sanju09 wrote:List T consists of 30 positive decimals none of which is an integer, and 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 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


By GMAC

Hi Sanju,
I got Answer as B. And this is how I solved it,

We Know: 30 Decimals. out of which 10 are Even and 20 are odd.
Actual Sum is S and Estimated Sum is E

To Find: E-S ?

For easy Calculation I assumed all the decimals are in the range between 0 and 1
So, we will get E = 10
S will be some positive number.

Definitely E-S will not be 10
now we can say Option II is possible

Hence eliminated C,D,E.

Similarly if we assume decimals between -1 and 0
Option I is possible.

So, both Option I and Option II is possible
Hence answer is B.

Regards,
Uva.

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.