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PS Inequalities

by gauravgundal » Mon Jul 12, 2010 5:54 pm
if 4< (7-x)/3, which of the following must be true?
1. 5 < x
2. |x+3|>2
3.-(x+5) is positive

A. 2 only
B. 3 only
C. 1 and 2 only
D. 3 and 2 only
E. 1 , 2 and 3 only

Solving the equation I get,
x < -5

1. Is not true

-----------------------------------------------------
x<-5
x+5 < 0 --- means x+5 is negative .Thus -(x+5) is positive.

3. Is true.
-----------------------------------------------------
2. |x+3|>2

x+3> 2 | -(x+3)>2
x>-1 | -x-3>2
| -5>x

This gives two inequalities
so how this equation can be true.

IS it the case that even if one of the solution satisfies the required solution i.e x<-5 ?The equation |x+3|>2 must be true for 4< (7-x)/3


expert please explain this.
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by amising6 » Mon Jul 12, 2010 6:23 pm
gauravgundal wrote:if 4< (7-x)/3, which of the following must be true?
1. 5 < x
2. |x+3|>2
3.-(x+5) is positive

A. 2 only
B. 3 only
C. 1 and 2 only
D. 3 and 2 only
E. 1 , 2 and 3 only

.
4< (7-x)/3
12<(7-x)
5<-x
-5>x

2 apeears most frequently so will start with option 2
|x+3|>2
-(x+3)>2 and +(x+3)>2
-x>5 and x>-1
x<-5 and x>-1
contradicting
3: -(x+5) is positive
now x will have to be less than <-5
so lets take x to be -5.1
so -_-5.1+5)=positive will always be positive

1. 5 < x
no not true as -5>x

so i guess answer will be B
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by barcebal » Mon Jul 12, 2010 8:19 pm
That's a crazy question!

I'm kind of confused too. But maybe this will help if we rephrase a few things.

After solving for 4< (7-x)/3 to arrive at x < -5 couldn't we say...

"if x<-5, which of the following must be true?"

1. is out because if x<-5 the inequality x>5 NEVER works.
3. is in because if x<-5, then -(x+5) is ALWAYS positive.

Now for the crazy one.

If x<-5, MUST |x+3|>2 ALWAYS yield an inequality that works? Yes it will, if FIRST off we know that x is already <-5. For any value where FIRST x<-5.... then |x+2| MUST ALWAYS be less than 2.

The reverse doesn't work though: for example, If |x+3|>2, MUST x<-5? No, because x can ALSO be >-1 which means that x MUST NOT ALWAYS be less than -5.

I choose D!
Last edited by barcebal on Mon Jul 12, 2010 8:20 pm, edited 1 time in total.
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by Stuart@KaplanGMAT » Mon Jul 12, 2010 8:20 pm
Let's just focus on (2), since that's the fun one.

|x+3|>2

As you noted, there are two solutions:

+(x+3) > 2 AND -(x+3) > 2

Solving:

+(x+3) > 2
x + 3 > 2
x > -1

-(x+3) > 2
x + 3 < -2
x < -5

Since the solutions are alternatives, we know that:

x < -5 OR x > -1

Is this a MUST BE TRUE?

No.. from the original, we know that x > -5, so x could be -3 or -2. Since both -3 and -2 don't agree with (2), (2) isn't a MUST be true.

Of course, rather than doing all the algebra, it would have been much quicker to test (2) with actual numbers. Just by looking at it, we should be able to see that if we pick x=-3, we get:

|0| > 2, which is clearly false. Since we're allowed to pick x=-3 (since we only have to satisfy x>-5), statement (2) need not be true.

Here's a general approach to picking numbers on roman numeral questions:

Step 1: characterize the choices. Here, the accredited statements MUST BE TRUE. The opposite of MUST BE TRUE is COULD BE FALSE.

Step 2: try to pick numbers that satisfy the original condition (e.g. x > -5) but don't satisfy the requirements for an accredited statement. For example, in this question we want to pick numbers that make the statement false. As soon as you find one number (or set of numbers, if there's more than 1 variable) that discredits a statement, you can eliminate all choices that include that statement.

Here are the 4 possible characterizations and their opposites:

MUST BE TRUE --> could be false
COULD BE TRUE --> must be false
MUST BE FALSE (sometimes written as CANNOT BE TRUE) --> could be true
COULD BE FALSE --> must be true
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by barcebal » Mon Jul 12, 2010 8:25 pm
Stuart,

You said,
No.. from the original, we know that x > -5, so x could be -3 or -2. Since both -3 and -2 don't agree with (2), (2) isn't a MUST be true.
But from the orginal we know that x < -5 not x>-5 as you posted.

In my take I said...
If x<-5, MUST |x+3|>2 ALWAYS yield an inequality that works? Yes it will, if FIRST off we know that x is already <-5. For any value where FIRST x<-5.... then |x+2| MUST ALWAYS be less than 2.

The reverse doesn't work though: for example, If |x+3|>2, MUST x<-5? No, because x can ALSO be >-1 which means that x MUST NOT ALWAYS be less than -5.
Doesn't this mean that D is correct?
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by Testluv » Mon Jul 12, 2010 9:45 pm
Yeah, it seems that Stuart may have misread x<-5 as x>-5.

Incidentally, Stuart's point about picking numbers rather than over-relying on algebra is great advice here.

The inequality in the question stem yields:

x<-5.

If you plug any number that satisfies this inequality into |x+3|, you will see that the result must be >2. For example, let x=-6. Then, |-6+3| = 3. Is 3>2? Yes. Let x=-5.1. Then, |-5.1+3| = 2.1. Is 2.1>2? Yes. We've convinced ourselves that given x<-5, roman numeral 2 must be true.

To answer the original poster's question, x is KNOWN to be <-5. So, we have to discard the inequality that has no overlap (x>-1). This is as in geometry where if we have a negative root, we discard it.
Last edited by Testluv on Tue Jul 13, 2010 5:47 pm, edited 2 times in total.
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by barcebal » Mon Jul 12, 2010 9:57 pm
So how, then, could the absolute value inequality have been a COULD BE FALSE? Suppose roman numeral 2 actually read:

|x+3|>1

Then we would have:

case 1:
x+3>1
x>-2

case 2:
-(x+3)>1
x<-4

In this case, we would discard case 1, as it is known to contradict what is known of x (that x<-5). For case 2, if x<-4, then we can have x=-4.5, in which case x is not <-5 or x=-100, in which case x is <-5. Then, it would be a COULD BE FALSE, and we would eliminate it.

In order for an absolute value inequality to be proven as MUST BE TRUE here, one of the solutions had to perfectly match x<-5. That's what happened here. (And obviously, it is impossible that both of the inequality solutions will match because they are going to be different inequality solutions).
I still think this case would be true.

Consider the question, if x<-5 THEN WHAT MUST BE TRUE?

Well x<-4 must be true if we know that x<-5. Now if you had a absolute value inequality that yielded x<-6 then it doesn't have to be true if the question stem requires that x<-5
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by Testluv » Mon Jul 12, 2010 10:00 pm
Well x<-4 must be true if we know that x<-5. Now if you had a absolute value inequality that yielded x<-6 then it doesn't have to be true is the question stem requires that x<-5
Yep, you caught my silly error too!

I was just about to edit but you caught it before I could!

If x<-5, then clearly it MUST BE TRUE that x<-4, x<0, etc, and no perfect match is required--just an inclusion in the appropriate range.

If x<-5, then it COULD BE FALSE that x<-6 because there is an exclusion of values that fall within the appropriate range (-5.5, etc.)

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by Ian Stewart » Tue Jul 13, 2010 3:38 am
gauravgundal wrote:if 4< (7-x)/3, which of the following must be true?
1. 5 < x
2. |x+3|>2
3.-(x+5) is positive

A. 2 only
B. 3 only
C. 1 and 2 only
D. 3 and 2 only
E. 1 , 2 and 3 only
We first want to rewrite the inequality (multiply by 3 on both sides, then add x to both sides, then subtract 12 from both sides) to get: x < -5. From here, 1) is clearly false, and 3) is clearly true.

There are a few approaches one can take when looking at 2). For an absolute value expression as simple as the one in this question, picking numbers is very fast - if you plug in any positive value for x, you can see that 2) does not need to be true. One can use the standard algebraic approach as well, as was done in explanations above. For more sophisticated absolute value questions on the GMAT, I prefer a number line approach (though it really isn't necessary for this particular question):

|a - b| is always equal to the distance between a and b on the number line. So, if we can get a minus sign between two terms inside an absolute value, we can interpret it as a distance. We can get a minus sign by replacing a '+' sign with two negative signs:

|x + 3| > 2
|x - (-3)| > 2

So this just says that 'the distance between x and -3 is greater than 2. Drawing a number line:

-----(-5)-----(-3)------(-1)--------

we can see that this means that x is either to the left of -5, or to the right of -1. So, since x might be to the right of -1, 2) does not need to be true.

EDIT - I should have looked back at the original question when I wrote the last sentence above - I reversed in my mind what the question was asking! The part I highlighted in red is clearly backwards - if we know that x < -5, then certainly it is true that the distance from x to -3 is more than 2. I've left what I wrote intact since people refer to it below. Thanks for the corrections!
Last edited by Ian Stewart on Tue Jul 13, 2010 8:22 pm, edited 1 time in total.
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by barcebal » Tue Jul 13, 2010 7:09 am
we can see that this means that x is either to the left of -5, or to the right of -1. So, since x might be to the right of -1, 2) does not need to be true.
Right, but the question is which of the following MUST be true if x<-5 (the original inequality simplified).

If x must be less than -5 then (2) must also be true for all cases where x < -5.
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by gauravgundal » Tue Jul 13, 2010 4:02 pm
I am little confused : The OG says that the answer is D .
if 2 . cannot be true then how the answer D is correct .Please explain me
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by barcebal » Tue Jul 13, 2010 4:12 pm
gauravgundal,

I think a lot of these posts have been really confusing. If we read the question really carefully, we understand that the questions is two parts

(1) IF 4< (7-x)/3...

We solve this part to be x<-5.

So now instead of saying IF 4< (7-x)/3 we can say if x < -5

(2) THEN which of the following must be true?

This is where wording is so important. If x is <-5 (don't argue with this or try to prove whether x<-5; it's in the question stem so it is true) then what else MUST be true.

Statement 2 must be true for all values where x<-5. It doesn't matter that statement 2 has additional solutions because IF FIRST, x<-5, then SECOND, statement two will ALWAYS work.
If x<-5, MUST |x+3|>2 ALWAYS yield an inequality that works? Yes it will, if FIRST off we know that x is already <-5. For any value where FIRST x<-5.... then |x+2| MUST ALWAYS be less than 2.

The reverse doesn't work though: for example, If |x+3|>2, MUST x<-5? No, because x can ALSO be >-1 which means that x MUST NOT ALWAYS be less than -5.
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by Testluv » Tue Jul 13, 2010 5:08 pm
I am little confused : The OG says that the answer is D .
if 2 . cannot be true then how the answer D is correct .Please explain me
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Looks like Ian did something wrong at the end. (2 is definitely a MUST BE TRUE.)

But his approach is still good. |x+3| can be rewritten as |x-(-3)|, which means "the distance between x and -3 on the number line."

IF x<-5, IS the distance between x and -3 on the number line >2 ?
Clearly, yes (look at it on the number line). Thus, 2 must be true.

But the easiest way BY FAR, here, is to pick numbers as Stuart said, and as I showed in my post above:
If you plug any number that satisfies this inequality into |x+3|, you will see that the result must be >2. For example, let x=-6. Then, |-6+3| = 3. Is 3>2? Yes. Let x=-5.1. Then, |-5.1+3| = 2.1. Is 2.1>2? Yes. We've convinced ourselves that given x<-5, roman numeral 2 must be true.
.
A third way of seeing this is to use algebra and a just a wee bit of formal logic (such a "wee" bit of formal logic, that you can even call it commonsense):

|x+3|>2

solves to :

x<-5 OR x>-1

For 2 to be necessarily true, one of these two inequalities must hold true (it's impossible for both of them to hold true simultaneously). So, you can read 2 as:

If x not <-5, then x>-1
and:
If x not >-1, then x<-5.

Because x<-5, we know that x is not>-1.
And we know from the above that if x is not >-1, then x<-5.
So, we are left with the quaint question: If x<-5, does it have to be true that x<-5?
Clearly, yes.
Thus, 2 must be true.
(This is also what barcebal is brushing on when she/he says when we know FIRST that x<-5...)

@barcebal: thanks for being patient with sleepy teachers.
Last edited by Testluv on Tue Jul 13, 2010 8:24 pm, edited 1 time in total.
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by Ian Stewart » Tue Jul 13, 2010 8:23 pm
Thanks for pointing that out - the very last sentence in my post above is backwards (I should have looked back at the question!). I've clarified in my post above.
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by Testluv » Tue Jul 13, 2010 8:25 pm
Ian,

every single one of the teachers in this thread made a silly error somehwere.

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