A train travels a distance of 300 km at a constant speed. If the speed of the train is increased by 5 km/h the journey would have taken 2 h less. The original speed of the train was |
A) 25 km/h
B) 20 km/h
C) 28 km/h
D) 30 km/h
Correct Answer: A
Solution :
Let the normal speed of train be x km/h. |
Let the normal time of train = T h |
\[\therefore \]\[\text{Time}=\frac{\text{Distance}}{\text{Speed}}\] |
So \[\frac{300}{x}=\text{T}\] (i) |
Similarly, \[\frac{300}{x+5}=\text{T}-2\] ... (ii) |
On solving Eqs. (i) and (ii), we get |
\[\frac{300}{x+5}=\frac{300}{x}-2\]\[\Rightarrow \]\[\frac{300}{x+5}=\frac{300-2x}{x}\] |
\[\Rightarrow \] \[300x=(x+5)(300-2x)\] |
\[\Rightarrow \] \[300x=300x-2{{x}^{2}}+1500-10x\] |
\[\Rightarrow \] \[-\,\,2{{x}^{2}}+1500-10x=0\] |
\[\Rightarrow \] \[{{x}^{2}}-750+5x=0\] |
\[\Rightarrow \] \[{{x}^{2}}+5x-750=0\] |
\[\Rightarrow \]\[{{x}^{2}}+30x-25x-750=0\] |
\[\Rightarrow \]\[x\,\,(x+30)-25\,\,(x+30)=0\] |
\[\therefore \] \[x=25,\]\[-\,\,30\] |
Since, speed of train cannot be negative, hence speed of train is 25 km/h, |
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