xy < zy < 0
Both xy and zy are negative.
Either y is negative, x and z both are positive and x > z.
Or y is positive, x and z both are negative and x < z.

Question: Is y positive?

(1) x < z
If x < z, x and z are both negative and y is positive. Sufficient.

(2) x is negative
If x is negative, z is also negative and then y must be positive. Sufficient.

Answer (D)
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Typical inequalities question where a step by step method is a must in order to reach the correct answer.

You are given xy<zy ---> y(x-z)<0 , 2 cases possible here

1. y<0 and x>z
2. y>0 and x<z

The question is asking us whether y >0

Per statement 1, x<z ---> case 2 comes into the picture, providing a definite answer to the question asked. Sufficient.

Per statement 2, x< 0 ---> use the given information that xy<0 --> y must be >0 for xy<0 with x<0. Thus sufficient as well.

Both statements are sufficient ----> D is the correct answer.

Hope this helps.
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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.

If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative


When you modify the original condition and the question, it becomes (x-z)y<0에서 y>0? --> x-z<0?. x<Z에서 1) is yes, which is sufficient.
For 2), if x<0, y>0 derived from xy<0, which is yes and sufficient. Therefore, the answer is D.


 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Can you just divide the original equation by Y?
XY < ZY < 0 --> X< Z < 0?

If not, why not?

Dividing an inequality by a variable whose sign is UNKNOWN is the biggest mistake you can make in inequality questions.

The sign of the inequality changes if you divide or multiply by a negative number. This is the reason why you CAN NOT divide by y.

If xy<zy ---> y*(x-z)<0 --> this means that either

Case 1: y>0 and x-z<0 OR
Case 2: y<0 and x-z>0

When you divide by y WITHOUT reversing the signs, you are inherently assuming that y>0. This is the very question asked.
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as per the question, it asks y(x-z)<0;
I agree for option 1 is sufficient on the basis that x<z;
for the second option, since x is -ve
x=-5 and z=-4 then y(-5-(-4))<0; hence y is +ve
x=-5 and z=-10 then y(-5-(-10))<0; here y should be -ve

there are conflicting solutions whether y is +ve/-ve

where am I wrong??

From Statement 2, we don't get any information for z & we are not bothered about z, all we need to figure out is whether y > 0

We know from question prompt that xy < 0, hence if x < 0, then y > 0

I hope that helps


Thanks,
GyM
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The #'s plugged in are not correct x < z, hence x=-5 and z=-10 is not valid
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xy<zy<0
1: x<z, then -x>-z, if we subtract it from above inequality we will get y>0
2: because xy<0 and x<0, therefore y>0

Target question: Is y positive?

Given: xy < zy < 0,
Let's focus on this part of the inequality: xy < zy
Subtract xy from both sides to get: 0 < zy - xy
Factor out the y to get: 0 < y(z - x) (this will come in handy later)

Statement 1: x < z
Subtract x from both sides to get: 0 < z - x
In other words, (z - x) is POSITIVE
This means we can take our given inequality 0 < y(z - x) and divide both sides by (z - x) to get: 0 < y
So, the answer to the target question is YES, y IS positive
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x is negative
The given information tells us that xy < zy < 0
This means xy < 0
In other words, xy is NEGATIVE
So, if x is negative, then it MUST be the case that y is POSITIVE
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

RELATED VIDEO FROM MY COURSE

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Official explanation

Solution: D

This difficult inequality problem requires that you properly leverage the information from the question stem. The second statement is relatively easy so you should start there – if x is negative and xy is negative then y must be positive. The second statement is sufficient. Since the second statement is relatively straightforward you should expect the 1st statement to be harder: from the question stem you know that xy < zy. If y were positive then you could divide by y and see that x < z. If y were negative then you could divide both sides to see that x > z. Since statement 1 proves that x < z then it must be the first case and y must be positive. Answer is D. Number picking is awkward and time consuming on first statement so don’t forget to use algebra as your first approach in any inequality problem.

xy<zy<0.

Tip 1: xy<zy, cancel y.. x<z.. dont do that. This is because you dont know what y is (positive or negative)

So, xy<zy,
or, xy-zy<0,
or, y(x-z)<0.

So, Either y<0, therefore in this case, (x-z)>0 .....(i)
OR, (x-z)<0, therefore in this case, y>0 ........(ii)

(1) x<z, which means (x-z)<0, which means equation (ii) is valid. y>0. SUFFICIENT
(2) x is negative, we know xy<0, if x negative, then y must be positive. SUFFICIENT

Answer: D

\(xy - zy < 0\)

\(y(x - z) < 0\)

\(y < 0 or x < z\)

Is \(y > 0?\)

(1) If \(x < z\), we can conclude \(y > 0\). SUFFICIENT.

(2) If \(x < 0\), and we know \(xy < 0\), then \(y > 0\). SUFFICIENT.

Answer is D.
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