List T consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The [/u]esimated 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 only
(B) I and II only
(C) I and III only
(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.
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