Question

(1) If the speed of an aeroplane is decreased by 120 km/h, it takes 18 minutes more<br />to travel some distance. If the speed is increased by 180 km/h, it takes 18 minutes<br />less to travel the same distance. Find the average speed of the aeroplane and the<br />distance.​

Answer 1

Step-by-step explanation:

distance= speed*time

total distance= 120*0.3 + 180*0.3

=90km

average speed= total distance/total time

= 90/0.3+0.3

=150km/h

Answer 2

The average speed of the aeroplane =720 km/h

The distance traveled by the aeroplane=1080 km

Step-by-step explanation:

Let x km/h be the average speed of aeroplane and s be the distance traveled by aeroplane.

Time=$\frac{Distance}{speed}$

According to question

$\frac{s}{x}+\frac{18}{60}=\frac{s}{x-20}$

$s(\frac{1}{x-120}-\frac{1}{x})=\frac{18}{60}$...(1)

From second condition

$\frac{s}{x}-\frac{18}{60}=\frac{s}{x+180}$

$s(\frac{1}{x}-\frac{1}{x+180})=\frac{18}{60}$....(2)

Equation (1) is divided by equation (2)

$\frac{\frac{1}{x-120}-\frac{1}{x}}{\frac{1}{x}-\frac{1}{x+180}}=1$

$\frac{1}{x-120}-\frac{1}{x}=\frac{1}{x}-\frac{1}{x+180}$

$\frac{x-x+120}{x(x-120)}=\frac{x+180-x}{x(x+180)}$

$\frac{120}{x(x-120)}=\frac{180}{x(x+180)}$

$\frac{x+180}{x-120}=\frac{180}{120}=\frac{3}{2}$

$2x+360=3x-360$

$3x-2x=360+360=0$

$x=720 km/h$

Substitute the value of x in equation then we get

$s(\frac{1}{720-120}-\frac{1}{720})=\frac{18}{60}$

$s(\frac{1}{600}-\frac{1}{720}=\frac{18}{60}$

$\frac{s}{60}(\frac{1}{10}-\frac{1}{12})=\frac{18}{60}$

$s(\frac{6-5}{60})=18$

$s(\frac{1}{60})=18$

$s=60\times 18=1080 km$

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