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

This topic has 1 expert reply and 2 member replies
alanforde800Maximus Senior | Next Rank: 100 Posts
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#### List T consist of 30 positive decimals, none of which is an

Sat Nov 19, 2016 4:36 am
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

I am totally horrified with the solution given in OG is there any alternate and best approach that can be applied to above problem??

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GMATGuruNY GMAT Instructor
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Sat Nov 19, 2016 7:44 am
I posted some advice and a solution here:
http://www.beatthegmat.com/og15-ps-218-t291281.html

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Marty Murray Legendary Member
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Sat Nov 19, 2016 5:12 am
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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Anaira Mitch Master | Next Rank: 500 Posts
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Sat Nov 19, 2016 4:42 am
Same here looking for some alternate approach that can be applied in 2 minutes.

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