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Is x^2 - 4/3*x + 5/12 < 0 ?

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Is x^2 - 4/3*x + 5/12 < 0 ?  [#permalink]

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New post 18 Mar 2015, 05:10
1
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A
B
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D
E

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  95% (hard)

Question Stats:

42% (02:14) correct 58% (02:20) wrong based on 132 sessions

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Is x^2 - 4/3*x + 5/12 < 0 ?  [#permalink]

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New post Updated on: 18 Mar 2015, 17:44
1
(1) Not sufficient. If we take several values of x such as x=1/2, 1, and 0, we get two different answers to the question.
(2) Sufficient. If x = 1, then we get that 1^2-4/3*1+5/12 is positive, hence the answer to the question is no. If x=-1, then (-1)^2-4/3*(-1)+5/12 will also give a positive number and the answer no to the question.

Answer B

Originally posted by viktorija on 18 Mar 2015, 16:31.
Last edited by viktorija on 18 Mar 2015, 17:44, edited 1 time in total.
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Is x^2 - 4/3*x + 5/12 < 0 ?  [#permalink]

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New post 18 Mar 2015, 17:37
1
(Factoring the equation)
x^2 - 4/3*x + 5/12 < 0?
x^2 - 16/12*x + 5/12 < 0?
12x^2 - 16x + 5 < 0?
(6x - 5)*(2x - 1) < 0?

(1) If we use any positive or negative integer, the signs will be the same for (6x - 5) and (2x - 1) and the equation will be greater than zero. However, if we use x = 1/2, we get -2*0 = 0, so the statement is insufficient.

(2) As mentioned above, using any integer will leave the roots with the same sign, so the product will always be positive. This statement is sufficient.

The correct answer is B.
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Re: Is x^2 - 4/3*x + 5/12 < 0 ?  [#permalink]

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New post 18 Mar 2015, 20:19
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Answer: B
x^2 - 4/3*x + 5/12 < 0 => (X-1/2)(X-5/6)<0 => 1/2<X<5/6)
(1) 0 <= x => Dont know positive or negative
(2) x is an integer => surely it is positive with all the value of x
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Re: Is x^2 - 4/3*x + 5/12 < 0 ?  [#permalink]

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New post 20 Mar 2015, 06:31
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Bunuel wrote:
Is x^2 - 4/3*x + 5/12 < 0 ?

(1) 0 <= x
(2) x is an integer


Kudos for a correct solution.


Solving x^2 - 4/3*x + 5/12 < 0
++++++1/2------5/6+++++++
1/2<X<5/6 , note that all the number in this range will be a fraction.

(1) 0 <= x Insufficient as X can be 0 in which case 5/12 is not less than 0. X can be 4/6 in which case x^2 - 4/3*x + 5/12 < 0.
(2) x is an integer . as X do not fall in the above range , the inequality will always yield a positive value.

Answer B
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Is x^2 - 4/3*x + 5/12 < 0 ?  [#permalink]

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New post 23 Mar 2015, 06:10
Bunuel wrote:
Is x^2 - 4/3*x + 5/12 < 0 ?

(1) 0 <= x
(2) x is an integer


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is a very tricky fraction with quadratics. We know that the lead coefficient, the coefficient of x-squared, is positive, so this is an “upward-facing” parabola, curved side pointing down and “arms” going up.

We can compute that at x = 0, y = +5/12, a positive number. At x = 1, we get y = 1 - 4/3 + 5/12 = 1/12.

At both x = 0 and x = 1, the output is positive. If a = 1 is the coefficient of x-squared, and ( – 4/3) is the coefficient of x, then the line of symmetry of the parabola is given by x = -b/(2a) = (4/3)/2 = 2/3.

The vertex, the lowest point on the parabola, would be at that position. The y-value there would be: y = (2/3)^2 - 4/3*2/3 + 5/12 = 5/12 - 4/3.

We don’t need to continue the calculation any further. A number less than one minus a fraction greater than one will be negative. At x = 2/3, there is a vertex with negative y-value, but the parabola curves up, and by the time it gets to x = 0 or x = 1, it’s already positive and above the x-axis. OK, now we can look at the statements.

Statement #1: 0 ≤ x

Well, if x = 1, then we get a “no” answer, but if x = 2/3, then we get a “yes” answer. Two different answers to the prompt are possible. This statement, alone and by itself, is not sufficient.

Statement #2: x is an integer

This is interesting and subtle. We need to think about the shape of a parabola. This parabola is negative at x = 2/3 and the immediate vicinity, but by the time we get over to either adjacent integer, x = 0 or x = 1, the parabola is already positive.
Attachment:
ghdmpp_img15.png
ghdmpp_img15.png [ 8.49 KiB | Viewed 1975 times ]


Once a parabola is going up, it keeps going, so for any integer to the left of x = 0, or any integer to the right of x = 1, it also will be positive. It is positive for every integer value. That means, we can give a definitive answer of “no” to the prompt question. Because we were able to give a definitive answer, this statement, alone and by itself, is sufficient.

Answer = (B)
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Is x^2 - (4/3)x + (5/12) < 0?  [#permalink]

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New post 25 Mar 2015, 10:52
Is \(x^2 - \frac{4}{3}x + \frac{5}{12} < 0\)?

Statement #1: \(0 \leq x\)
Statement #2: x is an integer

For a set of 14 challenging DS questions, including the OE to this particular question, see:
http://magoosh.com/gmat/2015/gmat-data- ... questions/

Mike :-)
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Is x^2 - (4/3)x + (5/12) < 0?  [#permalink]

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New post 25 Mar 2015, 16:08
mikemcgarry wrote:
Is \(x^2 - \frac{4}{3}x + \frac{5}{12} < 0\)?

Statement #1: \(0 \leq x\)
Statement #2: x is an integer

For a set of 14 challenging DS questions, including the OE to this particular question, see:
http://magoosh.com/gmat/2015/gmat-data- ... questions/

Mike :-)


I tested numbers to solve this one

Statement 1:
when x=1/2, 0<0 is false
when x=3/4, (-7/16)+(5/12) < 0 is true
Insufficient

Statement 2:
when x=1, \(x^2 - \frac{4}{3}x + \frac{5}{12} < 0\) is false
Plugging more integers in for x, \(x^2 - \frac{4}{3}x + \frac{5}{12} < 0\) seems to be false for all cases
Sufficient

Answer: B
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Re: Is x^2 - 4/3*x + 5/12 < 0 ?  [#permalink]

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Re: Is x^2 - 4/3*x + 5/12 < 0 ?   [#permalink] 31 Dec 2018, 22:44
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