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Sunday, December 9, 2007

Mathcounts Help Needed

Okay, my math kid has been working on his practice sheets for the upcoming Mathcounts competition next year. He seems to be doing great with the practice problems, but the warm-ups sometimes seem harder.

I'd love to help, but as many of you know, I am the resident Math Phobe of KTM. With the husband out of town, you are my only hope to helping him with this one particular problem. I have a feeling it is something really obvious.

Here it is:

"The sum of two numbers is 32, and the product of these two numbers is 48. What is the sum of the reciprocals of the two numbers? Express your answer as a common fraction."

I have the answer sheet, but no explanation on the best way to approach this.

Any and all help would be appreciated. Speak slowly, though....although the kid will probably understand what you're saying.

SusanS
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19 comments:

  1. SusanSDecember 9, 2007 at 1:49 PM

    I have no idea why it's so sprawled out, but..whatever. It's posted.

    ReplyDelete
  2. AnonymousDecember 9, 2007 at 2:14 PM

    I would start by trying to figure out the two numbers:

    Sum = 32 so ...

    X + Y = 32

    Product = 48 so ...

    X * Y = 48

    At this point I am inclined to try cheating because there aren't very many ways to multiply and get 48.

    48 breaks down into:

    2 * 2 * 2 * 2 * 3

    So my choices are:
    48 * 1
    24 * 2
    16 * 3
    12 * 4
    8 * 6
    6 * 8

    etc.

    The first observation is that I can't do this with integers (possibly because I've misunderstood the problem).

    I'll try one more thing:

    X + Y = 32 so X = (32 - Y)

    Then:

    X * Y => (32 - Y) * Y = 48
    32*Y - Y^2 = 48

    Now we can try polynomial factoring:

    Y^2 - 32*Y + 48 = 0

    But I can't find any integral factors :-(

    The solution will have to be of the form:

    (Y - a) * (Y - b)

    And 'a' and 'b' need to be factors of 48 ... but ... see the table above. Getting this equation to work seems problematic.

    Doing it with non-integers seems problematic because I'll need to add two 1/2s together to get a whole, but these won't multiply out to a whole number. Same for 1/3 and 2/3, etc.

    At this point, I'm stumped. Are the numbers base-10?

    -Mark Roulo

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  3. AnonymousDecember 9, 2007 at 2:21 PM

    We have two pieces of information

    x + y =32
    xy = 48

    Note that neither x nor y can be zero

    We have to find 1/x + 1/y

    Put these fractions onto their common denominator xy

    1/x = y/xy
    1/y = x/xy

    So 1/x + 1/y = y/xy + x/xy
    = (y+x)/xy
    = 32/48
    = 2/3


    Hope that makes it clear - the important point is that you never need to solve for x and y individually

    ReplyDelete
  4. SusanSDecember 9, 2007 at 2:26 PM

    Thanks Mark,

    I am thrilled to say that I did the factoring with the primes. The kid started out with the equations. We were stumped.

    The answer "2/3" on the answer sheet.

    ReplyDelete
  5. SusanSDecember 9, 2007 at 2:27 PM

    I'm always up for, "It's a typo," as the answer.

    ReplyDelete
  6. AnonymousDecember 9, 2007 at 2:28 PM

    PS You can also do it the clunky way by substitution and solving the resulting quadratic equation

    y = 32-x

    so xy = x(32-x) = 48

    so x^2 -32x + 48 = 0

    Work out x and y with the usual quadratic equation formula and then you can calculate the reciprocals and their sum.

    But this is hideous compared with the elementary manipulation of fractions in the first solution.

    ReplyDelete
  7. AnonymousDecember 9, 2007 at 2:28 PM

    "" the important point is that you never need to solve for x and y individually..."

    Does it matter that there don't seem to be any solutions for X and Y? Or is that not relevant here?

    -Mark R.

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  8. SusanSDecember 9, 2007 at 2:39 PM

    Amanda,

    Aha. That's why we got off. We were were so focused on finding x and y.

    ReplyDelete
  9. AnonymousDecember 9, 2007 at 2:46 PM

    Happy to help.

    x and y are 16 + 4 root 13 and 16 - 4 root 13 - not nice numbers to invert!

    ReplyDelete
  10. AnonymousDecember 9, 2007 at 2:48 PM

    "x and y are 16 + 4 root 13 and 16 - 4 root 13 - not nice numbers to invert!"

    Thanks!

    Thus also why the key is to avoid solving for them :-)

    -Mark Roulo

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  11. Catherine JohnsonDecember 9, 2007 at 3:13 PM

    Meanwhile I'm sitting here wanting to know why the difference of two squares is called the difference of two squares....

    ReplyDelete
  12. AnonymousDecember 9, 2007 at 4:37 PM

    I can kind of see now why it was used as a "warm-up." At first, I didn't get it, the actual problems seem easier.

    I might have to mess with my husband. Hopefully, he misses it and then I'll just pretend to figure it out. That would kill him.

    ReplyDelete
  13. AnonymousDecember 10, 2007 at 6:08 AM

    Catherine

    Look at it geometrically

    Draw your square with long side a
    Then inside at one corner put your smaller square with side b, extending its sides to create two rectangles with sides of length b and a-b and a square with sides of length (a-b)

    Then you can easily see that

    a^2 - b^2 = a^2 - 2ab + b^2

    ReplyDelete
  14. Catherine JohnsonDecember 10, 2007 at 7:15 AM

    I will do that!

    But FIRST I will write AT LEAST ONE PAGE about cows!

    ReplyDelete
  15. Catherine JohnsonDecember 10, 2007 at 7:15 AM

    Susan - sounds like a plan to me.

    ReplyDelete
  16. AnonymousDecember 10, 2007 at 12:28 PM

    Cows?????????????

    ReplyDelete
  17. Barry GarelickDecember 10, 2007 at 2:43 PM

    I might have to mess with my husband. Hopefully, he misses it and then I'll just pretend to figure it out. That would kill him.

    Have plenty of Prozac on hand!

    ReplyDelete
  18. AnonymousJune 30, 2008 at 9:14 AM

    the answer is 2/3 HA HA

    ReplyDelete
  19. AnonymousMarch 26, 2009 at 5:11 PM

    x+y=32
    xy=48
    48=1*48,2*24,3*16,4*12,6*8,8*6,12*
    ,24*2,48*1
    Ask a coach because there seems to be no integer solution.

    ReplyDelete