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Increasing Bacteria - Official Guide, #81, p. 163

Expert replies
by Ryan Ziemba » Sat Jul 10, 2010 7:16 pm
I'm not completely sure how to express what it is about this question
that I don't understand, but any insight beyond what the book provides
would be helpful.

I initially thought to approach this problem as one of exponential
growth simply because it deals with bacteria, but clearly the solution
deals with some other function of 'x'.

First off, I'm not familiar with the standard convention of writing
out function equations. Is there a typical formula or strategy I can
rely on?

The book explains that from the table we get 10.0(f)=x, but I'm
unclear how they derive that and then also how we can simply replace x
with 10.0 if the equation states that it is 10f that equals x, not 10
= x.

Any help would be much appreciated.
    Join the discussion
    Source: — Problem Solving |

    by amising6 » Sat Jul 10, 2010 7:32 pm
    dude please post the question also
    Ideation without execution is delusion
    Join the discussion

    by barcebal » Sat Jul 10, 2010 11:04 pm
    This question is very similar to compounded interest. I suggest that you review compounded interest as I think that it will help you understand what is being asked.

    In regards to setting up the equation.

    I

    1:00 time initial bacteria + the growth from 1:00 to 4:00 = 4:00 bacteria level.
    OR 10 + 10*(growth rate)= x

    growth rate= (x-10)/10

    You have two unknown variables which means you are going to set up another equation usually. Keep going.

    II

    4:00 bacteria level + the growth from 4:00 to 7:00 PM = 7:00 bacteria level.

    OR

    x + x*growth rate=14.4

    Now plug in the growth rate solved for in part I.

    x +x((x-10)/10)=14.4

    10x/10 + (x^2-10x)/10=14.4

    x^2/10=14.4

    x^2=144

    x=12

    OTHER METHOD FOR SOLVING

    If you know that the growth rate is similar to compound interest you KNOW that x cannot be the average of the 1:00 PM and 7:00 PM just because 4:00 (the time for x) is in the middle. The middle of the growth time will always be less than half of the total growth since the growth compounds with more time.

    In other words. The total range of growth is 2.4 grams. This means that LESS THAN 1.2 grams of growth occurred during the first half.

    With process of elimination you now know that C, D, E are out!

    Now just plug. Use option A first since it is easiest.

    By what percent (factor, according to the verbage in the question) did the bacteria growth from one to four PM assuming A is correct.

    20% (10*.2=2 grams of growth and a total of 12 grams of bacteria).

    Now multiply 12 grams by 20% again. Next amount of growth is 1.4 grams, which when added to 12 grams is 14.4 grams.

    Option A must be correct and no need to check B.
    Join the discussion