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Quads+Inequalities

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Source: — Data Sufficiency |

by mallika hunsur » Wed Apr 01, 2015 6:32 am
mallika hunsur wrote:Is x^2 - (4/3)x + (5/12) < 0?
Statement #1: 0 =< x
Statement #2: x is an integer
I think it's B..?

Anyone..?

Mallika
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by DavidG@VeritasPrep » Wed Apr 01, 2015 6:49 am
Looks like a good questions for picking numbers.

S1: If x = 0, we get 'Is 5/12 < 0?' So this gives us a NO
If x = 2/3, we get 'Is (4/9) - (8/6) + (5/12) <0?' This gives us a YES. Not Sufficient; (No need to calculate the exact value here: our positive terms (4/9 and 5/12 are both less than 1/2, so they'll sum to something less than 1. Our negative term (-8/6) is less than -1, so we'll end up with a negative value.)


S2: We already know x= 0 gives a NO. Any negative x will make each term in the expression positive, so that will also be a NO. x= 1, gives us 1 - 4/3 + 5/12, or 12/12 - 16/12 + 5/12, or 1/12, so that's a NO. As x gets bigger, the expression will become a larger and larger positive. The answer is always NO, so S2 is Sufficient.

Answer is B
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by GMATGuruNY » Wed Apr 01, 2015 1:33 pm
mallika hunsur wrote:Is x² - (4/3)x + (5/12) < 0?
Statement #1: 0 ≤ x
Statement #2: x is an integer
The graph of y = ax² + bx + c, where a>0, is shaped like a U.

For those adept at factoring, an alternate approach:

x² - (4/3)x + (5/12) < 0
12x² - 16x + 5 < 0
(6x - 5)(2x - 1) < 0.

x-intercept 1:
6x - 5 = 0
x = 5/6.

x-intercept 2:
2x - 1 = 0
x = 1/2.

Since the graph is U-shaped and crosses the x-axis at 1/2 and 5/6, it looks like this:
Image

As indicated by the graph, x² - (4/3)x + (5/12) < 0 when x is between 1/2 and 5/6.
Question stem, rephrased:
Is 1/2 < x < 5/6?

Statement 1: 0 ≤ x
If x = 0, then the answer to the rephrased question stem is NO.
If x = 2/3, then the answer to the rephrased question stem is YES.
INSUFFICIENT.

Statement 2: x is an integer
Since it cannot be true that 1/2 < x < 5/6, the answer to the rephrased question stem is NO.
SUFFICIENT.

The correct answer is B.
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