SOLUTION: " A motor boat travels upstream whose current is in 3 kph. The boat reaches a destination 54 km just to unload its cargo and returns immediately to its starting point. The boat is

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Question 483870: " A motor boat travels upstream whose current is in 3 kph. The boat reaches a destination 54 km just to unload its cargo and returns immediately to its starting point. The boat is back in 5 1/4 hours. Assuming that the unloading of cargo has a negligible time, what was the speed of the boat in still water? "
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Boat speed x
current speed 3

against current x- 3 kph
with current x+ 3 kph

Distance= 54 km

Time against + time with = 5.25 hours
t=d/r
54 /( x + 3 ) + 54 /(x - 3 ) = 5.25

LCD = (x - 3 ) ( x + 3 )
54 *( x - 3 ) + 54 (x + 3 ) = 5.25
54 x - 162 + 54 x + 162 = 5.25 ( x ^2 - 9 )
108 x = 5.25 x ^2 - 47.25
5.25 x ^2 - -108 x - 47.25

Find the roots of the equation by quadratic formula
a= 5.25 , b= -108 , c= -47.25

b^2-4ac=11664 + 992.25
b^2-4ac=12656.25
%09sqrt%28%0912656.25%09%29=%09112.5%09
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( 108 + 112.5 )/ 10.5
x1= 21
x2=( 108 -112.5 ) / 10.5
x2= -0.43
Ignore negative value
boat speed = 21 kph