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Confusing Algebra based question

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data

    If ADD = 9
    BAD = 7
    CAD = 8

    Find ADA .

    2. Relevant equations

    Algebra

    3. The attempt at a solution

    ADDBAD=9*7
    BADCAD=56
    ADABCC= 56*8/9
    B/C = 7/8

    I really don't know how to approach this problem, what must I eliminate and how should I do it ?
     
  2. jcsd
  3. Feb 2, 2013 #2

    Simon Bridge

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    Do we take it that each letter stands for a number, so that ADD=AD2, or are the whole words numbers?
    In which case you have 4 equations and 5 unknowns ... so you won't do it by simultaneous equations.

    I suspect there is some additional information to nail it since I can find quite a few combinations ... i.e.

    put D=3, then A=1, and AD=3 and ADA=3
    C=8/3, and B=7/3.

    put A=9, then D=1, and AD=9 and ADA=81 ... and you can still solve for C and B as above.

    I did notice that, since the order of multiplication dosn't matter, the rows look like:

    D(AD)=9
    C(AD)=8
    B(AD)=7
    A(AD)=?

    spot the pattern? ... well - maybe not.
    Unless the person who set the problem has a reputation for pulling this sort of trick?
     
    Last edited: Feb 2, 2013
  4. Feb 2, 2013 #3
    Can there be more than one correct solution ?
     
  5. Feb 2, 2013 #4

    Simon Bridge

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    Depends on your definition of "correct".
    I have supplied you with two sets of values for each letter which will make the system true.

    if we put X=ADA, then there are infinitely many values for X that will make the system of equations true. Just pick a few and see ... though it is easier to pick values for A and let the rest follow. (You could plot a graph of each of the other letters vs the value of A.)

    if there is supposed to be one correct solution - then there is some extra information you have not supplied to us.
    Where did you get the problem from?
     
  6. Feb 2, 2013 #5
    There is no extra information given, and I am not eliminating the possibility of more than one correct answer.

    Some things, however, trouble me :

    What if I took A=9 and D=1 .

    Then the value of ADA would be 81
    Would that be considered correct too ?

    Why did u state in post no. 2 that there are 5 variables and 4 equations ?
    Was that a typo ? aren't there 3 equations and 4 variables ?
     
  7. Feb 2, 2013 #6

    Ray Vickson

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    You can test A = 9, D = 1 and ADA = 81 yourself.
     
  8. Feb 2, 2013 #7

    Simon Bridge

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    Because I counted them. Didn't you?

    Don't forget to count the unnamed variable that the question wants you to find - helps if you give it a name.
    Define: "correct".
    That combination would satisfy the list of relations for particular values of B and C - does that make it "correct"?
    That there are infinite such combinations could just mean that there are no correct answers couldn't it?
    You will have to make that judgement call because only you know the context of the problem.
    Sometimes the extra information is implied in the question, or in the context in which you got it. i.e. if you got it as homework in a linear algebra course, then the way forward will probably have something to do with the topic just covered in class (or just about to be covered).

    Like I said - it could be a trick.
    We've been assuming that the different letters are separate variables usual to algebra.
    Then ADA=AAD=DAA ... and we expect them to have the usual algebraic relations in each statement. But if you look at the list in post #2, it would be consistent with the relations as written for ADA=6 ... but then that would be a trick.

    There is probably something you have missed that limits the possible values that X can take. Double check.
     
  9. Feb 4, 2013 #8
    Thanks for the clarification
    This Question was in one of my Math tests, and I remember it to be as concise as it is written above (in post no.1).

    I understand well now.
     
  10. Feb 4, 2013 #9

    Simon Bridge

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    Then it was probably part of the test was for you to get the context. The question was supposed to remind you of something in the coursework.
    No worries :) I'd ask the person who set the test, and/or other people who sat it.
     
  11. Feb 5, 2013 #10

    HallsofIvy

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    I have always wondered about the British usage "sat a test". The American usage would be "took a test". Would it count if you did the test standing up? Or isn't that allowed in Britain? On the other hand, in America, you are seldom allowed to "take" the test out of the room!
     
  12. Feb 5, 2013 #11

    Simon Bridge

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    English is a fun language. The UK usage comes from Old English.
    Both forms make sense in their contexts.

    Each kinda speaks to the sense of humor of the culture that originated the usage.
    The UK form comes from "set" in the sense of making ready. "set" and "sit" used to be the same word. So the tester "sets" the test and the testee "sits" it, afterwards, the test is "sat".
    Resonates with the common habit of sitting down to "sit" a test - so it's a pun.
    You can also take, do, complete, etc tests.

    In the US, the tester gives the testee the test, so the testee has to take it from them first, then attempt to complete the test. I understand there is also a special term in the US for when the test is over ... "Partay tahm" or something?
     
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