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Q and T

by kushal.adhia » Wed Oct 27, 2010 6:40 am
Is q > t ?

(1) qp^2 < tp^2

(2) qp^3 > tp^3

I know this is a simple question but i would really appreciate some help on this...

This is the way i worked it out:

Stmt 1: divide both sides by P^2;
Q>T
Answer is a definite no.

Stmt 2: divide both sides by p^3;
Q>T
Answer is a definite yes.

Hence i chose D... but OA is A

where did i go wrong??

Thanks

Kushal

Source: MGMAT
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Source: — Data Sufficiency |
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by kmittal82 » Wed Oct 27, 2010 6:48 am
Careful, don't just cancel out terms on both sides.

1) qp^2 - tp^2 < 0

p^2(q-t) < 0

p^2 can never be <0, thus q-t < 0 => q< t

Sufficient

2)

qp^3 - tp^3 < 0
p^3(q-t) < 0

If p^3 is < 0, then q > t
If p^3 is > 0, then q < t

Not sufficient
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by kapur.arnav » Wed Oct 27, 2010 9:57 am
kushal.adhia wrote:Is q > t ?

(1) qp^2 < tp^2

(2) qp^3 > tp^3

I know this is a simple question but i would really appreciate some help on this...

This is the way i worked it out:

Stmt 1: divide both sides by P^2;
Q>T
Answer is a definite no.

Stmt 2: divide both sides by p^3;
Q>T
Answer is a definite yes.

Hence i chose D... but OA is A

where did i go wrong??

Thanks

Kushal

Source: MGMAT
Ok, just remember this... while solving such questions remember to check for negative integers, fractions, zero etc before selecting your answer...
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by rkanthilal » Wed Oct 27, 2010 10:46 am
kmittal82 wrote:Careful, don't just cancel out terms on both sides.

1) qp^2 - tp^2 < 0

p^2(q-t) < 0

p^2 can never be <0, thus q-t < 0 => q< t

Sufficient

2)

qp^3 - tp^3 < 0
p^3(q-t) < 0

If p^3 is < 0, then q > t
If p^3 is > 0, then q < t

Not sufficient
Hi kmittal82,

I see how you arrived at your answer. Why is it incorrect to cancel out the p^2 or p^3 terms in either statement?
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by vijchid » Wed Oct 27, 2010 5:52 pm
Just to show why you cannot cancel the terms

2*-8> 3*-8

if you cancel out the terms -8 you will get 2>3...hope this helps...So when you are not sure about the sign of the number, you cannot cancel out. If you cancel out a negative number, you would have to reverse the inequality.
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by rkanthilal » Wed Oct 27, 2010 7:40 pm
vijchid wrote:Just to show why you cannot cancel the terms

2*-8> 3*-8

if you cancel out the terms -8 you will get 2>3...hope this helps...So when you are not sure about the sign of the number, you cannot cancel out. If you cancel out a negative number, you would have to reverse the inequality.
Thanks vijchid.
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