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by Uva@90 » Sat Jul 13, 2013 8:50 pm
List T consists of 30 positive decimals, none of which is an integer, and 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 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 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 only

Could you please explain this question in simpler way.

Thanks in advance.
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by GMATGuruNY » Sun Jul 14, 2013 3:39 am
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 decimal 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?

I. -16
II. 6
III. 10

A) I
B) I and II
C) I and III
D) II and III only
E) I, II and III
Make the problem CONCRETE by plugging in easy values.
10 of the values must have a tenths digit that is EVEN, while the other 20 values must have a tenths digit that is ODD.
To make the math easy, let's not consider decimals beyond the tenths place.
Try to MAXIMIZE E-S and MINIMIZE E-S.

E-S MAXIMIZED:
To MAXIMIZE the value of E-S, we must MINIMIZE the value of S.
To minimize S, we must ROUND UP the even decimals as MUCH as possible (from .2 to the next highest integer) and ROUND DOWN the odd decimals as LITTLE as possible (from .1 to the next smallest integer).
Let S = 10(.2) + 20(.1) = 4.
In E, .2 is rounded up to 1 and .1 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MAXIMUM possible value of E-S = 10-4 = 6.

E-S MINIMIZED:
To MINIMIZE the value of E-S, we must MAXIMIZE the value of S.
To maximize S, we must ROUND UP the even decimals as LITTLE as possible (from .8 to the next highest integer) and ROUND DOWN the odd decimals as MUCH as possible (from .9 to the next smallest integer).
Let S = 10(.8) + 20(.9) = 26.
In E, .8 is rounded up to 1 and .9 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MINIMUM possible value of E-S = 10-26 = -16.

Since the MAXIMUM difference is 6 and the MINIMUM difference is -16, only I and II are possible values of E-S.

The correct answer is B.
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by Uva@90 » Sun Jul 14, 2013 8:32 am
GMATGuruNY wrote:
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 decimal 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?

I. -16
II. 6
III. 10

A) I
B) I and II
C) I and III
D) II and III only
E) I, II and III
Make the problem CONCRETE by plugging in easy values.
10 of the values must have a tenths digit that is EVEN, while the other 20 values must have a tenths digit that is ODD.
To make the math easy, let's not consider decimals beyond the tenths place.
Try to MAXIMIZE E-S and MINIMIZE E-S.

E-S MAXIMIZED:
To MAXIMIZE the value of E-S, we must MINIMIZE the value of S.
To minimize S, we must ROUND UP the even decimals as MUCH as possible (from .2 to the next highest integer) and ROUND DOWN the odd decimals as LITTLE as possible (from .1 to the next smallest integer).
Let S = 10(.2) + 20(.1) = 4.
In E, .2 is rounded up to 1 and .1 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MAXIMUM possible value of E-S = 10-4 = 6.

E-S MINIMIZED:
To MINIMIZE the value of E-S, we must MAXIMIZE the value of S.
To maximize S, we must ROUND UP the even decimals as LITTLE as possible (from .8 to the next highest integer) and ROUND DOWN the odd decimals as MUCH as possible (from .9 to the next smallest integer).
Let S = 10(.8) + 20(.9) = 26.
In E, .8 is rounded up to 1 and .9 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MINIMUM possible value of E-S = 10-26 = -16.

Since the MAXIMUM difference is 6 and the MINIMUM difference is -16, only I and II are possible values of E-S.

The correct answer is B.
Hi GMATGuruNY,
I have one doubt regarding the rounding up. In question they mentioned E is obtained by
Each decimal 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.
But in your solution you have obtained S by rounding up
To minimize S, we must ROUND UP the even decimals as MUCH as possible (from .2 to the next highest integer) and ROUND DOWN the odd decimals as LITTLE as possible (from .1 to the next smallest integer).
Let S = 10(.2) + 20(.1) = 4.
In E, .2 is rounded up to 1 and .1 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MAXIMUM possible value of E-S = 10-4 = 6.
Could you explain why you did like this ?

Correct me if I am wrong.
Regards,
Uva
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by GMATGuruNY » Sun Jul 14, 2013 10:34 am
Uva@90 wrote: Hi GMATGuruNY,
I have one doubt regarding the rounding up. In question they mentioned E is obtained by
Each decimal 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.
But in your solution you have obtained S by rounding up
To minimize S, we must ROUND UP the even decimals as MUCH as possible (from .2 to the next highest integer) and ROUND DOWN the odd decimals as LITTLE as possible (from .1 to the next smallest integer).
Let S = 10(.2) + 20(.1) = 4.
In E, .2 is rounded up to 1 and .1 is rounded down to 0:
E = 10(1) + 20(0) = 10.
Thus, the MAXIMUM possible value of E-S = 10-4 = 6.
Could you explain why you did like this ?

Correct me if I am wrong.
Regards,
Uva
10 of the values must have an EVEN tenths digit:
0.2, 0.4, 0.6. or 0.8.
(Since we're considering decimals only up to the tenths place, I've excluded 0.0, which is an integer value.)

10 the values must have an ODD tenths digit:
0,1, 0.3., 0.5. 0.7, or 0.9.

When E-S is maximized and minimized, the value of E -- the ESTIMATED sum -- will be the same in each case.
The 10 even decimals will be rounded UP to 1, while the 20 odd decimals will be rounded DOWN to 0, yielding the following sum:
E = 10(1) + 20(0) = 10.

To MAXIMIZE the value of E-S, we must MINIMIZE the value of S -- the ACTUAL sum of the decimals.
To minimize the value of S, we must use the SMALLEST decimal in each list.
The smallest value in the first list is the even decimal that is rounded up the MOST:
0.2.
The smallest value in the second list is the odd decimal that is rounded down the LEAST:
0.1.
Thus:
Least possible value of S = 10(.2) + 20(.1) = 4.
Thus:
Maximum value of E-S = 10-4 = 6.

To MINIMIZE the value of E-S, we must MAXIMIZE the value of S -- the ACTUAL sum of the decimals.
To maximize the value of S, we must use the GREATEST decimal in each list.
The greatest value in the first list is the even decimal that is rounded up the LEAST:
0.8.
The greatest value in the second list is the odd decimal that is rounded down the MOST:
0.9.
Thus:
Greatest possible value of S = 10(.8) + 20(.9) = 26.
Thus:
Minimum value of E-S = 10-26 = -16.
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by Uva@90 » Sun Jul 14, 2013 7:31 pm
GMATGuruNY: Thanks a lot.
You made so clear.

Regards,
Uva.
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