Question

An aeroplane travelled a distance of 400 km/h at an average speed of x km/h. On the return journey, the speed was increased by 40 km/h . write down an expression for the time taken for (1) the onward journey , (2) the return journey . if the return journey took 30 minutes less than the onward journey , write down an equation in x and find its value.

Answer 1

Hey There!!


Here we can see that one thing remains constant in both forward and return journey:

Distance of the journey = 400 km


On the first journey, 

Average Speed = x km/hr

We know that:

$Speed = \frac{Distance}{Time} \\ \\ v=\frac{d}{t} \\ \\ \implies t=\frac{d}{v} \\ \\ \implies \boxed{t_1=\frac{400}{x}\,\,hours}$


On the return journey, Average Speed was increased by 40 km/hr. 

So, we have:

$x+40=\frac{400}{t_2} \\ \\ \\ \implies \boxed{t_2=\frac{400}{x+40}}$


We are given that the return journey took 30 minutes less than the onward journey. So we can say that the time difference is half an hour.

$t_1-t_2 = \frac{1}{2} \, \, hours \\ \\ \\ \implies \frac{400}{x} - \frac{400}{x+40} = \frac{1}{2} \\ \\ \\ \implies 400 \left(\frac{1}{x}-\frac{1}{x+40}\right)=\frac{1}{2} \\ \\ \\ \implies 800 (x+40 - x) = x(x+40) \\ \\ \\ \implies x^2+40x-32000=0 \\ \\ \\ \implies x^2+200x-160x-32000=0 \\ \\ \\ \implies x(x+200)-160(x+200)=0 \\ \\ \\ \implies (x+200)(x-160)=0 \\ \\ \\ \implies x=-200 \, \, OR \, \, x=160$

Since average speed cannot be negative, we have:

$\boxed{x=160\,\,km / hr}$

___________________________________________

So, finally your answers are:
$(1)\,\,t=\frac{400}{x}\,hr$


$(2)\,\,t=\frac{400}{x+40}\,hr$

$(3)\,\textrm{The equation is: }x^2+40x-32000=0 \\ \\ \\and \, \, x=160\,km / hr$


Hope it helps
Purva
Brainly Community

Answer 2

Answer:

X = 160 km/hr

Please refer to the above attachments

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