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Tough Number properties

Expert replies
by doclkk » Fri Jul 24, 2009 9:32 pm
Is A^2+A>-B^2?

1. A^2+B^2=1
2. A>0

(B)

My thoughts:

I'm trying to pick numbers but I'm not making sense of statement 1.

I skip to statement 2.

So I know that -B^2 is always negative. So This means B has to be true.

Answer is B/D

This is where I get stuck.

B^2 = 1 - A^2

Any thoughts?
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Source: — Data Sufficiency |

by rahulmehra13 » Sat Jul 25, 2009 12:19 am
IMO B

qn : is A^2 + A > - B^2

stmnt 1:

A^2 + B^2 = 1
=> A^2 - 1 = - B^2

Now to check whether A^2 + A > - B^2 we can check:
whether A^2 + A > A^2 - 1 ?
or A > -1 ?

as A could be any number, could be > -1 or < -1 or = -1

So stmnt 1 is insufficient.
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Re: Tough Number properties

by ankitns » Sat Jul 25, 2009 3:19 pm
So you solved for statement 2 and got stuck with either B or D.

For statement 1, go with simple numbers..
Say B^2 = 0, hence A^ = 1. A can be either -1 or 1

If A = -1

when we solve the equation A^2+A>-B^2 we get 0 > 0? No.

If A = 1

when we solve the equation A^2+A>-B^2 we get 2 > 0? Yes.

Hence NOT sufficient. So this will leave you with only B.
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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by navalpike » Fri Jul 31, 2009 10:13 am
Does the question say –(B^2) or (-B^2)? I am assuming it is the former.

Then S2 simply says that the left side of the equation is positive. In that case, no matter what B is, the right side will be negative or 0. Sufficient.
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by cubicle_bound_misfit » Sat Aug 01, 2009 6:06 am
simplify

the inequality becomes

is a2 +b2+a>0

we know a2 and b2 is always >0

hence question is asking is a>0 ?

that is what stmt 2 says.

hence B
Cubicle Bound Misfit
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by mohitsharda » Sat Aug 01, 2009 9:06 am
rahulmehra13 wrote:IMO B

qn : is A^2 + A > - B^2

stmnt 1:

A^2 + B^2 = 1
=> A^2 - 1 = - B^2

Now to check whether A^2 + A > - B^2 we can check:
whether A^2 + A > A^2 - 1 ?
or A > -1 ?

as A could be any number, could be > -1 or < -1 or = -1

So stmnt 1 is insufficient.

I disagree here 'coz A^2+B^2 = 1
as both A^2 and B^2 is positive... they need to be less than 1 or at max 1
So, A=< 1

So, 1 will be sufficient
MS
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by doclkk » Sun Aug 02, 2009 8:06 am
mohitsharda wrote:
rahulmehra13 wrote:IMO B

qn : is A^2 + A > - B^2

stmnt 1:

A^2 + B^2 = 1
=> A^2 - 1 = - B^2

Now to check whether A^2 + A > - B^2 we can check:
whether A^2 + A > A^2 - 1 ?
or A > -1 ?

as A could be any number, could be > -1 or < -1 or = -1

So stmnt 1 is insufficient.

I disagree here 'coz A^2+B^2 = 1
as both A^2 and B^2 is positive... they need to be less than 1 or at max 1
So, A=< 1

So, 1 will be sufficient
Rephrase the question

A^2 + B^2 + A > 0 (just adding B^2)

So at this point A could be > 0 or less than 0. You don't know. Right.

Think about A^2 + B^2 = 1. A could be positive -.707107 or positive
707107. The point is that squaring a number turns it into a positive so we don't know whether A > 0 or less than 0. Don't know at all.

The question reads: Is A^2+A>-B^2? We don't know if A is greater than 0 or less than 0 at this point. So if A =<1 then A is still insufficient because you can still answer Yes and No.

With statement B - after you rephrase the question A > 0. Bingo.
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