Sunday, May 17, 2009

I hate words problems - Anyone offer help?

This is way off topic and is something I would normally post on my personal blog. Frankly, more people read this one, so I am asking for help. I do not want to know the answer, but rather how to work the following problem (actually, I've edited it so that it is not the actual problem):

A cargo plane flew to a destination and back. It took one hour less time to get there than it did to get back. The average speed on the trip there was 350 mph. The average speed on the way back was 325 mph. How many hours did the trip take to get there?

As I said, I don't want to actual answer, but rather the "how do I solve this?" I am generally pretty good with the math homework my son brings my way, but word problems (or whatever they call them these days) drive me up the freakin wall.

I have no idea what I can offer in the way of a prize for helping, but I will come up with a game-used, auto, something or other....

Of course, I have to know by Monday night. Figgers. And THANKS to anyone that can explain this in English!

10 comments:

  1. Ok, I have no idea how to explain this, but I'm going to give it a shot.

    We are looking for the time in hours it took for the plane to get there. Distance (D) equals rate (R) times time (T). We know the rate and we know that the time it took to get there is an hour less than it took to get back.

    D = 350 * T = 325 * (T+1)

    We don't need distance, we need time.

    T = D / 350
    T +1 = D / 325

    Ok, here's where I forget my algebra formulas and start to cheat. I found the greatest common divisor for the rate. Both speeds are divisible by 25.

    350 / 25 = 14
    325 / 25 = 13

    There's some kind of algebraic rule that allows you to flip the denominator to find the answer, but I have no clue what the formula is. If you multiply 350 by 13 and 325 by 14 you'll get the same number which would be the distance. So basically I think I can give you the answer, but I can't give you the exact formula for getting there. I'm more the engineer type, I don't give a crap about the theory as long as it works.

    Hope that helps, don't worry about any prize, I just got a package from you anyway. Thanks!

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  2. Okay, but the first "=" in your equation is supposed to be a "+" right? So, my formula is:

    D = 350T + 325(T+1)

    So, once I have the cross-multiplied distance, substitute that back in for D and solve for T... Sound good?

    I'm glad you got your package! Hope you need some of the things in there....

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  3. You don't have to find a commen divisor just setit up like this.

    350T=D, 325(T+1)=D, so

    350T=325(T+1)

    solve for T

    T=13

    so

    350(13)=D and at the same time
    325(13+1)=D

    Now take the real problem out of the book and try that same formula.

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  4. I missed my chance to use my math teachering skills on a blog. AdamE set up the problem correctly and got the problem right. He left out a few steps about distributive property and such but you've got to contribute something to it, right?.

    Go math!!!

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  5. I stumble over math word problems. If you like them, I've got a book to suggest....

    The Man Who Counted by Malba Tahan

    Google Books linkAmazon.com link

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  6. Ah! Okay, I see how both solutions can get you there... Math, like most computer applications, provides more than one way to skin the ca... er, uh, find a solution.

    Thanks to everyone for your help!!

    Any/all responses before 7:00am CST on Monday morning get a prize!! That's just the way I roll...

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  7. i am in agreement with Adam and others: 350 T = 325 (T + 1)
    using the distributive property and tehn subtraction you get 25T = 325 then division T = 13 and T = 1 = 1
    In "english" if both ditances are the same then the time multilplied by the rate (speed) must equal each other; thereby saying D = 350 T AND D = 325 (T +1) and if D equals both the transitive property says they equal each other as well.
    A very popular style question especially for state standardized tests and college entrance exams.
    Have fun . . .

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  8. Looks like this problem has already been solved. The crazy thing about algebra is that often in real life we use it in our heads and have no idea that we are doing so.

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  9. This is why I'm a support tech and not a programmer :)

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  10. I remember the first time I was exposed to a word problem that had no answer... hence the correct being "can non solve." Every time I had trouble with a problem, this became my answer...

    So my best guess here is "Can not solve." Heh...

    ReplyDelete