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

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

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

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.

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)

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:
https://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

Taking 12 as the LCM of the denominators, the question stem can be rephrased as
Is \(\frac{(12 x^2 – 16x + 5) }{ 12 }\)< 0?

Since 12 is never less than 0, the question now becomes,
Is (12\( x^2\) – 16x + 5) <0?

This is a Quadratic inequality that can be solved by the Wavy curve method. For employing this method, we first factorise the Quadratic expression on the LHS.

12\(x^2\) – 6x – 10x + 5 = (6x – 5) (2x -1). The question can now be rephrased as,

Is (6x – 5) (2x – 1) < 0?

The zeroes of the Quadratic expression are (1/2) and (5/6).

When these are plotted on the number line, the parabola representing the expression will be below the water for all values of x in between (1/2) and (5/6).
For values greater than (5/6) or lesser than (1/2), the parabola will always be above the water (read as x-axis).

Therefore, the question can be rephrased as Is ½ < x < 5/6?

From statement I alone, 0≤ x or x ≥ 0.

Since we do not know what kind of a value x is, this could be in the target range or otherwise.
Statement I alone is insufficient. Answer options A and D can be eliminated.

From statement II alone, x is an integer.

This means that x is NEVER in the target range. This answers the question with a definite NO.

Statement II alone is sufficient. Answer options C and E can be eliminated.

The correct answer option is B.

Hope that helps!
Aravind BT

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