courtesy of a student:

If x and n are integers, is the sum of x and n less than zero?

(1) x + 3 < n – 1

(2) -2x > 2n

note that this problem is about the SUM of x and n (a COMBINATION). therefore, all of your initial efforts should be directed at isolating that combination (not solving for individual variables).

--

(1) alone
remember that your efforts are directed at isolating the combo x + n. to that end,
subtract 3 from both sides --> x < n - 4
add n --> x + n < 2n - 4
no way to determine whether x + n is sufficient or not
INSUFFICIENT

(2) alone
add 2x to both sides --> 0 > 2x + 2n
divide by 2 --> 0 > x + n
x + n is positive
SUFFICIENT

answer = b

Hi Ron,

Would you use a similar approach for the following question ... especially with regards to the first statement:

Is x + y > 0 ?

(1) x - y > 0

(2) x^2 - y^2 > 0

Statement 2 I could easily answer by x^2 - y^2 => (x + y) (x - y).
So (x + y) ( x - y) > 0 which means either (x + y) is pos and (x - y) is pos, or (x + y) is neg AND (x - y) is neg. so insuff.

How would you approach statement 1 ? thanks.

once you've seen enough of these, the correct approach to example 1 is something along the lines of, 'subtraction isn't the same as addition / a difference isn't the same as a sum, so, insufficient.' or, equivalently, you can realize that there's absolutely no way you're going to be able to rearrange (x - y) and get (x + y).

if you really don't think you can think along the lines of either of those**, then you can start plugging in different types of numbers that satisfy x - y > 0:
x = 3, y = 2: x + y > 0
x = -2, y = -3: x + y < 0
insufficient
note that it's important to pick different KINDS of numbers (positive and negative). if you just pick positive numbers all day, you'll get the mistaken impression that x + y has to be positive.

** you should really, really learn to think conceptually about the numbers in this kind of way. that's what number properties problems are all about: strange ways to twist the conceptual interpretation of a problem around. if your #1 response to all number properties problems is just to grind numbers without thinking about anything conceptual, then you are in for a hard time with all but the simplest ones.

quote]

courtesy of a student:

If x and n are integers, is the sum of x and n less than zero?

(1) x + 3 < n – 1

(2) -2x > 2n


once you've seen enough of these, the correct approach to example 1 is something along the lines of, 'subtraction isn't the same as addition / a difference isn't the same as a sum, so, insufficient.' or, equivalently, you can realize that there's absolutely no way you're going to be able to rearrange (x - y) and get (x + y).

if you really don't think you can think along the lines of either of those**, then you can start plugging in different types of numbers that satisfy x - y > 0:
x = 3, y = 2: x + y > 0
x = -2, y = -3: x + y < 0
insufficient
note that it's important to pick different KINDS of numbers (positive and negative). if you just pick positive numbers all day, you'll get the mistaken impression that x + y has to be positive.

** you should really, really learn to think conceptually about the numbers in this kind of way. that's what number properties problems are all about: strange ways to twist the conceptual interpretation of a problem around. if your #1 response to all number properties problems is just to grind numbers without thinking about anything conceptual, then you are in for a hard time with all but the simplest ones.

[/quote]

Hey Ron,

Here how I would solve this

We want to know if x+n<0. I set up all the possible scenarios

1)For x<0 & n<0 ( -2-1)=-3 answer is YES answer (sufficient)

2)For x>0 & n >0 answer is(2+1)=3 NO answer (sufficient)

3)For x >0 & n<0 answer is :
When we add a positive and negative number we must find whether the absolute value of the positive number x ( or distance from the 0 of on the x axis ) is larger that the one of the negative number n.
On that depends whether the x+n is > or < 0

3a) -for (x)<(-n) is ( 1+(-2))=-1 (sufficient) The absolute value of x here is (1) and smaller than the one of n which is (2).
The sum of x and n is less than zero- YES answer (sufficient)

3b)-for (x)>(-n) (2+(-1))= 1 The absolute value of x here is (2) and is larger than the one of n which is (1).
The sum of x and n is more than zero - No answer (sufficient)

4 For x <0 & n>0 answer is :

When we add a positive and negative number we must find whether the absolute value of the positive number n ( or distance from the 0 of on the x axis ) is larger that the one of the negative number x. On that depends whether the x+n is > or < 0

4a)-for (-x)<n i ( -2+1))=-1 (sufficient) The absolute value of x here is (2) is larger than the one of n which is (1). The sum of x and n is less than zero -YES answer (sufficient)

4b) -for (-x)>n (-1+2)= 1 (sufficient) The absolute of x here is (1) is smaller than the one of n which is (2).
The sum of x and n is more than zero- No answer (sufficient)

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Now let's look at the choices

1) x + 3 < n – 1 = x+4<n

This case applies in case 1) (both numbers are negative) which generates a YES answer and case 2) both numbers are positive (which generates NO) answer. This inconsistency generates a MAYBE , which is INSUFFICIENT

(2) -2x > 2n= -x>n

This is the case 4b and no other answer. Is a No answer and, unlike the choice 1 doesn't generate a MAYBE answer. SUFFICIENT

This was trying. I hope I got it right;)!

Ron, It took long time to build all this. Would you ever do it in a real test?

Thanks,

Ruben

i wouldn't dare, for at least two reasons:
1) as you acknowledged yourself, this method takes a positively ridiculous amount of time;
2) most of the build-up is completely irrelevant to the two statements at hand. specifically, you didn't use any of what you called case 3/3a/3b, ever, nor did you use case 4a. that means that a solid 50% of the build-up was a complete waste of time.
on a test like this one, on which time is a very precious commodity, you can ill afford to waste any time, let alone on irrelevant 'cases' as is done here.

so, here's the compromise:
it's ok to take a casewise approach, but you MUST restrict the cases to those that actually apply to the 2 given statements.