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Question Stats:
66% (01:47) correct
33% (00:44) wrong based on 1 sessions
If x is a positive integer, is \sqrt{x} \lt 2.5x - 5 ? 1. x \lt 32. x is a prime number Source: GMAT Club Tests - hardest GMAT questions
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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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I think serene is right.. root can be either +ve or -ve
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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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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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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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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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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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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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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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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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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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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 = 0If x = 3 then the inequality is satisfied, since \sqrt{3} is less than 2.5*3 - 5 = 2.5So Statement 2 alone is INSUFFICIENT.
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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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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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(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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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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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 -5In 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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A, since 1) x = 1 or 2 are both false. 2) x = 2 and 3 are false and true.
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