Q and T

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kushal.adhia: Senior | Next Rank: 100 Posts; Posts: 80; Joined: Thu Aug 12, 2010 11:47 pm; Location: chennai, india; Thanked: 2 times

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: Beat The GMAT — Data Sufficiency | Wed Oct 27, 2010 6:40 am

kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

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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kapur.arnav: Master | Next Rank: 500 Posts; Posts: 126; Joined: Mon Jul 19, 2010 9:36 am; Thanked: 6 times

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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rkanthilal: Master | Next Rank: 500 Posts; Posts: 154; Joined: Thu Aug 26, 2010 9:32 amFollowed by: 19 members; Location: Chicago,IL; Thanked: 46 times; GMAT Score:760

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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vijchid: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Wed Sep 29, 2010 8:14 am; Thanked: 1 times

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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rkanthilal: Master | Next Rank: 500 Posts; Posts: 154; Joined: Thu Aug 26, 2010 9:32 amFollowed by: 19 members; Location: Chicago,IL; Thanked: 46 times; GMAT Score:760

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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