GMATH practice exercise (Quant Class 14)
Is xy+zt+yz+tx positive?
(1) |x| = |y| = |z| = t
(2) x+y+z+t = 0
Answer: [spoiler]____(B)__[/spoiler]
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Is xy+zt+yz+tx positive?
This topic has expert replies
-
fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
-
fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
$$\underline {xy} + \underline{\underline {zt}} + \underline {yz} + \underline{\underline {tx}} \,\,\mathop > \limits^? \,\,0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y\left( {x + z} \right) + t\left( {z + x} \right) = \left( {x + z} \right)\left( {y + t} \right)\,\,\,\mathop > \limits^? \,\,0\,\,\,\,\,\left( * \right)\,$$fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 14)
Is xy+zt+yz+tx positive?
(1) |x| = |y| = |z| = t
(2) x+y+z+t = 0
$$\left( 1 \right)\,\,\,\left| x \right| = \left| y \right| = \left| z \right| = t\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y,z,t} \right) = \left( {0,0,0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y,z,t} \right) = \left( {1,1,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,\,x + y + z + t = 0\,\,\,\, \Rightarrow \,\,\,\,x + z = - \left( {y + t} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,?\,\,\,:\,\,\,\left( {x + z} \right)\left( {y + t} \right) = - {\left( {y + t} \right)^2} \le 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle $$
The correct answer is (B).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 14)
Is xy+zt+yz+tx positive?
(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
Brent Hanneson - Creator of GMATPrepNow.com
