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?

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Answer 1 · 2016-02-07T05:31:09

6.75 km

Let $x$ $x$ be the duration spent jogging.

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

$d = s t$ $d = st$

The distance covered by jogging is

$\Rightarrow d_{j} = 9 x$ $=> d_j = 9x$

Meanwhile, the distance covered by walking is

$d_{w} = 5 (1 - x)$ $d_w = 5(1 - x)$

Since the total distance covered is 8 km, we have

$\Rightarrow 9 x + 5 (1 - x) = 8$ $=> 9x + 5(1 - x) = 8$

$\Rightarrow 9 x + 5 - 5 x = 8$ $=> 9x + 5 - 5x = 8$

$\Rightarrow 4 x = 3$ $=> 4x = 3$

$\Rightarrow x = \frac{3}{4}$ $=> x = 3/4$

We want the distance covered by jogging,

$d_{j} = 9 x$ $d_j = 9x$

$d_{j} = 9 (\frac{3}{4})$ $d_j = 9(3/4)$

$d_{j} = \frac{27}{4} = 6 \frac{3}{4} = 6.75$ $d_j = 27/4 = 6 3/4 = 6.75$