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

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Math Expert
Joined: 02 Sep 2009
Posts: 52434
If a and b are real numbers such that a percent of (a − 2b) when added  [#permalink]

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19 Sep 2018, 22:58
00:00

Difficulty:

35% (medium)

Question Stats:

69% (01:51) correct 31% (02:12) wrong based on 49 sessions

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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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Manager
Joined: 07 Aug 2018
Posts: 110
Location: United States (MA)
GMAT 1: 560 Q39 V28
GMAT 2: 670 Q48 V34
If a and b are real numbers such that a percent of (a − 2b) when added  [#permalink]

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19 Sep 2018, 23:48
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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Intern
Joined: 23 Oct 2013
Posts: 14
Re: If a and b are real numbers such that a percent of (a − 2b) when added  [#permalink]

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20 Sep 2018, 04:45
a/100(a-2b)+b/100*b=0

(a-b)^2/100=0
a=b
Manager
Joined: 02 Aug 2015
Posts: 138
Re: If a and b are real numbers such that a percent of (a − 2b) when added  [#permalink]

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20 Sep 2018, 05:45
1
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!
Senior Manager
Joined: 15 Oct 2017
Posts: 312
GMAT 1: 560 Q42 V25
GMAT 2: 570 Q43 V27
GMAT 3: 710 Q49 V39
Re: If a and b are real numbers such that a percent of (a − 2b) when added  [#permalink]

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20 Sep 2018, 07:15
1
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.
Re: If a and b are real numbers such that a percent of (a − 2b) when added &nbs [#permalink] 20 Sep 2018, 07:15
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