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Re: List T consist of 30 positive decimals, none of which is an integer
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14 Dec 2017, 23:35
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 Check out our video solution to this very tricky problem: https://www.veritasprep.com/gmatsoluti ... olving_224
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Re: List T consist of 30 positive decimals, none of which is an integer
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18 Dec 2017, 19:49
Attached is a visual that should help. To keep it simple, I kept all my positive decimals between 0 and 1; since the question refers to the relative distance between the actual (S) and estimated (E) sets of sums, making them larger doesn't have any effect on the range of E  S. _
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Screen Shot 20171218 at 7.47.28 PM.png [ 203.95 KiB  Viewed 1162 times ]
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Re: List T consist of 30 positive decimals, none of which is an integer
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23 Dec 2017, 21:49
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 it is clear that in T 10 decimals have even tenths digit and 20 decimals have odd tenths digit. S is the sum of numbers in T Now let's try to find ES(max) and ES(min) to get the ranges. Here S is constant only E will change Now ES(max) will occur when there will be a maximum increment in E. This will happen when Even decimals are rounded up the max and odd decimals when rounded down has the least impact So lowest tenth digit Even decimals could be 0 for eg. 2.01 when rounded up becomes 3, an increment of 0.99 or 1 to be approx Hence increase from even decimals = 1*10=10 points Odd decimals has to be 0.1, when rounded down becomes 0, a decrease of 0.1 Hence decrease from Odd decimals = 0.1*20=2 Hence Net change i.e \(ES(max)=102=8\) Thus 6 is a possibility (if you take even 0.2 and odd as 0.1, you will get exact 6) and as 10>8 so 10 is not possible II holds trueNow ES(min) will be when even decimals increment is as low as possible and odd decimals are deceased the most So Even decimals has to be 0.899999 when rounded up becomes 1, an increase of 0.100001 or apprx 0.1 Hence increase from even decimals = 0.1*10=1 points Odd decimals has to be 0.99999, when rounded down becomes 0, a decrease of 0.999999 or approx 1 Hence decrease from Odd decimals = 1*20=20 Hence Net change i.e \(ES(min)=120=19\) so 16 is possible (if you take even to be 0.8 & odd to be 0.9, you will get exact 16) (I holds true)Hence Answer is B



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Re: List T consist of 30 positive decimals, none of which is an integer
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23 Dec 2017, 22:03
niks18 wrote: So Even decimals has to be 0.2 when rounded up becomes 1, an increment of 0.8
Hence increase from even decimals = 0.8*10=8 points The even decimals rounded up can actually equal (nearly) 10, not just 8, because 0.01 also has an even tenths digit (don't forget that zero, not 2, is the smallest even digit).
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Re: List T consist of 30 positive decimals, none of which is an integer
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23 Dec 2017, 22:45
mcelroytutoring wrote: niks18 wrote: So Even decimals has to be 0.2 when rounded up becomes 1, an increment of 0.8
Hence increase from even decimals = 0.8*10=8 points The even decimals rounded up can actually equal (nearly) 10, not just 8, because 0.01 also has an even tenths digit (don't forget that zero, not 2, is the smallest even digit). Hi mcelroytutoringYes 0.001 will have even tenths digit. actually the intention of my solution was to arrive at exact 6 & 16 to prove the option B is correct as this is a could be true question. I guess I should reword the solution to avoid ambiguity. Thanks for highlighting.



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Re: List T consist of 30 positive decimals, none of which is an integer
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23 Dec 2017, 23:12
Hi niks18, Sure, that makes senseboth strategies work.
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Re: List T consist of 30 positive decimals, none of which is an integer
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07 Jan 2018, 11:43
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= 2610 =16 MIN SE = 410= 6



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Re: List T consist of 30 positive decimals, none of which is an integer
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21 Feb 2018, 09:13
I can assume all of the 30 decimals are in the form of 0. something.
In this case, E = 10, S could never be 0 as all numbers are positive.
And S could be, 10 * 0,2 + 20 * 0.1 as a minimum. So ES = 6.
And S could be, 10 * 0,8 + 20 * 0,9. So ES = 16



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Re: List T consist of 30 positive decimals, none of which is an integer
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04 Mar 2018, 02:28
VeritasPrepKarishma wrote: 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
what is the best way to solve this question.
many thanks S 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. 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 actua0 is not a even number.. should'nt it be min 3.29999999



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Re: List T consist of 30 positive decimals, none of which is an integer
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04 Mar 2018, 03:09
qazi11 wrote:
0 is not a even number.. should'nt it be min 3.29999999
0 is an even integer. It is neither negative nor positive but it is even.
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Re: List T consist of 30 positive decimals, none of which is an integer
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07 Apr 2018, 07:44
Easier than it looks like:
E = 10 > 1/3 odd (rounded up)  1*10 + 2/3 even (rounded down) 0*20
ES: I)16 Possible > ES=10  10*0.6  20*0.5 = 16 II) 6 Possible > ES = 10  10*(0.2)  20*0.1 = 6 III) Not possible > ES = 10  a positive number = cannot be 10!!!
Hope it helps Matt



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Re: List T consist of 30 positive decimals, none of which is an integer
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07 Apr 2018, 10:27
Schawjibb wrote: 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/og13/218pro ... ition.htmlAlthough 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. thanks a lot for sharing the video explanation. Video explanations are a great resource for quicker preparation. I wish all 700+ level gmatclub questions have video explanations to them.



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Re: List T consist of 30 positive decimals, none of which is an integer
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14 Jun 2018, 17:25
a: the sum of 30 integers b: the sum of 30 decimals So : E=a+10,0 S= a,b ES= 10 + (0b)
b max: 08*10 + 0.9 *20 = 26 b min: 0.2*10 + 0.1 * 20 = 6
So the range of ES = [16,6]



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Re: List T consist of 30 positive decimals, none of which is an integer
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17 Jun 2018, 13:25
I've read through most of the solutions here, but nobody has really stated the obvious rule that actually makes this problem solvable:
1. The laws of this problem are absolute. If you find any solution for any set of numbers, it applies to every possible combination of numbers. Therefore, if you find that ES=10 for one set of example numbers, ES=10 is a solution for the entire problem space. No exceptions.
2. Therefore, since #1 is true, choose the simplest set of example numbers you can.
That's why the solution is easiest by choosing the following numbers: Round Up: 0.2, 0.4, 0.6, 0.8 Round Down: 0.1, 0.3, 0.5, 0.7, 0.9
Because whatever differences you find using the above numbers, will apply to every possible combination of other numbers! 1.1, 1000.1, 123413546.1  it doesn't matter!
If you can understand that, now solve the problem.



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Re: List T consist of 30 positive decimals, none of which is an integer
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22 Jun 2018, 23:03
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 \(E − S = 40 − 34 = 6\). Answer: B.
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Re: List T consist of 30 positive decimals, none of which is an integer
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11 Sep 2018, 11:16
Please explain why we didn't consider 0 (which is an even number) which calculating minimum E? As someone said earlier, the number could have been 1.03, in which case the maximum contribution for this would be 1*10 =10



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Re: List T consist of 30 positive decimals, none of which is an integer
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04 Oct 2018, 08:43
VeritasKarishma wrote: 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
what is the best way to solve this question.
many thanks S 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. are such type of questions solvable under the GMAT time constraints ? I think it will take 12 min to understand the question itself
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