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Homework Help: Word problem.

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that y is directly proportional to the cube of x and inversly proportional to the square of z, if x decreases by 23.3% and z increases by 45.3%, by what factor does y change?

    2. Relevant equations
    Couldn't make a equation but I tried like this:

    3. The attempt at a solution
    I couldn't get the question and where to start adding variables into the equation.
    I think the equation is wrong to me though. Someone help?
    (am I in the right thread for this question?)
    Also my English is sort of weak so please make statement simple as possible. :)
  2. jcsd
  3. Mar 24, 2010 #2


    User Avatar
    Homework Helper

    If c is proportional to a, and proportional to b, then

    c∝a and c∝b ⇒ c∝ab

    Try remaking your equation now.
  4. Mar 24, 2010 #3


    Staff: Mentor

    Start with y = Kx3/z2.
  5. Mar 24, 2010 #4
    Is the answer 21.37%?
  6. Mar 24, 2010 #5
    umm so I did it as y=Kx3/z2
    and so I got the value : y=K12649/2052.09.
    So do I replace the "y" with -23.3% and +45.3% then solve for "K"?
  7. Mar 24, 2010 #6
    umm I think so because the answer they left was in decimals so real answer is 0.214

    I don't what they're saying from the question "if x decreases by 23.3% and z increases by 45.3%, by what factor does y change?"
    Last edited: Mar 24, 2010
  8. Mar 24, 2010 #7
    x decreases BY 23.3% means x=100-23.3=76.6% So you have to sub. x=76.6. But if they had given decreases to 23.3% then x=23.3

    substituting this in the equation with k=1 we get x=21.37 in the answer which you have mentioned they have divided it by 100...
  9. Mar 24, 2010 #8
    hmmm I don't know how to get rid of "k" when I have y=K76.73/145.52
  10. Mar 24, 2010 #9


    Staff: Mentor

    Look at the ratio of the new y value to the old y value. In the ratio, the K's cancel, as do the x's and z's.

    BTW, 100-23.3[itex]\neq[/itex]76.6.
  11. Mar 24, 2010 #10
    Oops...Sorry Sir....
    its 76.7...
  12. Mar 24, 2010 #11
    yeah...Mark Sir is correct....

    Eqn(1) y=kx3/y2

    Eqn(2) y'=k(76.7)3/(145.5)2

    Eqn(2)/Eqn(1)=> y'=21.37y
  13. Mar 25, 2010 #12
    I think I got it. Thanks for help people :D
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