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First of all notice that since \(x\) is a positive number then \(\sqrt{x} \gt 0\).

(1) \(x \lt 3\). Since \(x\) is a positive integer then \(x=1\) or \(x=2\). For both those values, the right hand side (\(2.5x - 5\)) is less than or equal to zero, so it cannot be more than the left hand side (\(\sqrt{x}\)) which is positive. Hence the answer to the question is NO. Sufficient.

(2) \(x\) is a prime number. If \(x=2\) then the answer is NO but if \(x=11\) then the answer is YES. Not sufficient.

First of all notice that since \(x\) is a positive number then \(\sqrt{x} \gt 0\).

(1) \(x \lt 3\). Since \(x\) is a positive integer then \(x=1\) or \(x=2\). For both those values, the right hand side (\(2.5x - 5\)) is less than or equal to zero, so it cannot be more than the left hand side (\(\sqrt{x}\)) which is positive. Hence the answer to the question is NO. Sufficient.

(2) \(x\) is a prime number. If \(x=2\) then the answer is NO but if \(x=11\) then the answer is YES. Not sufficient.

Answer: A

Hey Bunuel,

Thanks for your help on this forum. Maybe I am wrong, but doesn't x>0 only tell me that the square root has a real value? So, as far as (1) why don't you consider negative value of \sqrt{x}? For example \sqrt{2}=+ or - 1.4 and this could be < or > 0.

First of all notice that since \(x\) is a positive number then \(\sqrt{x} \gt 0\).

(1) \(x \lt 3\). Since \(x\) is a positive integer then \(x=1\) or \(x=2\). For both those values, the right hand side (\(2.5x - 5\)) is less than or equal to zero, so it cannot be more than the left hand side (\(\sqrt{x}\)) which is positive. Hence the answer to the question is NO. Sufficient.

(2) \(x\) is a prime number. If \(x=2\) then the answer is NO but if \(x=11\) then the answer is YES. Not sufficient.

Answer: A

Hey Bunuel,

Thanks for your help on this forum. Maybe I am wrong, but doesn't x>0 only tell me that the square root has a real value? So, as far as (1) why don't you consider negative value of \sqrt{x}? For example \sqrt{2}=+ or - 1.4 and this could be < or > 0.

Thanks!

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 positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3; \(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice though, 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 \(x=\sqrt{9}=3\).
_________________

Question stem asks if √x <2.5x−5 This means, we need to prove if 2.5x - 5>0 => if x > 2? (we don't have equality sign because question mentions that x is a positive integer. This means >0).

Statement 1. x <3 As x is a positive integer, possible values of x are 1 and 2 and none of these is larger than 2. So answer to the question asked is NO. - Sufficient

Statement 2. x is a prime number. x can be any prime number including 2 so we don't know if x>2 or not. - Insufficient

Question stem asks if √x <2.5x−5 we know x is positive , so squaring both sides gives x< 6.25x^2 + 25 - 25x i.e 26x< 6.25x^2 + 25

so if put 1 answer is yes. for x = 2 answer is no.How can answer be A

Why are you squaring? We can only square inequality when we are certain that both sides are non-negative. We don't know that.

Also, you really do not need to square put x = 1 and x = 2 into √x <2.5x −5. What do you get?

Hey Bunuel- I know there is no need to square here as the stem is rather simple. But just to correct you, you can actually square in this case. LHS is a square root, which is definitely positive. RHS is given to be greater than LHS, so obviously RHS is greater than a positive number, so there is no problem with squaring this inequality. Thanks
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we shall fight on the beaches, we shall fight on the landing grounds, we shall fight in the fields and in the streets, we shall fight in the hills; we shall never surrender!

Question stem asks if √x <2.5x−5 we know x is positive , so squaring both sides gives x< 6.25x^2 + 25 - 25x i.e 26x< 6.25x^2 + 25

so if put 1 answer is yes. for x = 2 answer is no.How can answer be A

Why are you squaring? We can only square inequality when we are certain that both sides are non-negative. We don't know that.

Also, you really do not need to square put x = 1 and x = 2 into √x <2.5x −5. What do you get?

Hey Bunuel- I know there is no need to square here as the stem is rather simple. But just to correct you, you can actually square in this case. LHS is a square root, which is definitely positive. RHS is given to be greater than LHS, so obviously RHS is greater than a positive number, so there is no problem with squaring this inequality. Thanks

No, you are not right. If you tried and solve with and without squaring you'd get different ranges.

Simple example: \(\sqrt{x}<x\) --> x>1

But if you square you'll get: \(x<x^2\) --> x<0 or x>1.
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