A train travels 360, km at a uniform speed. I | vedclass

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Question

(A) Let the uniform speed of the train be $x\, km/h$.Time taken to cover $360\, km$ at speed $x$ is $t_1 = \frac{360}{x}\, h$.If the speed is increase

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Let the uniform speed of the train be $x\, km/h$.Time taken to cover $360\, km$ at speed $x$ is $t_1 = \frac{360}{x}\, h$.If the speed is increased by $5\, km/h$,the new speed is $(x + 5)\, km/h$.Time taken at the new speed is $t_2 = \frac{360}{x + 5}\, h$.According to the problem,the difference in time is $1\, hour$:$\frac{360}{x} - \frac{360}{x + 5} = 1$$360 \left( \frac{1}{x} - \frac{1}{x + 5} \right) = 1$$360 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 1$$360 \left( \frac{5}{x^2 + 5x} \right) = 1$$1800 = x^2 + 5x$$x^2 + 5x - 1800 = 0$Factoring the quadratic equation:$x^2 + 45x - 40x - 1800 = 0$$x(x + 45) - 40(x + 45) = 0$$(x + 45)(x - 40) = 0$This gives $x = -45$ or $x = 40$.Since speed cannot be negative,the speed of the train is $40\, km/h$.

$A$ train travels $360\, km$ at a uniform speed. If the speed had been $5\, km/h$ more,it would have taken $1\, hour$ less for the same journey. Find the speed of the train (in $km/h$).

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