#### Question

An Aero plane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. 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.

#### Solution

Distance = 400 km

The average speed of the aeroplane = x km/hr

Speed while returning = (x + 40) km/hr

1) We know : `"Time" = "Distance"/"Speed"`

Time taken for onward journey = 400/x hrs

2) Time taken for return journey = 400/(x + 40) hrs

From the given information we have

`400/x - 400/(x + 40) = 30/60`

`(400x + 16000 - 400x)/(x(x + 40)) = 1/2`

`16000/(x(x + 40)) = 1/2`

`x^2 + 40x - 32000 = 0`

`x^2 + 40x - 32000 = 0`

`x^2 + 200x - 160x - 32000 = 0`

x(x + 20) - 160(x + 2000) = 0

(x + 200)(x - 160) = 0

x = -200, 160

Since speed cannot be negative. Thus x = 160