Answers

2014-08-25T03:02:02+05:30
Let in time t ,I reach the bus stand with a speed v m/min. and distance between home and bus stand is x meter(m).
4 km/hr = 4*1000/60 = 200/3 m/min
5 km/hr = 250/3 m/min
t = x/(200/3) - 3
t = x/(250/3) + 3
x/(200/3) - 3 = x/(250/3) + 3
x = 2000m
time = 2000/(200/3) - 3 = 27min.
speed = 2000/27 m/min
hence distance between bus stand and home = 2000m
I hope you will understand. if not understand please ask. 
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2014-08-25T15:39:04+05:30
Let the distance be x km
Case (i)
Speed = 4 km/hr
Reached time is 3 minutes delay to the original time (T)
Distance = speed * time
x = 4(T +  \frac{3}{60} )
x = 4(T +  \frac{1}{20} )
x = 4T +  \frac{1}{5}       ......(1)

Case (ii)
Speed = 5 km/hr
Reached time 3 minutes before the scheduled time (T)
Distance = speed * time
x = 5(T - \frac{3}{60} )
x = 5(T - \frac{1}{20} )
x = 5T - \frac{1}{4}       ......(2)

Solving two equations, 

 4T + \frac{1}{5} = 5T - \frac{1}{4}

5T - 4T =  \frac{1}{5} +  \frac{1}{4}

T =  \frac{9}{20}

Substitute the value of T in any one of the equations

x = 5( \frac{9}{20} ) + \frac{1}{4}

x =  \frac{9}{4}  + \frac{1}{4}

x =  \frac{5}{2}

The bus stand is  2.5 km from home

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