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OG-11- DS-145

Expert replies
by ash_maverick » Mon Jul 13, 2009 3:52 am
Is 1/p >r/r^2+2 ?

1- p=r
2- r>0.

If i consider the stmt-1 then

1/r > r/r^2+2

In order to simplify it further, i can do a cross multlipication, that would result in
==> r^2 +2 >r^2

now looking at the above simplified inequality i can easily say that whether r>0 or r<0;

the above inequality will always be true..... So i chose optionn A. But the answer says C . Can somebody tells me what mistake i am doing? Is there any thing wrong in simplifying the inequality further.

Thanks in advance.
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Source: — Data Sufficiency |

Re: OG-11- DS-145

by nitya34 » Mon Jul 13, 2009 5:22 am
You cant.. becoz you are assuming p*r = +ve

say , 1/2 > 1/3

you can say 3>2(Cross Multiplication)


now say, -1/3>-1/2
=>Minus 0.333>minus 0.5
can you cancel out Minus sign and say 2>3?(NO)

remember, Inequality Will Change Sign when you Cancel Minus on Both Sides



ash_maverick wrote:Is 1/p >r/r^2+2 ?

1- p=r
2- r>0.

If i consider the stmt-1 then

1/r > r/r^2+2

In order to simplify it further, i can do a cross multlipication, that would result in
==> r^2 +2 >r^2

now looking at the above simplified inequality i can easily say that whether r>0 or r<0;

the above inequality will always be true..... So i chose optionn A. But the answer says C . Can somebody tells me what mistake i am doing? Is there any thing wrong in simplifying the inequality further.

Thanks in advance.
Many of the great achievements of the world were accomplished by tired and discouraged men who kept on working.
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by sreak1089 » Mon Jul 13, 2009 6:59 am
Is 1/p >r/r^2+2 ?

If p=r, then above eqn becomes 1/r > 1/r + 2? right in which case
stmt # 1 is sufficient. Is this correct or did I miss something?
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by hk » Mon Jul 13, 2009 10:04 am
sreak1089 wrote:Is 1/p >r/r^2+2 ?

If p=r, then above eqn becomes 1/r > 1/r + 2? right in which case
stmt # 1 is sufficient. Is this correct or did I miss something?
I think the question is not properly stated: it should be is 1/p>r/(r^2+2).

You cannot simply this futher without knowing the sign of R and P

1) Says p=r which does not help because, if both are positive then

1/r > r/ (r^2+2) and you can cross multiply and simplify.
But is both are negative then you need to invert the sign of inequality while cross multiplying.

2) r>0 No use as such without knowing the sign of p.

When combining both you know that both are positive hence we can cross multiply and simply. => Sufficient. Hence C!!!
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