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18 June, 14:23

Kiran drove from City A to City B, a distance of 228 mi. She increased her speed by 12 mi/h for the 400-mi trip from City B to City C. If the total trip took 14 h, what was her speed from City A to City B?

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  1. 18 June, 15:36
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    From city A to city B her speed was 38 mi/h

    Explanation:

    The traveled distance can be calculated using this equation:

    From city A to city B

    228 mi = v · t₁

    Where:

    v = velocity

    t₁ = time it took Kiran to travel the 228 mi from city A to city B

    From city B to city C

    400 mi = (v + 12 mi/h) · t₂

    We also know that the entire trip took 14 h, then:

    t₁ + t₂ = 14 h

    So, we have a system of three equations with three unknwons:

    228 mi = v · t₁

    400 mi = (v + 12 mi/h) · t₂

    t₁ + t₂ = 14 h

    Let's solve the third equation for t₁:

    t₁ = 14 h - t₂

    Now let's replace t₁ in the first equation and solve it for t₂

    228 mi = v · t₁

    228 mi = v · (14 h - t₂)

    228 mi/v - 14 h = - t₂

    t₂ = 14 h - 228 mi/v

    Now let's replace t₂ in the second equation:

    400 mi = (v + 12 mi/h) · t₂

    400 mi = (v + 12 mi/h) · (14 h - 228 mi/v)

    400 mi = 14 h · v - 228 mi + 168 mi - 2736 mi² / (v · h)

    400 mi = 14 h · v - 60 mi - 2736 mi² / (v · h)

    460 mi = 14 h · v - 2736 mi² / (v · h)

    Multiplicate by v both sides of the equation:

    460 mi · v = 14 h · v² - 2736 mi²/h

    0 = 14 h · v² - 460 mi · v - 2737 mi²/h

    Solving the quadratic equation:

    v = 38 mi/h

    (The other solution of the equation is negative, and therefore discarded)

    From city A to city B her speed was 38 mi/h
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