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# 700+ Quant DS.

Author Message
Intern
Joined: 26 Jan 2010
Posts: 7

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09 Feb 2010, 09:55
Hey Fellows,
Try this question.

Is a^2 + b^2 >= 130?

1. a/b = 3/11.
2. a < 3.

OA
[Reveal] Spoiler:
E
.

Last edited by punzoo on 09 Feb 2010, 10:43, edited 1 time in total.
Director
Joined: 27 Jun 2008
Posts: 542
WE 1: Investment Banking - 6yrs

### Show Tags

09 Feb 2010, 10:00
punzoo wrote:
Hey Fellows,
Try this question.

Is a^2 + b^2 >= 130?

1. a/b = 3/7.
2. a < 3.

OA
[Reveal] Spoiler:
E
.

E

1) It's a ratio, so a & b can be any number. Insuff
2) Not helpful; a = -infinity to +2. Insuff

combine, you get nothing.
Manager
Joined: 10 Feb 2010
Posts: 191

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12 Feb 2010, 20:00
Ans:E

1. a/b = 3/11.
when a=6, b=22 then a^2 + b^2>130
when a=.6, b=2.2 then a^2 + b^2<130
So insufficient

2. a < 3.
when a=.6, b=2.2 then a^2 + b^2<130
when a=-6, b=-22 then a^2 + b^2>130
So insufficient
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35

### Show Tags

13 Feb 2010, 00:16
punzoo wrote:
Hey Fellows,
Try this question.

Is a^2 + b^2 >= 130?

1. a/b = 3/11.
2. a < 3.

OA
[Reveal] Spoiler:
E
.

I m confused between C n E. pls letme know where I m wrong.

clearly 2nd eq is not sufficient.
Take 1st eq... take a=3x and b =11x

put in the question.....we get $$9x^2 + 121x^2 >= 130$$

=> $$130x^2 >=130$$
$$x^2>=1$$, thus we need to know whether this satisfy or not.
If we consider eq 2... a<3 that means x<1

Thus this should be C...but one case when x=1 we hav not consider this might take it to E.
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Math Expert
Joined: 02 Sep 2009
Posts: 39702

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13 Feb 2010, 06:29
gurpreetsingh wrote:
punzoo wrote:
Hey Fellows,
Try this question.

Is a^2 + b^2 >= 130?

1. a/b = 3/11.
2. a < 3.

OA
[Reveal] Spoiler:
E
.

I m confused between C n E. pls letme know where I m wrong.

clearly 2nd eq is not sufficient.
Take 1st eq... take a=3x and b =11x

put in the question.....we get $$9x^2 + 121x^2 >= 130$$

=> $$130x^2 >=130$$
$$x^2>=1$$, thus we need to know whether this satisfy or not.
If we consider eq 2... a<3 that means x<1

Thus this should be C...but one case when x=1 we hav not consider this might take it to E.

You did everything right, except the last step: $$x^2\geq{1}$$ --> $$x\leq{-1}$$ or $$x\geq{1}$$ --> $$a\leq{-3}$$ or $$a\geq{3}$$. So still E.

You can use number plugging when considering statements together: For $$a=0.3$$ and $$b=1.1$$ answer is NO, but for $$a=-3$$ and $$b=-11$$ answer is YES.

Hope it helps.
_________________
Re: 700+ Quant DS.   [#permalink] 13 Feb 2010, 06:29
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# 700+ Quant DS.

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