NIMCET 2009 Question Paper with solutions

Question : 71 of 120

Marks: +1, -0

A train after travelling

60 {km}

meets with an accident and then proceeds at

3 {/ }4

of its former rate and arrives at the terminus 40 min late. Had the accident happened

25 {km}

further on, it would have arrived

10 {min}

sooner. Find the speed of the train and the distance, respectively.

Solution:

Let

× {km} \/ {h}

be the speed of the train and

y {km}

be the distance,
Then,

\;{y-60}/{3 /4}=\;{25}/{x}+\;{y-85}/{3 x}+\;{10}/{60}

⇒25/{3/4x}=25/x+1/6

⇒25/{3x}=1/6

⇒ \;\; x=50 {km} \/ {h}

and

V=160 {km}

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