xz

Question

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|

chetan2u · Accepted Answer

HI  rahulkashyap ,
OA given is A, maximum students would miss out on a crucial point and ans C  but the answer should be E.

xz < 0 MEANS " Is x and z of OPPOSITE sign?"

1) |x+y| = |x| + |y| 
This tells us that x and y are of SAME sign..
But what about z?
Insuff

2) |y+z| = |y| + |z| 
This tells us that z and y are of SAME sign..
But what about x?
Insuff

combined..
 x,y and Z are of SAME sign, so xz>0..
so should be sufficient..

BUT what if y is 0 
1) |x+y| = |x| + |y|..... y = 0...... x can be POSITIVE or NEGATIVE
2) |y+z| = |y| + |z|..... y = 0...... z can be POSITIVE or NEGATIVE
 so x and z can be ANY sign 
Insuff

E

rahulkashyap · Answer

Can you please help solve this

TaN1213 · Answer

Amazing skill displayed by pointing out 'y=0' case.
+1.
Thanks  chetan2u

Plunkster82 · Answer

I got this wrong. I need to remember these problems can always be zero.

nishantmba · Answer

1) |x+y| = |x| + |y|
This tells us that x and y are of SAME sign..

Can you please explain, How x and y have same signs

MathRevolution · Answer

=>
Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.

Since we have 3 variables and 0 equation, E is most likely to be the answer.

Conditions 1) & 2)

From the condition 1) and 2),
|x+y| = |x| + |y| is equivalent to xy ≥ 0 and |y+z| = |y| + |z| is equivalent to yz ≥ 0. Then we have xy^2z ≥ 0 or xz ≥ 0.

x = 1, y = 0, z = 1 implies xz = 1 > 0.
x = -1, y = 0, z = 1 implies xz = -1 < 0.

Both conditions together are not sufficient.

The answer is 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, while C has 15% chance and A, B or D has 5% chance. Thus, E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously, there may be cases where the answer is A, B, C or D.

rahulkashyap · Answer

For the answer to be E, even if x or z=0, that would work, correct?

rahul16singh28 · Answer

Statmnt 1: This is possible only when x,y = +ve or x,y = -ve numer But we dont know about Z .. so insuff Statmnt 2: This is possible only when y,z = +ve or y,z = -ve But we dont know X.. so insuff.. Combining above 2 statmnts.. Insufff For Ex - when x,y is +ve... y,z can be +ve or - ve.. Please give +1 Kudos if you liked my Solution..

niks18 · Answer

Hi  nishantmba

RHS is the summation of two positive numbers i.e |x|+|y|, so LHS must be equal to this.
this is only possible when both are of equal sign i.e {+,+} or {-, -} or x=y=0. If either of the two numbers have different sign i.e{+,-}, then |x+y| will yield a less value and hence will not be equal to RHS

TaN1213 · Answer

Hey,

Would you please share your knowledge about "these" problems? What pattern in questions are you referring to?
I have got the idea but want to get extra sure from you.
Thank you

MathRevolution · Answer

The point of this question is the following property.

|x+y| = |x| + |y|  is equivalent to  xy ≥ 0 .

MathRevolution · Answer

|x+y| = |x| + |y| 
 ⇔ |x+y|^2 = (|x| + |y|)^2 
 ⇔ (x+y)^2 = |x|^2 + 2|x||y| + |y|^2 
 ⇔ x^2 + 2xy + y^2 = x^2 + 2|xy| + y^2 
 ⇔ 2xy = 2|xy| 
 ⇔ xy = |xy| 
 ⇔ xy ≥ 0

ArupRS · Answer

very good problem.

suchita2409 · Answer

I still have a doubt on this.

Stmt 1 says that the sign of x and y is same.
Stmt 2 says that the sign of y and z is same.

But every expert has considered y = 0 for solving .
My doubt is 0 has no sign ,and stmt 2 specifically says about the sign of y, so why should we take y to be 0.

Posted from my mobile device

Kinshook · Answer

Asked: xz < 0? 1) |x+y| = |x| + |y| x & y are of same sign Since z is unknown NOT SUFFICIENT 2) |y+z| = |y| + |z| z & y are of same sign Since x is unknown NOT SUFFICIENT (1) + (2) 1) |x+y| = |x| + |y| x & y are of same sign 2) |y+z| = |y| + |z| z & y are of same sign x, y & z are of same sign xz >=0 xz is NOT <0 But if y=0 1) |x+y| = |x| + |y|; x may be >0 or <0 2) |y+z| = |y| + |z|; z may be >0 or <0 xz may be >0 or <0 NOT SUFFICIENT IMO E

Archit3110 · Answer

target is xz<0
possible when either of x or z is -ve
#1
 |x+y| = |x| + |y|
possible when both x & y are of same sign 
z not known insufficient
#2
|y+z| = |y| + |z|
again same as #1 both y & z have to be of same sign
from 1 &2
x & z can be of same sign or can be opposite signs
insufficient
option E

xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z|

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