List T OG

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 95
Joined: Tue Sep 13, 2011 3:19 pm
Thanked: 3 times
GMAT Score:710

List T OG

by massi2884 » Sat Apr 07, 2012 7:21 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

OA is B but please explain.
Thanks

User avatar
GMAT Instructor
Posts: 1248
Joined: Thu Mar 29, 2012 2:57 pm
Location: Everywhere
Thanked: 503 times
Followed by:192 members
GMAT Score:780

by Bill@VeritasPrep » Sat Apr 07, 2012 8:28 am
I. If all 10 of the terms with even tenths digits have a tenths digit of 8, then rounding up to the nearest integer increases the term by .2; .2*10 gives a total increase of 2. If all 20 of the terms with odd tenths digits have a tenths digit of 9, then rounding down to the nearest integer decreases the term by .9; .9 * 20 gives a total decrease of 18. 2 - 18 = -16, which is the net decrease from the actual sum S to the estimated sum E. Thus, we can say that E - S could be -16.

II. If all 10 of the terms with even tenths digits have a tenths digit of 2, then rounding up to the nearest integer increases the term by .8; .8*10 gives a total increase of 8. If all 20 of the terms with odd tenths digits have a tenths digit of 1, then rounding down to the nearest integer decreases the term by.1; .1 * 20 gives a total decrease of 2. 8 - 2 = 6, which is the net increase from the actual sum S to the estimated sum E. Thus, we can say that E - S could be 6.

III. The situation in II shows the maximum increase possible, so there is no way to have E - S = 10.
Join Veritas Prep's 2010 Instructor of the Year, Matt Douglas for GMATT Mondays

Visit the Veritas Prep Blog

Try the FREE Veritas Prep Practice Test

User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 am
Location: Ahmedabad
Thanked: 31 times
Followed by:10 members

by ronnie1985 » Sat Apr 07, 2012 8:49 am
These type of questions appear in GMAT???
Follow your passion, Success as perceived by others shall follow you

Legendary Member
Posts: 1159
Joined: Wed Apr 16, 2008 10:35 pm
Thanked: 56 times

by raunekk » Sat Apr 07, 2012 10:33 am
@Bill

Can you please explain me the bold line. I am finding it difficult to figure how it would increase by 0.2

I. If all 10 of the terms with even tenths digits have a tenths digit of 8, then rounding up to the nearest integer increases the term by .2; .2*10 gives a total increase of 2. If all 20 of the terms with odd tenths digits have a tenths digit of 9, then rounding down to the nearest integer decreases the term by .9; .9 * 20 gives a total decrease of 18. 2 - 18 = -16, which is the net decrease from the actual sum S to the estimated sum E. Thus, we can say that E - S could be -16.


Thanks in advance!

User avatar
GMAT Instructor
Posts: 1248
Joined: Thu Mar 29, 2012 2:57 pm
Location: Everywhere
Thanked: 503 times
Followed by:192 members
GMAT Score:780

by Bill@VeritasPrep » Sat Apr 07, 2012 11:21 am
If we have a term of 7.8, for example, then rules state to round it up to the nearest integer, which is 8. 8 - 7.8 = .2
Join Veritas Prep's 2010 Instructor of the Year, Matt Douglas for GMATT Mondays

Visit the Veritas Prep Blog

Try the FREE Veritas Prep Practice Test

Senior | Next Rank: 100 Posts
Posts: 95
Joined: Tue Sep 13, 2011 3:19 pm
Thanked: 3 times
GMAT Score:710

by massi2884 » Sun Apr 08, 2012 8:21 am
It's from OG 13. Thanks Bill.

Senior | Next Rank: 100 Posts
Posts: 39
Joined: Mon Apr 06, 2009 1:25 pm
Thanked: 1 times

by yeloaw » Thu Apr 19, 2012 8:20 pm
My goodness, this problem took me 15 min.

It has to do with Max/min concepts.

Assume T = (1.a, 1.b,...etc) All units equal 1.xx.
E = 40 (due to rounding of ten even and 20 odd)
S max = 30 + 10(.8) + 20(.9) = 56
S min = 30 + 10(.2) + 20(.1) = 34

E-S min = 40 - 56 = -16
E-S max = 40 - 34 = 6
Thus, the min/max of E is -16 and 6, so I , II apply.

For (III. 10) to be true, S min/max in our equation above needs to equal 30. This is impossible since the question states T consists of 30 positive decimals. Therefore eliminate III.

User avatar
Senior | Next Rank: 100 Posts
Posts: 95
Joined: Sat Dec 28, 2013 8:53 am
Location: United States
Thanked: 2 times
Followed by:5 members

by tanvis1120 » Wed Sep 03, 2014 2:47 pm
My approach:

a. Since 1./3rd of the decimals in T have a tenths digit that is even, those will be rounded "up" to the nearest integer.
Example: 3.67 will be rounded up to 4; 5.28 to 6.


b. But since each decimal in T whose tenths digits is odd is rounded down to the nearest integer, it will look like:
5.37 rounded down to 5.
7.99 to 7.

from point (a), it is clear that the estimated sum of 1/3rd (10) of the decimals in the list T will exceed by 10, thus increasing the actual summation by 10.
So, E is always greater that S (E>S) by 10, i.e. E-S>0 Always

This made me eliminate the negative result for E-S.
So, the answer can be 6 and 10.