This is how I would solve it:

Even tenth digit - Round up - 10 numbers
Odd tenth digit - Round down - 20 numbers

E - S can take many values so how do we figure which ones it cannot take? We need to find the range of E - S - the minimum value it can take and the maximum value it can take.

Minimum value of E - S => E is much less than S. How do we make E much less than S?
By doing 2 things:

1. When I round up, the difference between actual and estimate should be little. Say the numbers are something like 3.8999999 (very close to 3.9) and they will be rounded up to 4 i.e. the estimate gains 0.1 per number. Since there are 10 even tenth digit numbers, the estimate will be apprx .1*10 = 1 more than actual
2. When I round down, the difference between actual and estimate should be huge. Say the numbers are something like 3.999999 (very close to 4) and they will be rounded down to 3 i.e. the estimate loses apprx 1 per number. Since there are 20 such numbers, the estimate is 1*20 = 20 less than actual.
Overall, the estimate will be apprx 20 - 1 = 19 less than actual

E - S = -19

Maximum value of E - S => E is much greater than S. How do we make E much greater than S?
By doing 2 things:

1. When we round up, the difference between actual and estimate should be very high. Say the numbers are something like 3.000001 (very close to 3) and they will be rounded up to 4 i.e. the estimate gains 1 per number. Since there are 10 even tenth digit numbers, the estimate will be apprx 1*10 = 10 more than actual
2. When we round down, the difference between actual and estimate should be very little. Say the numbers are 3.1. They will be rounded down to 3 i.e. the estimate loses apprx 0.1 per number. Since there are 20 such numbers, the estimate is 0.1*20 = 2 less than actual.

Maximum value of E - S = 10 - 2 = 8

So 10 cannot be the value of E - S.
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Ok, so I don't normally post here, but one of my private tutoring students pointed me to this thread.

So yeah, there are a lot of words in this problem, and it looks complicated. And if you actually start thinking about "E" and "S" separately -- as most people on this thread seem to be doing -- then it becomes quite complicated indeed.
But ... we don't care about E, and we don't care about S. We care about E - S. That's a HUGE difference. (Analogy: I drive my car 54 miles. Do you know the starting odometer mileage? Nope. Do you know the ending odometer mileage? Nope. Do you know the difference between them? Yep, 54 miles.)

What affects E - S? To figure that out, temporarily forget that there are 30 numbers, and pretend there's just 1 number.
* If the number is rounded up, then E - S is positive. It's the amount by which the number is rounded up. (If you round 14.87 up to 15, then E - S is 0.13.)
* If the number is rounded down, then E - S is negative. It's the amount by which the number is rounded down. (If you round 14.77 down to 14, then E - S is -0.77.)
Thinking this way, we can see that it's a waste of time to consider the "integer part" at all. If you round 14.87 up to 15, or 99.87 up to 100, or anything point 87 up to the next integer, you're still looking at +0.13.

Now that we know that, things are more transparent.

* 10 numbers are rounded up.
The most by which they can be rounded up is just barely less than 1 each (if you start with x.00001 type thing).
The least by which they can be rounded up is just barely more than 0.1 each (if you start with x.89999 type thing).

* 20 numbers are rounded down.
The most by which they can be rounded down is just barely less than 1 each (if you start with x.99999 type thing).
The least by which they can be rounded down is 0.1 each (if you start with x.1).

To make E - S as big as possible, make the positive contributions as big as possible: 10 * (approximately +1) = approximately +10. Make the negative contributions as small as possible: 20 * -0.1 = -2. So, the maximum value of E - S is somewhere around 8.

To make E - S as small as possible, make the positive contributions as small as possible: 10 * (approximately 0.1) = approximately +1. Make the negative contributions as big as possible: 20 * (approximately 1) = approximately -20. So, the minimum value of E - S is somewhere around -19.

There you go.

The point here is that you should think about whatever the problem actually asks you for. In those terms, it sounds stupidly simple, but look at what's happening in this thread -- everyone is thinking separately about E and S, even though we only care about E - S. (Think about the odometer.) Oops.
Focus!

This idea is even more important on data sufficiency problems. (If a data sufficiency problem asks for something minus something else, and you try to find the individual values of "something" and "something else" rather than just finding the difference, you're practically certain to get the problem wrong, even if your math is all correct.)


--

Ron Purewal
ManhattanGMAT

Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

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IMO, the best solution is shown by M Dabral of GMAT Quantum in one of its video explanations.

Here is the link: http://www.gmatquantum.com/og13/218-pro ... ition.html

Although NOT most of GMAT Quantum's video explanations/solutions are up to the par, I found that this one along with some others (e.g., PS 178) is the best video explanation floating out there in the internet.

Assume T = (1.a, 1.b,...etc) All units equal 1.xx.
E = 40 (due to rounding of ten even and 20 odd)
S max = 30 + 10(.8) + 20(.9) = 56
S min = 30 + 10(.2) + 20(.1) = 34

E-S min = 40 - 56 = -16
E-S max = 40 - 34 = 6
Thus, the min/max of E is -16 and 6, so I , II apply.

For (III. 10) to be true, S min/max in our equation above needs to equal 30. This is impossible since the question states T consists of 30 positive decimals. Therefore eliminate III.
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To clarify, when you got E = 40
1.xxx rounded up, 2 x 10 (even) = 20
1.xxx rounded down, 1 x 20 (odd) = 20

is that correct?

I found the responses in this question very helpful, but I think I found another way to explain it that I can understand better... hopefully someone finds this useful someday.

**
As the question ask for the difference between S and E, the actual "integer" portions are negligible as they will cancel each other out in the minus (-) operation (e.g. 15.9 - 15 = 0.9) . If it helps in understanding, we can also assume that all numbers in the list T are 0.xxxx (as the minus operation cancels the "integer" portion anyway).

Let's recap:
S = Real sum
E = Estimated sum, 10 are rounded up, 20 are rounded down

We can also first establish that there is only one value of E, as we are given in the question that 10 of them WILL be rounded up, and 20 WILL be rounded down. There is no Emax or Emin. To solve for E and assuming that numbers in list T are 0.xxx as per above point,
E = 10 x (T values rounded up) + 20 x (T values rounded down)
= 10 x (0.2) + 20 x (0.9) --> doesnt matter what the values are, they can be any even / odd combinations at the tenth
= 10 x (1) + 20 x (0), after performing round up/down operations
= 10

There are however Smax and Smin, as these are possible ranges prior to the round up/down operations.

Smax = 10 x (maximum even tenth) + 20 x (maximum odd tenth)
= 10 x (0.8) + 20 x (0.9)
= 8 + 18
= 26

Smin = 10 x (minimum even tenth) + 20 x (minimum odd tenth)
= 10 x (0.2) + 20 x (0.1)
= 2 + 2
= 4

The largest range of E - S is therefore
E - Smin = 10 - 4
= -6

The lowest range of E - S is therefore
E - Smax = 10 - 26
= -16



Unlike some contributions here, I found it absolutely necessary to solve for E and S. The previous solution by RonPurewal looks sound, but when you look closer, it assumes that E - S = (10 round up values - 20 round down values), which is not true, and therefore the answers are not exactly to the range given by the question. The full algebraic equation of his solution should have been:

- Largest range of E - S = (10x round up + 20x round down) - (10x Minimum round up + 20x minimum round down)
- Lowest range of E - S = (10x round up + 20x round down) - (10x Maximum round up + 20x maximum round down)

Essentially calling for the need to resolve for E and S individually, eventually.
I apolgise if I sound rude - its 130am in the morning and I had just spent an hour just analyzing just this question alone, and my heart hurts! Irregardless, I have learn plenty just by analyzing everyone's solution to help me arrive to this way of looking into this problem. Many thanks!

The two extremes can be calculated as:
1. Highest tenths place 0.9 and 0.8

so E = S +(10*0.2 - 20*0.9) >>>> Change from 0.1 to 1 is 0.2(increase) & 0.9 to 0 is 0.9(decreased)
= S +(2-18)
E-S = -16

2. Lowest tenths place 0.1 & 0.2 >>>> Change from 0.2 to 1 is 0.8 & 0.1 to 0 is 0.1
so E = S+(10*0.8 - 20*.2)
E -S = (8-2)
E-S = 6

Hence answer is B!!

Let the number with even tenths place be denoted by x and the number with odd tenths place digit by y.

We can further split these decimals into their integer and decimal parts denoted by x(i) and x(d) respectively.
Similarly, y can be split and written as y(i) and y(d).

Now, according to question,
-------------->>>>> S = x(i) + x(d) + y(i) + y(d) ........................... (1)

And, since E comprises only of the integer values
-------------->>>> E = x(i) +10 + y(i) ............................(2) { because, the number with even tenths place digit is rounded up
and the number with odd tenths place digit is rounded down }

Now, from (1) and (2)
E - S = x(i) +10 + y(i) - { x(i) + x(d) + y(i) + y(d) }
= 10- { x(d) + y(d) }

Therefore, the final value is always going to be less than 10, and there are only two options in the question that support this result .


Please let me know if there is any problem with this approach.

You're missing something here, Souvik. First of all, I approached this differently than you, although that doesn't matter. I first looked at the differential irrespective of S. E is either going to be a net (+) or (-) based on the observations' tenth's-digits, and the magnitude of gain due to rounding.

BUT! (x.2) doesn't represent the maximum net gain from an even tenth's-digit rounded up to the nearest integer, (x.0) does.
Thus,
delta-E(max) = 10(1) - 20(.1) = 8 (=E-S max)
delta-E(min) = 10(.2) - 20(.9) = (-16) (=E-S min)

Inconsequential in the scheme of the answer choices, but something worth noting.

Cheers!

At first glance question may look difficult but in reality it can be solved in 2 mins ..

\(E_m_a_x = S + ( 20*(-0.1) + 10*0.8) = S+6\) so E-S=6
\(E_m_i_n = S + ( 20*(-0.9) + 10*0.2) = S-16\) so E-S=-16

Answer B.

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 --> 4 so we get 0,8*10=8
Min: 3,8 --> 4 so we get 0,2*10=2

Round down - Odd
Max: 3,9 --> 3 so we get -0,9*20=-18
Min: 3,1 --> 3 so we get -0,1*20=-2

Now we can manipulate those numbers:
I. -16 --> -18 + 2 = -16 OK
II. 6 --> -2 + 8 = 6 OK
III. 10 X You can not get 10 because the max positive Value is 8, Hence, correct answer is (B)

Note that the only thing that effects the difference E-S are the possible values of E, as S is the set of stationary fixed values. Thus, the max and min value changes of E are all that we care about.

Max
If the true number, S, of even tenths decimals end in X.0000000001, the even decimals in E would get rounded up to effectively +1 more, or +1*10=+10 away from the true value of S.
If the true number, S, of odd tenths decimals end in X.100000, the odd decimals in E would each lose -.1, or -0.1*20= -2 from the true value of S.
Net max value for E-S = 8

Min
If the true number, S, of even tenths decimals end in X..89999999, the even decimals in E would get rounded up to effectively +0.1 more, or +0.1*10=+1 away from the true value of S.
If the true number, S, of odd tenths decimals end in X.999999999, the odd decimals in E would each effectively move -1, or -1*20= -20 from the true value of S.

Net min value of E-S = -19

Thus, the possible range of E-S values is -19 to 8.

Below is my approach -

10 numbers are with even tenth's digit: Let them be (n + d1) each [Ex: - 1.22 = 1 + 0.22]
20 numbers are with odd tenth's digit: Let them be (n + d2) each [Ex: - 1.23 = 1 + 0.23]

When rounded off - they become :
10*(n+1) and
20*(n) respectively.

Hence, E-S = [10*(n+1) + 20*(n)] - [10*(n+d1) + 20*(n+d2)]
= 10 - 10*d1 -20*d2
As, d1 and d2 can range between 0 and 1. It implies - 0<10*d1<10 AND 0<20*d2<20
Therefore, we can conclude - -20<E-S<10

Answer - B

MATH REVOLUTION VIDEO SOLUTION:


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Dont know what's the issue here but I can see atleast 3 good solutions with Karishma's the best one , IMO

Refer to the posts : http://gmatclub.com/forum/list-t-consis ... l#p1081121
Ron Purewal's: http://gmatclub.com/forum/list-t-consis ... l#p1292975
http://gmatclub.com/forum/list-t-consis ... l#p1090856

The concepts tested here are max/min , rounding of decimals and number properties (odd/even). All these word problems become complicated when you do not break the problem down into manageable chunks.

To make the calculations simpler, you can assume that the decimals are

0.1 and 0.2 such that 0.2 0.2 0.2 .... 10 times and 0.1 0.1 0.1 .....20 times.

Thus, S = 0.2*10+0.1*20 = 4 and E = 1*10+0 = 10. Thus, E-S = 6. You will see that this is the most you will get for E-S.

Now, consider the other end of the spectrum: 0.8 0.8 0.8 ....10 times and 0.9 0.9 0.9 ...20 times

Thus, S = 0.8*10+20*0.9 = 26 and E = 1*10+0=10, Thus E-S = -16. Whatever values you play with, you will never get to 10.

Look at the solutions above and let me know if you still have questions. If you do, mention the doubt/question as well.

Hope this helps.
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I have not gone through the entire thread and would explain you the best way I can...
Since I am not aware what has been explained earlier, the way and solution I am giving may be NEW or already discussed..

Lets first take the INFO from the Q--

1) there are 30 decimals, NONE of them is an INTEGER..

2) decimals with even TENTHS is moved to upper integer..
which means any decimal .2,.4,.6,.8 moves to 1..

3) decimals with odd TENTHS is moved to lower integer..
which means any decimal .1,.3,.5,.7,.9 moves down to 0..

4) We are NOT concerned with what is to LEFT of decimal as that will get cancelled out in E-S

5) 10 are moving up (1/3 rd even decimals), whereas the other 20 move down

SOLUTION-

In such Qs, the best way is to look at the least and max values..
so lets take the smallest and biggest even and odd decimals..
EVEN--
smallest-0.2
EFFECT= 10 is moving up But in actual .2*10=2 is already moved up in S..
so effect on E-S= 10-2= +8
Largest-0.8
EFFECT= 10 is moving up But in actual .8*10=8 is already moved up in S..
so effect on E-S= 10-8= +2

ODD--
smallest-0.1
EFFECT= .1*20= 2moved down= -2
Largest-0.9
EFFECT= 0.9*20= 18 moved down = -18..

so now we can take combinations of effect of even and effect of moving down..
even= +2 and +8
odd= -2 and -18..
least (opposite effect)= -18+2=-16
biggest effect= 8-2=6..

so E-S will lie between -16 and 6, both inclusive..
so 10 is not possible
ans B
Hope it helps

n1+n2+n3...n30 = S

Estimated Value = E

*To maximize one quantity, minimize the others
To minimize one quantity, maximize the others*

(E-S) max occurs when gain from rounding off is maximum . Therefore we assume each even number in Set T to have tenths digit as 2. And we minimize the loss from rounding off by assuming odd numbers in set T to have tenths digit 1. Therefore gain = (0.8*10) & loss = (0.1*20). Therefore 8-2 = 6

So option II is possible whereas option III is not.

(E-S)min occurs when you reverse the conditions , i.e. minimise gain and maximise loss. Therefore even digits have tenths digit 8 whereas odd numbers have tenths digit 9. Therefore gain = (0.2*10) & loss = (0.9*20). Therefore 2-18 = -16

So option I is feasible.

Major question.

We are quick to assume that if 10 numbers are even, the rest are odd.
What if one of the digits is 0: neither even nor odd?

Why have we not assumed that one of the digits could be 0?

VeritasPrepKarishma