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If a and b are real numbers such that a percent of (a − 2b) when added

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Bunuel
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If a and b are real numbers such that a percent of (a − 2b) when added [#permalink] New post   19 Sep 2018, 23:58
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If a and b are real numbers such that a percent of (a − 2b) when added to b percent of b, the value obtained is 0, then which of the following statements must be true?

I. a = b
II. a + b = 0
III. a − b = 1

(A) Only I
(B) Only II
(C) Only III
(D) Only I and III
(E) Only II and III
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A
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T1101
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If a and b are real numbers such that a percent of (a − 2b) when added [#permalink] New post   20 Sep 2018, 00:48
I came up with \(\frac{a-2b}{100}\)+\(\frac{{b}^{2}}{100}\) = 0

\(b^2\)-2b+a=0

\(b^2\)=2b-a

And then by testing values only II is correct, hence B

But I am unsure if this is correct... Can anyone help?
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Re: If a and b are real numbers such that a percent of (a − 2b) when added [#permalink] New post   20 Sep 2018, 05:45
a/100(a-2b)+b/100*b=0

(a-b)^2/100=0
a=b
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Re: If a and b are real numbers such that a percent of (a − 2b) when added [#permalink] New post   20 Sep 2018, 06:45
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Bunuel wrote:
If a and b are real numbers such that a percent of (a − 2b) when added to b percent of b, the value obtained is 0, then which of the following statements must be true?

I. a = b
II. a + b = 0
III. a − b = 1

(A) Only I
(B) Only II
(C) Only III
(D) Only I and III
(E) Only II and III


a%(a-2b) + b%(b) = 0

1/100[a^2 -2ab+b^2]=0

a^2 -2ab+b^2 = 0

(a-b)^2=0

a=b.

Hence A.

Cheers!
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Re: If a and b are real numbers such that a percent of (a − 2b) when added [#permalink] New post   20 Sep 2018, 08:15
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Given: a*(a-2b)/100 + (b/100)*b=0. Therefore, a^2 -2ab + b^2=0
(a-b)^2=0, (a-b)=0, hence a=b.

IMO A.
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Re: If a and b are real numbers such that a percent of (a − 2b) when added [#permalink]
20 Sep 2018, 08:15
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