Since this is a round-trip, Christy covered the same distance in both journey.
Let the distance covered be
[tex]d[/tex]
When she left the house, she travelled at
[tex]24 \: kph[/tex]
We were given the average speed to be
[tex]20 \: kph[/tex]
Meaning if we add the speed in the return journey to 24 kph and find the average , we must obtain 20 kph.
Let
[tex]x[/tex]
be the speed in the return trip.
Then
[tex] \frac{24 + x}{2} = 20[/tex]
We solve for x to obtain;
[tex]24 + x = 40[/tex]
[tex]x = 40 - 24 = 16 \: kph[/tex]
We were also told that she took half an hour longer in the return trip.
If we let
[tex]t[/tex]
represent the time the first journey took, then the return trip will take
[tex](t + 0.5) \: hours[/tex]
Now we have the following:
[tex]<b><u>First Journey</u></b>[/tex]
[tex]speed = \frac{d}{t} [/tex]
[tex]24 = \frac{d}{t} [/tex]
[tex] \Rightarrow d = 24t - - - (1)[/tex]
[tex]<b><u>Second Journey</u></b>[/tex]
[tex]16 = \frac{d}{t + 0.5} [/tex]
[tex]
\Rightarrow d = 16(t + 0.5) - - - (2)[/tex]
Now let us equate the two equations to obtain,
[tex]
16(t + 0.5)= 24t [/tex]
We expand and simplify to obtain;
[tex]
16t + 8= 24t [/tex]
This implies that,
[tex]
8= 24t - 16t[/tex]
.
[tex]
8= 8t[/tex]
[tex]
t = 1[/tex]
Hence the return trip took,
[tex]t + 0.5 = 1 + 0.5 = 1.5[/tex]
hours.
