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An Aeroplane Left 50 Minutes Later than Its Scheduled Time, and in Order to Reach the Destination, 1250 Km Away, in Time, It Had to Increase Its Speed by 250 Km/Hr from Its Usual Speed. Find Its Usual Speed. - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

Question

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

Solution

Let the usual speed of aero plane be x km/hr. Then,

Increased speed of the aero plane = (x + 250) km/hr

Time taken by the aero plane under usual speed to cover 1250 km = 1250/xhr

Time taken by the aero plane under increased speed to cover 1250 km = 1250/(x +250)hr

Therefore,

1250/x-1250/(x+250)=50/60

(1250(x+250)-1250x)/(x(x+250))=5/6

(1250x+312500-1250x)/(x^2+250x)=5/6

312500/(x^2+250x)=5/6

312500(6) = 5(x2 + 250x)

1875000 = 5x2 + 1250x

5x2 + 1250x - 1875000 = 0

5(x2 + 250x - 375000) = 0

x2 + 250x - 375000 = 0

x2 - 500x + 750x - 375000 = 0

x(x - 500) + 750(x - 500) = 0

(x - 500)(x + 750) = 0

So, either

x - 500 = 0

x = 500

Or

x + 750 = 0

x = -750

But, the speed of the aero plane can never be negative.

Hence, the usual speed of train is x = 500 km/hr.

Is there an error in this question or solution?

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Solution An Aeroplane Left 50 Minutes Later than Its Scheduled Time, and in Order to Reach the Destination, 1250 Km Away, in Time, It Had to Increase Its Speed by 250 Km/Hr from Its Usual Speed. Find Its Usual Speed. Concept: Solutions of Quadratic Equations by Factorization.
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