An aeroplane takes 1hour less for journey of 1200km if it's speed is increased by 100 kmhr from its usual speed find the usual speed
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let it's original speed be x km/h
ATQ
1200/x -1200/x+100 =1
1/x-1/x+100 =1/1200
x+100-x/x(x+100) =1/1200
100/x(x+100)=1/1200
x^2+100x-120000=0
x^2 -300x+400x-120000=0
x(x-300)+400(x-300)
(x-300)(x+400)
x=-400 (neglecting because speed can't be -ve)
x=300 Ans
ATQ
1200/x -1200/x+100 =1
1/x-1/x+100 =1/1200
x+100-x/x(x+100) =1/1200
100/x(x+100)=1/1200
x^2+100x-120000=0
x^2 -300x+400x-120000=0
x(x-300)+400(x-300)
(x-300)(x+400)
x=-400 (neglecting because speed can't be -ve)
x=300 Ans
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