If ab ≠ 0 and ax – by < 0 , is x < y ?

[yo4561]

If ab ≠ 0 and ax – by < 0 , is x < y ?
1) a = b
2) a^3 > 0


Hi! I am a first time poster (apologies if I did something wrong here). I am not sure if this a CAT problem, but it was written by Manhattan Prep (I am assuming that it is because it was not in the book, it was an exercise emailed from the course instructor). I am confused as to why statement 1 is not sufficient. If you know that ax<by and a=b, isn't that the same as saying a and b are basically constants and that is enough to know x<y? I am just not sure why you would need to divide the constants, as the answer mentions below.

"Problem 14 Solution: C
It makes sense to manipulate the inequality in the question stem before addressing the statements. Adding by to both sides, this inequality becomes ax <
by.

Statement 1: If a = b, then the inequality from the question stem becomes ax< ay. However, dividing by a yields two potential outcomes: either x < y or x > y, depending on the sign of a. Statement 1 is not sufficient.

Statement 2: If a^3 > 0, then a > 0. However, this indicates nothing about b, so depending on the value b takes on we could get two different answers to the question in the stem. Statement 2 is not sufficient. Together: Combining the information allows us to rewrite the inequality in
the question stem as ax < ay, and then to divide by a (which is positive). The result is x < y, and together the statements are sufficient'

[esledge]

If ab ≠ 0 and ax – by < 0 , is x < y ?
1) a = b
2) a^3 > 0

Hi! I am a first time poster (apologies if I did something wrong here). I am not sure if this a CAT problem, but it was written by Manhattan Prep (I am assuming that it is because it was not in the book, it was an exercise emailed from the course instructor). I am confused as to why statement 1 is not sufficient. If you know that ax<by and a=b, isn't that the same as saying a and b are basically constants and that is enough to know x<y? I am just not sure why you would need to divide the constants, as the answer mentions below.

Hi! You are forgetting that if those constants were negative, the inequality sign would have to flip when you divide ax<by by the constant a=b.

I'd probably do case testing on paper in my workspace for stmt (1):

Case 1: ax < by
and a = b = 2
... so 2x < 2y, which means x < y and the answer is Yes.

Case 2: ax < by
and a = b = -3
... so -3x < -3y, which means x > y and the answer is No.

A Yes and a No mean Maybe = Insufficient

More generally, you'll probably see this "exception" tested again. So when you see inequalities, think "what about negatives? are they allowed?" Here, there were two other clues that sign could be relevant:
1. The ab ≠ 0 in the question stem prevents a=b=0, but still allows positives and negatives.
2. The a^3 > 0 in stmt (2) rephrases to "a is positive."

So here's one other DS tip: When working with Statement (1) alone, make sure to check what happens when you deliberately violate Statement (2). And vice-versa: if you were testing cases for (2), you must plug a positive value for variable a, but don't plug the same number in for variable b.

[yo4561]

If ab ≠ 0 and ax – by < 0 , is x < y ?
1) a = b
2) a^3 > 0

Hi! I am a first time poster (apologies if I did something wrong here). I am not sure if this a CAT problem, but it was written by Manhattan Prep (I am assuming that it is because it was not in the book, it was an exercise emailed from the course instructor). I am confused as to why statement 1 is not sufficient. If you know that ax<by and a=b, isn't that the same as saying a and b are basically constants and that is enough to know x<y? I am just not sure why you would need to divide the constants, as the answer mentions below.

Hi! You are forgetting that if those constants were negative, the inequality sign would have to flip when you divide ax<by by the constant a=b.

I'd probably do case testing on paper in my workspace for stmt (1):

Case 1: ax < by
and a = b = 2
... so 2x < 2y, which means x < y and the answer is Yes.

Case 2: ax < by
and a = b = -3
... so -3x < -3y, which means x > y and the answer is No.

A Yes and a No mean Maybe = Insufficient

More generally, you'll probably see this "exception" tested again. So when you see inequalities, think "what about negatives? are they allowed?" Here, there were two other clues that sign could be relevant:
1. The ab ≠ 0 in the question stem prevents a=b=0, but still allows positives and negatives.
2. The a^3 > 0 in stmt (2) rephrases to "a is positive."

So here's one other DS tip: When working with Statement (1) alone, make sure to check what happens when you deliberately violate Statement (2). And vice-versa: if you were testing cases for (2), you must plug a positive value for variable a, but don't plug the same number in for variable b.

Thank you very much Emily, I appreciate it

[esledge]

You're welcome!