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Math Help - Fractions

  1. #1
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    Fractions

    Hi.
    Can you help me solve this question?
    Thanks in advance!


    One layer of tinting material on a window cuts out 1/5 of the sun's UV rays.
    (a) What fraction would be cut out by using two layers?
    (b) How many layers would be required to cut out at least 9/10 of the sun's UV rays?
    Last edited by mr fantastic; November 30th 2009 at 02:08 AM. Reason: Changed post title
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  2. #2
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    Hello ipokeyou
    Quote Originally Posted by ipokeyou View Post
    Hi.
    Can you help me solve this question?
    Thanks in advance!


    One layer of tinting material on a window cuts out 1/5 of the sun's UV rays.
    (a) What fraction would be cut out by using two layers?
    (b) How many layers would be required to cut out at least 9/10 of the sun's UV rays?

    (a) If one layer cuts out \frac15 of the UV, then it must allow \frac45 through. Of this, \frac45 will be allowed through by a second layer. So, of the original quantity of UV the fraction allowed through by two layers will be:
    \frac45\times\frac45 = \frac{16}{25}
    Therefore 1 - \frac{16}{25}= \frac{9}{25} will have been cut out.

    (b) Similarly, if there are n layers, the fraction allowed through will be \left(\frac45\right)^n. And the fraction cut out will be
    1-\left(\frac45\right)^n
    So we want the smallest integer value of n for which
    1-\left(\frac45\right)^n\ge\frac{9}{10}

    \Rightarrow \left(\frac45\right)^n\le\frac{1}{10}
    If you understand how to take logs of both sides, you can solve this inequality that way. Or you can simply use your calculator until you get the answer. Either way, I think it's 11 layers.

    Grandad
    Last edited by mr fantastic; November 30th 2009 at 02:11 AM.
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