Problem Solving

Hi ,

I was wondering if there is an easy way to solve this question, the OG 13 explanation is long and not very clear,

here is the question

List T consists of 30 positive decimals, none of which is an integer, and the sum of 30 decimals is S, the estimated sum of the 30 decimals E is defined as follows. each decimal in T whose tenths digit is even is rounded up to the nearest integer, each integer in decimal T whose tenth digit is odd is rounded down to the nearest integer, E is sum of resulting integers, . IF 1/3 of decimals in T have tenths digits that are even, which of follwoing is possible value of E - S ?


-16
6
10

A I only
B I and II
c I and III
D II and III
E I , II and III

List T consists of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digit is odd is rounded down to the nearest integer; E is the sum of the resulting integers. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E - S?

I. -16
II. 6
III. 10

A) I
B) I and II
C) I and III
D) II and III only
E) I, II and III

Make the problem CONCRETE by plugging in easy values.
10 of the values must have a tenths digit that is EVEN, while the other 20 values must have a tenths digit that is ODD.
To make the math easy, let's not consider decimals beyond the tenths place.
Try to MAXIMIZE E-S and MINIMIZE E-S.

E-S MAXIMIZED:
To MAXIMIZE the value of E-S, we must MINIMIZE the value of S.
To minimize S, we must ROUND UP the even decimals as MUCH as possible (from .2 to the next highest integer) and ROUND DOWN the odd decimals as LITTLE as possible (from .1 to the next smallest integer).
Let S = 10(.2) + 20(.1) = 4.
In E, .2 is rounded up to 1 and .1 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MAXIMUM possible value of E-S = 10-4 = 6.

E-S MINIMIZED:
To MINIMIZE the value of E-S, we must MAXIMIZE the value of S.
To maximize S, we must ROUND UP the even decimals as LITTLE as possible (from .8 to the next highest integer) and ROUND DOWN the odd decimals as MUCH as possible (from .9 to the next smallest integer).
Let S = 10(.8) + 20(.9) = 26.
In E, .8 is rounded up to 1 and .9 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MINIMUM possible value of E-S = 10-26 = -16.

Since the MAXIMUM difference is 6 and the MINIMUM difference is -16, only I and II are possible values of E-S.

The correct answer is B.

Thanks alot, this is helpful