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m05 q15

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m05 q15 [#permalink] New post 28 Jan 2009, 05:26
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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

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Re: m05 q15 [#permalink] New post 28 Jan 2009, 11:40
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] New post 18 Feb 2009, 13:33
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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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Re: m05 q15 [#permalink] New post 19 Feb 2009, 08:22
I think serene is right.. root can be either +ve or -ve
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Re: m05 q15 [#permalink] New post 24 Aug 2009, 16:18
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] New post 24 Aug 2009, 20:06
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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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Re: m05 q15 [#permalink] New post 24 Aug 2009, 21:34
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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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Re: m05 q15 [#permalink] New post 25 May 2010, 08:24
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] New post 25 May 2010, 08:53
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] New post 25 May 2010, 18:04
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] New post 26 May 2010, 04:09
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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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Re: m05 q15 [#permalink] New post 26 May 2010, 09:41
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] New post 26 May 2010, 13:08
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] New post 26 May 2010, 13:15
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] New post 28 May 2010, 07:39
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] New post 13 Jun 2010, 03:22
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] New post 27 May 2011, 04:45
(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] New post 27 May 2011, 05:59
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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Re: m05 q15 [#permalink] New post 27 May 2011, 10:07
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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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Re: m05 q15 [#permalink] New post 29 May 2011, 03:01
A, since
1) x = 1 or 2 are both false.
2) x = 2 and 3 are false and true.
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Re: m05 q15   [#permalink] 29 May 2011, 03:01
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