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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
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slingfox wrote:
Is x > y?

1. 9x = 4y
2. x = -y


Thank you for this question. here is my solution.

1. \(9x=4y\) also means:
(a) \(\frac{x}{y}=\frac{4}{9}\) which means y > x NO!
(b) \(\frac{x}{y}=\frac{{-4}}{{-9}}\) which means x > y YES!
INSUFFICIENT!

2. \(\frac{x}{y}=-1\) this means:
(a) they have the same number but of different sign, one is negative the other is positive BUT we don't know which one is positive or negative
(b) x = y = 0
INSUFFICIENT!

Combine both:
\(9(-y) = 4y\)
\(13y = 0\)
\(y=0\) and \(x=0\)

Answer: C
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
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slingfox wrote:
Is x > y?

1. 9x = 4y
2. x = -y

Another simpler approach could be...

Statement 1-

Say x= 4, & y=9 => x<y => NO.
But if x= -4, & y=-9 => x>y => Yes.

Insufficient

Statement 2-

Say x= 1, & y=-1 => x>y => Yes.
But if x= -1, & y=1 => x<y => No.

Insufficient

Statement 1, & 2 together-

Only one case is possible => x=y=0. Answer C.

-Shalabh Jain
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
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GMATT73 wrote:
Is x > y?

(1) 9x = 4y
(2) x = -y



1) NS
if x=4 & y=9 the statement is sufficient;
but if x=9 & y=4 the statement is insufficient;

2) NS
if \(x=-y\) is \(-y>y\)?
It depends on whether y is negative.

T) S
Is y negative? \(-y > y\),
substitute in the second prompt and we get \(\frac{x}{y}=\frac{4}{9}\) -> \(\frac{-y}{y}=\frac{4}{9}\) -> \(-1=\frac{4}{9}\)
-> Impossible: y is never negative and x>y is never true.

Answer C.
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
Expert Reply
Is x > y?

(1) 9x = 4y
(2) x = -y

If we observe both equations,
from 1st equation, both x and y have same sign
from 2nd equation, both x and y have opposite sign.
hence this is possible only when both x and y = 0

hence Combination of both statements is the sufficient, and Answer to question stem (Is x > y?) is No,

Answer C
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
GMATT73 wrote:
Is x > y?

(1) 9x = 4y
(2) x = -y


1 & 2 Alone are not sufficient.. Combining the two, we get -

\(9(x+y) = 13y\)

From Statement II, \(x + y = 0\)

So, \(9 * 0 = 13y\)

\(y =0, x = 0.\)

Hence, C.

Originally posted by rahul16singh28 on 02 Jun 2018, 04:19.
Last edited by rahul16singh28 on 02 Jun 2018, 08:42, edited 1 time in total.
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
Expert Reply
hi

\(9(x+y) = 13y\)
what makes u conclude \(y =0, x = 0.\)
Reasoning is more important than the answer.

rahul16singh28 wrote:
GMATT73 wrote:
Is x > y?

(1) 9x = 4y
(2) x = -y


1 & 2 Alone are not sufficient.. Combining the two, we get -

\(9(x+y) = 13y\)
\(y =0, x = 0.\)

Hence, C.
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
gmatbusters wrote:
hi

\(9(x+y) = 13y\)
what makes u conclude \(y =0, x = 0.\)
Reasoning is more important than the answer.

rahul16singh28 wrote:
GMATT73 wrote:
Is x > y?

(1) 9x = 4y
(2) x = -y


1 & 2 Alone are not sufficient.. Combining the two, we get -

\(9(x+y) = 13y\)
\(y =0, x = 0.\)

Hence, C.


Hi gmatbusters

From Statement 1, we can write -
\(9(x+y) = 13y\)

From Statement B -
\(x + y = 0\)

So, we have by combining two statements -

\(9(x+y) = 13y\) or
\(9*0 = 13y\)

\(y = 0, x = 0\)
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
Expert Reply
yes, this is a complete explanation. :thumbup:

rahul16singh28 wrote:
hi

\(9(x+y) = 13y\)
what makes u conclude \(y =0, x = 0.\)
Reasoning is more important than the answer.

rahul16singh28 wrote:
GMATT73 wrote:
Is x > y?

(1) 9x = 4y
(2) x = -y


1 & 2 Alone are not sufficient.. Combining the two, we get -

\(9(x+y) = 13y\)
\(y =0, x = 0.\)

Hence, C.


Hi gmatbusters

From Statement 1, we can write -
\(9(x+y) = 13y\)

From Statement B -
\(x + y = 0\)

So, we have by combining two statements -

\(9(x+y) = 13y\) or
\(9*0 = 13y\)

\(y = 0, x = 0\)[/quote]
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Re: Is x > y? (1) 9x = 4y (2) x = -y [#permalink]
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