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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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03 Jun 2016, 07:34
what if we assume the digits to be: 4.02 (0 is even) and 3.9 (9 is odd). Then in this case, ES comes to 8 which is absurd.
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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03 Jun 2016, 07:45
kanav06 wrote: what if we assume the digits to be: 4.02 (0 is even) and 3.9 (9 is odd). Then in this case, ES comes to 8 which is absurd.
Please help Hi 4.02... O is even, it moves up an dbecomes 5... 3.9 ... .9 is odd, it moves down and becomes 3.. ans ES =5+3  (4.02+3.9) = 87.92 = 0.8..
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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03 Jun 2016, 07:49
4.02 cannot be considered because technically it is an integer if rounded till the tenth place. And the question states none of the 30 positive decimals in the set are integers. Isn't that the case ? chetan2u
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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03 Jun 2016, 09:41
rishi02 wrote: 4.02 cannot be considered because technically it is an integer if rounded till the tenth place. And the question states none of the 30 positive decimals in the set are integers. Isn't that the case ? chetan2uHi, the Q states that  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 we apply it to 4.02.. decimal is .02, which has 0 in tenths place and 0 is EVEN.. It doesn't matter the tenths place is 0,2,4,6,8 all will lead to next higher integer..so .02 will take 4.02 to be rounded UP to nearest integer, which would be 5 in this case If we take 4.00, it is an integer but if it is 4.0987645... it will go up to 5
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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05 Jun 2016, 21:44
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? VeritasPrepKarishmaOn 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 integer [#permalink]
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15 Jul 2016, 04:46
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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. The answer is B.
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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19 Nov 2016, 06:10
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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 ns (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 ES= (20n+10(n+1))(10(n+0.1)+20(n+0.1n)= 20n+10n+1010nn20n2n=103n So ES=103n We know that n is a positive decimal and it cannot be an integer, so use the answer choices: Check 16 > 103n=16 > n=26/3, it works! Check 6 > 103n=6 > n=4/3, it works! Check 10 > 103n=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 integer [#permalink]
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19 Nov 2016, 06:26
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. The correct answer is .
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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19 Nov 2016, 06:40
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 ns (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 ES= (20n+10(n+1))(10(n+0.1)+20(n+0.1n)= 20n+10n+1010nn20n2n=103n So ES=103n We know that n is a positive decimal and it cannot be an integer, so use the answer choices: Check 16 > 103n=16 > n=26/3, it works! Check 6 > 103n=6 > n=4/3, it works! Check 10 > 103n=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 ns (no limitations in the question to do this):
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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19 Nov 2016, 07:20
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anairamitch1804Maybe 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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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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07 Dec 2016, 06:46
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 > 10D > 1030 => 10 > 10D > 20
so the desired range is 20 < 10D < 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 integer [#permalink]
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10 Dec 2016, 11:40
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 ES lies within (104) to (1026) 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 integer [#permalink]
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16 Feb 2017, 11:54
Option BRoundingoff(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, roundingoff(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 roundingoff exercise. :One way is to maximize and minimize the possible difference created by the roundingoff 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 integer [#permalink]
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16 Feb 2017, 11:58
E  S = a*(almost 1)  b*(almost 1)
with a in (0...10) and b in (0..20)
Max value of ES is 10*(almost 1) Min value of ES is 20*(almost 1)
Therefore, 19.999 <= ES <= 9.999
I and II are in that range, so it's letter B



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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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19 Apr 2017, 07:55
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 = +82 = 6 Even + Odd = +218 = 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 integer [#permalink]
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18 May 2017, 11:06
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
The answer is B.



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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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23 May 2017, 20:06
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 subscenarios: 1.1 and 2.1) ES=Gain+Loss=82=6 1.1 and 2.2) ES=Gain+Loss=818=16 1.2 and 2.1) ES=Gain+Loss=22=0 1.2 and 2.2) ES=Gain+Loss=218=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 integer [#permalink]
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17 Jul 2017, 20:05
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 ES. 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. ES can be in the range from just numerically less than 20 to just less than 10. So B is the answer
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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26 Jul 2017, 22:15
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 The solution is as mentioned in the picture attached.
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Re: List T consist of 30 positive decimals, none of which is an integer [#permalink]
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14 Nov 2017, 05:14
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 ES, so maximum value of the ES can be [max(increase)+min(decrease)] i.e. 8+(2)=6 so we can rule out value 10. Similarly minimum value of ES= [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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