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IrinaOK
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 02:22
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A 87/20
B 63/20
C 47/20
D 15/4
E 14/5



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Juaz
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 02:31
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IrinaOK
A 87/20
B 63/20
C 47/20
D 15/4
E 14/5
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A.
(x+3) - (y-1)
=> 15/4 + 3/5 = 87/20
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Ravshonbek
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 02:43
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IrinaOK
A 87/20
B 63/20
C 47/20
D 15/4
E 14/5
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hmm, did u post it for fun :) or is it tricky :)

sqrt(x+3)^2 - sqrt(y-1)^2=x+3-y+1=x+y+4=3/4-2/5+4=(15-8+80)/20=87/20
A
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IrinaOK
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 02:51
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IrinaOK
A 87/20
B 63/20
C 47/20
D 15/4
E 14/5
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Nope, I really can`t understand OA and OE.

AO is B.
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Ravshonbek
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:04
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IrinaOK
A 87/20
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tricky and brutal for me requires better thinking heads :)

we have not tested it to sqrt(-(x+3))^2 - sqrt(-(y-1))^2 which changes the answer

i think we should plug the numbers without further simplifying the equation.sqrt(x^2+6x+3) - sqrt(y^2-2y+1)
this will yield us 63/20
is it level 3 question?
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IrinaOK
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:07
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IrinaOK
A 87/20
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OE is

There are always two roots, one positive and one negative.
X+3 is the positive root, no doubt.
Y-1 is the negative root, (y-1 = -3/5 ; 1 -y = 3/5)
You need both positive roots, try and calculate:
(X + 3) - (1-Y) = X + 3 - 1 + Y = 3/4 +2 +2/5 = 63/20

can someone explain text in bold, please?
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IrinaOK
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:15
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Ravshonbek

is it level 3 question?
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level 3, does not mean the question requires knowledge of complex math concepts; level 3 indicates that test taker has higher chance of getting it wrong on the GMAT :btw
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:18
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B it is :)

sqrt(x^2 + 6*x + 9) - sqrt(y^2-2*y+1)
= sqrt((x+3)^2) - sqrt((y-1)^2)
= |x+3| - |y-1| as a definition of absolute value is |b| = sqrt(b^2)

Then, if x=3/4 and y=2/5, we have:
sqrt(x^2 + 6*x + 9) - sqrt(y^2-2*y+1)
= |3/4 + 3| - |2/5-1|
= |15/4| - |-3/5|
= 15/4 - 3/5
= (75-12)/(20)
= 63/20
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:19
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IrinaOK
Ravshonbek

is it level 3 question?

level 3, does not mean the question requires knowledge of complex math concepts; level 3 indicates that test taker has higher chance of getting it wrong on the GMAT :btw
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challenges perhaps? one can learn a lot from them
need to get them as well, later.
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:20
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IrinaOK
IrinaOK
A 87/20
B 63/20
C 47/20
D 15/4
E 14/5

OE is

There are always two roots, one positive and one negative.
X+3 is the positive root, no doubt.
Y-1 is the negative root, (y-1 = -3/5 ; 1 -y = 3/5)
You need both positive roots, try and calculate:
(X + 3) - (1-Y) = X + 3 - 1 + Y = 3/4 +2 +2/5 = 63/20

can someone explain text in bold, please?
Show more


The bold reflects my absolute values.... sqrt has to be always done on positive number (or 0 of course) :)
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IrinaOK
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:22
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Fig
B it is :)

sqrt(x^2 + 6*x + 9) - sqrt(y^2-2*y+1)
= sqrt((x+3)^2) - sqrt((y-1)^2)
= |x+3| - |y-1| as a definition of absolute value is |b| = sqrt(b^2)

Then, if x=3/4 and y=2/5, we have:
sqrt(x^2 + 6*x + 9) - sqrt(y^2-2*y+1)
= |3/4 + 3| - |2/5-1|
= |15/4| - |-3/5|
= 15/4 - 3/5
= (75-12)/(20)
= 63/20
Show more


Fig,

I will ask them to post your explanation as alternative OE, if you do not mind, it is much easier to comprehend.

Thank you. :flower:
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A 87/20 B 63/20 C 47/20 D 15/4 E 14/5 07 Oct 2007, 03:23
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IrinaOK
Fig
B it is :)

sqrt(x^2 + 6*x + 9) - sqrt(y^2-2*y+1)
= sqrt((x+3)^2) - sqrt((y-1)^2)
= |x+3| - |y-1| as a definition of absolute value is |b| = sqrt(b^2)

Then, if x=3/4 and y=2/5, we have:
sqrt(x^2 + 6*x + 9) - sqrt(y^2-2*y+1)
= |3/4 + 3| - |2/5-1|
= |15/4| - |-3/5|
= 15/4 - 3/5
= (75-12)/(20)
= 63/20

Fig,

I will ask them to post your explanation as alternative OE, if you do not mind, it is much easier to comprehend.

Thank you. :flower:
Show more


Go for it & your welcome :)



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!