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An equation for an algebra word problem

  1. May 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose that b is 20% more than a, c is 25% more than b, and d is k% less than c. Find k such that a=d




    2. Relevant equations
    a=d; b=1.2a; c=1.25b=1.5a


    3. The attempt at a solution
    1. a=##\frac{c}{15 }##=d=##\frac{100c-k}{100}##
    2. By multiplication of extremes and means we get 100c=150c-15.k
    3. 50c=1.5k
    4. Again simplifying k=33.3c

    I have tried above solution but I think something is incorrect because k should be a number not a ratio in terms of c.
    I don't latex well enough and it seems that step 1 is not seeming properly, too small symbols.
    Source: Algebra and Trigonometry by Keedy Bittinger.
    Thank you.
     
  2. jcsd
  3. May 8, 2017 #2

    SammyS

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    You have a typo and an error in multi-equation #1.

    For one, ##\ \displaystyle a=\frac{c}{1.5} \ ##. (Missed the decimal point.)

    More importantly the last part is in error.
    ##\ \displaystyle \frac{100c-k}{100} \ ##​
    is incorrect. ##\ k\ ## must also multiply ##\ c\ ##.
     
  4. May 8, 2017 #3
    But if ##d## is ##k%## less than ##c##, does not this mean ##\frac{c}-k}{100} ## and isn't it ##\frac{100c-k}{100}##?

    And would you please explain why percent sign not appear in this code?
    upload_2017-5-8_18-11-8.png



    Thank you.
     

    Attached Files:

    Last edited: May 8, 2017
  5. May 8, 2017 #4

    Ray Vickson

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    You need
    $$d =\left( 1 - \frac{k}{100} \right) c$$
     
  6. May 8, 2017 #5
    $$d =\left( 1 - \frac{k}{100} \right) c$$=##\frac{c}{1.5}##

    ##50c=15ck##

    and k=3.333

    Would you please write above in the proper latex form?

    Thank you.
     
  7. May 8, 2017 #6
    @mech-eng, I suggest you check your answer of ##k = 3.333\%## against your initial set of equations ##a=d; b=1.2a; c=1.25b##.

    Also, although percentages proper are expressed as ##x/100##, it's easier & faster during calculation to assume 1 = 100%; thus instead of $$d = \left(1 - \frac{k}{100}\right) c$$ you can work more simply with ##d = (1 - k)c##. For that matter you can substitute a temporary value for that expression and do the subtraction later on.
     
    Last edited: May 8, 2017
  8. May 8, 2017 #7

    SammyS

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    That looks proper to me.

    Mathematically, it's equivalent to

    ##\ \displaystyle d = \left(\frac{100-k}{100}\right) c ##
     
  9. May 8, 2017 #8
    I think the OP was referring to the part of his comment that wasn't already formatted in LaTeX.
     
    Last edited: May 8, 2017
  10. May 8, 2017 #9
    ##\ \displaystyle d = \left(\frac{100-k}{100}\right) c=\frac c {15} ##

    I added the right side but is it proper. I didn't used symbols such as \left and \right.
    The following is what I added there. But it is not the same when it is on the right side. Would you please explain this?
    ## \frac c {15}##

    Thank you.
     
  11. May 8, 2017 #10

    SammyS

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    Proper in what way?

    You dropped the decimal point from 1.5 again, otherwise, it looks right math-wise .

    By the way: for % symbol in LaTeX, use \ suffix as in \%.

    ## 7\% ## or ##k\%##
     
  12. May 8, 2017 #11
    Sorry. Question is solved. But look the Latex. Their size of the c/15 parts are different.

    Thank you.
     
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