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68% (00:41) correct 32% (00:51) wrong based on 215 sessions
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16 Sep 2014, 00:44



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21 Oct 2015, 07:26
I think this is a highquality question and I agree with explanation. This and M0407 are duplicates



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28 Oct 2015, 00:54
How can a negative number be prime? Statement II should limit value to x=1 for prime number 2, I think.. please explain



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28 Oct 2015, 05:16



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17 Apr 2017, 06:02
what about the rule : 'When the GMAT provides square root sign for an even root,such as x^(1/2) or x^(1/4), then the only accepted answer is the positive root''? I am asking this because i used this rule to find only one solution (1) in the second statement. Is the rule only talking about x^(even power) ? thanks



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17 Apr 2017, 06:23



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14 Jun 2017, 08:29
I think this is a highquality question and I don't agree with the explanation. Since a prime number can't be negative, why we are considering x = 1? this should always be x =1. isn't it?



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14 Jun 2017, 08:36



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09 Jul 2018, 16:04
BunuelSince \sqrt{x} = lxl i.e. x can be +ve as well as ve, can you please make me understand why \sqrt{(1) (1)} is not equals to (1). We can take one negative value out of root, and multiple of 2*(1) would be negative i.e. nonprime number. Why we have to calculate \sqrt{(1) (1)} before taking value out of the root? I know this way it would be 2 and a prime number.



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09 Jul 2018, 19:28
Avinash2303 wrote: BunuelSince \sqrt{x} = lxl i.e. x can be +ve as well as ve, can you please make me understand why \sqrt{(1) (1)} is not equals to (1). We can take one negative value out of root, and multiple of 2*(1) would be negative i.e. nonprime number. Why we have to calculate \sqrt{(1) (1)} before taking value out of the root? I know this way it would be 2 and a prime number. When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the nonnegative root. That is: \(\sqrt{9} = 3\), NOT +3 or 3; \(\sqrt[4]{16} = 2\), NOT +2 or 2; Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and 3. Because \(x^2 = 9\) means that \(x =\sqrt{9}=3\) or \(x=\sqrt{9}=3\).
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08 Jan 2019, 06:08
why is it wrong to simplify the second statement to 2*x since the sq.root and the square cancel each other? If i do that then for the second statement to be true, only when x=1 will the result be a prime.



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08 Jan 2019, 06:40
ameralhajj wrote: why is it wrong to simplify the second statement to 2*x since the sq.root and the square cancel each other? If i do that then for the second statement to be true, only when x=1 will the result be a prime. Because \(\sqrt{x^2}=x\), not x. Try plugging in a negative numbers to check. Why does \(\sqrt{x^2}=x\)? The point here is that square root function cannot give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=x\)
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M2848
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09 Jan 2019, 00:14
Bunuel wrote: If \(x\) is an integer, what is the value of \(x\)?
(1) \(23x\) is a prime number.
(2) \(2\sqrt{x^2}\) is a prime number. #1 : x can be +/1 so insufficeint #2: again x can be +/1 in sufficeint IMO E
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