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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
55% (01:36) correct
45%
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abikumar
The red part is not correct.
The point here is that square root function cannot give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{25}=5\) (not +5 and -5). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.
Hope it's clear.
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LM
Is \(\sqrt{x}\) a prime number?
1) \(|3x-7|=2x+2\) --> we have one check point 7/3 (check point - the value of x for which an expression in absolute value equals to zero):
A. \(x\leq{\frac{7}{3}}\) --> \(3x-7\leq{0}\) hence \(|3x-7|=-(3x-7)\) --> \(-(3x-7)=2x+2\) --> \(x=1\) --> \(\sqrt{1}=1\neq{prime}\).
B. A. \(x>{\frac{7}{3}}\) --> \(3x-7>0\) hence \(|3x-7|=3x-7\) --> \(3x-7=2x+2\) --> \(x=9\) --> \(\sqrt{9}=3=prime\).
Two different answer. Not sufficient.
2) \(x^2=9x\) --> \(x(x-9)=0\) --> \(x=0\) or \(x=9\) --> \(\sqrt{0}=0\neq{prime}\) or \(\sqrt{9}=3=prime\). Not sufficient.
(1)+(2) Intersection of values from (1) and (2) is \(x=9\) --> \(\sqrt{9}=3=prime\). Sufficient.
Answer: C.
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General Discussion
You guys have taken Sqrt(9) as 3 where as it should be plus or minus 3. In this case, statements (1) and (2) taken together wont be sufficient hence the answer should be E. Please explain. I know OA is C but it may be wrong.
Note that from statement 1: x>7/3 but root of x is not required to be greater than 7/3 or in fact sqrt(x) has no conditions on it so that logic wont work too.
Note that from statement 1: x>7/3 but root of x is not required to be greater than 7/3 or in fact sqrt(x) has no conditions on it so that logic wont work too.
ohh yes. Thanks Bunuel. I read other articles and realized what you have said is followed by GMAT. Anyways, i appreciate the explanation, thanks for the information.
Please correct me if i am wrong
x\sqrt{2}=9x
if we divide both sides by x then we get x= 9 which makes B sufficient.
isnt it??
x\sqrt{2}=9x
if we divide both sides by x then we get x= 9 which makes B sufficient.
isnt it??
manjusu
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.
So, if you divide (reduce) \(x^2=9x\) by \(x\), you assume, with no ground for it, that \(x\) does not equal to zero thus exclude a possible solution (notice that both x=9 AND x=0 satisfy the equation).
Hope it's clear.
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As from the first statement x >-1 so it can be 0 too.
Second statement gives values 0 and 3
Even after combining both statements, As we are not sure of answer ( 0 or 3 )
i.e why I concluded E.
I know am skipping few imp. concepts here please help.
Second statement gives values 0 and 3
Even after combining both statements, As we are not sure of answer ( 0 or 3 )
i.e why I concluded E.
I know am skipping few imp. concepts here please help.
Apex231
x > -1 implies that whatever value of x will satisfy this equation, it will be greater than -1. It does not mean that every value greater than -1 will satisfy it. You cannot take one part of an equation in isolation and solve from it.
|3x-7|=2x+2
Point is that no value of x less than -1 can satisfy this equation. But, it doesn't mean that every value greater than or equal to -1 will satisfy it.
When you solve this equation, you get x = 1 or 9 (both greater than -1). No other value of x satisfies this equation. If you put x = 0, you get 7 = 2 which is not true. So x cannot be 0.
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LM
Hey Bunuel KarishmaB, please correct my reasoning in statement 2:
given:- x^2 = 9x
if i take root both sides, then
\sqrt{x^2} can be written as |x|
so the equation becomes,
|x| = \sqrt{9x}
hence,
1] x = 3\sqrt{x}
2] x = -3\sqrt{x}
from 1]
\sqrt{x}*\sqrt{x} = 3\sqrt{x}
cancel \sqrt{x} both sides, we get \sqrt{x} = 3
from 2]
\sqrt{x}*\sqrt{x} = -3\sqrt{x}
cancel \sqrt{x} both sides, we get \sqrt{x} = -3
But since root value cannot be negative, hence only 1] remain, and from this i concluded that B is sufficeint.
DonBosco7
Your doubt is addressed here: https://gmatclub.com/forum/is-root-x-a- ... l#p1133668
Basically, you cannot reduce by \(\sqrt{x}\) because it can be 0.
P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.
DS questions in GMAT Focus encompass various types of word problems, such as:
- Word Problems
- Work Problems
- Distance Problems
- Mixture Problems
- Percent and Interest Problems
- Overlapping Sets Problems
- Statistics Problems
- Combination and Probability Problems
While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."
Check GMAT Syllabus for Focus Edition
You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.
So, you can ignore this question.
Hope it helps.
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