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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). _________________

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.

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!

Question: \(\sqrt{x} < (2.5x - 5)\) and x is a positive integer By squaring on both sides, it can be simplified as \(x < (2.5x-5)^2\). 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:

\(\sqrt{x}<{2.5*x-5}\) ------------1 For x=1 \(\sqrt{1}<{2.5*1-5}\) No.

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 ............. _________________

Sun Tzu-Victorious warriors win first and then go to war, while defeated warriors go to war first and then seek to win.

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.

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