If xyz < 0, is x < 0 ?

Statement one :

x-y<0
x<y

Here x can take value of a positive or a negative number. x can be a negative number and y and z can be positive number or x and y can be positive numbers and z can take a negative value. Statement is insufficient.

Statement two:
x-z<0
x<z

This statement is insufficient to determine sign of x. x,y and z all can be negative or x and z can take positive values and y can be a negative number.

Both statements combined together, x is the smallest number and product of x,y and z is a negative value. so x has to be a negative number. Either x is a negative number and y and z are positive or all the three numbers are negative. In both the cases, x is a negative number. Hence Ans=C

My answer : C

x.y.z < 0

This is possible when :
All 3 (x,y,z) are <0
OR
One of x,y,z < 0

1.
x-y<0
-> x<y
Both x & y can be positive OR x can be negative
Cannot say anything

2.
x-z<0
-> x<z
Again, Both x & z can be positive OR x can be negative
Cannot say anything

1+2
On combining x<y & x<z
This means x is the smallest of all three
therefore it must be negative for x.y.z <0 to hold true.

MAKE A line number to see that the matter is easy

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 xyz<0 , is x<0 ?

(1) x−y<0

(2) x−z<0

There are 3 variables (x,y,z) and one equation (xyz<0) in the original condition, 2 equations in the given conditions, so there is high chance (C) will be our answer.
Looking at the conditions together,
x<y and x<z.
If x>0, then yz>0, so this does not satisfy xyz<0 (out of scope). Hence x has to be x<0 . This answers the question 'yes' and is therefore sufficient, making the answer (C).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% 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 C 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, D or E.

Hi,

This method is little longer, but just though to put out so that some users can understand what cases should be considered for such type of questions.

we have given a base condition that xyz<0
we can infer two things form this
either x<0&y<0&z<0 or one of x,y,z is <0.

Now Statement 1:
x-y<0
So we have three cases for this
x<0 & y<0 and |x|>|y|
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & y>0 and |x|>|y|
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & y>0 and |y|>|x|
But since xyz<0 from this we have x<0,y>0,z>0

Now Statement 1:
x-y<0
So we have three cases for this
x<0 & y<0 and |x|>|y| ------(a)
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & y>0 and |x|>|y| ------(b)
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & y>0 and |y|>|x|------(c)
But since xyz<0 from this we have x>0,y>0,z<0

Now Statement 2: x-z<0
So we have three cases for this
x<0 & z<0 and |x|>|z| -----(d)
But since xyz<0 from this we have x<0,y<0,z<0
x<0 & z>0 and |x|>|z|-----(e)
But since xyz<0 from this we have x<0,y>0,z>0
x>0 & z>0 and |z|>|x|-----(f)
But since xyz<0 from this we have x>0,y<0,z<0

Combing 1 and 2 we have
cases a, b, d,e we see that

we get either x<0, y<0, z<0
or
x<0 , y>0, z>0

But in both the cases x<0

Probus

1) x < y
Not sufficient
2) x < z
Not sufficient

1+2) x < y, z
Then x has to be negative.

ANSWER: C

Analyse four cases when xyz<0:

x y z
- + + -----(a)
+ - + -----(b)
+ + - -----(c)
- - - -----(d)

1) x<y ---> a,c,d all possible. Not SUFFICIENT
2) x<z ---> Similar to above, a,b,d all possible. Not SUFFICIENT

1 & 2 combined ---> a and d both possible. And in both cases x<0. SUFFICIENT.

We can simply imagine a number line and solve this question in moments.

Given xyz < 0 which means that either exactly 1 of them is negative or all three are negative. None of them is 0.
Question: "Is x negative?"

Statement 1: x - y < 0
means x < y
So x is to the left of y on the number line (check the Number Line module)
Is x negative? It may or may not be. All three could be negative so x may be. Only z could be negative so x may not be. Not Sufficient

Statement 2: x - z < 0
x < z
So x is to the left of z too.
Is x negative? It may or may not be. All three could be negative so x may be. Only y could be negative so x may not be. Not Sufficient

Together I know that x is the smallest one. So if all three are negative, x is negative. If only one of them is negative, then it must be x which is negative because it must be the smallest.

Answer (C)

Check this post on how to use the Number Line: https://youtu.be/3gxVx3Y9xJA