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April 20, 2014

April 20, 2014

Posted by **Anonymous** on Tuesday, August 7, 2012 at 6:41pm.

- averages -
**Anonymous**, Tuesday, August 7, 2012 at 8:00pmwell, a+b+c+d+e<500

to minimize a, e=109.999, d=109.998, c=109.997, b=109.996 you get the drift, so

a=500-about 110*5=550, so a could be a minimum of about -50, or if you did the math above, about -49.99

- averages -
**Ana**, Saturday, September 8, 2012 at 6:29amThe answer is 70.

First off if we want to know the LEAST possible value for a, then b, c, d, e have to be the MAXIMUM value. In this case it would be e=109, d=108, c=107, and b=106.

Now that we have values for all the variables except a, we can solve for a. To solve for a, we use the average that is given. Because the average of the 5 variables is 100, we know that the addition of all 5 variables is 500.

(a+b+c+d+e)/5=100

Therefore, a+b+c+d+e =500

Now that we know what b,c,d,e is, we can solve for a.

a=500-b-c-d-e

a=500-106-107-108-109

a=70

- averages -
**Sanket**, Sunday, February 3, 2013 at 12:27amThis question appears in power prep 2 CD. Actually the question is as follows:

a<=b<=c<=d<=e<=110;

for a to be minimum, b,c,d,e must be maximum. Assume b,c,d and e to be 110 (Yes, the sum has less-than-or-equal-to signs) that way you get the answer as:

a+440=100*5;

a=60;

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