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ds divisble [#permalink]
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04 Sep 2008, 13:02
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This topic is locked. If you want to discuss this question please repost it in the respective forum. if \(x^2=y+5\) and \(y=z2\) and z=2x , is \(x^3+y^2+z\) divisible by 7? 1) x>0 2) y=4 Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient If we substitute z and y with x in the three equations we have
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Re: ds divisble [#permalink]
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04 Sep 2008, 13:18
D is the answer. St.1 By substitutions one can get the equation, x^2  2x3 =0 Solving for x, x= +3 or 1; If x=3, then y=4 and z=6 Putting these values in, we get the result of 49 which is divisible by 7. However, if x= 1, we get 13 as the result and it is not divisible by 7. Therefore, St.1 is required and sufficient.
St. 2 By putting the values in, we get 49 that is divisible by 7. So, sufficient.



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Re: ds divisble [#permalink]
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04 Sep 2008, 13:48
i'd go for b.
The stem's equations can be used to put the equation of the question in terms of x onl. So x^3+4X^26X+4.
St 1) insuff try x=1 answer no x=3 ans yes St 2) we can get the value of x using one of the eq in the stem and replace. Suff.



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Re: ds divisble [#permalink]
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04 Sep 2008, 14:19
will go for B
for 1 reduce equation in terms of x
for 2 reduce the equation in terms of y



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Re: ds divisble [#permalink]
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04 Sep 2008, 17:34
x2suresh wrote: if \(x^2=y+5\) and \(y=z2\) and z=2x , is \(x^3+y^2+z\) divisible by 7?
1) x>0 2) y=4
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient If we substitute z and y with x in the three equations we have What is the OA dude ? B is very tempting.. but when i try stem 1 with x=1,2,3,4, and 5.. only x=3 seems to satisfy all the 3 equations in the question. So D is looks realistic. what is the source and OE ?



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Re: ds divisble [#permalink]
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04 Sep 2008, 18:06
I vote for D.
X^2=Y+5, Y=Z2, Z=2X
A. X>0
X^2=2X2+5 X^22X+3=0, solving for X gives X=3
Solving for Y and Z gives, Y=4, Z=6... Sufficient
B. Y=4
Solving for X and Z gives, X=3, Z=6... Sufficient



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Re: ds divisble [#permalink]
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04 Sep 2008, 20:04
OA is D This is from GMATCLUB challenges.
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Re: ds divisble [#permalink]
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05 Sep 2008, 22:14
x2suresh wrote: if \(x^2=y+5\) and \(y=z2\) and z=2x , is \(x^3+y^2+z\) divisible by 7?
1) x>0 2) y=4
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient If we substitute z and y with x in the three equations we have (1)here solving eqns we get x=1,3 => 3 gives div by 7 but when x=1 the expr is noit div => INSUFFI (2)gives x=3 and hence when substituted the expr is div by 7 SUFFI IMO B
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Re: ds divisble [#permalink]
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05 Sep 2008, 22:16
spriya wrote: x2suresh wrote: if \(x^2=y+5\) and \(y=z2\) and z=2x , is \(x^3+y^2+z\) divisible by 7?
1) x>0 2) y=4
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient If we substitute z and y with x in the three equations we have (1)here solving eqns we get x=1,3 => 3 gives div by 7 but when x=1 the expr is noit div => INSUFFI (2)gives x=3 and hence when substituted the expr is div by 7 SUFFI IMO B i made silly mistake overlooked x>0 oh god!!!
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Re: ds divisble [#permalink]
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06 Sep 2008, 02:47
IMO D
x^2=y+5, y=z2, z=2x Now, y=z2 => z=2+y but, z=2x, means 2x=2+y => x=1+y/2 Putting x in x^2=y+5 => (1+y/2)^2=y+5 Solving further gives y=4 or 4
Now eliminating y from x^2=y+5 and y=z2 gives x^2=z+3 putting z=2x gives, x^2=2x+3 Solving further gives x=1 or 3
Now Option (1) says x>0, means x=3 and z=6 & y=4 => SUFFI to answer
Option (2) says y=4, thus gives z=6 & x=3 again => SUFFI to answer
thus D










