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Inequalities - Range

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Inequalities - Range

by bml1105 » Sat Apr 12, 2014 8:48 am
I generally understand how they got the ranges (although the 0s confuse me a little), but the explanation in the guide just confused me about how I got to the ranges on my own.

Problem #1 (#3 - MGMAT Algebra - p 169)

If 4/x < 1/3, what is the possible range of values for x?

Problem #2 (#4 - MGMAT Algebra - p 169)

If 4/x < -1/3, what is the possible range of values for x?


Any help with the explanation would be greatly appreciated!
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by [email protected] » Sat Apr 12, 2014 11:55 am
Hi bml1105,

It's important to remember that the rules of math do not change, even if a situation is "complex looking."

The GMAT will test you on rules that you know, but in ways that you're sometimes not used to thinking about. Here, we're dealing with inequalities and fractions. You probably know lots about each individual category, but now you have to combine these math factoids and answer a question.

Sometimes that "math" comes down to how you "see" it.

In the first example (4/X < 1/3), we're asked for the possible range of values for X.

I "see" a fraction that is less than another fraction, so I'm going to think about 4/X....

ANY NEGATIVE value for X will make 4/X < 1/3, but what about the POSITIVE values?

4/4 = 1
4/8 = 1/2
4/12 = 1/3

As the denominator gets BIGGER, the fraction gets SMALLER.

4/12 = 1/3 but we need it to be LESS than 1/3, so X has to get BIGGER.

In this question, the range of values is X > 12 or X is ANY NEGATIVE value.

Here, the final range is X < 0 or X > 12
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In the second example (4/X < -1/3), we're also asked for the range of values of X.

Again, I "see" a fraction less than another fraction, BUT there's a negative sign, so I have to account for that.

Since we're looking for values that are less than -1/3, I know that we'll have to be dealing with other negatives, so X MUST be NEGATIVE.

4/-4 = -1
4/-8 = -1/2
4/-12 = -1/3

As the denominator becomes MORE NEGATIVE, the fraction gets CLOSER TO 0. So we need a negative denominator, but we're facing a limitation....

X > -12 BUT it must still be negative, so X is also < 0

Here, the final range is -12 < X < 0

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by GMATGuruNY » Sat Apr 12, 2014 12:06 pm
bml1105 wrote: If 4/x < 1/3, what are the possible range of values for x?
Determine the CRITICAL POINTS.
Critical points occur when:
Case 1: The given expression is UNDEFINED.
Case 2: The left-hand side is EQUAL to the right-hand side.

Case 1: 4/x is undefined when x=0.
Case 2: 4/x = 1/3
Cross-multiplying, we get:
x=12.

The critical points are x=0 and x=12.
To determine the range(s) where 4/x < 1/3, test one value TO THE LEFT and one value TO THE RIGHT of each critical point.

x<0:
Substituting x=-1 into 4/x < 1/3, we get:
4/-1 < 1/3
-4 < 1/3.
This works.
Thus, x<0 is a valid range.

0<x<12:
Substituting x=1 into 4/x < 1/3, we get:
4/1 < 1/3
4 < 1/3.
Doesn't work.
Thus, 0<x<12 is not a valid range.

x>12:
Substituting x=16 into 4/x < 1/3, we get:
4/16 < 1/3
1/4 < 1/3.
This works.
Thus, x>12 is a valid range.

Result:
4/x < 1/3 when x<0 and when x>12.
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by Matt@VeritasPrep » Sat Apr 12, 2014 1:22 pm
One last approach:

4/x < 1/3

Start by considering three cases: x is positive, x is zero, and x is negative.

If x is negative, then 4/x is negative, so 4/x will always be less than 1/3. So x can be any negative number.

If x is even, we have 4/0, which isn't allowed, so x cannot be 0.

If x is positive, we can rearrange the inequality. (Since we're assuming x is positive in this case, we can multiply both sides by x and not worry about having to flip the inequality.) Crossmultiplying gives us 4/(1/3) < x, or 12 < x.

So we've got our valid ranges: either x is any negative number (i.e. x < 0) or 12 < x.

Now let's do the same thing for 4/x < -1/3.

Suppose that x is positive. We'll first multiply both sides by -3, which gives us (4*-3)/x > 1. (Notice how multiplying both sides by a negative number FLIPS the inequality sign.) Now we multiply both sides by x, which is positive, to get -12 > x. But no positive number is less than -12! So there are NO positive solutions. (That should be intuitive: how could 4/positive be less than -1/3?)

Suppose that x is zero. This again gives us 4/0, so x is not 0.

Suppose that x is negative. We'll multiply both sides by -3, again getting (-12)/x > 1, then we'll multiply both sides by x. Since x is NEGATIVE, we again FLIP the inequality, giving us -12 < x. Since we assumed x is negative, we really have -12 < x < 0, as x must be less than 0 to be negative. That's our inequality, and we're done!
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by bml1105 » Sat Apr 12, 2014 4:58 pm
Thanks! Those are the answers I got, and I got it the same way as Rich. For some reason, when I was reading the explanation about flipping the signs, I got confused and thought I was doing it wrong because I hadn't flipped any signs.

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by [email protected] » Sat Apr 12, 2014 11:39 pm
Hi bml1105,

The GMAT tends to reward flexible thinkers. Most Quant questions are written in such a way that there are multiple ways to get to the correct answer. Clever Test Takers know all of the options and which is fastest for any given situation, so there's no "wrong way" to approach a question as long as you're getting the correct answer and not spending too much time on any one question. Most print materials offer just one explanation (due to size restrictions of the printed document), so you should keep an open mind about the alternative approaches that you might come up with (or that you'll find explained by the experts here).

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