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

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

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05 Jun 2016, 20:44
kanav06 wrote:
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?

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On GMAT, all integers have an even/odd designation.

Even: ... -4, -2, 0, 2, 4 ...
Odd: ... -3, -1, 1, 3...
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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15 Jul 2016, 03:46
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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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List T consist of 30 positive decimals, none of which is an [#permalink]

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01 Aug 2016, 11:58
vsrsankar wrote:
First let us find the value of E:
"1/3 of the decimals in T have a tenths digit that is even" => 10 numbers have an even tenths digit and remaining 20 numbers have an odd tenths digit.
=> E = (sum of all 30 integer parts) +10(1)-20(1) = (sum of all 30 integer parts)-10

Now let us come to "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
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

E-S ranges between (E-Smax) to (E-Smin) = -36 to -14

IMO, the answer option "A" is correct, as -16 only lies in this range.

The answer for me would be "B", and my reasoning was the following:

Let's take extreme cases of decimals with even and odd tenth units:

Even: 0.20 (lowest) - 0.8999... (greatest)
Odd: 0.10 (lowest) - 0.9999... (greatest)

Let's use for the first case the lowest odd and the lowest even numbers

S = 20 x (0.10) + 10 x (0.2) = 4
E = 0 + 10 x 1 = 10

So in this case E - S = 6. So we already know that 6 is a possibility and so we can drop choices "A" and "C".

Now let's use the greatest odd and the greatest even numbers:

S = 20 x (0.999...) + 10 x (0.8999....) = 29 approximately
E= 0 + 10 x 1 (by rounding each 0,1 to 1) = 10
E - S = 10 - 29 = -19

So -19 < S-E <= 6

So -16 is also in the interval. Hence answer is B.

Hope it is clear!

Best,

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

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19 Nov 2016, 04:45
Is there any alternate approach that can be applied to above problem because none of solutions posted seems to be cover problem in 2 minutes.
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List T consist of 30 positive decimals, none of which is an [#permalink]

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19 Nov 2016, 05:10
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Hi anairamitch1804, I don't normally post here, but I read all the explanations to this question and it just seems that it is pretty hard get your head around this question and solve it in under 2 minutes.
So, the following worked for me to solve it in the time limit:

We have 10 (1/3 times 30) even and 20 (2/3 times 30) odd numbers and we know that the 10th digit is either rounded up (if even) or down (if odd).

Just assume for simplicity, we have 2 digit decimal n-s (no limitations in the question to do this):

So the actual sum is expressed in the following way: S=10(n+0.1n)+20(n+0.1n)

We know that in estimated sum E, evens are rounded up and odds rounded down, so it follows:
If tenth digit is even: n+0.1n≈n+1, and we have 10 of these --> 10(n+1)
If tenth digit is odd: n+0.1n≈n, and we have 20 of these --> 20n
So E=10(n+1)+20n

So now E-S= (20n+10(n+1))-(10(n+0.1)+20(n+0.1n)= 20n+10n+10-10n-n-20n-2n=10-3n
So E-S=10-3n

We know that n is a positive decimal and it cannot be an integer, so use the answer choices:
Check -16 --> 10-3n=-16 --> n=26/3, it works!
Check 6 --> 10-3n=6 --> n=4/3, it works!
Check 10 --> 10-3n=10 --> this means that n is 0, which cannot be the case, so it does not work!

Hence, I & II work only, Answer B
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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19 Nov 2016, 05:26
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.

[Reveal] Spoiler:
B
.
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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19 Nov 2016, 05:40
iako27 wrote:
Hi anairamitch1804, I don't normally post here, but I read all the explanations to this question and it just seems that it is pretty hard get your head around this question and solve it in under 2 minutes.
So, the following worked for me to solve it in the time limit:

We have 10 (1/3 times 30) even and 20 (2/3 times 30) odd numbers and we know that the 10th digit is either rounded up (if even) or down (if odd).

Just assume for simplicity, we have 2 digit decimal n-s (no limitations in the question to do this):

So the actual sum is expressed in the following way: S=10(n+0.1n)+20(n+0.1n)

We know that in estimated sum E, evens are rounded up and odds rounded down, so it follows:
If tenth digit is even: n+0.1n≈n+1, and we have 10 of these --> 10(n+1)
If tenth digit is odd: n+0.1n≈n, and we have 20 of these --> 20n
So E=10(n+1)+20n

So now E-S= (20n+10(n+1))-(10(n+0.1)+20(n+0.1n)= 20n+10n+10-10n-n-20n-2n=10-3n
So E-S=10-3n

We know that n is a positive decimal and it cannot be an integer, so use the answer choices:
Check -16 --> 10-3n=-16 --> n=26/3, it works!
Check 6 --> 10-3n=6 --> n=4/3, it works!
Check 10 --> 10-3n=10 --> this means that n is 0, which cannot be the case, so it does not work!

Hence, I & II work only, Answer B

Thanks for solutions this seems to be very practical but I am stuck with below line :

Just assume for simplicity, we have 2 digit decimal n-s (no limitations in the question to do this):
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List T consist of 30 positive decimals, none of which is an [#permalink]

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19 Nov 2016, 06:20
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anairamitch1804

Maybe wording is not clear enough, sorry. As we are only concerned with the tenth digit of the decimal, n+0.1n would be enough to write vs. writing a decimal with greater digits after 0... hope it is clear
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List T consist of 30 positive decimals, none of which is an [#permalink]

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04 Dec 2016, 02:57
anairamitch1804 wrote:
Is there any alternate approach that can be applied to above problem because none of solutions posted seems to be cover problem in 2 minutes.

Check out the Approach taken by chetan2u here =>
list-t-consist-of-30-positive-decimals-none-of-which-is-an-131755-20.html#p1654863

I think he has Explained it in the Simplest Possible way.

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

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07 Dec 2016, 05:46
Guys,
I used following method to solve this, the explanation makes it look long but I was able to solve this in no time.

Lets say N is a decimal number, so we can write
N = integer part of N + decimal part of N
For simplicity N = I + d for any decimal

and 0 < d < 1

When we round up N to its nearest Integer, we make d = 1. This means for each rounding up we gain 1 over the integer value of respective number
e.g.
N = 1.2 = 1 + 0.2
In this case d = 0.2.
On round up N we get 2 which is 1 + 1 and d = 1.

When we round down N to its nearest Integer, we make d = 0. This mean for each rounding down we get 0 over integer value of respective decimal.
e.g.
N = 1.3 = 1 + 0.3
In this case d = 0.3
On rounding down N, we get 1 which is I + 0 and d = 0.

Coming back to the question,
S = Sum of integer part of all numbers (A) + Sum of decimal part of all numbers (D) = A + D.

note that 0 < D < 30 because decimal parts of all 30 integers have a range of 0 < D < 1

E = A + gain for numbers which were rounded up + gain for numbers which were rounded down
E = A + 1*10 + 0*20 (because 1/3 were rounded up and 2/3 were rounded down)
E = A + 10

So E- S = A + 10 - A - D = 10 - D, so essentially we need to find the range of 10 - D

But we know that 0 < D < 30 => 0 > -D > -30 => 10 +0 > 10-D > 10-30 => 10 > 10-D > -20

so the desired range is -20 < 10-D < 10.

Now we can clearly see that only 6 and -16 fall in this range and hence B is the correct answer.
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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10 Dec 2016, 10:40
Top Contributor
My Assumption,

Each decimal whose tenths digit is even =0.X and whose tenths digit is odd=0.Y,Here(X=2,4,6 or 8 and Y=1,3,5,7 or 9)

---------------------------------

So,For any value of X & Y, E=10(0.X) +20(0.Y)=10(1)+20(0)=10

Minimum value (when X=2 and Y=1 ) of S=10(0.2)+20(0.1)=4
Maximum value (when X=8 and Y=9) of S=10(0.8)+20(0.9)=26

So Range of possible value of E-S lies within (10-4) to (10-26) or 6 to -16

Only a & b are within this range

So correct answer is a & b only or B
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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16 Feb 2017, 10:54
Option B

Rounding-off(up/down) a decimal ending with an odd number will have max. impact if it ends with 1 or 9, i.e., the number should look like - Y.1 or Y.9
Similarly, rounding-off(up/down) a decimal ending with even number will have max. impact if it ends with 2 or 8, i.e., the number should look like - Z.2 or Z.8

Q Stem: Identify the possible values/range for E - S ?

:In short, identify the difference created by the rounding-off exercise.
:One way is to maximize and minimize the possible difference created by the rounding-off exercise.
:There are 10 numbers like Z.2 or Z.8 and 20 numbers like Y.1 or Y.9.
:Numbers like Z.2 or Z.8 are rounded UP - so difference created will be 0.8 (Z+1 - Z.2 = 0.8) OR 0.2 (Z+1 - Z.8 = 0.2)
:Numbers like Y.1 or Y.9 are rounded DOWN - so difference created will be -0.1 (Y - Y.1 = -0.1) OR -0.9 (Y - Y.9 = -0.9)

To maximize the difference, we should add the max. possible POSITIVE value and least possible NEGATIVE value = 10*(0.8) + 20*(-0.1) = 8 - 2 = 6
To minimize the difference, we should add the max. possible NEGATIVE value and LEAST possible POSITIVE value = 10*(0.2) + 20*(-0.9) = 2 - 18 = -16

Hence, all the possible values of E - S should lie in the range (Both values inclusive) -16 to 6. And, from given options only 10 lies outside this range.
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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16 Feb 2017, 10:58
E - S = a*(almost 1) - b*(almost 1)

with a in (0...10) and b in (0..20)

Max value of E-S is 10*(almost 1)
Min value of E-S is -20*(almost 1)

Therefore, -19.999 <= E-S <= 9.999

I and II are in that range, so it's letter B
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List T consist of 30 positive decimals, none of which is an [#permalink]

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19 Apr 2017, 06:55
Solved in UNDER A MINUTE:

10 Even and 20 Odd
Max Even value: +8 (0.8*10, considering all were 0.2 we made all 0.2 to 1)
Min Even value: +2 (0.2*10, considering all were 0.8 and we made all 0.8 to 1)

Max Odd value: -18 (Same logic as positive ones)
Mix even value: -2

Even + Odd = +8-2 = 6
Even + Odd = +2-18 = -16
Value CANNOT be 10 coz max value cannot exceed 6

Hence B
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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18 May 2017, 10:06
T = ( X1, X2, X3 ……… X30)
All the numbers that T is consisting of are positive DECIMALS. None of them are integers.
Sum = X1+ X2 + X3 + ……… + X30 = S
We have 10 even and 20 odd numbers.

The ESTIMATED sum = E and is defined:
________________________________________________
Each even decimal is rounded $$UP$$ to nearest integer:
2.00000002 = 3 => here we approximately adding 1
2.89999999 = 3 as well => here we approximately adding 0.1 =>
the difference between estimated sum E and real sum E must be in range 0.1 to 1
0.1 <E - S< 1
________________________________________________________
Each odd decimal is rounded DOWN to the nearest integer:

2.1111111 = 2 => here we approximately subtracting 0.1
2.9999999 = 2 as well => we approximately subtracting 1.
the difference between estimated sum E and real sum E must be in the range -1 to -.01

-1 <E - S<-0.1
_________________________________________________________
since we have 10 even numbers and 20 odd the range is as follows:

for even numbers:

10*0.1 <E - S< 1*10
1 < E - S < 10

for odd numbers:
20*(-1) <E - S<20* (-0.1)
-20 < E –S < - 2

To find total difference between estimated and real sum, we add the odd and even numbers differences =>

-19 < E – S < 2

So all possible values must be in the range of -19 and 2.

- 16 is in range
- 6 is in range
- 10 – is not in range

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List T consist of 30 positive decimals, none of which is an [#permalink]

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23 May 2017, 19:06
I don't know if my approach helps or whether is correct, but I had to come up with it since I couldn't understand other people's approaches to this question.

We basically have have 10 decimals with even tenths digits and 20 decimals with odd tenths digits.

If we only take extremes, we can define four scenarios:

1) For the 10 decimals with even tenths digits we would have:
1.1) 10 digits ending in x.2 which are rounded up to (x+1).0. Gain is +0.8(10)=8
1.2) 10 digits ending in x.8 which are rounded up to (x+1).0. Gain is +0.2(10)=2

2) For the 20 decimals with odd tenths digits we would have:
2.1) 20 digits ending in x.1 which are rounded up to (x).0. Loss is -0.1(20)=-2
2.2) 10 digits ending in x.9 which are rounded up to (x).0. Loss is -0.9(20)=-18

We can now combine the four sub-scenarios:
1.1 and 2.1) E-S=Gain+Loss=8-2=6
1.1 and 2.2) E-S=Gain+Loss=8-18=-16
1.2 and 2.1) E-S=Gain+Loss=2-2=0
1.2 and 2.2) E-S=Gain+Loss=2-18=-16

Since we took the extremes, we know that the possible differences are in the range [-16,6], and options outside this range aren't valid. Alternatively, options within the range should be valid as well, but since the only options mentioned in the question stem are 6 and -16 (voilà, the extremes we calculated!), therefore the correct answer is B.

Hope it helps.
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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17 Jul 2017, 19:05
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.

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

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26 Jul 2017, 21:15
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

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Re: EFFICIENCY IN PROBLEM SOLVING [#permalink]

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14 Nov 2017, 04:14
waitherakariuki wrote:
Hi guys,

I need your help with answering question below from the GMAT Test bank in a less time consuming/efficient way.

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

I got the correct answer, but took me 4 minutes.
As 1/3 rd numbers are even only 10 decimals will be rounded off to next integer. So at max the difference can be 0.8 for any decimal.
So for 10 decimals at max this value will be 0.8*10=8 = max(increase)
& min(increase)=0.2*10=2

Similarly, for odd decimals, minimum difference can be -0.1*20=-2 = min(decrease)
& maximum difference can be -0.9*20=-18 = max(decrease)

And we need to find out E-S, so maximum value of the E-S can be [max(increase)+min(decrease)] i.e. 8+(-2)=6

so we can rule out value 10. Similarly minimum value of E-S= [min(increase) + max(decrease) ] i.e. 2+(-18) = -16

So Only I & II follows. Option B is the answer.

Anyone with a better approach?
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Re: List T consist of 30 positive decimals, none of which is an [#permalink]

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09 Dec 2017, 06:14
kelvinp wrote:
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!

Thank you! The best possible solution I was able to understand.
Re: List T consist of 30 positive decimals, none of which is an   [#permalink] 09 Dec 2017, 06:14

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