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Thread: Find Rate Against Wind

  1. #1
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    Find Rate Against Wind

    To train for the bicycle portion of the race, Jerri rides 24 miles out a straight road, then turns around and rides 24 miles back. The trip is out against the wind, whereas the trip back is with the wind. If she rides 10mph faster with the wind then she does against the wind, and the complete trip out and back takes 2 hours, how fast does she ride when traveling against the wind?

    I know this problem is a distance problem involving D = rt.

    With the wind:

    rate = x + 10

    Time = 1 hour

    Distance = 24 miles

    Against the wind:

    rate = x

    Time = 1 hour

    Distance = 24 miles

    Here is my equation set up:

    x + (x + 10) = 48

    Is this correct so far?
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  2. #2
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    Re: Find Rate Against Wind

    No, it's not. If x add x+ 10 are speeds then their sum is a speed, not the total distance. I notice that you say "Time = 1 hour" so I presume that you mean that x(1)+ (x+ 10)(1)= 48 which would be correct units. But the "1 hour" cannot be correct! Certainly there is nothing in the problem that says the time taken on each leg is 1 hour. Further, if you are going the same distance at two different speeds, they cannot take the same time.

    That is your error- the problem says that the total time taken is two hours and you incorrectly assume that each leg takes half of that time.

    Instead, let "t" be the time, in hours, taken on the first leg, where you are going against the wind. Then the time on the second leg is 2- t hours. Using the "distance is time times speed" that is t(x+ 10)= 24 and (2- t)x= 24. Solve those two equations for x and t.
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  3. #3
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    Re: Find Rate Against Wind

    I find these problems easier to do if you set up a table:

    ............................Speed...........Distan ce.................Time
    Against wind.............x..................24............ ...........24/x
    With wind..............x+10.............. 24....................24/(x+10)

    The table is set up in the following way:

    Let x be the speed in mph when travelling against the wind (this is what you are required to find).
    Therefore x+10 is the speed with the wind.

    Distance in both cases is the same (24 miles).

    Time = distance/speed

    Now you know the total time is 2 hours, so \frac{24}{x} + \frac{24}{x+10} =2.

    Then solve for x.
    Last edited by Debsta; Jan 10th 2017 at 01:09 PM.
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  4. #4
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    Re: Find Rate Against Wind

    Always a good idea to "make one up" yourself in order
    to understand what's going on.
    Make the regular speed 15, wind speed 5, distance 60;
    then speed against the wind = 10, with the wind = 20.

    @10................60...................>6 hr
    3 hr<....................................@20
    Total of 9 hours.

    Easy to jump to the conclusion that if there was no wind,
    the trip would also take 9 hours.....surprise: NO!

    @15.................60......................>4 hr
    4 hr<........................................@15
    Total of 8 hours !

    Get my drift? (or should I say "my wind")
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  5. #5
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    Re: Find Rate Against Wind

    Quote Originally Posted by HallsofIvy View Post
    No, it's not. If x add x+ 10 are speeds then their sum is a speed, not the total distance. I notice that you say "Time = 1 hour" so I presume that you mean that x(1)+ (x+ 10)(1)= 48 which would be correct units. But the "1 hour" cannot be correct! Certainly there is nothing in the problem that says the time taken on each leg is 1 hour. Further, if you are going the same distance at two different speeds, they cannot take the same time.

    That is your error- the problem says that the total time taken is two hours and you incorrectly assume that each leg takes half of that time.

    Instead, let "t" be the time, in hours, taken on the first leg, where you are going against the wind. Then the time on the second leg is 2- t hours. Using the "distance is time times speed" that is t(x+ 10)= 24 and (2- t)x= 24. Solve those two equations for x and t.
    Nicely explained. I did assume that two hours should be divided in two-one hour times. Setting up the problem equation for word problems has been a challenge for me for many years. Many students hate word problems for that specific reason.
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  6. #6
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    Re: Find Rate Against Wind

    Quote Originally Posted by DenisB View Post
    Always a good idea to "make one up" yourself in order
    to understand what's going on.
    Make the regular speed 15, wind speed 5, distance 60;
    then speed against the wind = 10, with the wind = 20.

    @10................60...................>6 hr
    3 hr<....................................@20
    Total of 9 hours.

    Easy to jump to the conclusion that if there was no wind,
    the trip would also take 9 hours.....surprise: NO!

    @15.................60......................>4 hr
    4 hr<........................................@15
    Total of 8 hours !

    Get my drift? (or should I say "my wind")
    What is the best way to master word problems? I think solving word problems is the most important math skill to learn. What do you say?
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  7. #7
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    Re: Find Rate Against Wind

    Well...got no idea WHY they're called word problems;
    basically:
    teacher asks Joe: what's 3 + 5?
    another teacher asks Jim: if you have 3 apples and 5 bananas,
    how many fruits have you got?
    2nd one is a word problem?!!

    That's over-simplifying, I agree.

    Take your problem; could be worded:
    travels from A to B at v-5 mph,
    then from B to A at v+5 mph.
    WHY bring in the wind!
    Thanks from mathdad1965
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  8. #8
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    Re: Find Rate Against Wind

    Quote Originally Posted by DenisB View Post
    Well...got no idea WHY they're called word problems;
    basically:
    teacher asks Joe: what's 3 + 5?
    another teacher asks Jim: if you have 3 apples and 5 bananas,
    how many fruits have you got?
    2nd one is a word problem?!!

    That's over-simplifying, I agree.

    Take your problem; could be worded:
    travels from A to B at v-5 mph,
    then from B to A at v+5 mph.
    WHY bring in the wind!
    To apply with the wind or against the wind, with the current or against the current involves physics.
    Last edited by USNAVY; Jan 11th 2017 at 09:56 AM.
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  9. #9
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    Re: Find Rate Against Wind

    Quote Originally Posted by Debsta View Post
    I find these problems easier to do if you set up a table:

    ............................Speed...........Distan ce.................Time
    Against wind.............x..................24............ ...........24/x
    With wind..............x+10.............. 24....................24/(x+10)

    The table is set up in the following way:

    Let x be the speed in mph when travelling against the wind (this is what you are required to find).
    Therefore x+10 is the speed with the wind.

    Distance in both cases is the same (24 miles).

    Time = distance/speed

    Now you know the total time is 2 hours, so \frac{24}{x} + \frac{24}{x+10} =2.

    Then solve for x.
    Nice set up. I must learn how to do this with similar problems.
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  10. #10
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    Re: Find Rate Against Wind

    Quote Originally Posted by mathdad1965 View Post
    Nicely explained. I did assume that two hours should be divided in two-one hour times. Setting up the problem equation for word problems has been a challenge for me for many years. Many students hate word problems for that specific reason.
    Yes, they require thinking rather than just memorizing formulas.
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  11. #11
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    Re: Find Rate Against Wind

    Quote Originally Posted by HallsofIvy View Post
    Yes, they require thinking rather than just memorizing formulas.
    This is the most important math skill to know. I recall in 2006, I took a test to work as a Financial Asvisor for Bank One. Most of the test covered word problems.

    I know that working for Bank One would have surely changed my life for the better, financially speaking. It was then when I realized that my life was going to be a rugged road. I became fanatical about math and started going through every textbook question I could find only to find myself getting stuck when facing word problems.
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  12. #12
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    Re: Find Rate Against Wind

    Quote Originally Posted by mathdad1965 View Post
    I recall in 2006, I took a test to work as a Financial Asvisor for Bank One.
    Is that right?!
    I had only one job: 40 years (1961 to 2001) with CS-Coop Credit Society Ltd.,
    now the Alterna Community Savings and Credit Union.
    Ended up as Financial Analyst...so had to practice my maths!
    Thanks from mathdad1965
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  13. #13
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    Re: Find Rate Against Wind

    Quote Originally Posted by DenisB View Post
    Is that right?!
    I had only one job: 40 years (1961 to 2001) with CS-Coop Credit Society Ltd.,
    now the Alterna Community Savings and Credit Union.
    Ended up as Financial Analyst...so had to practice my maths!
    Congratulations! I love math. I really want to learn as much as possible. I do not care what my age is, honestly.

    Yes, my story concerning Bank One is true. I really tried to pass but was not able to. The bank representative told me that my score was the lowest he had seen in many years. I walked out of the bank depressed and feeling sorry for myself.
    Thanks from DenisB
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