Question

Please help with the question.

Answer 1

Step-by-step explanation:

let the original speed =xkm/hr

: normal time flight = 2800/x

speed = (x -100) km/hr : time of flight ,

when speed is reduced = 2800 /x-100 hours

A/q to given eq :-

2800 /x-100 - 2800 -x = 30 /60

1 /2 = 2800x - 2800 ( x-100) / x ( x-100) [ cross multiple]

= x² -100x =560000

x² -100x -560000 =0

x² -800x +700x -560000 =0

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

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

x =800 ; x = -700

originally speed of aircraft = 800

originally duration of flight = 2800 /800 =28/8 km /hr

hopefully helping you.......