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# decimals

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Manager
Joined: 04 Sep 2006
Posts: 113

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31 May 2009, 05:14
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T is a set of 30 decimals, sum of which is S. All the decimals are classified to two groups: if the tenth¡¯ digit is even, the decimal rounds to the immediate greater integer (For example, 2.2=>3); if the tenth' digit is odd, the digits to the right of the decimal point are abandoned (For example, 2.1=>2). The sum of these integers is E. If 1/3 of the decimals have a even tenth's digit, which of the following could not be the value of E-S?
I. -16
II. 6
III. 10
Manager
Joined: 08 Feb 2009
Posts: 144
Schools: Anderson

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31 May 2009, 06:25
III, 10 cannot be the value of E-S.

$$\frac{1}{3}$$ of the decimals, or 10 decimals, have even tenth digit $$\Rightarrow$$ their actual value has been increased. i.e. 3.6 = 4
$$\frac{2}{3}$$ of the decimals, or 20 decimals, have odd tenth digit $$\Rightarrow$$ their actual value has been decreased. i.e. 3.5 = 3

When E has a maximum value, Maximum Positive value of E-S will result.

Lets assume the even decimals end in 0.2, which is rounded to the next highest integer $$\Rightarrow$$ there is a gain of 0.8 in each integer m]\Rightarrow[/m] Gain = 0.8 x 10 = +8
Lets assume the odd decimals end in 0.1, which is rounded to the next lowest integer $$\Rightarrow$$ there is a loss of 0.1 in each integer m]\Rightarrow[/m] Loss = 0.1 x 20 = -2
Therefore, net Gain = 8-2 = 6 = E-S.

When E has a Lowest value, Lowest value of E-S will result.

Lets assume the even decimals end in 0.8, which is rounded to the next highest integer $$\Rightarrow$$ there is a gain of 0.2 in each integer m]\Rightarrow[/m] Gain = 0.2 x 10 = +2
Lets assume the odd decimals end in 0.9, which is rounded to the next lowest integer $$\Rightarrow$$ there is a loss of 0.9 in each integer m]\Rightarrow[/m] Loss = 0.9 x 20 = -18
Therefore, net Loss = -18+2 = -16 = E-S.

The value of E-S has to be between 6 & -16.
Thus 10 is ruled out.
Manager
Joined: 04 Sep 2006
Posts: 113

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31 May 2009, 07:00
except this

\frac{1}{3} of the decimals, or 10 decimals, have even tenth digit \Rightarrow their actual value has been increased. i.e. 3.6 = 4 ( how did u get ? )
\frac{2}{3} of the decimals, or 20 decimals, have odd tenth digit \Rightarrow their actual value has been decreased. i.e. 3.5 = 3
other part of the solution sounds nice .
Manager
Joined: 08 Feb 2009
Posts: 144
Schools: Anderson

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31 May 2009, 07:08
vcbabu wrote:
except this

\frac{1}{3} of the decimals, or 10 decimals, have even tenth digit \Rightarrow their actual value has been increased. i.e. 3.6 = 4 ( how did u get ? )
\frac{2}{3} of the decimals, or 20 decimals, have odd tenth digit \Rightarrow their actual value has been decreased. i.e. 3.5 = 3
other part of the solution sounds nice .

I made changes to the solution I posted. Please go through the changes.
Senior Manager
Joined: 16 Jan 2009
Posts: 355
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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31 May 2009, 14:45
goldeneagle94 wrote:
III, 10 cannot be the value of E-S.

$$\frac{1}{3}$$ of the decimals, or 10 decimals, have even tenth digit $$\Rightarrow$$ their actual value has been increased. i.e. 3.6 = 4
$$\frac{2}{3}$$ of the decimals, or 20 decimals, have odd tenth digit $$\Rightarrow$$ their actual value has been decreased. i.e. 3.5 = 3

When E has a maximum value, Maximum Positive value of E-S will result.

Lets assume the even decimals end in 0.2, which is rounded to the next highest integer $$\Rightarrow$$ there is a gain of 0.8 in each integer m]\Rightarrow[/m] Gain = 0.8 x 10 = +8
Lets assume the odd decimals end in 0.1, which is rounded to the next lowest integer $$\Rightarrow$$ there is a loss of 0.1 in each integer m]\Rightarrow[/m] Loss = 0.1 x 20 = -2
Therefore, net Gain = 8-2 = 6 = E-S.

When E has a Lowest value, Lowest value of E-S will result.

Lets assume the even decimals end in 0.8, which is rounded to the next highest integer $$\Rightarrow$$ there is a gain of 0.2 in each integer m]\Rightarrow[/m] Gain = 0.2 x 10 = +2
Lets assume the odd decimals end in 0.9, which is rounded to the next lowest integer $$\Rightarrow$$ there is a loss of 0.9 in each integer m]\Rightarrow[/m] Loss = 0.9 x 20 = -18
Therefore, net Loss = -18+2 = -16 = E-S.

The value of E-S has to be between 6 & -16.
Thus 10 is ruled out.

Nice expln goldeneagle94.
IMO '10'
_________________

Lahoosaher

Intern
Joined: 12 May 2009
Posts: 46
Location: Mumbai

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01 Jun 2009, 04:37
IMO 10

E-S =>
(i) (10 * 0.8) - (20 * 0.1) = 8-2 = 6
(ii) (10 * 0.2) - (20 * 0.9) = 2 - 18 = -16,

Out of the options 10 is left out and 10 cannot be the value of E-S

Thanks,
Senior Manager
Joined: 08 Jan 2009
Posts: 324

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01 Jun 2009, 21:34
Great explaination goldeneagle94
Re: decimals   [#permalink] 01 Jun 2009, 21:34
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