = > 2 {x}^{2} + 10x - 1500 = 0 \\ \\ = > {x}^{2} + 5x - 750 = 0 \\ \\ = > {x}^{2} + 30x - 25x - 750 = 0 \\ \\ = > x(x + 30) - 25(x + 30) = 0 \\ \\ = > (x + 30)(x - 25) = 0$
So , x = -30 & 25

Speed can never be in negative

So,The Usual speed of Train is 25 km/hr✔
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