# A student walks and jogs to college each day. The student averages 5 km/h walking and 9 km/h jogging. The distance from home to college is 8 km, and the student makes the trip in 1 hour. How far does the student jog?

Feb 7, 2016

6.75 km

#### Explanation:

Let $x$ be the duration spent jogging.

Since the total travel duration is 1 hour, $1 - x$ is the duration spent walking

$d = s t$

The distance covered by jogging is

$\implies {d}_{j} = 9 x$

Meanwhile, the distance covered by walking is

${d}_{w} = 5 \left(1 - x\right)$

Since the total distance covered is 8 km, we have

$\implies 9 x + 5 \left(1 - x\right) = 8$

$\implies 9 x + 5 - 5 x = 8$

$\implies 4 x = 3$

$\implies x = \frac{3}{4}$

We want the distance covered by jogging,

${d}_{j} = 9 x$

${d}_{j} = 9 \left(\frac{3}{4}\right)$

${d}_{j} = \frac{27}{4} = 6 \frac{3}{4} = 6.75$