After covering a distance of 30 km with a uniform speed there is some defect in a train engine and therefore its speed gets reduced to 4/5 of its original speed ,Consequently , the train reaches its destination late by 45 minutes . If the defect in the engine would have occurred after covering 18 km more distance , the train would have reached 9 minutes earlier . Find the original speed of train.
Answers
Answer:
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Step-by-step explanation:
Let the speed of the train=x km\hr
Length of the journey=y km.
Time taken=yx
According to the first condition.
Speed for the first 30 km = x km/hr
speed for the remaining (y−30) km = 54km/hr
∴ time taken to cover 30 km = x30
time taken to cover (y−30)=54xy−30=4x5(y−30)
According to the problem
⇒x30+4x5(y−30)=xy+6045
⇒x30+4x5y−150=xy+43
⇒120+5y−150=4y+3x
⇒y−3x=30⇔3x−y=−30 ....(1)
According to the second condition.
Speed for the first 48 km = x km/hr
speed for the remaining (y−48) km = 54km/hr
∴ time taken to cover 48 km = x48
time taken to cover (y−48)=4x/5y−48=4x5(y−48)
According to the problem
⇒x48+4x5(y−48)=xy+6036
⇒x30+4x5y−150=xy+53
⇒4x192+5y−240=5x5y+3x
⇒4x5x−48=5x5y+3x
⇒25y−240=20y+12x
⇒−12x+5y=240⇔12x−5y=−240 ........(2)
multiply eq 1 by 5
⇒15x−5y=−150 ........(3)
Substract eq2 and eq3
⇒3x=90⇒x=30
put x=30 in eq1
⇒3×30−y=−30
⇒y=120
Let the speed of the train=30 km/hr
Length of the journey=120 km.