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10pt for nasty word problem? help ;(?

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A man travels from Town X to Town Y at an average rate of 50 mph and returns at an average rate of 40 mph. He takes a 1/2 hour longer than he would take if he made the round trip at an average of 45 mph. What is the distance from Town X to Town Y?

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  1. My teacher gave my class two websites that help you ALOT 8)

    www.hotmath.com

    www.smarthinking.com

    hope i helped :)

    IF those dont work out--

    try googling a free tutoring website,where someone could help you LIVE xD just find a safe one :) the above two,are tutoring websites also,and are very safe (:


  2. Make some variables, and things start to get less nasty.

    Let:

    d = distance from X to Y (the thing we're trying to answer)

    t_xy = time to get from X to Y

    t_yx = time to get from Y to X

    > "...travels from Town X to Town Y at an average rate of 50 mph..."

    This means:

    d / t_xy = 50 mph

    > "...returns at an average rate of 40 mph..."

    This means:

    d / t_yx = 40 mph

    > "He takes a 1/2 hour longer than he would take if he made the round trip at an average of 45 mph."

    This is a long sentence, so let's break it down.

    "He takes a 1/2 hour longer.."

    They're talking about his normal round trip, which take a time of t_xy + t_yx.  So they're saying that "t_xy + t_yx is 1/2 hour longer than" something.

    "...he would take if he made the round trip at an average of 45 mph."

    So "t_xy + ty_x is 1/2 hour longer than T", where T is the length of this hypothetical 45mph trip.  (We just made another variable).

    So that sentence in math is:

    t_xy + t_yx = 1/2 + T

    Now, before we go farther, notice that:

    (2d) / T = 45 mph

    or:

    T = (2d) / 45

    So combine that with the previous equation:

    t_xy + t_yx = 1/2 + (2d) / 45

    Next, remember these from above:

    d / t_xy = 50 mph

    d / t_yx = 40 mph

    So we can substitute "d / 50" for t_xy and "d / 40" for t_yx:

    d/50 + d/40 = 1/2 + (2d) / 45

    Now there's just one variable "d" in that equation; so solve for "d".

  3. 120

    speed x time = distance

    40=x(distance)/t+30(time + 30 because it takes him an extra half an our tan 45mph)

    45=x/t

    45t=x

    40=45t/t + 30

    40t + 120 = 45t

    t=24

    plug 24 in for t in any equation and solve for d

    d=120

  4. Let  the distance from Town X to Town Y be d.

    time = distance / speed

    t=d /50 + d/40

    Half an hour more with 45 mph  both ways.

    t+1/2 = 2d / 45

    2d/45 +1/2 = d/50+d/40

    4000d+45000=1800d+2250d

    -50d=-45000

    d=900 miles

  5. wait a minute, if you go there at 50 and come back at 40 doesn't that average out to 45 round trip anyways?

  6. Distance= Rate x Time, so Time=Distance/Rate

    D/50+D/40=(2D/45)+.5

    D=900 miles

  7. Hi,

    distance/rate to + distance/rate from - ½ = distance/ave rate

    x... ....x.... 1..  ..2x

    ---.+.----.-.--- = -------

    50.. .40.. .2.. ...45

    Multiply every term by 1800 to cancel out denominators. then solve for x.

    36x + 45x - 900 = 80x

    81x - 900 = 80x

    x = 900

    It is 900 miles from town X to town Y. <==ANSWER

    I hope that helps!! :-)

  8. x/50 + x/40 = 2x/45 + 1/2

    x + 50x/40 = 100x/45 + 25

    40x + 50x = 4000x/45 + 1000

    1800x + 2250x = 4000x + 45000

    50x = 45000

    x = 900 miles

  9. d1 = d2

    d1 = 50t1

    d2 = 40t2

    t2 = (1/2) + t1

    Avg. = (d2 - d1)/(t2 - t1)

    t2 = (5/4)t1

    45 = [50(5/4)t1 - 50t1]/(1/2)

    90 = (1/4)(50t1)

    360 = 50t1

    7.2 = t1

    d1 = 50(7.2) = 360

    Therefore, the distance is 360 miles.


  10. time=distance x rate

    I'd guess I'd set it up like this, I'll keep trying to solve if there isn't an answer.

    t=2d*45

    (d*50)+(d*40)=1.5t

    (d*50)+(d*40)=1.5(2d*45)

    d*50+d*40=3d+67.5

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