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 m

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

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Ngọc Hoa 4 years 2021-08-21T17:16:00+00:00 2 Answers 13 views 0

Answers ( )

  1. minhkhoi
    0
    2021-08-21T17:17:27+00:00
    Reply

    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

  2. minhkhue
    0
    2021-08-21T17:17:47+00:00
    Reply

    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.

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