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anybody plz help in TSD quest...arun sharma

Submitted by kisan_comp on Thu, 2007-08-30 14:35

arun sharma tsd lod II

q 48-

two ghats r located on a river bank& r 21 km apart.leaving one of the ghats for the other , a motor boat returns to the 1st ghat in 270 min. spending 60 min of that time in taking the pasengers at the 2nd ghat.Find the sped of the boat in still water if the speed of the river flow is 2.5 km\h??

a)10.4km/h

b)12.5 km/h

c)22.5 km/h

d)11.5 km/h

 

 

plz yaar solve it 4 me

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Posted on: Thu, 2007-08-30 18:04 #1
Let the speed of the boat in
Let the speed of the boat in still water = u
            speed of the river flow = v
 
Time taken to cross the river while moving with the flow = 21/(u + v)
Time taken to cross the river while moving against the flow = 21/(u - v)
 
Total time given is 270 min = 3.5 hr
 
So 3.5 = 21 [ 1 / (u + v ) + 1 / (u - v ) ]
 
=> 1/6 = 2u / ( u2 – v2 )           
 
=>  u2 – v2  – 12u =0   given v = 2.5
 
=> u2 – 12u  - 6.25 =0  
 
So u = [ 12 ± sqrt (144 + 25) ] /2
 
u = (12 ± 13 ) / 2
 

= 15.5 or -0.5

 

__________________

n/a

Posted on: Thu, 2007-08-30 23:54 #2
hey...thanx buddy....u made

hey...thanx buddy....u made this quest a dam easy one.....thanx 4 ur effort

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