Source: Official Guide for the GMAT 13th Ed. Problem Solving; #218 Official Guide for the GMAT 2015 14th Ed. Problem Solving; #218

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# List T consists of 30 positive decimals

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 decimals 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?

### 6 Explanations

3

Jose Maria Cao

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 E-S = 6.

And S could be, 10 * 0,8 + 20 * 0,9. So E-S = -16

May 22, 2016 • Comment

Cydney Seigerman, Magoosh Tutor

Nice work. Thanks for sharing your solution! :)

5

Debarati Das Gupta

I think of this as a tough Min Max Problem. Well as a student taking the test, I look at this question in one way i.e. where all the decimal numbers are different. I mean they can all be different i.e. 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.1, 2.2, 2.3, 2.8, 2.5, 2.6, 2.7, 2.8, 2.9, 3.1, 3.2, 3.3, 3.4, 3.5, 3.7, 3.9, 4.1, 4.3, 4.5, 4.7, 4.9

Here, for the distinct set, we've 10 even tenth decimal digit numbers, i.e,
1.2, 1,4, 1.6, 1.8 who will be rounded up to 2*4 = 8
2.2, 2.8, 2.6, 2.8 who will be rounded up to 3*4 = 12
3.2, 3.4 who will be rounded up to 4*2 = 8

the distinct odd integers are:
1.1, 1.3, 1.5, 1.7, 1.9 who will be rounded down to 1*5 = 5
2.1, 2.3, 2.5, 2.5, 2.9 who will be rounded down to 2*5 = 10
3.1, 3.3, 3.5, 3.7, 3.9 who will be rounded down to 3*5 = 15
4.1, 4.3, 4.5, 4.7, 4.9 who will be rounded down to 4*5 = 20

Now S = 83
and E = 78
which gives us E-S = -5
Now the problem with distinct decimal numbers is that it is cumbersome to take everything into account. During the GMAT we just have 1:20mins to 2:00mins to solve. So we can't afford to take distinct values.
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Alternate Route: Take one value for even and one value for odd that is repetitive.

Suppose 2.1 is taken as the odd tenth place decimal value 20 times.
and 2.2 is taken as the even place decimal value 10 times.

but again the qn arises : This will also give me only one value, then what should i do?
think like this: the left side of the decimal really doesnt matter, it can be anything 2 or 3 or 8 or whatever. its what is on the right of the decimal that counts.... So I need a RANGE of E-S.

So for that I'll pick 2.1 and 2.9 AND 2.2 and 2.8

Case 1: 2.2 10 times = 22
Rounded off will give me 30

2.1 20 times = 42
Rounded off will give me 40
E-S = 70 - 64 = 6 (one of the options!!)

Case 2: 2.8 10 times = 28
Rounded off will give 30

2.9 20 times = 58
Rounded off will give me 40

E-S = 70 - 86 = -16 (Again one of the options!)

So E-S can have I. and II. value not III (10)
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I've written it like this because as a student and fellow companion of every student like me prepping for GMAT, it is important to show that no thought process is wrong. The trick is to bring it on track. Many like me would have thought "oh!, I can take distinct values" but wait! Is the GMAT testing me for working out cumbersome numbers? NO!
So what can i do? Take repeated numbers! -> but that gives me only one value! So take the least and highest of the range (2.1...2.9) and (2.2....2.8) and work it out!

Hope this helps (someone who is like me, and thinks a little too much while solving problems, lol :) )

Jun 10, 2015 • Comment

Jonathan , Magoosh Tutor

Debarati, thanks for your message. You're right: you can also take specific numbers to find the minimum and maximum differences. 10 can be excluded by using (2.0+) and (2.1).

1 Jonathan , Magoosh Tutor

Hi Jon,
That's right -- as shown in the video, numbers with tenths digit of 0 would be rounded up. 0 is indeed an even integer :)

Mar 29, 2015 • Comment

2

siddhartha Bhardwaj

S= e + o (where e is the sum of even fractions and o is the sum of odd fractions)

E = 10 (as all odds omitted and all evens are turned to integers, which are 1/3 of 30 =10)
E-S= 10- (e+o)
We will find the min and max value of (e+o) to put E-S into interval of max and min.
The max even is 0.8 and max odd is 0.9
The min even is 0.2 and min odd is 0.1

To find max value of e+o assume all e to be max and all o to be max. so 10*0.8+20*0.9=26

To find min value of (e+o) assume all e to be min and all o to be min. so 10*0.2+20*0.1=4
Hence 4<= o+e<=16
(e+o) = 10- (E-S)

So 4<= 10- (E-S)<=16 is equal to 6=> E-S=>-16
Hence B

Aug 31, 2014 • Comment

Jon Kostem

Actually, just a friendly heads up, the lowest even decimal value should be .0, as zero is an even number :)

3

Isha ajmani

The difference between the Estimated and the actual Sum will be the sum of the gains and losses from the round ups and round downs respectively. So,
Round UP Round Down

Max 0.8 -(0.9)

Min 0.2 -(0.1)

For 10 even Max - 8, Min - 2
For 20 odd Max - (-18), Min - (-2)

Possibilities for (E-S)
= 8+ (-2) = 6 (II - Yes)

= 2 + (-18) = -16 (I - Yes)

But any combination of the sum of the gain from the 10 even round ups and the losses from the 20 odd round downs does not yields 10. Also, the max gain from rounding up is 0.8 for any term (8 for the 10 terms), so even if there is no loss from rounding down, 10 cannot be a possibility.

Jul 21, 2014 • Comment

7 Mike McGarry, Magoosh Tutor

Dec 28, 2013 • Comment