Question

When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks. Of her average walking speed is 3 mph and her average biking speed is 12 mph, how far is it from her apartment to the classroom building

Answer 1

Answer:

$d=2.4\ miles$

Explanation:

Given:

• average walking speed, $v_w=3\ mph$

• average biking speed, $v_b=12\ mph$

According to given condition:

$t_w=t_b+\frac{36}{60}$

where:

$t_w=$ time taken to reach the building by walking

$t_b=$ time taken to reach the building by biking

We know that,

$\rm time=\frac{distance}{speed}$

so,

$\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}$

$\frac{d}{3}=\frac{d}{12} +\frac{3}{5}$

$d=2.4\ miles$

Answer 2

Answer:

The distance from her apartment to the classroom building is 2.4 miles.

Explanation:

Given that,

Time  = 36 min

Walking average speed of her = 3 m/h

biking average speed of her = 12 m/h

If she takes n minutes to ride, then if the distance is d miles,

We need to calculate the distance

Using formula of time

$t=t_{1}+t_{2}$

$t=\dfrac{d}{v_{1}}+\dfrac{d}{v_{2}}$

Put the value into the formula

$\dfrac{36}{60}=\dfrac{d}{3}-\dfrac{d}{12}$

$d=\dfrac{36\times12}{60\times3}$

$d=2.4\ miles$

Hence, The distance from her apartment to the classroom building is 2.4 miles.