List T OG

[massi2884]

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

[Bill@VeritasPrep]

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.

[ronnie1985]

These type of questions appear in GMAT???

[raunekk]

@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!

[Bill@VeritasPrep]

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

[massi2884]

It's from OG 13. Thanks Bill.

[yeloaw]

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.

[tanvis1120]

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.