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Question

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

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Solution

Let the usual speed of aeroplane be x km/hr,

Then the increased speed of the aeroplane is = (x + 250) km/hr

Time taken by the aeroplane under usual speed to cover 1250 km =1250 over xhr

Time taken by the aeroplane under increased speed to cover 1250 km = fraction numerator 1250 over denominator x plus 250 end fractionhr

Therefore,

1250 over x � fraction numerator 1250 over denominator left parenthesis x plus 250 right parenthesis end fraction equals 50 over 60
fraction numerator 1250 left parenthesis x plus 250 right parenthesis � 1250 x over denominator x left parenthesis x plus 250 right parenthesis end fraction equals 5 over 6
fraction numerator 1250 x plus 312500 � 1250 x over denominator x squared plus 250 x end fraction equals 5 over 6
1875000 space equals space 5 x squared space plus space 1250 x space space 5 x squared space plus space 1250 x space � space 1875000 space equals space 0 space space 5 left parenthesis x squared space plus space 1250 x space � space 375000 right parenthesis space equals space 0 space space x squared space plus space 250 x space � space 375000 space equals space 0 space space x squared space � space 500 x space plus space 750 x space � space 375000 space equals space 0 space space x left parenthesis x space � space 500 right parenthesis space plus space 750 left parenthesis x space � space 500 right parenthesis space equals space 0 space space left parenthesis x space � space 500 right parenthesis left parenthesis x space plus space 750 right parenthesis space equals space 0 space space x space equals space 500 space o r space space x space equals space � space 750

Since, the speed of the aeroplane can never be negative,

Therefore, the usual speed of the train = 500 km/hr.


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