Posted by Anonymous on Tuesday, August 7, 2012 at 6:41pm.
well, 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
The 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
This 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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