SOLUTION: Betty and Carl live in the same apartment building, work at the same office, and set off for work in the morning at the same time. They must each travel 42 km, and they arrive at w

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Question 1099768: Betty and Carl live in the same apartment building, work at the same office, and set off for work in the morning at the same time. They must each travel 42 km, and they arrive at work at the same time. Betty travels by car at 60 km/h, parks at the car park, then walks at 4 km/h the rest of the way to work. Carl takes the bus, which travels at 40 km/h, to the same car park as Betty uses, and then rides his bike the rest of the way at 12 km/h. If they both leave home at 6 a.m., at what time do they arrive at work?
Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
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1.  I am glad that after my notice at

        https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1099270.html

    you changed the problem formulation.

    Now the problem is correctly posed and is solvable.



2.  Let x be the distance from the apartment building to the parking lot, and 
    let y the distance from the parking lot to the office.


    Then you have these two equations

    x + y = 42,            (1)     (total distance)

    x%2F60 + y%2F4 = x%2F40 + y%2F12     (2)    ("time" equation")


    Multiply eq(2) by 120 (both sides). You will get

    2x + 30y = 3x + 10y,   or

    x = 20y.              (3)


    Now express y = 42-x from eq(1)  and substitute it into eq(3). You will get

    x = 20*(42-x),

    x = 20*42 - 20x  ====>  21x = 20*42  ====>  x = %2820%2A42%29%2F21 = 20*2 = 40.


    Thus, the distance x is 40 kilometers.  Then the distance  y is 42-40 = 2 km.

    And the total time of the Carl's travel is  40%2F40+%2B+2%2F12 = 1 hour and 10 minutes.


    So, they both arrive to the office at  7:10 am.

Solved.