Data Sufficiency

Hi,

I'm having a difficult time with number properties DS questions, like the following:

Is xy > 0 ?

1) x - y > -2

2) x - 2y < -6

OA is C

Ron Purewell from MGMAT showed how to do this problem on the MGMAT Forum, but I'm unclear as to how he decided upon which numbers to pick. I'm always worried I'm going to miss something. For example, "If you select integers, would the answer come out differently if you selected fractions?" Or, "What if X is larger than Y"... "What if one is negative and one is positive?" By the time I prove each statement sufficient or insufficient, I've completely lost my place. If both are insufficient, I find it nearly impossible to decide between C & E, because the information is so garbled in my mind.

Here's how Ron Purewell explained how to do the above problem:

ok, well, first establish that (a) and (b) <i>individually</i> are insufficient by testing cases:
(1)
x = y = 1 --> yes
x = y = 0 --> no
insufficient

(2)
x = y = 100 --> yes
x = -10, y = 10 --> no
insufficient

so think about the two statements together:
SWITCH THE SIGNS of the latter inequality, so that the signs face the same way:
x - y > -2
-x + 2y > 6

add them (note that you can add inequalities whose signs are facing the same way - a useful fact):
y > 4
this means y is positive.
back to the first inequality
x > -2 + y
since y is more than 4, this means that x is more than 2
so x is positive

so x and y are both positive.

How did he decide upon the different sets of numbers used to prove/disprove statement #1 and statement #2? How did he know not to test fractions? What about if X were bigger than Y and visa versa, why didn't he test those scenarios? What if one number were negative and the other were positive?

You see, by the time you think all these things through, you would be out of time (by a long shot). I'm thinking Ron knew which smart numbers to pick based on the information given. I would've tried to test every possible scenario, and never would've finished the question in time.

Can anyone offer some insight?

Hi Stockmoose16,

What Ron did was simply to prove that statments 1 and 2 are not sufficent on their own. To do this, he simply proved that you can answer the question stem differently (i.e. Yes AND No) with values that definitely satisfy the statements.

To prove that xy>0 is TRUE, all you need to prove is EITHER (x>0 AND y>0) OR (x<0 AND y<0).

To prove that xy>0 is FALSE, all you need to prove is EITHER (x<0 AND y>0) OR (x>0 AND y<0) OR (x=0 AND/OR y=0).

The point to remember here is this: For any given set of x and y values that you use to prove the question stem to be TRUE or FALSE, it MUST satisfy the statement that you are considering at that time.

If the above is clear, then the next step is to pick numbers (which I believe is your main concern):

When evaluating statement 1:
Step 1: Try to see if you can pick a set values that have the same sign which MUST satisfy statement 1. You are doing this to prove the question stem TRUE.

Ron picked x=y=1 (both +ve) which yields x - y = 0 which is larger than -2. So question stem is TRUE.

You can also pick x =5, y =1 (both +ve) which yields x-y=4 (which is > -2) or you can pick x = -1, y = -3 (both -ve) which yields x - y= 2 (which is > -2). Again, these sets of values all prove the question stem to be TRUE.

Step 2: Try to see if you can pick a set of values that have opposite signs or contain the value 0 which MUST satisfy statement 1. You are doing this to prove the question stem FALSE.

Ron picked x=y=0 (both zero) which yields x-y=0, which is > -2. The question stem is now FALSE.

You can also pick x=1, y=-1 (opposite signs) which yields x-y=2, which is > -2. Again, this combination also proves the question stem to be FALSE.

Once you are able to prove the question stem to be TRUE AND FALSE, then statement 1 is insufficient.

The same process is carried out for statement 2 as well.

What I'm trying to assert here is that picking numbers is only the secondary step in solving this question. The main thing you need to do here is to parapharase the question stem which will help you narrow your scope of what to evaluate/look out for.

I hope this will be of use to you, and sorry for the long explanation.

-BM-

bluementor wrote:Hi Stockmoose16,

What Ron did was simply to prove that statments 1 and 2 are not sufficent on their own. To do this, he simply proved that you can answer the question stem differently (i.e. Yes AND No) with values that definitely satisfy the statements.

To prove that xy>0 is TRUE, all you need to prove is EITHER (x>0 AND y>0) OR (x<0 AND y<0).

To prove that xy>0 is FALSE, all you need to prove is EITHER (x<0 AND y>0) OR (x>0 AND y<0) OR (x=0 AND/OR y=0).

The point to remember here is this: For any given set of x and y values that you use to prove the question stem to be TRUE or FALSE, it MUST satisfy the statement that you are considering at that time.

If the above is clear, then the next step is to pick numbers (which I believe is your main concern):

When evaluating statement 1:
Step 1: Try to see if you can pick a set values that have the same sign which MUST satisfy statement 1. You are doing this to prove the question stem TRUE.

Ron picked x=y=1 (both +ve) which yields x - y = 0 which is larger than -2. So question stem is TRUE.

You can also pick x =5, y =1 (both +ve) which yields x-y=4 (which is > -2) or you can pick x = -1, y = -3 (both -ve) which yields x - y= 2 (which is > -2). Again, these sets of values all prove the question stem to be TRUE.

Step 2: Try to see if you can pick a set of values that have opposite signs or contain the value 0 which MUST satisfy statement 1. You are doing this to prove the question stem FALSE.

Ron picked x=y=0 (both zero) which yields x-y=0, which is > -2. The question stem is now FALSE.

You can also pick x=1, y=-1 (opposite signs) which yields x-y=2, which is > -2. Again, this combination also proves the question stem to be FALSE.

Once you are able to prove the question stem to be TRUE AND FALSE, then statement 1 is insufficient.

The same process is carried out for statement 2 as well.

What I'm trying to assert here is that picking numbers is only the secondary step in solving this question. The main thing you need to do here is to parapharase the question stem which will help you narrow your scope of what to evaluate/look out for.

I hope this will be of use to you, and sorry for the long explanation.

-BM-

Bluementor,

That was a very helpful explanation. Can you further explain what happens once you prove both statements insufficient (as above)? How can you quickly figure out whether the combined statements are sufficient? How do you know not to test fractions and negative fractions, etc, just to make sure you didn't leave anything out?

When you look at both statements together, you get a pair of inequalities that you can use to solve for x and y. You can do this by adding both inequalities as long as the inequality signs face the same way (as mentioned by Ron):

(1) x-y > -2
(2) x-2y < -6

You can't quite solve them yet since both inequalities have their signs in opposite ways. So first, we need to get them aligned. Lets choose to do this by manupulating inequality (1) (You can also choose to do this for inequality (2) instead). We can choose to do this in two ways; both will come down to the same thing:

-You can simply flip the whole inequality to get: -2 < x -y (1a)

-Or you can multiply -1 on both sides to get the sign flipped : y - x < 2 (1b)


Now you can solve the inequalities to obtain the inequalities for x and y respectively. You do this by adding both inequalities (lets use 1b and 2):

(y - x) + (x-2y) < 2 -6
-y < -4
y > 4

You can add the inequality y>4 to inequality (1), since the signs both face the same way:

y + (x -y) > 4 + (-2)
x > 2

So now you have x>2 and y>4. Both are definitely positive. Therefore xy > 0 is TRUE, C.

bluementor wrote:When you look at both statements together, you get a pair of inequalities that you can use to solve for x and y. You can do this by adding both inequalities as long as the inequality signs face the same way (as mentioned by Ron):

(1) x-y > -2
(2) x-2y < -6

You can't quite solve them yet since both inequalities have their signs in opposite ways. So first, we need to get them aligned. Lets choose to do this by manupulating inequality (1) (You can also choose to do this for inequality (2) instead). We can choose to do this in two ways; both will come down to the same thing:

-You can simply flip the whole inequality to get: -2 < x -y (1a)

-Or you can multiply -1 on both sides to get the sign flipped : y - x < 2 (1b)


Now you can solve the inequalities to obtain the inequalities for x and y respectively. You do this by adding both inequalities (lets use 1b and 2):

(y - x) + (x-2y) < 2 -6
-y < -4
y > 4

You can add the inequality y>4 to inequality (1), since the signs both face the same way:

y + (x -y) > 4 + (-2)
x > 2

So now you have x>2 and y>4. Both are definitely positive. Therefore xy > 0 is TRUE, C.

Sweet! Thanks for the explanation.

How can you quickly figure out whether the combined statements are sufficient? How do you know not to test fractions and negative fractions, etc, just to make sure you didn't leave anything out?

When you have combined both inequalities, there is no need to test any numbers since you have proven both x and y to be positive, which is sufficient to answer the question stem.

-BM-

bluementor wrote:

How can you quickly figure out whether the combined statements are sufficient? How do you know not to test fractions and negative fractions, etc, just to make sure you didn't leave anything out?

When you have combined both inequalities, there is no need to test any numbers since you have proven both x and y to be positive, which is sufficient to answer the question stem.

-BM-

Wow. Brilliant explanation. Thank you very much!

lunarpower wrote:

Stockmoose16 wrote:How do you know not to test fractions and negative fractions, etc, just to make sure you didn't leave anything out?

you don't need to test fractions here because there aren't any signals associated with fractions in the problem.

in general, fractions come into play when you're comparing powers of the same variable. this is so because increasing powers of a fraction get smaller (e.g., fraction^3 < fraction^2), while increasing powers of "normal" numbers get bigger.

fractions also come into play when you perform multiplication and division, because they also act weird under those operations: multiplying by a fraction makes numbers smaller, while dividing by a fraction makes numbers bigger.

since none of these things is happening in this problem - all you're doing is adding and subtracting - there's no need to consider fractions.

Correct me if I'm wrong, but aren't you multiplying in statement #2 (i.e. 2y)?