List T OG

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.

These type of questions appear in GMAT???

@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!

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

It's from OG 13. Thanks Bill.

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.

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.