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

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Solution

## Step 1Let the original speed of train$=skm/h$We know that$Time=\frac{dis\mathrm{tan}ce}{speed}$According to the question, we have,Time is taken by train $=\frac{360}{s}hour$And, Time taken by train its speed increase $5km/h=360/\left(s+5\right)$Step 2Now, It is given that,Time taken by train in first – time taken by train in 2nd case =$48min=48/60hour$$360/s–360/\left(s+5\right)=48/60=4/5\phantom{\rule{0ex}{0ex}}360\left(1/s–1/\left(s+5\right)\right)=4/5\phantom{\rule{0ex}{0ex}}360×5/4\left(5/\left(s²+5s\right)\right)=1\phantom{\rule{0ex}{0ex}}450×5=s²+5s\phantom{\rule{0ex}{0ex}}s²+5s-2250=0\phantom{\rule{0ex}{0ex}}s=\left(-5±\surd \left(25+9000\right)\right)/2\phantom{\rule{0ex}{0ex}}=\left(-5±\surd \left(9025\right)\right)/2\phantom{\rule{0ex}{0ex}}=\left(-5±95\right)/2\phantom{\rule{0ex}{0ex}}=-50,45\phantom{\rule{0ex}{0ex}}Buts\ne -50becausespeedcannotbenegative\phantom{\rule{0ex}{0ex}}So,s=45km/h$Hence, original speed of train $=45km/h$  Suggest Corrections  0      Similar questions  Explore more