Question

A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speeds of the two trains.

Answer 1

Given: Total distance of a journey= 600 km
Let the speed of the slow train be X km/h  and the speed of the fast train is (x+10) km)h.

Time taken by the slow train to cover 600 km= 600/x hrs
Time taken by the fast train to cover 600 km= 600/(x +10) hrs

[ Time = Distance/speed]

600/(x +10) = 600/ x - 3
600/ x -  600/(x +10) = 3
600(x+10) -60x /(x(x+10) = 3
600x+6000 - 60x /(x(x+10) = 3
6000 / x² +10x = 3
3(x² +10) = 6000
3x² +30 = 6000
3x² +30 - 6000= 0
3 (x² +10x - 2000) = 0
x² +10x - 2000 = 0
x² +50x -40x -2000= 0
x(x+50) - 40 ( x +50)= 0

(x-40) (x+50)= 0

x = 40 or x= - 50 (speed cannot be negative)

speed of the slow train be = x = 40 km/h
speed of the fast train be = (x+10)= 40+10 =  50 km/h

Hence the speeds of two trains are 40 km/h and 50 km/h.

HOPE THIS WILL HELP YOU...

Answer 2

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A fast train takes three hours less than the slow train for a journey of 600 km.
 If the speed of slow train is 10km/hr less than that of fast train.
 Find the speed of the two trains. : Let s = slow train speed then (s+10) = fast train

 : Write a time equation
: Time {{{dist/speed}}}
 ; Slow train time - Fast train speed = 3 hrs {{{600/s}}} - {{{600/((s+10))}}} = 3
 Multiply each term by s(s+10); results 600(s+10) - 600s = 3s(s+10)
: 600s + 6000 - 600s = 3s^2 + 30s : Arrange as a quadratic equation

, (600s's cancel) 3s^2 + 30s - 6000
= 0 Simplify, divide by 3 s^2 + 10s - 2000 = 0
 Factors to (s = 50)(s - 40) = 0 positive solution
 s = 40 km/hr speed of the slow train then 50 km/hr speed of the fast train 
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