A train,travelling at a uniform speed for 360 | vedclass

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(B) Let the original speed of the train be $x\, km/h$.Then,the increased speed of the train is $(x+5)\, km/h$.Distance $= 360\, km$.Time taken at orig

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Let the original speed of the train be $x\, km/h$.Then,the increased speed of the train is $(x+5)\, km/h$.Distance $= 360\, km$.Time taken at original speed $= \frac{360}{x}\, h$.Time taken at increased speed $= \frac{360}{x+5}\, h$.Given that the difference in time is $48\, min = \frac{48}{60}\, h = \frac{4}{5}\, h$.According to the condition: $\frac{360}{x} - \frac{360}{x+5} = \frac{4}{5}$.$\Rightarrow 360 \left( \frac{1}{x} - \frac{1}{x+5} ight) = \frac{4}{5}$.$\Rightarrow 360 \left( \frac{x+5-x}{x(x+5)} ight) = \frac{4}{5}$.$\Rightarrow \frac{360 	imes 5}{x^2+5x} = \frac{4}{5}$.$\Rightarrow 1800 	imes 5 = 4(x^2+5x)$.$\Rightarrow 9000 = 4x^2 + 20x$.$\Rightarrow 4x^2 + 20x - 9000 = 0$.Dividing by $4$: $x^2 + 5x - 2250 = 0$.Solving by factorization: $x^2 + 50x - 45x - 2250 = 0$.$x(x+50) - 45(x+50) = 0$.$(x+50)(x-45) = 0$.Since speed cannot be negative,$x = 45$.Therefore,the original speed of the train is $45\, km/h$.

$A$ train,travelling at a uniform speed for $360\, km$,would have taken $48\, min$ less to travel the same distance,if its speed were $5\, km/h$ more. Find the original speed of the train (in $km/h$).

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