Question

An aeroplane left 30 minutes later than its scheduled time. In order to reach the destination 2800km 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?

Steps should be mentioned

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. ✅✅

____________________________________

$\huge \boxed{ \mathbb{THANKS}}$

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

Answer 2

Let the usual speed be x km/hr.
Actual speed = ( x + 100 ) km/hr.
Time taken at usual speed = 2800/x h.
Time taken at actual speed =2800/x+100h.

Difference between the two times taken =1/2 h.

2800/x -2800/x+ 100 = 1/2.

=> 2800 ( 1/x - 1/x+ 100 ) = 1/2.

=> x+100-x/x(x+100) = 1/5600 .

=> 100/ x^2+100x) = 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.

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

[ => Speed cannot be negative . ]