If a + b 0 and (a+b) b, which of the following must be true?

Question

If a + b > 0 and √(a+b) > b, which of the following must be true? (A) a > 0 (B) b > 0 (C) a < 0 (D) b < 0 (E) None of these

chetan2u · Accepted Answer

Since there are no restrictions on values of a and b, we can have various combinations that will fit in. So E should be the answer. But let us solve it. \sqrt{a+b}>b Square both sides, a+b>b^2..........b^2-b or < 0, so let us show that a and b can take value of 0. 1) If b=0, then 0(0-1)0.......So a can be 2 and b can be 0, a+b=2+0=2>0. Eliminate options b<0 and b>0 and a<0. 2) Only option left is a<0. b(b-1)0.....let a=0 b+0>0.... So 0

CrackverbalGMAT · Answer

Since this is a must be true question and we have a couple of inequalities to test out, plugging in simple values can be a good strategy to answer this question.

Let a = 2 and b = 1. This satisfies both the inequalities since 3>0 and √3 > 1. We can eliminate answer options C and D since, with this case, we have proven them otherwise.

Let a = 2 and b = -1. This satisfies the inequalities since 1>0 and √1 > -1. We can eliminate answer option B.

Let a = -1 and b = 1.1. This satisfies the first inequality since 0.1>0 but it doesn’t satisfy the second since the left hand side will be smaller than the right hand side. This means that a CANNOT be a negative value.

Let a = 0 and b = 1. Again, this satisfies the first inequality but doesn’t satisfy the second. This means that a CANNOT be ZERO. 
Therefore, it is definitely true that a HAS to be a positive number i.e. a>0 must be true.

The correct answer option, IMHO, is A.

Hope that helps!
Aravind BT

LaveenaPanchal · Answer

Hi

Is there any method other than number plugging of values possible here. Can we figure out using number line or defining ranges..

LaveenaPanchal · Answer

Hi,

Can you please post the Official Answer.

Emdad · Answer

Official answer is given E and the explanation is "By trial and error method, we can see that a and b can both be positive or negative
and still satisfy the equation."
Thanks

chetan2u · Answer

You are missing out on a=0 and b=1/4..
0+(1/4)=1/4>0
 \sqrt{a+b}>b

\sqrt{0+\frac{1}{4}}>\frac{1}{4}

\sqrt{\frac{1}{4}}>\frac{1}{4}

\frac{1}{2}>\frac{1}{4}

So both inequalities satisfied.

Thus E is the answer.

santosh93 · Answer

Given \sqrt{a+b}>b Square both sides, a+b>b^2 but a+b>0 implies 0>b^2 So -10 and -11 Hence the answer is Option E PS : Option A) a>0 is not always true for given conditions, consider a=0.5 and the above equations fall apart.

CrackverbalGMAT · Answer

chetan2u ,

Thanks for the correction! I did miss out on Proper fractions indeed (probably because I focussed excessively on a = 0)

For proper fractions,  \sqrt{x}  > x. My bad! Correcting my answer.

The correct answer option is E.

Aravind B T

DrVanNostrand · Answer

I don't think we can square both the sides of the inequality as we don't know whether both the sides are non-negative or not.

chetan2u · Answer

Yes, we can’t square an inequality when we don’t know the sign.

This is a must be true question. 
So when we do not know anything about the variables and I am able to prove that a can be 0 for some value of b and b also can be 0 for some value of a, all the options fail. 
Since, we were able to eliminate all options, we don’t require to check for various combinations for  |b|>\sqrt{a+b}  when b<0, that is  a+b<b^2 .

DrVanNostrand · Answer

I don't think we can square both the sides of the inequality as we don't know whether both the sides are non-negative or not.[/quote]

Yes, we can’t square an inequality when we don’t know the sign.

This is a must be true question.
So when we do not know anything about the variables and I am able to prove that a can be 0 for some value of b and b also can be 0 for some value of a, all the options fail.
Since, we were able to eliminate all options, we don’t require to check for various combinations for \(|b|>\sqrt{a+b}\) when b<0, that is \(a+b<b^2\).[/quote]

Thanks for your various answers. I learn a lot from them. However, I don't get what you mean to say here. We can conclude that a>b and b can be
(-a<b<infinity). Can't I say that I still don't know the sign of B?

Crytiocanalyst · Answer

let us assume a=0 then b=1 then the equations still satisfies so we eleminate A and C

Similarly b=0 and a=4 still both the equations are satisfied therefore let us eleminate B and D

Therefore the only option left is E IMO E since all the posiibilities are accounted when the variables are taken to be zero

nasimbijoy · Answer

'' let us assume a=0 then b=1 then the equations still satisfies so we eleminate A and C''
if we do so, then the 2ns condition here which is √(a+b)>b, become 1>1?
then how does it satisfy?

iliagigauri · Answer

Thank you for the explanation, I agree that plugging in simple numbers is the best strategy for the question! it saves time and it's easily understandable, but the numbers you used to plug in are limited. with better examples you could've arrived to answer E.

If a + b > 0 and (a+b) > b, which of the following must be true?

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