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Math Help - Take a Hike

  1. #1
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    Take a Hike

    Some friends on a hike leave at 9 a.m. and return at 2 p.m. They spend one-quarter of the total distance walking uphill, one-half walking on level ground, and one-quarter walking downhill. If their speed is 4 mph on level ground, 2 mph uph ill, and 6 mph downhill, approximately how far did they walk?
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    Quote Originally Posted by MathMage89 View Post
    Some friends on a hike leave at 9 a.m. and return at 2 p.m. They spend one-quarter of the total distance walking uphill, one-half walking on level ground, and one-quarter walking downhill. If their speed is 4 mph on level ground, 2 mph uph ill, and 6 mph downhill, approximately how far did they walk?
    d = v*t for each leg of the journey.

    Call tu the time taken to walk uphill, td the time taken to walk downhill, and tl the time taken to walk level.

    Then
    (1/2)d = 4tl ==> tl = (1/8)d
    (1/4)d = 2tu ==> tu = (1/8)d
    (1/4)d = 6td ==> td = (1/24)d

    and
    tl + tu + td = 5 (hours)

    (1/8)d + (1/8)d + (1/24)d = 5

    (7/24)d = 5

    d = 5*(24/7) = 17.1429 (or so) miles.

    -Dan
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  3. #3
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    Take a hike

    Ok you have a hike that lasts 5 hours (9am-2pm)

    1/4 of that time uphill at 2 miles per hour (1/4x)
    1/2 of that time flat at 4 miles per hour (1/2x)
    1/4 of that time downhill at 6 miles per hour (1/4x)

    1/4x + 1/2x + 1/4x = 5 hours

    1/4x = 1.25 hours
    1/2x = 2.5 hours
    1/4x = 1.25 hours

    so 1.25 hours times 2 miles per hour = 2.5 miles uphill
    2.5 hours times 4 miles per hour = 10 miles on flat ground
    1.25 hours down times 6 miles per hour = 7.5 miles downhill

    add all three together to get a walk that was 20 miles in length.
    Last edited by rose234; February 27th 2007 at 03:17 AM. Reason: I did the math wrong
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    Super Member Aryth's Avatar
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    Quote Originally Posted by rose234 View Post
    Ok you have a hike that lasts 5 hours (9am-2pm)

    1/4 of that time uphill at 2 miles per hour (1/4x)
    1/2 of that time flat at 4 miles per hour (1/2x)
    1/4 of that time downhill at 6 miles per hour (1/4x)

    1/4x + 1/2x + 1/4x = 6 hours

    1/4x = 2.25 hours
    1/2x = 4.5 hours
    1/4x = 2.25 hours

    so 2.25 hours times 2 miles per hour = 4.5 miles uphill
    2.5 hours times 4 miles per hour = 10 miles on flat ground
    2.25 hours down times 6 miles per hour = 13.5 miles downhill

    add all three together to get a walk that was 28 miles in length.
    There's something seriously wrong with this calculation. For one it says 5 hours at the top and then 6 hours shortly after. Then it says 4.5 hours for one and then 2.5 a little while afterwards and 1/4 of 5 is 1.25 not 2.25. If you correct these errors, then you get:

    1/4x + 1/2x + 1/4x = 5 hours

    1/4x = 1.25 hours
    1/2x = 2.5 hours
    1/4x = 1.25 hours

    1.25 x 2 = 2.5
    2.5 x 4 = 10
    1.25 x 6 = 7.5

    2.5 + 10 + 7.5 = 19.5 miles.
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  5. #5
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    Quote Originally Posted by MathMage89 View Post
    Some friends on a hike leave at 9 a.m. and return at 2 p.m. They spend one-quarter of the total distance walking uphill, one-half walking on level ground, and one-quarter walking downhill. If their speed is 4 mph on level ground, 2 mph uph ill, and 6 mph downhill, approximately how far did they walk?
    They were walking for 5 hours.
    So they walked for 5/4 hours uphill covering a distance of (5/4)*2 uphill,
    So they walked for 5/2 hours on level covering a distance of (5/2)*4 on level
    So they walked for 5/4 hours down hill covering a distance of (5/4)*6 down hill.

    Total distance = (5/4)*2 + (5/2)*4 + (5/4)*6 = 20 miles.

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MathMage89 View Post
    Some friends on a hike leave at 9 a.m. and return at 2 p.m. They spend one-quarter of the total distance walking uphill, one-half walking on level ground, and one-quarter walking downhill. If their speed is 4 mph on level ground, 2 mph uph ill, and 6 mph downhill, approximately how far did they walk?
    I would just like to point out to rose234, Aryth, and CaptainBlack that the OP states that we are talking about fractions of the distance walked, not fractions of the time taken.

    -Dan
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