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If \(x\) is a positive integer, is \(\sqrt{x} \lt 2.5x  5\) ? 1. \(x \lt 3\) 2. \(x\) is a prime number Source: GMAT Club Tests  hardest GMAT questions SOLUTION: m05q157517220.html#p1091067



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24 Aug 2009, 22:34
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ddtiku wrote: I think serene is right.. root can be either +ve or ve Thats not correct. In fact thats misleading statement. If root is already given without ve sign, then it is already +ve. For ex: If sqrt (x) = sqrt2, then it is +ve. If square is given and you are taking sqrt, then only one sqrt is +ve and another is ve and vice versa. For ex: If x^2 = 2, then x = either sqrt2 or (sqrt2).
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sher676 wrote: But the equation does not hold for x=1,2
x=1
sq1< 2.5(1) 5 1 <2.5, which is not correct
x=2
sq2<2.5(2) 5
sq2<0, again the equation does not hold
Am i missing somethin? You calculation is correct. and it means 1) can prove that "No, √x < 2.5x 5 is not correct", therefore choose A
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27 May 2011, 11:07
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thuylinh wrote: I go for E. 1.x<3 > x= 1 or 2. But both the two values don't match the first inequation > wrong. 2.Assume a and b is the root of an inequation. The inequation has two ranges of number satisfied: x<a or x>b, there for there are plenty of number can satisfied. Therefore, both are insufficient Please see subhashghosh's solution above. that is correct. "A" is indeed correct. Data Sufficiency is all about proving whether a given statement is sufficient to answer the question asked. This is a classic example of Yes/No type data sufficiency: Question asked: Is \(\sqrt{x} < 2.5x 5\) In words: Is root of x less than 2.5 times x minus 5? St1 is sufficient to answer the question asked with a definitive No. No, root of x IS NOT less than 2.5 times x minus 5. Sufficient. St2: answer may be Yes or No. Not definite. Not Sufficient. Ans: "A"
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18 Feb 2009, 14:33
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Isnt root of x either negative or positive? for x=2, the (root)of 2 < 0 which is true.



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26 May 2010, 05:09
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Quote: It's not.
Square of any value is positive. A value is square of its root. So, the value you find from its square root is always positive. So, x is 4 if sqrt(x) is 2 or 2. It can never be 4.
At the same time, if x = 4, x = 2^2 or x = 2^2. So, square root of x is 2 or 2. This says, square root of some value can be either positive or negative. Your logic is perfect, but this is a matter of convention. In GMAT math, the roots of x are expressed as \(+\sqrt{x}\) and \(\sqrt{x}\) \(\sqrt{x}\) itself is ALWAYS positive. I completely understand your logic, but if you don't accept this as a convention, you are bound to either get DS sums wrong or frown on several PS sums. For example, the solutions for \(x\) in the equation \(x^2 = 25\) are \(x=5\) and \(x=5\). HOWEVER, if \(x = 25\), then \(\sqrt{x} = 5\). PERIOD. Remember, convention not logic! Good luck!



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30 May 2011, 04:23
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seku wrote: Question: \(\sqrt{x} < (2.5x  5)\) and x is a positive integer By squaring on both sides, it can be simplified as \(x < (2.5x5)^2\). This is not always correct.
statement 1. x < 3, which means the possible values are 1 or 2 only substitute 1 in the simplified equation, \(1 < (2.5 (1)  5)^2 => 1 < (2.5)^2 => 1 < 6.25\) True substitute 2 in the simplified equation, \(2 < (2.5 (2)  5)^2 => 2 < (0)^2 => 2 < 0\) False NOT SUFFICIENT
statement 2. x is prime #, which means the possible values are 2,3,5,7 etc substitute 2 in the simplified equation, \(2 < (2.5 (2)  5)^2 => 2 < (0)^2 => 2 < 0\) False substitute 3 in the simplified equation, \(3 < (2.5 (3)  5)^2 => 3 < (2.5)^2 => 3 < 6.25\) True NOT SUFFICIENT
So, my pick was E. Any comments ? \(100<1\) \((100)^2>1^2\) \(0.1<1\) \((0.1)^2<1^2\) \(1<2\) \(1^2<2^2\) Thus, squaring both sides in inequality may give undesired result, esp when we don't know the signs of the expression on both sides. Something similar happened here: \(\sqrt{x}<{2.5*x5}\) 1 For x=1 \(\sqrt{1}<{2.5*15}\) No. \((\sqrt{1})^2<(2.5*15)^2\) 2 \(1<6.25\) Yes. Does this make statement 1 insufficient? No. It just proves the following: if \(a<b\) then, \(a^2<b^2\) may not be true.
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28 Jan 2009, 12:40
ConkergMat wrote: If x is a positive integer, is √x < (2.5x 5)?
1. x < 3 2. x is a prime number 1: x could be 1 or 2. suff. A.
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Re: m05 q15 [#permalink]
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19 Feb 2009, 09:22
I think serene is right.. root can be either +ve or ve



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24 Aug 2009, 17:18
But the equation does not hold for x=1,2
x=1
sq1< 2.5(1) 5 1 <2.5, which is not correct
x=2
sq2<2.5(2) 5
sq2<0, again the equation does not hold
Am i missing somethin?



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Re: m05 q15 [#permalink]
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25 May 2010, 09:24
GMAT TIGER wrote: ddtiku wrote: I think serene is right.. root can be either +ve or ve Thats not correct. In fact thats misleading statement. If root is already given without ve sign, then it is already +ve. For ex: If sqrt (x) = sqrt2, then it is +ve. If square is given and you are taking sqrt, then only one sqrt is +ve and another is ve and vice versa. For ex: If x^2 = 2, then x = either sqrt2 or (sqrt2). Consider X^2=4 , roots are x=+2 case2, if x= 4 & we have to find \(\sqrt{x}\) , so according to above quote if x is known to be +ve , value of \(\sqrt{x}\) shall be +2 only.... Generalizing : if X^even , then roots are =+x However if \(\sqrt{x}\) is asked and x is +ve , then we have only one root =+\(\sqrt{x}\) Is it right , i don't know .............
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Re: m05 q15 [#permalink]
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25 May 2010, 09:53
Yes, by definition square root function only yields positive numbers.
However, as stated above, if you are solving for roots you have to consider all cases.



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Re: m05 q15 [#permalink]
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25 May 2010, 19:04
gsothee wrote: GMAT TIGER wrote: ddtiku wrote: I think serene is right.. root can be either +ve or ve Thats not correct. In fact thats misleading statement. If root is already given without ve sign, then it is already +ve. For ex: If sqrt (x) = sqrt2, then it is +ve. If square is given and you are taking sqrt, then only one sqrt is +ve and another is ve and vice versa. For ex: If x^2 = 2, then x = either sqrt2 or (sqrt2). Consider X^2=4 , roots are x=+2 case2, if x= 4 & we have to find \(\sqrt{x}\) , so according to above quote if x is known to be +ve , value of \(\sqrt{x}\) shall be +2 only.... Generalizing : if X^even , then roots are =+x However if \(\sqrt{x}\) is asked and x is +ve , then we have only one root =+\(\sqrt{x}\) Is it right , i don't know ............. It's not. Square of any value is positive. A value is square of its root. So, the value you find from its square root is always positive. So, x is 4 if sqrt(x) is 2 or 2. It can never be 4. At the same time, if x = 4, x = 2^2 or x = 2^2. So, square root of x is 2 or 2. This says, square root of some value can be either positive or negative.



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Re: m05 q15 [#permalink]
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26 May 2010, 10:41
I found (1) is sufficient by testing 0,1, and 2 and for each the answer is No. Therefore it is sufficient.
For statement (2) there are prime integers that can yield a yes or a no and is therefore insufficient.



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Re: m05 q15 [#permalink]
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26 May 2010, 14:08
marcusaurelius wrote: I found (1) is sufficient by testing 0,1, and 2 and for each the answer is No. Therefore it is sufficient.
For statement (2) there are prime integers that can yield a yes or a no and is therefore insufficient. I chose D. Please can you give examples of prime numbers that give a "yes" and those that give a "no" in statement (2)
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Re: m05 q15 [#permalink]
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26 May 2010, 14:15
gmatbull wrote: marcusaurelius wrote: I found (1) is sufficient by testing 0,1, and 2 and for each the answer is No. Therefore it is sufficient.
For statement (2) there are prime integers that can yield a yes or a no and is therefore insufficient. I chose D. Please can you give examples of prime numbers that give a "yes" and those that give a "no" in statement (2) If \(x = 2\) then the inequality is not satisfied, since \(\sqrt{2}\) is not less than \(2.5*2  5 = 0\) If \(x = 3\) then the inequality is satisfied, since \(\sqrt{3}\) is less than \(2.5*3  5 = 2.5\) So Statement 2 alone is INSUFFICIENT.



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Re: m05 q15 [#permalink]
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28 May 2010, 08:39
I think its E
1. Statement 1: x<3 Hence x=1 or x=2
If x=1, 1<2.5(1)5 1<2.5, we have a unique answer
If x=2, √2=2.5(2)5 A square root will have a +ve value & a ve value So √2= +1.4 or 1.4 Hence, we have two answers: 1.4>0 & 1.4<0 So A is out.
2. Statement 2 will have multiple answers due to +ve & ve value of square roots So B &D are out.
3. If we combine both statements, we again arrive at x=2 & 1.4>0 & 1.4<0, So C is out.
Hence answer is E.
Please correct me if I'm wrong. Thanks.



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Re: m05 q15 [#permalink]
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13 Jun 2010, 04:22
I jumped to C, forgetting that there are only 3 positive integers < 3, which give same answer. Now I get i. Thanks.



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Re: m05 q15 [#permalink]
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27 May 2011, 05:45
(1) x can be 1 and 2 only, and the answer is definitive no for both. Sufficient. (2) for x = 2, answer is No, for x = 5, answer is yes Not sufficient. Answer  A
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Re: m05 q15 [#permalink]
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27 May 2011, 06:59
I go for E. 1.x<3 > x= 1 or 2. But both the two values don't match the first inequation > wrong. 2.Assume a and b is the root of an inequation. The inequation has two ranges of number satisfied: x<a or x>b, there for there are plenty of number can satisfied. Therefore, both are insufficient







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