Question

An aeroplane left 30 minutes later than its scheduled time. In order to reach his destination 2800 km away in time, the pilot decided to increase its speed by 100 km/hr from its usual speed. What was the usual speed of the aeroplane?

Also mention the steps.

Answer 1

Hey there !!

→ Let the usual speed be x km/hr.

→ Actual speed = ( x + 100 ) km/hr.

→ Time taken at usual speed = $(\frac{2800}{x}) hr.$

→ Time taken at actual speed =$( \frac{2800}{x+100}) hr$

→ Difference between the two times taken = $\frac{60}{30} = \frac{1}{2} hr.$

=> $\frac{2800}{x} - \frac{2800}{x + 100} = \frac{1}{2} .$

=> 2800 $( \frac{1}{x} - \frac{1}{x+ 100} ) = \frac{1}{2} .$

=> $\frac{x + 100 - x}{x ( x + 100) }= \frac{1}{5600} .$

=> $\frac{100}{ {x}^{2} + 100x ) }= \frac{1}{5600} .$

=> x² + 100x - 560000 = 0.

=> x² + 800x - 700x - 560000 = 0.

=> x( x + 800 ) -700( x + 800 ) = 0.

=> ( x - 700 ) ( x + 800 ) = 0.

=> x - 700 = 0. | => x + 800 = 0.

=> $\huge \boxed{ x = 700 .}$| => x = -800.

[ => Speed cannot be negative . ]

✔✔ Hence, the usual speed of the aeroplane was 700 km/hr. ✅✅

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$\huge \boxed{ \mathbb{THANKS}}$ 

$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

Answer, Let the usual time taken by the aeroplane = x km/hr Distance to the destination = 1500 km Case (i) Speed = Distance / Time = (1500 / x) Hrs Case (iI) Time taken by the aeroplane = (x - 1/2) Hrs Distance to the destination = 1500 km Speed = Distance / Time = 1500 / (x - 1/2) Hrs Increased speed = 250 km/hr ⇒ [1500 / (x - 1/2)] - [1500 / x] = 250 ⇒ 1/(2x2 - x) = 1/6 ⇒ 2x2 - x = 6 ⇒ (x - 2)(2x + 3) = 0 ⇒ x = 2 or -3/2 Since, the time can not be negative, The usual time taken by the aeroplane = 2 hrs and the usual speed = (1500 / 2) = 750 km/hr.