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A train covers a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken four hours less than the scheduled time. And if the train were slower by 6 km/hr, the train would have taken six hours more than the scheduled time. Find the length of the journey in kilometers.

 Jan 31, 2021
 #1
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Equating times  we  have this system

 

D/ ( R + 6)   =   D/R -  4

D/ (R - 6) =  D/R + 6                       simplify

 

D / (R + 6)  =  (D  - 4R)  / R

D/ ( R -6)   =  ( D + 6R) / R

 

DR  =  (R +6) ( D - 4R)

DR  = (R - 6) ( D + 6R)

 

DR  = DR  +  6D - 4R^2  - 24R

DR  = DR - 6D + 6R^2   -36R

 

0  =  -4R^2 + 6D  -  24R

0  =   6R^2  - 6D  -36R             add  these

 

0 = 2R^2  + 60R

2R ( R - 30)  = 0

 

Setting the second factor to  0   and  solving for R   produces  the  normal  rate of     30 Km/ hr

 

And

 

-4 (30)^2  +  6D  - 24 (30)   = 0

 

-3600 +  6D  - 720   = 0

 

6D  =  3600 +  720

 

6D  =  4320

 

D = 4360  /  6  =   720  km =   distance of the journey

 

cool cool cool         

 Jan 31, 2021

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