# Question: v-w

## Comment on v-w

### If |x| denotes the least

If |x| denotes the least integer greater than or equal to x,
is |x|=0?
(1) –1 < x < 1
(2) x < 0

hi Brent can u pls explain above problem ?

### Given: [x] denotes the least

Given: [x] denotes the least integer greater than or equal to x.
So, for example, [1.3] = 2, since 2 is the smallest INTEGER that's greater than 1.3
Likewise, [8.8] = 9, since 9 is the smallest INTEGER that's greater than 8.8
[-3.5] = -3, since -3 is the smallest INTEGER that's greater than -3.5
[-0.9] = 0, since 0 is the smallest INTEGER that's greater than -0.9

Target question: Is [x] = 0?
In order for [x] to equal 0, x must be greater than -1 and less than or equal to zero.

So... REPHRASED target question: Is -1 < x ≤ 0?

Statement 1: –1 < x < 1
This is not enough information to answer the REPHRASED target question.
case a: x = -0.5, in which case -1 < x ≤ 0
case b: x = 0.5, in which case 0 < x
Since we cannot answer the REPHRASED target question, statement 1 is NOT SUFFICIENT

Statement 2: x < 0
This is not enough information to answer the REPHRASED target question.
case a: x = -0.5, in which case -1 < x ≤ 0
case b: x = -2, in which case x < -1
Since we cannot answer the REPHRASED target question, statement 2 is NOT SUFFICIENT

Statements 1 and 2 COMBINED
Statement 1 tells us that -1 < x
Statement 2 tells us that x < 0
When we combine these inequalities, we get: -1 < x < 0
PERFECT.
This is EXACTLY what the REPHRASE target question is asking.

Since we can answer the REPHRASED target question, the COMBINED statements are SUFFICIENT

Here's a related question to practice: http://www.beatthegmat.com/number-property-question-t271768.html

### When answering the question I

When answering the question I plugged in numbers in the answer's equations and I had the same answer. Is this is a valid approach?

### Plugging in values yields

Plugging in values yields conclusive results IF doing so yields conflicting answers to the target question, in which case the statement is NOT sufficient.

However, in cases where a statement is sufficient, plugging in values will help us feel more confident about the sufficiency of the statement, but it doesn't yield conclusive results.

More on this here: http://www.gmatprepnow.com/articles/data-sufficiency-when-plug-values

### hi Brent, I'm confused.

hi Brent, I'm confused.
can you please confirm if plugging in values for this question is right or not?

### Testing values is a strategy

Testing values is a strategy that is best used when you have a feeling that a statement is not sufficient, in which case you need only find two contradictory cases in order to be certain the statement is, indeed, insufficient.

However, if a statement is sufficient, then testing values will never yield conclusive results, since the answer to the target question will always be the same (when the statement is sufficient).

More on this concept here: http://www.gmatprepnow.com/articles/data-sufficiency-when-plug-values

### Hi Brent,

Hi Brent,

Is it possible to do the following:

Leave the question as is and do the following to the statements:

statement 1: v > x and w < y

flip them so that the inequality faces the same way: x < v and w < y
Add them: x + w < v + y
Manipulate them so that it reads: v - w > x - y

Of course the method you have shown is quicker and I'll be using that in the future. I wanted to know if I'm breaking any rules.

Thanks again for your quick responses.

Your solution is perfect! Nice work.

Cheers,
Brent

### HI Brent ,

HI Brent ,

for the question @ https://www.beatthegmat.com/is-b-0-t298460.html

i saw your explanation and have one doubt.
While combining the statements the result comes out to be
b³ - b² < 0
on further simplifying
b²(b-1)>0

means b>1 since square of a number can not be negative
so ultimate outcome will be b>1 which is not sufficient to comment that b<0
let me know where i am missing the trick.

That's a great idea. However, you made one small error.
It's true to write: b³ - b² < 0
But, when you factored it, you got: b²(b-1) > 0 (notice that you accidentally reversed the direction of the inequality symbol)

ASIDE: If we had been able to conclude that b>1, then that would be SUFFICIENT to answer the target question (i.e., if b>1, then b is definitely not less than 0)

Cheers,
Brent

### 'w' and then 'y' (or 'y' and

'w' and then 'y' (or 'y' and then 'w') were added so all the negative signs would disappear, and arrive at the new target question: 'v+y > x+w'?

Can (and how can) the old inequalities be answered correctly without combining them? Or, is it recommended to always combine inequalities to make them positive, and is it the fastest way?

Thanks!

### The strategies we use for

The strategies we use for these kinds of questions must be determined on a question-by-question manner.

When we scan the two statements, we see that each statement provides two separate inequalities.
Since we can only add two inequalities (we can't subtract them), it's useful to rephrase the target question in such a way that we can take advantage of our ability to add two inequalities.

That said, we also have the option of keeping the target question as it is and then manipulating the two statements. Here's what I mean:

TARGET QUESTION: Is v - w > x - y?

STATEMENT 1: v > x and w < y
Take the second inequality and multiply both sides by -1 to get: -w > -y
Aside: Since I multiplied both sides by a negative value, I reversed the direction of the inequality symbol.
We now have:
v > x
-w > -y
When we ADD the two inequalities we get: v + (-w) > x + (-y)
Simplify to get: v - w > x - y
Perfect. The answer to the target question is YES.
Statement 1 is sufficient.

STATEMENT 2: w < v and x < y
If w < v, then we know that v - w = some POSITIVE number
If x < y, then we know that x - y = some NEGATIVE number
This means we can be certain that v - w > x - y
Once again, the answer to the target question is YES.
Statement 2 is sufficient.

Does that help?

Cheers,
Brent

### Yes, the explanation helps.

Yes, the explanation helps. It appears that manipulating the statements right away actually saves 1 step (the combining of the inequalities to make them positive). It's great to know how to combine (by adding) inequalities. However, for this question, it seems that evaluating the statements right away saves a bit of time. Thanks!

### hi Brent,

hi Brent,

how would you solve if the question was like ' is v*w > x*y ?

### Let's find out...

Let's find out...

Target question: Is vw > xy?
(1) v > x and w < y
(2) w < v and x < y

Since no relationship with ZERO is provided, it seems that there can be many different cases in which some values are negative, zero values are negative, and all values are negative.

Given all of this, go straight to the two statements COMBINED and test some values.
There are several sets of values that satisfy all of the given conditions. Here are two:

CASE I: v = -2, w = -10, x = -4 and y = 3
In this case, vw = (-2)(-10) = 20 and xy = (-4)(3) = -12
So the answer to the target question is "YES, vw is greater than xy"

CASE II: v = 2, w = 0, x = 1 and y = 3
In this case, vw = (2)(0) = 0 and xy = (1)(3) = 3
So the answer to the target question is "NO, vw is not greater than xy"

Since we get conflicting answers to the target question, the correct answer is E

### I got it, thanks. Did you

I got it, thanks. Did you mean answer 'E' ?

### Oops. Yes I meant to say the

Oops. Yes I meant to say the answer is E. I changed that part. Thanks for the heads up!