List \(T\) consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is \(S.\) The

[VJesus12]

List \(T\) consist 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 digits is odd is rounded down to the nearest integer. 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

Answer: B

Source: Official Guide

[Brent@GMATPrepNow]

List \(T\) consist 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 digits is odd is rounded down to the nearest integer. 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

Answer: B

Source: Official Guide

Given: 10 of the values must have an EVEN tenths digit, and the remaining 20 values must have an ODD tenths digit.

Let's try to determine the MAXIMUM value of E - S and the MINIMUM value of E - S
Once we do that, we'll know the range of possible values of E - S

IMPORTANT: To make things easier, let's consider decimals in the form 0.something

MAXIMUM value of E - S
In order to MAXIMIZE the value of E - S, we must MINIMIZE the value of S
We can do this by making the decimals with an ODD tenths digit 0.1, and by making the decimals with an EVEN tenths digit 0.01
So List T consists of twenty 0.1's and ten 0.01's
This means S = 20(0.1) + 10(0.01) = 2.1

When we complete our ROUNDING, we get: twenty 0's and ten 1's
So, E = 20(0) + 10(1) = 10

So, the MAXIMUM value of E - S = 10 - 2.1 = 7.9


MINIMUM value of E - S
In order to MINIMIZE the value of E - S, we must MAXIMIZE the value of S
We can do this by making the decimals with an ODD tenths digit 0.99999..., and by making the decimals with an EVEN tenths digit 0.89999...
So List T consists of twenty 0.9999....'s and ten 0.89999....'s
This means S ≈ 20(1) + 10(0.9) ≈ 29
ASIDE: We need not get super crazy about how many 9's we add to our decimals. Let's just look for an APPROXIMATE value.

When we complete our ROUNDING, we get: twenty 0's and ten 1's
So, E = 20(0) + 10(1) = 10

So, the MINIMUM value of E - S = 10 - 29 = -19

So, the value of E - S can range from (approximately) -19 to 7.9

Since -16 and 6 fall within this range, the correct answer is B

Cheers,
Brent

[swerve]

List \(T\) consist 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 digits is odd is rounded down to the nearest integer. 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

Answer: B

Source: Official Guide

We have whole numbers in the answer choices, so we don't need to go through all the \(3,0000001\) cases. We limit here to the tenth!

Round Up -Even
Max: \(3,2 \Rightarrow 4\) so we get \(0,8\cdot 10=8\)
Min: \(3,8 \Rightarrow 4\) so we get \(0,2\cdot 10=2\)

Round down - Odd
Max: \(3,9 \Rightarrow 3\) so we get \(-0,9\cdot 20=-18\)
Min: \(3,1 \Rightarrow 3\) so we get \(-0,1\cdot 20=-2\)

Now we can manipulate those numbers:
I. \(-16 \Rightarrow -18 + 2 = -16\) OK \(\color{green}\checkmark\)
II. \(6 \Rightarrow -2 + 8 = 6\) OK \(\color{green}\checkmark\)
III. \(10 X\) You can not get \(10\) because the max positive Value is \(8.\)

Therefore, the correct answer is B