List T consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The [/u]esimated 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 only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III
It's number 218 in OG on page 183.
I need help with this and the explanation plea.
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bnpetteway
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GMATGuruNY
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bnpetteway wrote:List T consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The [/u]esimated 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 only
(B) I and II only
(C) I and III only
(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.
Last edited by GMATGuruNY on Fri Jan 18, 2013 12:48 am, edited 1 time in total.
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vishugogo
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GmatGuruNY....not able to understand the following statement
To maximize the value of E, we want to 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).
To maximize the value of E, we want to 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).
GMATGuruNY wrote:bnpetteway wrote:List T consist of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The [/u]esimated 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 only
(B) I and II only
(C) I and III only
(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, we want to 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, we want to 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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Tommy Wallach
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Hey Vishugogo,
GMATGuru's explanation is spot on, but he did incorrectly write "maximize the value of E" in his explanation. What he meant is "maximize the value of E-S". In fact, E is never going to change. If we know there are going to be 10 numbers with an even tenths digit and 20 numbers with an odd tenths digits, based on the rules of rounding given, there will be 10 numbers that round to 1 and 20 numbers than round to 2, creating an E of 10.
So if we want to maximize E-S, we need the smallest possible S (so Guru chose .2 and .1, which are the smallest decimals that meet the tenths digit criteria). To minimize E-S, we need the biggest possible S. To get that, we need to choose the biggest possible numbers for our two decimals (technically, Guru did not choose the biggest possible decimals, but it makes no difference): .899999 and .999999:
S = 10(.8999999) + 20(.999999) = 9 + 20 = 29 (approximately)
10 - 29 = -19
This means -16 is in the range.
Hope that helps!
-t
GMATGuru's explanation is spot on, but he did incorrectly write "maximize the value of E" in his explanation. What he meant is "maximize the value of E-S". In fact, E is never going to change. If we know there are going to be 10 numbers with an even tenths digit and 20 numbers with an odd tenths digits, based on the rules of rounding given, there will be 10 numbers that round to 1 and 20 numbers than round to 2, creating an E of 10.
So if we want to maximize E-S, we need the smallest possible S (so Guru chose .2 and .1, which are the smallest decimals that meet the tenths digit criteria). To minimize E-S, we need the biggest possible S. To get that, we need to choose the biggest possible numbers for our two decimals (technically, Guru did not choose the biggest possible decimals, but it makes no difference): .899999 and .999999:
S = 10(.8999999) + 20(.999999) = 9 + 20 = 29 (approximately)
10 - 29 = -19
This means -16 is in the range.
Hope that helps!
-t
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just a quick clarification needed
I used 0.01 as lowest even decimal. In this case get the same answer just slightly different range.
please clarify which is correct the lowest even decimal number 0.2 or 0.01
I used 0.01 as lowest even decimal. In this case get the same answer just slightly different range.
please clarify which is correct the lowest even decimal number 0.2 or 0.01
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The least possible even tenths digit is 0, so decimals such as .01 and .001 -- even .00000000001 -- should be considered when we try to maximize E-S.J N wrote:just a quick clarification needed
I used 0.01 as lowest even decimal. In this case get the same answer just slightly different range.
please clarify which is correct the lowest even decimal number 0.2 or 0.01
I plugged in .2 to simplify the arithmetic.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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