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

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

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18 Mar 2015, 05:10
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Difficulty:

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 ?

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

Kudos for a correct solution.

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

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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.

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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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.

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

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18 Mar 2015, 20:19
1
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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20 Mar 2015, 06:31
1
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.

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

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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 [ 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.

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

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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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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

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

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