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

Answers (1)

Given , Total distance $=360\; km$

Let's normal speed of train be $x$

In normal conditions

$v=\frac{d}{t}\Rightarrow t=\frac{d}{v}\Rightarrow t_{x}=\frac{360}{x}$ ______(1)

When speed is increased

$t_{x+5}=\frac{360}{x+5}$ _________(2)

According to question

$t_{x}-t_{x+5}=1\; hr$

from eq (1) and (2)

$\frac{360}{x}-\frac{360}{x+5}=1$

$\Rightarrow \frac{360}{x}-1=\frac{360}{x+5}$

$\Rightarrow \frac{360-x}{x}=\frac{360}{x+5}$

$\Rightarrow (360-x)(x+5)=360\times x$ (cross multiplication)

$\Rightarrow 360x+360\times 5-x^{2}-5x=360\; x$

$\Rightarrow x^{2}+5x-1800=0$ (rearranging equation)

$\Rightarrow x^{2}+(45-40)x-1800=0$

$\Rightarrow x^{2}+45x-40x-1800=0$

$\Rightarrow (x+45)(x-40)=0$

$\Rightarrow x=-45,\; 40$

$x\neq -45$ (because speed can not be negative)

hence $\Rightarrow x=40km/hr$

speed of train $=40\; km/hr$

Posted by

Safeer PP

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