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

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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: https://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.
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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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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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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shamanth25 wrote:
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

When reading through this question, notice that we are asked which of the following is a POSSIBLE value of E – S. This tells us that we will not be looking for a definite answer here.

We are given that list T has 30 decimals, and that the sum of this list is S.

Next we are given that each decimal whose tenths digit is even is rounded up to the nearest integer and each decimal whose tenths digit is odd is rounded down to the nearest integer.

We are next given that E is the sum of these resulting integers.

Finally, we are given that 1/3 of the decimals in list T have a tenths digit that is even. This this means that 2/3 have a tenths digit that is odd. This means we have 10 decimals with an even tenths digit and 20 decimals with an odd tenths digit.

This is very helpful because we are going to use all this information to create a RANGE of values. We will calculate both the maximum value of E – S and the minimum value of E – S.

Another way to say this is that we want the maximum value of the sum of our estimated value in list T minus the sum of the actual values in list T and also the minimum value of the sum of the estimate values in list T minus the sum of the actual values in list T. Thus, we need to determine the largest estimated values and the smallest estimated values for the decimals in list T.

To do this, let’s go back to some given information:

Each decimal whose tenths digit is even is rounded up to the nearest integer.

Each decimal whose tenths digit is odd is rounded down to the nearest integer.

Let’s first compute the maximum estimated values for the decimals in list T. To get this, we want our 10 decimals with an even tenths digit to be rounded UP as MUCH as possible and we want our 20 decimals with an odds tenths digit to be rounded DOWN as LITTLE as possible. Thus, we can use decimals of 1.2 (for the even tenths place) and 1.1 (for the odd tenths place). Let’s first calculate S, or the sum of the 30 decimals. We know we have 10 decimals of 1.2 and 20 decimals of 1.1, so we can say:

S = 10(1.2) + 20(1.1) = 12 + 22 = 34

Now we can round up 1.2 to the nearest integer and round down 1.1. We see that 1.2 rounded up to the nearest integer is 2, and 1.1 rounded down to the nearest integer is 1. So now we can calculate E, or the sum of the resulting integers. We can say:

E = 10(2) + 20(1) = 20 + 20 = 40

Thus, the maximum value of E – S, in this case, is 40 – 34 = 6.

Let’s next compute the minimum estimated values for the decimals in list T. To get this we want our 10 decimals with an even tenths digit to be rounded UP as LITTLE as possible and we want out 20 decimals with an odds tenths digit to be rounded DOWN as MUCH as possible. Thus, we can use decimals of 1.8 (for the even tenths place) and 1.9 (for the odd tenths place). Let’s first calculate S, or the sum of the 30 decimals. We know we have 10 decimals of 1.2 and 20 decimals of 1.1, so we can say:

S = 10(1.8) + 20(1.9) = 18 + 38 = 56

Now we can round up 1.8 to the nearest integer and round down 1.9 to the nearest integer. 1.8 rounded up to the nearest integer is 2, and 1.9 rounded down to the nearest integer is 1. So now we can calculate E, or the sum of the resulting integers, so we can say:

E = 10(2) + 20(1) = 20 + 20 = 40

Thus, the minimum value of E – S, in this case, is 40 – 56 = -16.

Thus, the possible range of E – S is between -16 and 6, inclusive. We see that I and II fall within this range.

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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!
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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shamanth25 wrote:
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

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,
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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
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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Key point:

1/3 of the decimals in T have a tenths digit that is even.

There are 30 decimals in T. So 10 have even tenths digits and ten have odd tenths digits.

Notice:

When you round down one of the decimals, you are reducing E. So you are reducing E - S.

When you round up one of the decimals, you are increasing E. So you are increasing E - S.

So according to the question we will be rounding up and increasing E ten times and rounding down and reducing E 20 times.

Possible even decimals are .2, .4, .6 and .8. So rounding up adds .8, .6, .4 or .2.

Possible odd decimals are .1, .3, .5, .7 and .9. So rounding down subtracts .1, .3, .5, .7 or .9.

So really the question is can (10 values from the adds list) - (20 values from the subtracts list) equal one of the given answers.

Check the values:

I. -16 is pretty low. So to get it we need to do some serious subtraction and not much addition.

Let's try minimizing E by minimizing the addition, by choosing the smallest number from the adds list, and maximizing the subtraction, by choosing the largest number from the subtracts list.

(10 x .2) - (20 x .9) = 2 - 18 = -16

Value I works.

II. 6 is between -16 and 10. So I am going to skip it for now. If 10 works, I think 6 is going to as well.

III. 10 is pretty high. So let's maximize the addition and minimize the subtraction.

(10 x .8) - (20 x .1) = 8 - 2 = 6.

So 10 does not work, but 6 does.

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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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shamanth25 wrote:
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

Official Explanation:

Since 1/3 of the 30 decimals in T have an even tenths digit, it follows that 1/3*(30)=10 decimals in T have an even tenths digit. Let $$T_E$$ represent the list of these 10 decimals, let $$S_E$$ represent the sum of all 10 decimals in $$T_E$$, and let $$E_E$$ represent the estimated sum of all 10 decimals in $$T_E$$ after rounding. The remaining 20 decimals in T have an odd tenths digit. Let $$T_O$$ represent the list of these 20 remaining decimals, let $$S_O$$ represent the sum of all 20 decimals in $$T_O$$, and let $$E_O$$ represent the estimated sum of all 20 decimals in $$T_O$$ after rounding. Note that $$E = E_E + E_O$$ and $$S = S_E + S_O$$ and hence $$E − S = (E_E + E_O) − (S_E + S_O) = (E_E − S_E) + (E_O − S_O)$$.

The least values of $$E_E − S_E$$ occur at the extreme where each decimal in TE has tenths digit 8. Here, the difference between the rounded integer and the original decimal is greater than 0.1. (For example, the difference between the integer 15 and 14.899 that has been rounded to 15 is 0.101.) Hence, $$E_E − S_E > 10(0.1) = 1$$. The greatest values of $$E_E − S_E$$ occur at the other extreme, where each decimal in $$T_E$$ has tenths digit 0. Here, the difference between the rounded integer and the original decimal is less than 1. (For example, the difference between the integer 15 and 14.001 that has been rounded to 15 is 0.999.) Hence, EE − SE < 10(1) = 10. Thus, $$1 < E_E − S_E < 10$$.

Similarly, the least values of EO − SO occur at the extreme where each decimal in TO has tenths digit 9. Here, the difference between the rounded integer and the original decimal is greater than −1. (For example, the difference between the integer 14 and 14.999 that has been rounded to 14 is −0.999.) Hence EO − SO > 20(−1) = −20. The greatest values of EO − SO occur at the other extreme where each decimal in TO has tenths digit 1. Here, the difference between the rounded integer and the original decimal is less than or equal to −0.1. (For example, the difference between the integer 14 and 14.1 that has been rounded to 14 is −0.1.) Hence, $$E_O − S_O ≤ 20(−0.1) = −2$$. Thus, $$−20 < E_O − S_O ≤ −2$$.

Adding the inequalities $$1 < E_E − S_E < 10$$ and $$−20 < E_O − S_O ≤ −2$$ gives $$−19 < (E_E − S_E) + (E_O − S_O) < 8$$. Therefore, $$−19 < (E_E + E_O) − (S_E + S_O) < 8$$ and $$−19 < E − S < 8$$. Thus, of the values −16, 6, and 10 for E − S, only −16 and 6 are possible.

Note that if T contains 10 repetitions of the decimal 1.8 and 20 repetitions of the decimal 1.9, $$S = 10(1.8) + 20(1.9) = 18 + 38 = 56$$, $$E = 10(2) + 20(1) = 40$$, and $$E − S = 40 − 56 = −16$$. Also, if T contains 10 repetitions of the decimal 1.2 and 20 repetitions of the decimal 1.1, $$S = 10(1.2) + 20(1.1) = 12 + 22 = 34$$, $$E = 10(2) + 20(1) = 40$$, and $$E − S = 40 − 34 = 6$$.

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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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The list T consists of 30 positive decimals, of which 10 decimals (1/3rd of the 30 decimals) have their tenths place as an even digit and the remaining 20 decimals (remaining 2/3rd of the 30 decimals) have their tenths place as an odd digit.
Let us recall that, if the general representation of a decimal is
0.pqrst…….,
p = tenths place, q = hundredths place, r = thousandths place and so on.

Note: Remember that the decimal can also be 1.pqrst or 2.pqrst and so on.

In our case, we will have to focus on the tenth’s place because it is based on this that we are rounding off the decimal, either to a higher integral value or to a lower integral value.
For example, if the decimal is 0.2qrst…, then this has to be rounded up to the next higher integer that is 1; if the decimal is 1.2qrst…., then this has to be rounded up to the next higher integer, 2; and so on. I hope this is clear.

Similarly, if the decimal is 0.1qrst….., then this has to be rounded down to the next lower integer, which in this case is obviously ZERO; if the decimal is 1.1qrst…., then this has to be rounded down to the next lower integer, 1; and so on. I hope this is clear as well.

Now, let us address the actual problem at hand. Since we are trying to find out the impossible values of E-S, which is a difference of two sums, it is a good idea to find out the range of values that E-S can possibly take, which will help us figure out which value is not possible.
The minimum value of E-S can be obtained by making E as small as possible while making S as big as possible. S can be made as large as possible by taking the tenth’s place as a bigger digit like 8 or 9 and the contrary is true to make is as small as possible.
The following tables should make it clearer:

Kindly go through the pic attached.

Let us compute all possible values of S and E from the table above by taking different combinations:
If S = 0.1 +2 = 2.1 and E = 10 + 0 = 10, then E-S = 8.
If S= 0.1 + 19.8 = 19.9 and E = 10 + 0, then, E –S = -9.9
If S = 8.9 + 2 = 11.9 and E = 10 +0 = 10, then, E – S = -1.9
If S = 8.9 + 19.8 = 28.7 and E = 10 + 0, then E = -18.7
Therefore, the range of values is between -18 and 8. Hence, 10 is not a possible value for E-S.
An important point that you have to note here is the fact that I have taken the smallest possible positive decimal in 0.01 and therefore, with decimals like 1.01 or 1.21, the upper range of 8 will only reduce, but will never increase beyond 8 and hence, we can safely say that 10 is definitely not a possible option

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General Discussion
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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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

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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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shamanth25 wrote:
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

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

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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)
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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.
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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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 : https://gmatclub.com/forum/list-t-consis ... l#p1081121
Ron Purewal's: https://gmatclub.com/forum/list-t-consis ... l#p1292975
https://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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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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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.
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
shamanth25 wrote:
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

1.Consider the extreme cases so that you know the maximum and minimum values possible and therefore can find the range. of E-S.
2. In the case of the tenths digits even, maximum of just less than 10 can be gained.
3. In the case of tenths digits odd, maximum of just less than 20 can be lost i.e, -20
4. E-S can be in the range from just numerically less than -20 to just less than 10.