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You went on a 360km train ride but you are late! If the train had just been 5 km/hr faster, then the train ride would have been one hour shorter.  How fast was the train going?

 Jul 8, 2020
 #1
avatar+1013 
+3

so we have (360/x)-(360/(x+5))=60 so we multiply all terms to get 360(x+5)+−360x=1x(x+5) and then −x^2−5x+1800=0 where x=-45 or x=40 but it can't be negative so x=40 and the speed is 40kph

 Jul 8, 2020
 #3
avatar+6 
+2

Correct

Pro12345  Jul 8, 2020
 #2
avatar+25880 
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x = k/hr

 

360 / x    -1   = 360 / (x+5)    solve for x

360/x - 360/(x+5)  = 1

360(x+5) - 360x = x (x+5)

1800= x^2+5x

x^2+5x-1800 = 0      Quadratic formula shows x = 40  k/hr   or   -45(throw  out)

 Jul 8, 2020

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