Question

Christy went on a round-trip bicycle ride starting at her house. When she left, she traveled downhill at a rate of 24 kilometers per hour (kph). She rested for 30 minutes, and then returned to her house. Since the return trip was partly uphill, it took half an hour longer and Christy averaged only 20 kph. How long did the return trip take?

Answer 1

Since this is a round-trip, Christy covered the same distance in both journey.

Let the distance covered be
$d$

When she left the house, she travelled at
$24 \: kph$

We were given the average speed to be
$20 \: kph$

Meaning if we add the speed in the return journey to 24 kph and find the average , we must obtain 20 kph.

Let
$x$
be the speed in the return trip.

Then
$\frac{24 + x}{2} = 20$

We solve for x to obtain;

$24 + x = 40$

$x = 40 - 24 = 16 \: kph$

We were also told that she took half an hour longer in the return trip.

If we let
$t$
represent the time the first journey took, then the return trip will take

$(t + 0.5) \: hours$

Now we have the following:

$First Journey$

$speed = \frac{d}{t}$

$24 = \frac{d}{t}$

$\Rightarrow d = 24t - - - (1)$

$Second Journey$

$16 = \frac{d}{t + 0.5}$

$\Rightarrow d = 16(t + 0.5) - - - (2)$

Now let us equate the two equations to obtain,

$16(t + 0.5)= 24t$

We expand and simplify to obtain;

$16t + 8= 24t$

This implies that,

$8= 24t - 16t$

.
$8= 8t$

$t = 1$

Hence the return trip took,

$t + 0.5 = 1 + 0.5 = 1.5$

hours.