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If r = q^2 and r does not equal 0, is r > q? (1) -(r*q)/2

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If r = q^2 and r does not equal 0, is r > q? (1) -(r*q)/2 [#permalink] New post 13 Nov 2005, 20:54
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D
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If r = q^2 and r does not equal 0, is r > q?

(1) -(r*q)/2 < 1
(2) 1/r < 1/q
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Re: DS: r > q? [#permalink] New post 13 Nov 2005, 22:19
B. r is +ve. q could be both.

from i, if r = 0.25 then q=0.5 and q>r. if r =4, q=-2 then r>q. so not suff.

from ii, g<r. so enough because r is +ve so it doesnot matter whether q is +ve or -ve. q is always smaller than r. so suff...
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 [#permalink] New post 14 Nov 2005, 02:22
lest see.

r=q^2 that means r is positive r!= 0

statement 1 says -r*q/2<1
or -rq<2

because r is positve so q will also be positive from here only way r can be less than q is when q is more than 1. becasue -rq is less than 2 that means

rq> 2. so there is no way q can be less than 1.

so this statement is sufficient to answer that r>q

satement 2 says

1/r < 1/q

unless r is negative ( which r is not)

r>q from this equation.

so B is sufficient.

so our answr choice is D
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 [#permalink] New post 14 Nov 2005, 03:21
nakib77 wrote:
lest see.

r=q^2 that means r is positive r!= 0

statement 1 says -r*q/2<1
or -rq<2
because r is positve so q will also be positive from here only way r can be less than q is when q is more than 1. becasue -rq is less than 2 that means

rq> 2. so there is no way q can be less than 1.

so this statement is sufficient to answer that r>q

satement 2 says

1/r < 1/q

unless r is negative ( which r is not)

r>q from this equation.

so B is sufficient.

so our answr choice is D


buddy, the second bold part should be rp>-2. Btw, I agree with Himalaya that OA is B.
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Re: DS: r > q? [#permalink] New post 14 Nov 2005, 03:58
duttsit wrote:
If r = q^2 and r does not equal 0, is r > q?

(1) -(r*q)/2 < 1
(2) 1/r < 1/q


From 1 we get -rq < 2 or rq > -2 . r=q^2 substituting we get q^3>-2, we can find values for Q which satisfies and which also not satisfies this property. Hence A is not sufficient.

1/r < 1/Q => r > Q Hence statement 2 is sufficient.

Hence B
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 [#permalink] New post 14 Nov 2005, 04:18
laxieqv wrote:
nakib77 wrote:
lest see.

r=q^2 that means r is positive r!= 0

statement 1 says -r*q/2<1
or -rq<2

s


buddy, the second bold part should be rp>-2. Btw, I agree with Himalaya that OA is B.


:oops: oops.... I agree it should b B.
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 [#permalink] New post 14 Nov 2005, 07:39
If r = q^2 and r does not equal 0, is r > q?

(1) -(r*q)/2 < 1
r =q^2 so r is positive....so q can be positive or negiative ..but gives no Information about the r> q so Not Suff




(2) 1/r < 1/q
as r Not = 0, multiply by r
1 < r/q

r> q Sufficient


B
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 [#permalink] New post 14 Nov 2005, 09:14
Good job guys. OA is B.
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 [#permalink] New post 16 Nov 2005, 17:09
Ok late on this but I get B also...

r=q^2....r>q?

look at (2) first cause it looks simple..

we know that r is positive since q^2 is positive (non-zero)

so from (2) we know that r is greater than q..suff

(1)
-(q^2*q)/2 <1

well...lets see if -q^3/2 <1 couple of things here...if q is 1, then -1/2 <1, r=q...

if q=0.5 , then q^2=0.25 then q^3=.125/2 <1; r <q....
if q =2, then q^2=4, q^3=8/2=4 *-1=-4<1; r>q

not sufficient...

B it is..
  [#permalink] 16 Nov 2005, 17:09
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