Is xy+zt+yz+tx 0?

Question

GMATH  practice exercise (Quant Class 14)

Is  xy+zt+yz+tx > 0\,  ?

(1)  |x| = |y| = |z| = t 
(2)  x+y+z+t = 0

rvarora · Answer

Is xy+zt+yz+tx>0 ?

(1) |x|=|y|=|z|=t
(2) x+y+z+t=0

Simplifying the question

xy+zt+yz+tx>0 --> y(x+z)+ (t(x+z)>0
Now the question becomes  IS (x+z) (y+t)>0 
Taking statement 2
x+z= -(y+t)
Sufficient
Taking Statement 1
x,y,z,t can take up any value positve or negative or both so not sur --> insufficient

Hence , B

fskilnik · Answer

\underline {xy} + \underline{\underline {zt}} + \underline {yz} + \underline{\underline {tx}} \,\,\mathop > \limits^? \,\,0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y\left( {x + z} ight) + t\left( {z + x} ight) = \left( {x + z} ight)\left( {y + t} ight)\,\,\,\mathop > \limits^? \,\,0\,\,\,\,\,\left( * ight)\,

\left( 1 ight)\,\,\,\left| x ight| = \left| y ight| = \left| z ight| = t\,\,\,\,\left\{ \matrix{
 \,{m{Take}}\,\,\left( {x,y,z,t} ight) = \left( {0,0,0,0} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{m{NO}}} ightangle \,\, \hfill \cr 
 \,{m{Take}}\,\,\left( {x,y,z,t} ight) = \left( {1,1,1,1} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{m{YES}}} ightangle \,\, \hfill \cr} ight.

\left( 2 ight)\,\,\,x + y + z + t = 0\,\,\,\, \Rightarrow \,\,\,\,x + z = - \left( {y + t} ight)\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,?\,\,\,:\,\,\,\left( {x + z} ight)\left( {y + t} ight) = - {\left( {y + t} ight)^2} \leqslant 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{	ext{NO}}} ightangle

The correct answer is (B).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

BrentGMATPrepNow · Answer

Target question :    Is xy + zt + yz + tx positive?

Statement 1 : |x| = |y| = |z| = t  
Let's TEST some values.
There are several values of x, y, z and t that satisfy statement 1. Here are two: 
 Case a : x = y = z = t = 1. In this case, xy + zt + yz + tx = (1)(1) + (1)(1) + (1)(1) + (1)(1) = 4. So, the answer to the target question is  YES, xy + zt + yz + tx IS positive 
 Case b : x = y = z = t = 0. In this case, xy + zt + yz + tx = (0)(0) + (0)(0) + (0)(0) + (0)(0) = 0. So, the answer to the target question is  NO, xy + zt + yz + tx is NOT positive 
Since we cannot answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : x + y + z + t = 0  
Take: x + y + z + t = 0
Subtract y from both sides to get: x + z + t = -y
Subtract t from both sides to get: x + z = -y - t
Rewrite as follows:  x + z = -(y + t)

Now take the expression in the target question:  xy + zt + yz + tx 
Rearrange the terms as follows:  xy + yz + tx + zt 
Factor the terms in PAIRS:  y(x + z) + t(x + z) 
Simplify to get:  (y + t)(x + z)

Since we already know that  x + z = -(y + t) , we can replace (x + z) with -(y + t) to get:  (y + t)  
Notice that, if (y + t) is POSITIVE, then -(y + t) is NEGATIVE, which means  (y + t)  is NEGATIVE 
Similarly, if (y + t) is NEGATIVE, then -(y + t) is POSITIVE, which means  (y + t)  is NEGATIVE 
Finally, if (y + t) = 0, then -(y + t) = 0, which means  (y + t)  is ZERO 
IMPORTANT: Notice that, in all 3 possible cases above, the answer to the target question is  NO, xy + zt + yz + tx is NOT positive 
Since we can answer the  target question  with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

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Is xy+zt+yz+tx > 0?

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