List T consist of 30 positive decimals, none of which is an integer

Total 30.
the value of E:
1/3 of the decimals in T have a tenths digit that is even so
10 numbers have an even tenths digit
20 numbers have an odd tenths digit.

now for any decimal with a even tenths digits would be 0.2,.04,0.6, 0.8, 1.2, 1.4, 1.6, 1.8 and so on!

for evens we round up! so they'll become 1,1,1,1,2,2,2 and so on!

now for any decimal with a odd tenths digits would be 0.1,0.3,0.5, 0.7, 0.9, 1.1, 1.3, 1.5, 1.7 and so on!

for odds we round down! so they'll become 0,0,0,0,0,1,1,1,1 and so on!


For E :


E = (sum of all 30 integer parts) +10(1)-20(1)
=(sum of all 30 integer parts)-10

or simply 10!
(if you consider all decimals are between 0 and 1 then sum of evens = 10(1) and sum of odds become 20(0)

Now for "S":
The maximum possible value of S occurs when ten numbers have '8' as tenths digit and remaining 20 numbers have '9' as tenths digit.
Smax = (sum of all 30 integer parts) +10(0.8)+20(0.9)
= (sum of all 30 integer parts)+26

or simply .8(10)+0.9(20) = 26


The minimum possible value of S occurs when ten numbers have '2' as tenth digit and remaining 20 numbers have '1' as tenth digit.
Smin = (sum of all 30 integer parts) +10(0.2)+20(0.1)
= (sum of all 30 integer parts)+4

or simple 0.2(10)+0.1(20) = 4

MAX S - E= 26-10 =16
MIN S-E = 4-10= -6

SIMPLE SOLUTION TO THIS ANSWER.

SOLVE FOR E-S !!

LETS TRY TO FIND THE RANGE OF E-S

FIRST MAXIMUM

FOR THIS E SHOULD BE GREATEST POSIBLE ,S SHOULD BE SMALLEST.

E WILL BE HIGH WHEN IT IS ROUNDED UPWARDS E.G 1.01 BECOMES 2 ( GAIN OF 0.99).

S WILL BE SMALL WHEN THE NUMBER IS VERY LOW LIKE 1.01 , NOW

NOW FOR 10 DIGITS SIMILAR WE HAVE ( 0.99 *10) =9.9+ OTHER 20 DIGITS ODD NUMBERS WHICH ARE DOWNGRADED (THESE NEED TO BE VERY LESS) LIKE 1.10 ( .10*20)=2

THEREFORE MAX E-S WILL BE 10-2= 8 ( 10 EVEN TENTH NO. + 20 ODD TENTH NO.)


NOW FOR MIN E-S


AS YOU CAN GUESS E MIN AND S MAX.


LIKE FOR 10 EVEN 10 NO .( 1.89) = 2 NET GAIN OF 0.11 = + 1.1 FOR 10 NO.

FOR 20 ODD TENTH DIGIT NUMBERS (1.99) = 1 = NET DOWN OF = 0.99 , FOR 20 NO. IS -19.80
THEREFORE E-S = +1.1-19.80 = -18.7

WE DONT CALCULATE S FOR BOTH CASES AS WE ARE TAKING NET DIFFERENCE


HOPE IT IS CLEAR !! :please

I made a blunder in this one even after spending a lot of time. Although i reached to get a figure of 18, 8 and 2, I subtracted 8 from 18 and rest, anyone can guess. Anyway, on retrospection, I figured out certain things that I'm sharing here.

First, that I was overwhelmed by the question for what it was about. I unnecessarily made my life miserable by overthinking as I was thought of 30 huge numbers - about the integer part and decimal part. I went into gimmicky approach as I thought integer part would change, impacting massively the result of E - S. And finally that I need the rough values of E and S.
However, the question talks only about the positive decimals. There are 10 decimals with even digit at tenths place that are rounded up and 20 with odd digit at tenths place that are rounded down. As we are looking for E - S, we know that the integer part would not get affected at all. Okay, that's not right perfectly since 0.2 becomes 1.0, changing the integer part.

But on a closer look we can calculate the total approximate change and not bother about the integer part at all. Whatever integer we get as result of either rounding up or rounding down, we are taking care of that by calculating - it not being carry forwarded.
Here's how it is:To know what possible values E - S can have, we need to first get the range of its values for which we need to calculate the maximum and minimum value of E-S.
\((E - S)_{max} = E_{max} - S = Δ_{E_{max}}\) AND
\((E - S)_{min} = E_{min} - S = Δ_{E_{min}}\)
(Note: E-S can have negative values depending on each value)

\(Δ_{E_{max}} = E_{Emax} + E_{Omin}\) - Eqn. 1 AND
\(Δ_{E_{min}} = E_{Emin} + E_{Omax}\) - Eqn. 2

Now, let the 30 decimals are(extreme possibilities)
0.00001, 0.00001 ....... (10 0.00001 values) and 0.9999, 0.9999 .... (20 0.9999 values)
0.00001 round up to 1.0, total Max change becomes ~10*0.9999 = 10 AND 0.9999 becomes 0.0, total Max change becomes ~0.9999*20 = 20
Precisely, the rounding up becomes max. a little < 10 and rounding down becomes max. a little < -20. For ease let's take 10 and -20('-' for rounding down).

0.8999, 0.8999 ..... (10 0.8999 values) and 0.1000, 0.1000 .... (20 0.1000 values)
0.8999 round up to 1.0, total Min change becomes ~10*0.1 = 1 AND 0.1 becomes 0.0, total Min change becomes 0.1*20 = 2
Precisely, the rounding up becomes min. a little > 1 and rounding down becomes min. a little = 2. For ease let's take 1 and -2('-' for rounding down).

Thus,
\(Δ_{E_{max}} = 10 - 2 = 8\) AND
\(Δ_{E_{min}} = 1 - 20 = -19\)

Hence only I and II are possible i.e. 6 and -16.

Subtlety in Question(or may be answer):
On the otherhand, GMAC would have been notoriously insane had it any one option with any of the possibilities viz. 7, 8, -17, -18 or -19. Here's how:
Under time pressure, normally people could take T as 0.2(smallest tenths digit) OR 0.8(largest tenths digit) for 10 decimals whose tenths digits is even and 0.1(smallest tenths digit) OR 0.9(largest tenths digit) for 20 decimals whose tenths digits is odd.

Maximum change: Rounding up becomes max. change = 0.8*10 = 8 and Rounding down becomes max. change = 0.9*20 = 18(actually '-18' for rounding down)

Minimum change: Rounding up becomes min. change = 0.2*10 = 2 and Rounding down becomes min. change = 0.1*20 = 2(actually '-2' for rounding down)

Thus,
\(Δ_{E_{max}} = 8 - 2 = 6\) AND
\(Δ_{E_{min}} = 2 - 18 = -16\)
Hope you got my point.

Finally, we can make our life easier if we consider all the 30 decimals as 0.xxxx OR at best each of the 10 as 0.01/0.89 and each of the 20 as 0.1/0.99.
Answer B.

HTH.

Often when we are asked for possible values, the problem is testing the min/ max content area. This type of problem can be straight forward and 600 level problems, or require us to apply a definition or test examples to see some pattern that happens from maxing a quantity.

Since we are asked the possible difference between E and S, E-S, lets think about maxing E

We maximize E when we gain the most from rounding up and lose the least from rounding down. This occurs when all of our even 10th place decimals are 0 and all of our even 10th place decimals are 1

Now, while none of our numbers can be zero, we can let our number with 0 tenth place decimal get so close to zero, that we are essentially gaining 1 by rounding it, and the minimum lost to rounding occurs when we let all our numbers with 1 tenth place decimal = .1

Now, in terms of S, E = S + gains from rounding. There will be 1/3 (30) = 10 even decimals and 20 odd decimals

So, Max E=S + 10(1) - 20(.1) = S+10-2 = S+8. Therefore Max E-S = S+8-S=8, so any value up to 8 can be the difference of the two sets. We know that III is not possible, therefore our only answer choice that works is B (I and II). Notice, you dont even have to show I to be true to get this problem correct. by eliminating III and seeing II: 6<8, we know the choice has to be B as it is the only option that works for us.

Let's make it easy--

if tenths digits 2,4,6,8= then increase to next integer
if tenths digits 1,3,5,7,9= then decrease to next integer

let assume the cases for even tenths digits
case (1) 0.2*10= 2 (increase over original value)
case (2) 0.8*10=8 (increase over original value)

let assume the cases for odd tenths digits
case (3) 0.1*20= -2 (decrease over original value)
case (4) 0.9*20= -18 (decrease over original value)

now, lets check the different possible combinations of even and odd values-
1) case 1 + case 3= 2+(-2)= 0
2) case 2 + case 4= 8+(-18)= -10
3) case 1 + case 4= 2+(-18)= -16
4) case 2 + case 3= 8+(-2)= 6

3 and 4 holds for option B (Min and max values)

GMAT Quant is designed such that smart people who haven’t done math in awhile can still do well on the test. Given ANY bizarro question like this, the smart non-math approach is to pick some basic and specific numbers, just to see what’s happening.

The specific numbers should be “bounded”: smaller numbers, larger numbers, numbers close together, and numbers for the large range.

1/3 of the 30 decimals are even. 0.2 is even. 0.1 is odd.

So start with ten 0.2s and twenty 0.1s. Using these numbers, follow the directions of the question and see what happens. You should end up with one of the Roman Numeral options(!)

Since 0.1 and 0.2 represent the lower bound and are close together, the next set of numbers should have a larger range.

Pick twenty 0.1s and ten 0.8s. Using these numbers, follow the directions the question and see what happens. The result should NOT be found among the Roman Numerals (darn!)

BUT, assuming you kept your work high and tight (that is, legible), compare your work to the remaining Roman numerals. You might see that one of the Roman numerals will never work under any circumstances. And then you’re done.

If you’re still stuck, pick the final set of numbers: larger numbers with a small range.

Pick ten 0.8s and twenty 0.9s. Follow the directions. Another Roman Numeral option should appear in your solution. Additionally, you should see why the final remaining Roman is is impossible. So you’re done.

NOTE: fully aware that I did not provide the actual math here. Other replies have done so. But it’s all about the approach and not the actual math, right?

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Hi GMATters,

Here's a video of my solution to this problem:

my solution is plain and simple: try to make your own analogy
Let's go to MONEY!

E-S

Emily's money - Sandy's money? What's the range ?

Assume Emily is super rich, she got 2*10+1*20 dollars. While sandy is a poor student she only has 1.2*10+1.1*20
We have their wealth GAP which is 6!

But wait a second, what if the situation is totally reversed. Emily now is the poor one and vice versa. She only got 2*10+1*20, but sandy got 56.
Then here we go we have their wealth GAP again, - 16!

Hi BrentGMATPrepNow, how do we know that we are looking for Range of E-S here? E-S mean Range or looking at answer choices have both positive and negative values?

Hi souvik101990, could you elaborate this part a bit more please of what number do you use to get 40 ? "E = 40 (due to rounding of ten even and 20 odd)"
Also in E-S min, not sure why you are using 56 from S max? Thanks

The question asks. ". . . which of the following is a possible value of E - S?"

Noted thanks BrentGMATPrepNow

Ignore the integer part of these decimals, only care for the after- the- comma part in determining the (E-S)
So for easier calculation and less confusing, I rename a little bit
Suppose S = total of decimals part before before "estimated"
and E = total of decimals part after "estimated", i.e. round up or round down

So we have E = 10 (10 decimals round up to 1 and 20 decimals round down to 0)
Then we will look for max and min of (E-S) or in other words minS and maxS respectively
S max when each decimal max (~ 0.9999999) so S max ~ 30 (noted that S can up very close to 30 but can not equal to 30)
S min when each decimal min (~0.00...0001) so S min ~ 0 (noted that S can down very close to 0 but can not equal to 0)

Therefore
(E-S) min ~ 10 - 30 = -20 (can down very close to -20 but can not equal to -20)
(E-S) max ~ 10 - 0 = 10 (can up very close to 10 but can not equal to 10)
Thus only I and II are possible, III cant.

d_k = round_down(d_k)+{d_k} => round_down(d_k) = d_k - {d_k} (I)
d_k = round_up(d_k)-1+{d_k} => round_up(d_k) = d_k + 1 - {d_k} (II)

E = sum f(d_k) = sum(round_down(d_i))+sum(round_up(d_j)), 1<=i<=10, 11<=j<=30
S = sum (d_k) = sum(d_i)+sum(d_j), 1<=i<=10, 11<=j<=30

E - S = [ sum(round_down(d_i))+sum(round_up(d_j)) ] - [sum(d_i)+sum(d_j)], 1<=i<=10, 11<=j<=30
E - S = sum[(round_down(d_i))-sum(d_i)] - sum[(d_j)-(round_up(d_j))], 1<=i<=10, 11<=j<=30
Using (I) and (II):

E - S = sum[1-{d_i}] - sum[-{d_j}], 1<=i<=10, 11<=j<=30
E - S = 10-sum[{d_k}], 1<=k<=30
Since the question says about d_k that "none of which is an integer" <=> {d_k} not equal to zero <=> 0<{d_k}<1 :
0 < sum[{d_k}]={d_1}+{d_2}+...+{d_30} < 30
-20 < E - S < 10

I. -16 -> ok
II. 6 -> in
III. 10 -> out

Correct option is B

Given: 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.

Asked: 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 ?

Let \(T = {t_1, t_2, .... , t_{30}}\)
\(t_1 + t_2 + ... + t_{30} = S\)

10 decimals in T have even tenth digit and 20 decimals in T have odd tenth digit

At most increase in E = 10
At most decrease in E = 20

Change in E may vary from -20 to 10 with extremes not possible since the numbers are not integers.

I. -16 : Possible
II. 6 : Possible
III. 10 : Not possible

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

IMO B

This is a beast of a problem. Here is a way I work it out as an optimiziation problem:

https://youtu.be/yY947L2IhpU

Example of the question I would skip in the real exam

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