If a and b are constants, is the expression (x + b)/(x + b)^(1/2)

Question

If a and b are constants, is the expression  \frac{x+b}{\sqrt{x+a}}  defined for x = –2 ?

(1) a = 5

(2) b = 6

DS24571.01

BrentGMATPrepNow · Accepted Answer

Target question :    Is the expression defined for x = –2?  
This is a great candidate for  rephrasing the target question .

If x = -2, then the expression becomes  \frac{(-2)+b}{\sqrt{(-2)+a}}

There are two ways in which the expression  \frac{(-2)+b}{\sqrt{(-2)+a}}  is NOT defined:

case i) If a = 2, then the fraction's denominator is 0, which means the entire rational expression is NOT defined. 
case ii) If a < 2, then the value inside the square root sign is NEGATIVE, in which case the denominator is NOT defined. 
When we combine both cases, we see that the expression is NOT defined when a ≤ 2. .. 
Or we can say,  the expression IS defined when a > 2

REPHRASED target question :    Is a > 2?

Aside: the video below has tips on rephrasing the target question

Statement 1 : a = 5  
The answer to the REPHRASED target question is  YES, a IS greater than 2 
Since we can answer the  REPHRASED target question  with certainty, statement 1 is SUFFICIENT

Statement 2 : b = 6 
This does not help us answer the  REPHRASED target question .
Statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

RELATED VIDEO
 https://www.youtube.com/watch?v=MyG1K3ee69w

EMPOWERgmatRichC · Answer

Hi rebinh1982,

For the given fraction to be 'defined', the denominator CANNOT be 0. In addition, since we have a square-root in the denominator, we also cannot have a negative total (since that would create an imaginary number). As such, since we're given a value for X (re: -2), we need to know the value for A to answer this question. In simple terms, if A is less than or equal to 2, then the answer to the question is NO. If A is greater than 2, then the answer to the question is YES.

GMAT assassins aren't born, they're made,
Rich

globaldesi · Answer

Value of expression depends entirely onvalue of a
since a is what makes the expression either infinite, non real or real

\frac{x+b}{\sqrt{x+a}}

Stmt one tells a =5
which makes denominator real 
and value of expression becomes
  \frac{-2+b}{\sqrt{-2+5}}  =  \frac{-2+b}{\sqrt{3}}

hence sufficient

Stmt II gives no info for a
hence insufficient
thus answer A

alpacino · Answer

Necessary condition is (1) as allows to define the numerator (making sure no negative root arises). The denominator is already defined no matter what. Even if b=2, the whole fraction would result in 0 which is defined.
Condition (2) isn't useful as does not tell us anything about the denominator. Denominator could still be root of a negative number.

EMPOWERgmatRichC · Answer

Hi All,

We're told that A and B are constants. We're asked if the expression (X+B)/(√(X+A)) is "defined" for x = -2. This is a YES/NO question and is a great 'concept' question - meaning that you don't have to do much math if you understand the concepts involved. For the given fraction to be 'defined', the denominator CANNOT be 0. As such, since we're given a value for X (re: -2), we need to know the value for A to answer this question.

1) A = 5

Fact 1 gives us the value of A. With the given value of X, we know that the denominator will be a POSITIVE number, so the fraction will be defined and the answer to the question is ALWAYS YES. 
Fact 1 is SUFFICIENT

2) B = 6

Fact 2 gives us information to define the numerator of the fraction (re: -2 + 6 = +4), but not the denominator. 
IF.... 
A = +2, then the answer to the question is NO 
A = +3, then the answer to the question is YES 
Fact 2 is INSUFFICIENT

Final Answer:  A

GMAT assassins aren't born, they're made, 
Rich

rebinh1982 · Answer

Can some one please explain what ''defined'' mean in this context or rephrase this question in an easier way?

Does defined mean terminating? Does the question mean whether this fraction becomes a terminating decimal?

Thanks in advance!

Ekland · Answer

In layman arthmetic "Undefined" in math means that the math CANNOT be solved. If you put it in your calculator for instance it says "math error". "Defined" means that it's solvable.

Examples of undefined or senseles or unmathematical expression are any number divided by zero, square root of any negative number, etc...

So the question is asking you, if X = -2, and 1) a = 5
2) b = 6, is the expression solvable mathematically?

It's basically asking you "Can you tell if this is sensible or senseless?"

Remember that any math has to be logical. You can't multiply any negative number by itself and still get negative.so you can't say square root of a negative number is equal to something.

In 1) we have enough info to tell that it's solvable or not. 
"a' has to be greater than 2 for the denominator to be solvable mathematically otherwise you will have a zero or negative inside the square root sign. Sqrt of zero is zero and zero as a denominator is unsolvable bcos you cant divide anything by nothing. Senseless. So it's sufficient.

2) is not sufficient bcos we don't know the value of "a" in the denominator. So it can be solvable or not.

A is answer

Posted from my mobile device

MHIKER · Answer

The expression will be defined if the denominator will not be "0"

(1) Given a = 5, x = -2 , the total denominator will be something nonzero. So, Option 1 is Sufficient.

(2) Only b's value is given. if a=2 then denominator will be "zero" and the expression will not be defined but if as like previous a = 5 then the expression will be defined. Insufficient.

Ans. A.

Mazba95 · Answer

We cannot calculate the square root of negative number or 0. To know whether the expression is defined or not when the value of x is -2, we need to know the value of a. If the value of a is less than or equal to 2, then the expression cannot be defined as the value of (x+a) or (-2+a) will be 0 or negative. So, we need to know the value of a. Value of b does not matter in this question.

Statement 1 is sufficient as it specify the value of a.

Statement 2 is not sufficient as we are not concerned about the value of b.

So, Answer is [A]

Statement 1 is sufficient as it specify the value of a.

Statement 2 is not sufficient as we are not concerned about the value of b.

So, Answer is

Fighter15 · Answer

Do we treat all imaginary numbers as 'undefined' on the GMAT? i.e. if the denominator is root of a negative number, the expression becomes undefined...

Bunuel · Answer

We cannot take square roots of negatives on the GMAT:  \sqrt{negative}  is undefined. All numbers on the GMAT are always real numbers by default.

ScottTargetTestPrep · Answer

Solution:

Question Stem Analysis:

We need to determine whether the expression (x + b) / √(x + a) is defined for x = -2. Notice that the expression is defined when x + a is positive. Therefore, as long as (-2 + a) is positive, the expression will be defined.

Statement One Alone:

Since a = 5, then -2 + a = 3, which is positive. So the expression is defined. Statement one alone is sufficient.

Statement Two Alone:

Since we don’t know anything about the value of a, we can’t determine whether the expression is defined or not. Statement two alone is not sufficient.

Answer: A

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=S6ofBeRPTSY

GMATinsight · Answer

Answer: Option A

Video solution by  GMATinsight

https://www.youtube.com/watch?v=v8ahZImj1qU

wickedvikram · Answer

What if value of b is 2? expression will be 0, since 0 by imaginary num or infinity is 0 (defined). So, I don't agree with the solution from OG. I think it should be C

Bunuel · Answer

If OG says that OA is A, then it's A. I remember only 1 OG quant question (among thousands) which had wrong OA.

For  \frac{-2+b}{\sqrt{-2+a}}  to be defined, -2+a must be positive, which means that -2 + a > 0 must be true so a > 2 must be true. (1) says that a = 5 > 2. So, the expression is defined.

(For (1) if b = 2, then we'd have  \frac{0}{\sqrt{3}} = 0 , which means that the expression is also defined for b = 2.)

KanikaJain · Answer

If a and b are constants, is the expression (x+b)/(x+a)^1/2 defined for x = –2 ?

(1) a = 5
(2) b = 6

for this to be a valid expression, x+a>0.

thus Statement 1.
hence A

bumpbot · Answer

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If a and b are constants, is the expression (x + b)/(x + b)^(1/2)

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If a and b are constants, is the expression (x + b)/(x + b)^(1/2)

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