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A superfast train runs having the speed15 kms / hr more than that of an express train. Leaving same station the superfast train reached at a station of 180 kms distance 1 hour before than the express train. Determine the speed of the superfast train in km/ hr. |
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Answer» Solution :Let the speed of the superfast train was x KM/hr. `:.` the speed of the express train was (x-5)km/hr. `:.` to travel 180 km, time taken by the superfast train is `(180)/(x)` hr and by express train is `(180)/(x-15)` hr As per questions, `(180)/(x-15)-(180)/(x)=1` or, `180((1)/(x-15)-(1)/(x))=1or,(x-x+15)/((x-15)x)=(1)/(180)` or, `(15)/(x^(2)-15x)=(1)/(180)` or, `x^(2)-15x=2700` or, `x^(2)-15x-2700=0...............(1)` Comparing (1) with `ax^(2)+bx+c=0,(ANE0)` we get, `a=1,b=-15 andc=-2700` `:.` by Sreedhar Acharya's formula we get, `x=(-(-15)pmsqrt((15)^(2)-4xx1xx(-2700)))/(2xx1)` or, `x=(15pmsqrt(225+10800))/(2)` or, `x=(15pmsqrt(11025))/(2)or,x=(15pm105)/(2)` `:.x=(15+105)/(2)` (taking +sign) and `:.x=(15-105)/(2)` (taking -sign) `=(120)/(2)andx=(-90)/(2)` =60 and x=-45 But the speed of a train connot be negative, `:.x=60`. Hence the speed of the superfast train was 60 km/hr. |
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