Is z less than 0?

Question

Is z less than 0 ?

(1) xz > yz
(2) xy > yz

iamgame · Answer

Should be E.
We dont have any information about x,y or z to start with.
1) xz>yz
=> z <> 0 (since if it is = 0 this inequality doesnt hold good), we also get x > y => insufficient
2) xy>yz
=> y<> 0 (since if it is = 0 this inequality doesnt hold good), we also get x>z => insufficient

Combining both we get tget relationship x>y and x>z
we cannot get any further information.

Hence E

IanStewart · Answer

The conclusions you've drawn above (highlighted in red) are not quite right. If you know from Statement 1 that xz > yz we cannot just divide by z on both sides to get x > y. We don't know if z is positive or negative (that's the question after all!). If z is positive, we could divide by z to find that x > y, but if z is negative, when we divide by z on both sides we need to reverse the inequality; we would find that x < y. As for the question, Is z < 0? (1) xz > yz (2) xy > yz it is easy to come up with a scenario where all three letters are positive since x is on the left side of each inequality - we just need to make x large. So if x = 100, y = 1 and z = 1, then the statements will be true and z > 0. We can also come up with a scenario where z < 0. If, say, z = -1, then for Statement 1 to be true, x will need to be smaller than y. If we make x and y positive and z negative, Statement 2 will be true (the left side will be positive and the right side will be negative), so we could have x = 1, y = 2 and z = -1. So the answer is E.

danzig · Answer

Is z less than 0 ?

(1) xz > yz
(2) xy > yz

What should be the approach to solve this question?, picking numbers, algebra, etc?

Thanks!

EvaJager · Answer

Having 3 variables, I feel I have too much freedom, so I would start directly testing (1) and (2) together. None of the variables can be 0. A nice number to plug in would be y = 1. 0, 1, -1 are nice choices to plug in for testing. If 0 is not acceptable, next choice would be 1. I am using such choices frequently. Sometimes, numbers greater than 1 or smaller than one, depending on the data. From (1), we get xz > z, or z(x - 1) > 0, and from (2) we obtain x > z. If x > 1, so x - 1 > 0, then z must be positive. We have plenty of choices for x and z such that also x > z. For example x = 2, z = 1. If x < 1, so x - 1 < 0, then z must be negative. Again, no problem to find values for which x > z. For example x = -1, z = -2. Not sufficient. You can try instead of y = 1, taking x = 1 and work out some examples to prove that (1) and (2) together are not sufficient. Answer E.

Bunuel · Answer

Merging similar topics. Please refer to the solutions above. Similar question to practice: is-z-less-than-129362.html Hope it helps.

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is z less than 0 ?

(1) xz > yz
(2) xy > yz
 
There are 3 variables (x,y,z) but only 2 equations are given by the conditions, so there is high chance (E) will become the answer.
Looking at the conditions,
from 1) x(y-z)>0 2) y(x-z)>0, the answer to the question becomes 'no' for x=y=2, z=1
but 'yes' for x=y=2, z=-1, so the conditions are insufficient, and the answer therefore becomes (E).

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

bumpbot · Answer

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Is z less than 0?

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