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Is root{x} a prime number?
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Is \(\sqrt{x}\) a prime number? (1) \(3x7=2x+2\) (2) \(x^2=9x\)
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Originally posted by LM on 26 Jan 2012, 05:41.
Last edited by Bunuel on 26 Jan 2012, 06:29, edited 1 time in total.
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Is root{x} a prime number?
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17 Oct 2012, 14:17
abikumar wrote: 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. 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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Re: Square root is prime?
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26 Jan 2012, 06:28
LM wrote: Is \(\sqrt{x}\)
(1) \(3x7=2x+2\)
(2) \(x^2=9x\) Is \(\sqrt{x}\) a prime number? 1) \(3x7=2x+2\) > we have one check point 7/3 (check point  the value of x for which an expression in absolute value equals to zero): A. \(x\leq{\frac{7}{3}}\) > \(3x7\leq{0}\) hence \(3x7=(3x7)\) > \((3x7)=2x+2\) > \(x=1\) > \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) > \(3x7>0\) hence \(3x7=3x7\) > \(3x7=2x+2\) > \(x=9\) > \(\sqrt{9}=3=prime\). Two different answer. Not sufficient. 2) \(x^2=9x\) > \(x(x9)=0\) > \(x=0\) 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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Re: Is root{x} a prime number?
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17 Oct 2012, 13:11
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.



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Re: Is root{x} a prime number?
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17 Oct 2012, 21:34
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.



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Re: Is root{x} a prime number?
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20 Oct 2012, 17:49
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??



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Re: Is root{x} a prime number?
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Re: Is root{x} a prime number?
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18 Nov 2012, 21:46
Hi Bunuel, There is one confusion .In many of your posts you have suggested whenever we have modulus at one side (Foreg: LHS in the first statement here). Why can't we compare the RHS as below 2x +2 >=0(LHS absolute value so always above or equal zero) x >1 In this case answer could be different ie E Please suggest where I am doing wrong
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Re: Is root{x} a prime number?
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19 Nov 2012, 03:00



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Re: Is root{x} a prime number?
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20 Nov 2012, 08:09
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.
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Re: Is root{x} a prime number?
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03 Jan 2013, 22:12
solving 3x7 = 2x + 2 > x = 1 or 9 > Not Suff solving x^2 = 9x > x = 0 or 9 > Not Suff Together suff



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Re: Is root{x} a prime number?
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17 Jan 2013, 00:21
Apex231 wrote: prinkashar wrote: 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. any explanation for this? From statement 1, you only have two roots, x = 1 or 9. Its not a range of numbers.
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Re: Is root{x} a prime number?
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17 Jan 2013, 18:26
PraPon wrote: Apex231 wrote: prinkashar wrote: 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. any explanation for this? From statement 1, you only have two roots, x = 1 or 9. Its not a range of numbers. I am referring to one of the posts above which mentions following for stmt 1 2x +2 >=0(LHS absolute value so always above or equal zero) x >1 In this case answer could be different ie E



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Re: Is root{x} a prime number?
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17 Jan 2013, 19:58
Apex231 wrote: prinkashar wrote: 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. any explanation for this? 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. 3x7=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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Re: Is root{x} a prime number?
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20 Jan 2013, 21:24
VeritasPrepKarishma wrote: Apex231 wrote: prinkashar wrote: 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. any explanation for this? 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. 3x7=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. simply... for condition 1...square both sides... x=9 or x=1 we get two solutions for cndition 2..we get x=9 therefore, both statements are reqd.



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Re: Square root is prime?
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29 Nov 2013, 22:52
Bunuel wrote: LM wrote: Is \(\sqrt{x}\)
(1) \(3x7=2x+2\)
(2) \(x^2=9x\) Is \(\sqrt{x}\) a prime number? 1) \(3x7=2x+2\) > we have one check point 7/3 (check point  the value of x for which an expression in absolute value equals to zero): A. \(x\leq{\frac{7}{3}}\) > \(3x7\leq{0}\) hence \(3x7=(3x7)\) > \((3x7)=2x+2\) > \(x=1\) > \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) > \(3x7>0\) hence \(3x7=3x7\) > \(3x7=2x+2\) > \(x=9\) > \(\sqrt{9}=3=prime\). Two different answer. Not sufficient. 2) \(x^2=9x\) > \(x(x9)=0\) > \(x=0\) 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. Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done



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Re: Square root is prime?
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30 Nov 2013, 03:12
mohnish104 wrote: Bunuel wrote: LM wrote: Is \(\sqrt{x}\)
(1) \(3x7=2x+2\)
(2) \(x^2=9x\) Is \(\sqrt{x}\) a prime number? 1) \(3x7=2x+2\) > we have one check point 7/3 (check point  the value of x for which an expression in absolute value equals to zero): A. \(x\leq{\frac{7}{3}}\) > \(3x7\leq{0}\) hence \(3x7=(3x7)\) > \((3x7)=2x+2\) > \(x=1\) > \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) > \(3x7>0\) hence \(3x7=3x7\) > \(3x7=2x+2\) > \(x=9\) > \(\sqrt{9}=3=prime\). Two different answer. Not sufficient. 2) \(x^2=9x\) > \(x(x9)=0\) > \(x=0\) 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. Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done Both x=1 and x=9 are valid for (1). Please elaborate what you mean?
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Re: Is root{x} a prime number?
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24 Dec 2015, 23:24
Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done[/quote]
Both x=1 and x=9 are valid for (1). Please elaborate what you mean?[/quote]
Bunel, I too have the same doubt in my mind. I will try to explain it.
By taking condition, X>0, i got the value of X as 9
And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.
So i took X=9 only and got the answer as A.
Pls explain how X=1 with the condition X<0 is valid.
Thanks in advance



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Re: Is root{x} a prime number?
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24 Dec 2015, 23:29
mohnish104 wrote: Bunuel wrote: LM wrote: Is \(\sqrt{x}\)
(1) \(3x7=2x+2\)
(2) \(x^2=9x\) Is \(\sqrt{x}\) a prime number? 1) \(3x7=2x+2\) > we have one check point 7/3 (check point  the value of x for which an expression in absolute value equals to zero): A. \(x\leq{\frac{7}{3}}\) > \(3x7\leq{0}\) hence \(3x7=(3x7)\) > \((3x7)=2x+2\) > \(x=1\) > \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) > \(3x7>0\) hence \(3x7=3x7\) > \(3x7=2x+2\) > \(x=9\) > \(\sqrt{9}=3=prime\). Two different answer. Not sufficient. 2) \(x^2=9x\) > \(x(x9)=0\) > \(x=0\) 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. Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been doneBunel, I too have the same doubt in my mind. I will try to explain it. By taking condition, X>0, i got the value of X as 9 And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition. So i took X=9 only and got the answer as A. Pls explain how X=1 with the condition X<0 is valid.Thanks in advance



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Re: Is root{x} a prime number?
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27 Dec 2015, 03:31
ArunpriyanJ wrote: mohnish104 wrote: Bunuel wrote: Is \(\sqrt{x}\) a prime number?
1) \(3x7=2x+2\) > we have one check point 7/3 (check point  the value of x for which an expression in absolute value equals to zero):
A. \(x\leq{\frac{7}{3}}\) > \(3x7\leq{0}\) hence \(3x7=(3x7)\) > \((3x7)=2x+2\) > \(x=1\) > \(\sqrt{1}=1\neq{prime}\). B. A. \(x>{\frac{7}{3}}\) > \(3x7>0\) hence \(3x7=3x7\) > \(3x7=2x+2\) > \(x=9\) > \(\sqrt{9}=3=prime\). Two different answer. Not sufficient.
2) \(x^2=9x\) > \(x(x9)=0\) > \(x=0\) 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. Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been doneBunel, I too have the same doubt in my mind. I will try to explain it. By taking condition, X>0, i got the value of X as 9 And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition. So i took X=9 only and got the answer as A. Pls explain how X=1 with the condition X<0 is valid.Thanks in advance Why are you talking about x<0 and x>0 when the transition point is 7/3, not 0? You should brush up fundamentals on absolute value:
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Re: Is root{x} a prime number? &nbs
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