if a<b<c<d<e<110 and the average (arithmetic mean) of a,b,c,d,e is 100 what is the lease possible value of a

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;

To determine the least possible value of 'a,' we need to consider the given information.

Given:
a < b < c < d < e < 110
Average (arithmetic mean) of a, b, c, d, e = 100

Since the numbers are in increasing order, and the average is closer to a than to any other number, we can determine the least possible value of 'a' by finding the lowest value among a, b, c, d, e that is still less than or equal to 100.

Since we know that the average of a, b, c, d, e is 100, the sum of these numbers would be 100 multiplied by 5 because there are 5 numbers in total. So the sum of a, b, c, d, e is 500.

Now, let's consider the lowest possible value for e, which is 109, as the maximum value is given as 110. We can find the remaining values using the given information.

e = 109
d < e, so the maximum value of d is one less than e: d = 108
c < d, so the maximum value of c is one less than d: c = 107
b < c, so the maximum value of b is one less than c: b = 106
a < b, so the maximum value of a is one less than b: a = 105

So, the least possible value of 'a' is 105 when the given conditions are met.