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Is xy+zt+yz+tx > 0?

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Is xy+zt+yz+tx > 0?  [#permalink]

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New post 25 Feb 2019, 15:28
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33% (02:10) correct 67% (02:29) wrong based on 48 sessions

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GMATH practice exercise (Quant Class 14)

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

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

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Re: Is xy+zt+yz+tx > 0?  [#permalink]

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New post 25 Feb 2019, 17:43
1
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
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Re: Is xy+zt+yz+tx > 0?  [#permalink]

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New post 25 Feb 2019, 17:56
fskilnik wrote:
GMATH practice exercise (Quant Class 14)

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

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

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


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

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


The correct answer is (B).


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

Regards,
Fabio.
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Re: Is xy+zt+yz+tx > 0?  [#permalink]

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New post 26 Feb 2019, 07:46
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fskilnik wrote:
GMATH practice exercise (Quant Class 14)

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

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


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)[-(y + t)]
Notice that, if (y + t) is POSITIVE, then -(y + t) is NEGATIVE, which means (y + t)[-(y + t)] is NEGATIVE
Similarly, if (y + t) is NEGATIVE, then -(y + t) is POSITIVE, which means (y + t)[-(y + t)] is NEGATIVE
Finally, if (y + t) = 0, then -(y + t) = 0, which means (y + t)[-(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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Re: Is xy+zt+yz+tx > 0?   [#permalink] 26 Feb 2019, 07:46
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